4 Quantum mechanics

Only the basics of quantum mechanics (QM) are explained here; for a more detailed explanation see [20]. In QM, the trajectory $\mathbf{r}\left(t\right)$ is replaced by the wave function $\psi\left(\mathbf{r}, t\right)$ to determine the state of a particle without spin. $\psi$ is in general complex-valued and $\left|\psi\left(\mathbf{r}, t\right)\right|^2$ is the probability density of finding the particle at $\mathbf{r}$ at time $t$. For functions $f, g:\mathbb{R}^3\to\mathbb{C}$ one defines the unitary product $\langle|\rangle$ by

Since the particle is somewhere, one has

The Schrödinger equation (SE) holds

with the Hamilton operator

Here $\hbar \coloneqq \frac{h}{2\pi}$ with Planck's constant $h$. An operator turns a function into a new function and is denoted by a hat symbol. Two further axiomatic assumptions of QM are the de Broglie relation

for the connection between momentum $\mathbf{p}$ and circular wave number $\mathbf{k}$ of a particle as well as the Planck-Einstein relation

for the connection between energy $E$ and angular frequency $\omega$. A plane wave can therefore be written as

\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \psi_0\exp\left(i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)\right) = \psi_0\exp\left(\frac{i}{\hbar}\left(\mathbf{p}\cdot\mathbf{r} - Et\right)\right) \end{align} \]

Here one finds that

\[ \begin{align} i\hbar\frac{\partial}{\partial t}\psi\left(\mathbf{r}, t\right) = E\psi\left(\mathbf{r}, t\right) \end{align} \]

holds, so the right-hand side of the SE brings the energy to the front. Analogously, for the left-hand side one obtains

\[ \begin{align} - \frac{\hbar^2}{2m}\Delta\psi\left(\mathbf{r}, t\right) = \frac{p^2}{2m}\psi\left(\mathbf{r}, t\right) = E_{\text{kin}}\psi\left(\mathbf{r}, t\right) \end{align} \]

with the non-relativistic energy-momentum relationship $E_{\mathrm{kin}} = \frac{p^2}{2m}$. The SE is therefore the law of conservation of energy for a plane probability wave. Because of

\[ \begin{align} - i\hbar\nabla\psi\left(\mathbf{r}, t\right) = \mathbf{p}\psi\left(\mathbf{r}, t\right) \end{align} \]

one defines

as the momentum operator. For wave functions $\psi\left(\mathbf{r}, t\right)$ whose time dependence is trivial and can be separated from the spatial dependence, i.e.

\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \psi\left(\mathbf{r}\right)\exp\left(-i\frac{E}{\hbar}t\right), \end{align} \]

the partial-time-derivative term $i\hbar\frac{\partial}{\partial t}$ becomes

\[ \begin{align} i\hbar\frac{\partial}{\partial t} = E. \end{align} \]

The Schrödinger equation then becomes the stationary Schrödinger equation

This is an eigenvalue problem, where $E$ is the eigenvalue and $\psi\left(\mathbf{r}, t\right)$ is the eigenvector.

4.1 Free particle

A free particle of mass $m$ moves in the absence of a potential $\newhat{v}_h = 0$ according to the equation

\[ \begin{align} - \frac{\hbar^2}{2m}\Delta \psi = i\hbar\frac{\partial\psi }{\partial t}. \end{align} \]

If one again makes the ansatz of a plane wave for $\psi\left(\mathbf{r}, t\right)$

\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \psi_0\exp\left(i\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right)\right) = \psi_0\exp\left(\frac{i}{\hbar}\left(\mathbf{p}\cdot\mathbf{r} - Et\right)\right), \end{align} \]

the dispersion relation follows from this:

\[ \begin{align} \frac{\hbar^2\mathbf{k}^2}{2m} = E = \hbar\omega\Leftrightarrow\omega\left(\mathbf{k}\right) = \frac{\hbar\mathbf{k}^2}{2m}.\tag{4.17}\label{eq:disp_rel_qm_frei} \end{align} \]

Eq. (4.17) is called the dispersion relation. The phase velocity $c_{\mathrm{ph}}$ of a matter wave thus becomes

\[ \begin{align} c_{\text{ph}} = \frac{\omega}{k} = \frac{\hbar k}{2m}, \end{align} \]

while for the group velocity $c_{\mathrm{gr}}$ one has

\[ \begin{align} c_{\text{gr}} = \sum_{i = 1}^{3}\frac{\partial \omega}{\partial k_i}\mathbf{e}_i = \frac{\hbar\mathbf{k}}{m}. \end{align} \]

Unlike electromagnetic waves in a vacuum, matter waves are not dispersion-free. The group velocity is twice as large in magnitude as the phase velocity.

4.2 Potential well

The potential well describes a potential $V\left(x\right)$ of the form

\[ \begin{align} V\left(x\right) \coloneqq \begin{cases} 0, \:0\leq x\leq L,\\ \infty, \:\text{otherwise} \end{cases} \end{align} \]

with $L>0$. The Hamiltonian $\newhat{H}$ of a particle in this potential is

\[ \begin{align} \newhat{H} = -\frac{\hbar^2}{2m}\Delta + V\left(x\right). \end{align} \]

The boundary conditions arise from the fact

\[ \begin{align} \psi\left(x\right) = 0\text{ for }x\not\in\left(0, L\right). \end{align} \]

One makes the ansatz

\[ \begin{align} \psi\left(x\right) = C\exp\left(ik_1x\right) + C_2\exp\left(ik_2x\right) \end{align} \]

with $k_1, k_2>0$. From the boundary condition

\[ \begin{align} \psi\left(0\right) = 0 \end{align} \]

follows

\[ \begin{align} C = -C_2, \end{align} \] so define $C \coloneqq C$; then the ansatz reads

\[ \begin{align} \psi\left(x\right) = C\left(\exp\left(ik_1x\right) - \exp\left(ik_2x\right)\right). \end{align} \] From the boundary condition

\[ \begin{align} \psi\left(L\right) = 0 \end{align} \] follows

\[ \begin{align} \exp\left(ik_1L\right) = \exp\left(ik_2L\right). \end{align} \]

In general, the two solution components $\exp\left(ik_1x\right)$ and $\exp\left(ik_2x\right)$ may be linearly independent; therefore one has

\[ \begin{align} k_1 L = k_2L + 2n\pi \end{align} \]

with $n\in \mathbb{Z}$, so

\[ \begin{align} k_1 - k_2 = \frac{2n\pi}{L}. \end{align} \]

So set

\[ \begin{align} k_1 = n_1\frac{\pi}{L}, & {} & k_2 = n_2\frac{\pi}{L} \end{align} \]

with $n_1 - n_2\in \mathbb{Z}$ even. The case $k_1 = k_2$ must be excluded, since otherwise the wave function vanishes. The normalization condition holds

\[ \begin{align} 1&\hastobe \int_{0}^{L}\left|\psi\left(x\right)\right|^2dx = \left|C\right|^2\int_0^L 2 - \exp\left(i\left(k_1 - k_2\right)x\right) - \exp\left(i\left(k_2 - k_1\right)x\right)dx\nonumber\\ &\Leftrightarrow \left|C\right|^22L\Leftrightarrow \left|C\right| = \frac{1}{\sqrt{2L}}. \end{align} \]

The overall solution for the wave function therefore reads, under the assumption $C>0$ with $n_1, n_2\in \mathbb{Z}$, $n_1\not = n_2$ and $n_1 - n_2$ even,

For the energy eigenvalues $E_{n_1, n_2}$ one obtains

The energy $E = 0$ is not possible because $n_1$ and $n_2$ cannot be zero at the same time. This is different in classical mechanics.

4.3 Potential step

The following potential is given

\[ \begin{align} V\left(x\right) = \begin{cases} V_0, \text{ }0\leq x\leq L,\\ 0, \text{ otherwise} \end{cases} \end{align} \]

with $V_0, L>0$. If a classical particle with an energy $0<E<V_0$ is in the range $x<0$, it can never reach the range $x>L$. In quantum mechanics this is different; this is called the tunnel effect because, figuratively speaking, the particle digs a tunnel through the potential barrier.

This will also be discussed here, but for the simpler case $L\to\infty$. From now on, an index $1$ stands for the range $x<0$, while an index $2$ stands from now on for the range $x\geq0$. The potential $V\left(x\right)$ is given by

\[ \begin{align} V\left(x\right) = \begin{cases} 0, \:x<0\\ V_0, \:x\geq 0 \end{cases} \end{align} \]

with $V_0>0$. In the two regions $1$ and $2$, two different stationary Schrödinger equations hold

\[ \begin{align} - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_1\left(x\right) &= E\psi_1\left(x\right),\\ - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_2\left(x\right) &= \left(E - V_0\right)\psi_2\left(x\right) \end{align} \]

for the two wave functions $\psi_1\left(x\right)$, $\psi_2\left(x\right)$; let $E<V_0$ be the energy of the particle. For $\psi_1\left(x\right)$ one makes an ansatz

\[ \begin{align} \psi_1\left(x\right) = C\exp\left(ikx\right) + C_2\exp\left(-ikx\right) \end{align} \]

with $k > 0$. The first term corresponds to the part of the wave arriving to the right, while the second term corresponds to the reflection. This time $C, C_2$ are not normalization constants, since one cannot normalize on unbounded sets. For $\psi_2\left(x\right)$ one makes an ansatz

\[ \begin{align} \psi_2\left(x\right) = C_3\exp\left(-\lambda x\right) \end{align} \]

with $C_3, \lambda > 0$.

The stationary Schrödinger equation applies to $\psi$, so the second derivative of $\psi$ has the same continuity properties as the potential $V$. In this case the second derivative is discontinuous, but $\psi$ is still continuously differentiable. This results in the following two connection conditions:

\[ \begin{align} C + C_2 &= C_3,\\ ikC - ikC_2 &= -\lambda C_3 \end{align} \]

Substituting the first equation into the second one gives

\[ \begin{align} ikC - ikC_2 &= -\lambda C - \lambda C_2\Leftrightarrow C_2\left(\lambda - ik\right) = -C\left(\lambda + ik\right)\nonumber\\ \Leftrightarrow C_2 &= -\frac{\lambda + ik}{\lambda - ik} = -\exp\left(2\phi\right) \end{align} \]

with

\[ \begin{align} \phi = \arctan\left(\frac{k}{\lambda}\right). \end{align} \]

From this it follows

\[ \begin{align} C_3 = C - \exp\left(2\phi\right). \end{align} \]

$k$ is obtained from the Schrödinger equation for region 1:

\[ \begin{align} \frac{\hbar^2k^2}{2m} = E \Rightarrow k = \frac{1}{\hbar}\sqrt{2mE} \end{align} \]

$\lambda$ is obtained from the Schrödinger equation for region $2$:

\[ \begin{align} - \frac{\hbar^2}{2m}\lambda^2 = E - V_0 \Rightarrow \lambda = \frac{1}{\hbar}\sqrt{2m\left(V_0 - E\right)} \end{align} \]

The solutions are summarized again:

$C$ is not determined by the Schrödinger equation, but by a normalization that is not possible here. In the case $L<\infty$ one can calculate transmission and reflection coefficients of the probability current density at the potential threshold.

The result of this section and sections 4.1 and 4.2 can be stated:

• In the case $E>V_0 = \text{ const}.$ there is a continuous energy spectrum.

• In the case $E<V_0 = \text{ const}.$ the particle can propagate into this region in the form of an exponential decay.

• If the particle's motion is bounded, a discrete energy spectrum exists.

4.4 Product wave functions

Let a one-dimensional potential $V_x\left(x\right)$ be given with an associated Hamiltonian

\[ \begin{align} \newhat{H}_x = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V_x\left(x\right), \end{align} \]

the solutions $\left(\psi_{n_x}\left(x\right), E_{n_x}\right)$ of the eigenvalue problem

\[ \begin{align} \newhat{H}_x\psi_{n_x}\left(x\right) = E_{n_x}\psi_{n_x}\left(x\right) \end{align} \]

are known. Now one extends the problem to three dimensions, i.e.

\[ \begin{align} V\left(x, y, z\right) = V_x\left(x\right) + V_y\left(y\right) + V_z\left(z\right), \end{align} \]

where, analogous to the x component, the solutions of the Schrödinger equation belonging to $V_y\left(y\right)$ and $V_z\left(z\right)$ are also known. However, it does not have to be $V_x = V_y$. The Hamiltonian $\newhat{H}$ of the three-dimensional system becomes

\[ \begin{align} \newhat{H} = \newhat{H}_x + \newhat{H}_y + \newhat{H}_z \end{align} \]

Now one makes a product ansatz for the solution $\psi\left(x, y, z\right)$

\[ \begin{align} \psi\left(x, y, z\right) = \psi_{n_x}\left(x\right)\psi_{n_y}\left(y\right)\psi_{n_z}\left(z\right), \end{align} \]

then one has

\[ \begin{align} \newhat{H}\psi\left(x, y, z\right) &= \left(\newhat{H}_x + \newhat{H}_y + \newhat{H}_z\right)\psi_{n_x}\left(x\right)\psi_{n_y}\left(y\right)\psi_{n_z}\left(z\right)\nonumber\\ &= \left(E_{n_x} + E_{n_y} + E_{n_z}\right)\psi_{n_x}\left(x\right)\psi_{n_y}\left(y\right)\psi_{n_z}\left(z\right). \end{align} \]

The solutions constructed in this way $\psi\left(x, y, z\right)$ form the complete solution set of

\[ \begin{align} \newhat{H}\psi\left(x, y, z\right) = E\psi\left(x, y, z\right). \end{align} \]

With this knowledge, one can easily solve, for example, the Schrödinger equation for a particle in a three-dimensional harmonic oscillator or in a three-dimensional potential well if the one-dimensional solutions are known.

4.5 Harmonic oscillator

The Hamiltonian $\newhat{H}$ of the harmonic oscillator is

\[ \begin{align} \newhat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2x^2, \end{align} \]

thus the stationary Schrödinger equation is

\[ \begin{align} - \frac{\hbar^2}{2m}\psi'' + \frac{1}{2}m\omega^2x^2\psi = E\psi. \end{align} \]

This eigenvalue problem must be solved. One can write it as

\[ \begin{align} \psi'' - \frac{x^2}{b^4}\psi = - \frac{2mE}{\hbar^2}\psi \end{align} \]

with $b \coloneqq \sqrt{\frac{\hbar}{m\omega}}$. Define $u \coloneqq \frac{x}{b}$; then

\[ \begin{align} \frac{d}{dx} = \frac{du}{dx}\frac{d}{du} = \frac{1}{b}\frac{d}{du}, & {} & \frac{d^2}{dx^2} = \frac{1}{b^2}\frac{d^2}{du^2}. \end{align} \]

With the substitution

\[ \begin{align} \psi\left(x\right)\to\psi\left(u\right) \end{align} \]

it becomes

\[ \begin{align} \psi'' - u^2\psi = -\frac{2E}{\hbar\omega}\psi.\tag{4.62}\label{eq:sg_harm_osz} \end{align} \]

At infinity this becomes

\[ \begin{align} \psi'' - u^2\psi\approx 0. \end{align} \]

For this, use the ansatz

\[ \begin{align} f\left(u\right) = \exp\left(-\frac{u^2}{2\sigma^2}\right) \end{align} \]

with $\sigma>0$. Differentiating twice yields

\[ \begin{align} \frac{d^2}{du^2}\exp\left(-\frac{u^2}{2\sigma^2}\right) = \frac{d}{du}\left(-\frac{u}{\sigma^2}\exp\left(-\frac{u^2}{2\sigma^2}\right)\right) \approx \frac{u^2}{\sigma^4}\exp\left(-\frac{u^2}{2\sigma^2}\right) \end{align} \]

for large $u$. Substituting this gives

\[ \begin{align} \frac{u^2}{\sigma^4} - u^2\hastobe0\Rightarrow\sigma = 1. \end{align} \]

For $\psi$, use the ansatz

\[ \begin{align} \psi\left(u\right) = P\left(u\right)\exp\left(-\frac{u^2}{2}\right). \end{align} \]

The second derivative is given by

\[ \begin{align} \frac{d^2\psi}{du^2} &= \frac{d}{du}\left[P'\exp\left(-\frac{u^2}{2}\right) - uP\exp\left(-\frac{u^2}{2}\right)\right] = \exp\left(-\frac{u^2}{2}\right)\left[P'' - 2uP' - P + u^2P\right]. \end{align} \]

Eq. (4.62) becomes, with $\newtilde{E} = \frac{2E}{\hbar\omega}$,

\[ \begin{align} P'' - 2uP' - P = -\newtilde{E}P\Leftrightarrow P'' - 2uP' + P\left(\newtilde{E} - 1\right) = 0.\tag{4.69}\label{eq:dgl_p} \end{align} \]

For $P\left(u\right)$, use a power-series ansatz

\[ \begin{align} P\left(u\right) = \sum_{i = 0}^{\infty}a_iu^i \end{align} \]

This yields

\[ \begin{align} P'\left(u\right) = \sum_{i = 0}^{\infty}ia_iu^{i - 1} \end{align} \]

and

\[ \begin{align} P''\left(u\right) = \sum_{i = 0}^{\infty}\left(i - 1\right)ia_iu^{i - 2} = \sum_{i = 0}^{\infty}\left(i + 2\right)\left(i + 1\right)a_{i + 2}u^i. \end{align} \]

Substituting into Eq. (4.69) gives

\[ \begin{align} \sum_{i = 0}^{\infty}\left(i + 2\right)\left(i + 1\right)a_{i + 2}u^i - 2\sum_{i = 0}^{\infty}ia_iu^{i} + \left(\newtilde{E} - 1\right)\sum_{i = 0}^{\infty}a_iu^i = 0. \end{align} \]

For this to hold for all $u\in \mathbb{R}$, the individual coefficients must vanish:

\[ \begin{align} \left(i + 2\right)\left(i + 1\right)a_{i + 2} - 2ia_i + \left(\newtilde{E} - 1\right)a_i = 0, \end{align} \]

hence the recursion formula for $a_i$ is

\[ \begin{align} a_{i + 2} = a_{i}\frac{2i - \newtilde{E} + 1}{\left(i + 2\right)\left(i + 1\right)}. \end{align} \]

Since the harmonic series $\sum_{n = 1}^{\infty}n^{-1}$ diverges, $\sum_{i = 0}^{\infty}a_iu^i$ also diverges unless the series terminates. Therefore, there must be an $n\in\mathbb{N}$ with

These are the oscillator's energy eigenvalues. $n$ defines a state. The states are therefore quantized, their energies are equally spaced with spacing $\hbar\omega$, and the ground-state energy $E_0>0$ is positive.

Rearranging the recursion formula gives

\[ \begin{align} a_{i} = a_{i + 2} \frac{\left(i + 2\right)\left(i + 1\right)}{2\left(i - n\right)}, \tag{4.77}\label{eq:hermite_poly_rek} \end{align} \]

The polynomials $P_n\left(u\right)$ are determined by $a_{n}$ and $a_{n + 1} = 0$; they are called Hermite polynomials $H_n\left(u\right)$, see Sect. C.3.

The normalization condition is

\[ \begin{align} \int_{ - \infty}^{\infty}\left|\psi\left(y\right)\right|^2dy = \int_{ - \infty}^{\infty}c_n^2\exp\left(-\frac{y^2}{b^2}\right)H_n^2\left(\frac{y}{b}\right)dy = \int_{ - \infty}^{\infty}c_n^2b\exp\left(-u^2\right)H_n^2\left(u\right)du = 1, \end{align} \]

thus, with Eq. (C.124),

\[ \begin{align} c_n^2b = \frac{1}{\sqrt{\pi}2^nn!}. \end{align} \]

Hence, the eigenstates of the harmonic oscillator are

with

\[ \begin{align} b = \sqrt{\frac{\hbar}{m\omega}}, & {} & c_n = \frac{1}{\sqrt{b\sqrt{\pi}2^nn!}}, & {} & u = \frac{x}{b}. \end{align} \]

Finally, two important relations are noted:

\[ \begin{align} u\psi_n\left(u\right) &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:herm_pol_prop_2}{\text{Eq. (C.123)}}}{=} n\frac{1}{\sqrt{2n}}\psi_{n - 1}\left(u\right) + \frac{1}{2}\sqrt{2\left(n + 1\right)}\psi_{n + 1}\left(u\right) = \sqrt{\frac{n}{2}}\psi_{n - 1}\left(u\right) + \sqrt{\frac{n + 1}{2}}\psi_{n + 1}\left(u\right),\tag{4.82}\label{eq:harm_osz_zust_prop_1}\\ \frac{d\psi_n}{du}\left(u\right) &\stackrel{\href{ch-41-orthogonal-function-systems.html#eq:herm_pol_prop_1}{\text{Eq. (C.122)}}}{=} 2n\frac{1}{\sqrt{2n}}\psi_{n - 1}\left(u\right) - u\psi_n\left(u\right) = \sqrt{2n}\psi_{n - 1}\left(u\right) - u\psi_n\left(u\right)\nonumber\\ &= \sqrt{2n}\psi_{n - 1}\left(u\right) - \sqrt{\frac{n}{2}}\psi_{n - 1}\left(u\right) - \sqrt{\frac{n + 1}{2}}\psi_{n + 1}\left(u\right) = \sqrt{n}\left(\sqrt{2} - \frac{1}{\sqrt{2}}\right)\psi_{n - 1}\left(u\right) - \sqrt{\frac{n + 1}{2}}\psi_{n + 1}\left(u\right)\nonumber\\ &= \sqrt{n}\left(\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\psi_{n - 1}\left(u\right) - \sqrt{\frac{n + 1}{2}}\psi_{n + 1}\left(u\right) = \sqrt{\frac{n}{2}}\psi_{n - 1}\left(u\right) - \sqrt{\frac{n + 1}{2}}\psi_{n + 1}\left(u\right)\tag{4.83}\label{eq:harm_osz_zust_prop_2} \end{align} \]

4.6 Hilbert space and operators

A particle is described in QM by its generally time-dependent wave function $\psi\left(t\right):\mathbb{R}^3\to\mathbb{C}$. This wave function is also referred to as the state of the particle.

4.6.1 Hilbert space

The set of all continuously differentiable functions $\mathbb {R}^3\to \mathbb {C}$ is called Hilbert space $H$ in QM. The dimension of $H$ is infinite. One can specify an infinite number of different bases $\left(f_1, f_2, \dotsc\right)$ with $f_i\in H$ for all $i\in \mathbb{N}$ with $i\geq 1$ of $H$. Usually one chooses orthonormal bases, i.e. bases with

\[ \begin{align} \langle f_i|f_j\rangle = \delta_{i, j} \end{align} \]

for all $i, j\in \mathbb {N}$ with $i, j\geq 1$. One can thus write for a state $\psi\left(\mathbf{r}, t\right)$

\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \sum_{i = 1}^{\infty}a_i\left(t\right)f_i\left(\mathbf{r}\right). \end{align} \]

For the $a_i$, with $k\in \mathbb {N}$, $k\geq 1$, one has

\[ \begin{align} \langle f_k|\psi\rangle = \int_{\mathbb{R}^3}f_k^\star\left(\mathbf{r}\right)\left(\sum_{i = 1}^{\infty}a_if_i\left(\mathbf{r}\right)\right)d^3r = \sum_{i = 1}^{\infty}a_i\int_{\mathbb{R}^3}f_k^\star\left(\mathbf{r}\right)f_i\left(\mathbf{r}\right)d^3r = \sum_{i = 1}^{\infty}a_i\delta_{ki} = a_k, \end{align} \]

so

\[ \begin{align} a_i = \langle f_i|\psi\rangle.\tag{4.87}\label{eq:formel_ent_koeff} \end{align} \]

If the order of the $f_i$ is chosen appropriately, $\lim\limits_{i\to\infty}a_i = 0$, so in practice the above sum can be truncated somewhere depending on the desired precision. The vector $\left(a_i\right)$ thus determines the state $\psi\left(\mathbf{r}\right)$ of a particle. One can transform to another basis $\left(\newtilde{f}_i\right)$; the transformation results from

\[ \begin{align} \psi\left(\mathbf{r}, t\right) = \sum_{i = 1}^{\infty}a_i\left(t\right)f_i\left(\mathbf{r}\right) = \sum_{i = 1}^{\infty}\newtilde{a}_i\left(t\right)\newtilde{f}_i\left(\mathbf{r}\right). \end{align} \]

In this case, the state $\psi\left(\mathbf{r}, t\right)$ is determined by $\left(\newtilde{a}_i\left(t\right)\right)$. An example of such a transformation is the Fourier transform.

If one has solved the Schrödinger equation for a system and found a quantization of the states, one can specify a state $\psi\left(\mathbf{r}\right)$ by $N\geq 1$ quantum numbers $n_1, \dotsc, n_N$, $\psi\left(\mathbf{r}, t\right) = \psi\left(n_1, \dotsc, n_N,\mathbf{r},t\right)$. There are thus infinitely many ways to specify a state of a particle. If one does not wish to commit to such a possibility, one simply writes

\[ \begin{align} |\psi\rangle. \end{align} \]

These can be, for example, spectral coefficients $a_i$ $|\psi\rangle = \left(a_i\right)$ or a quantum number $n$ as in the one-dimensional harmonic oscillator, $|\psi\rangle = |n\rangle$. This notation is also known as Dirac notation. If one needs two quantum numbers $n, m$ to specify a state, one writes $|\psi\rangle = |n, m\rangle$. One defines

\[ \begin{align} \langle\psi| \coloneqq |\psi\rangle^\star. \end{align} \]

This means: if $|\psi\rangle = \psi\left(\mathbf{r}, t\right)$ holds for the state $|\psi\rangle$ in position space, then $\langle\psi| = \psi^\star\left(\mathbf{r}, t\right)$ holds. Derived from the English word bracket, $|\psi\rangle$ is also called a ket state or simply ket, and $\langle\psi|$ is called a bra state or simply bra.

An operator $\newhat{O}$ turns a state into a new state (see also Sect. A.8), so it is a mapping on Hilbert space. One writes

\[ \begin{align} |\chi\rangle = \newhat{O}|\psi\rangle \end{align} \]

for the state $|\chi\rangle$, which arises when the operator $\newhat{O}$ acts on the state $|\psi\rangle$. If $\newhat{O}$ is linear (all operators used in QM are linear), then $\newhat{O}$ is a matrix if one expands the states with respect to some orthonormal basis of $H$. This matrix has an infinite number of rows and columns, unless one uses, approximately, only a finite number of basis elements. The operator adjoint to $\newhat{O}$ is called $\newhat{O}^+$ and is defined by the relation

\[ \begin{align} \left\langle\psi|\newhat{O}\chi\right\rangle = \left\langle\newhat{O}^+\psi|\chi\right\rangle, \end{align} \]

which is to hold for all $\psi, \chi\in H$. This determines $\newhat{O}^+$. In linear algebra, a matrix $A$ over $\mathbb{C}$ is called unitary, if

\[ \begin{align} A^+ = A^{-1} \end{align} \]

holds. Here $A^{-1}$ is the inverse matrix of $A$, i.e.

\[ \begin{align} A^{-1}A = 1 \end{align} \]

and $A^+$ is the adjoint matrix

\[ \begin{align} A^+= \left(A^\star\right)^T. \end{align} \]

Writing, for a state,

\[ \begin{align} |\psi\rangle = \sum_{i = 1}^{\infty}a_i|\psi_i\rangle, \end{align} \]

one has

\[ \begin{align} 1 &= \langle\psi|\psi\rangle = \left\langle\sum_{i = 1}^{\infty}a_i^\star f_i^\star\left(\mathbf{r}\right)\newvline \sum_{j = 1}^{\infty}a_if_i\left(\mathbf{r}\right)\right\rangle\nonumber\\ &= \int_{\mathbb{R}^3}\left(\sum_{i = 1}^{\infty}a_i^\star f_i^\star\left(\mathbf{r}\right)\right)\left(\sum_{j = 1}^{\infty}a_if_i\left(\mathbf{r}\right)\right)d^3r = \sum_{i, j = 1}^{\infty}\int_{\mathbb{R}^3}a_i^\star f_i^\star\left(\mathbf{r}\right)a_jf_j\left(\mathbf{r}\right)d^3r\nonumber\\ &= \sum_{i, j = 1}^{\infty}a_i^\star a_j\delta_{i, j} = \sum_{i = 1}^{\infty}\left|a_i\right|^2. \end{align} \]

The probability density in spatial space is given by $\rho = \psi\psi^\star = \sum_{i, j = 1}^{\infty}a_i^\star a_j f_i^\star\left(\mathbf{r}\right)f_j\left(\mathbf{r}\right)$. The probability $P_i$ of encountering the particle in a state $f_i$ is given by

\[ \begin{align} P_i = \left|\left\langle f_i|\psi\right\rangle\right|^2 = \left|a_i\right|^2. \end{align} \]

If every state $i$ is accompanied by an energy $E_i$, then the expectation value of the energy $\langle E\rangle$ is given by

\[ \begin{align} \langle E\rangle = \sum_{i = 1}^{\infty}P_iE_i = \sum_{i = 1}^{\infty}\left|\langle f_i|\psi\rangle\right|^2 E_i. \end{align} \]

If the equation $\newhat{O}|f_i\rangle = o_i|f_i\rangle$ holds for an operator $\newhat{O}$, then the expectation value $\langle o\rangle$ of the quantity $o$ is given by

\[ \begin{align} \langle o\rangle &= \sum_{i = 1}^{\infty}P_io_i = \sum_{i = 1}^{\infty}\left|\langle f_i|\psi\rangle\right|^2o_i = \sum_{i = 1}^{\infty }\langle \psi|f_i\rangle o_i\langle f_i|\psi\rangle = \sum_{i = 1}^{\infty}\langle\psi|o_if_i\rangle a_i\nonumber\\ &= \sum_{i = 1}^{\infty}\int_{\mathbb{R}^3}\psi^\star o_ia_if_id^3r = \int_{\mathbb{R}^3}^{}\psi^\star \sum_{i = 1}^{\infty}o_ia_if_id^3r = \int_{\mathbb{R}^3}^{}\psi^\star \sum_{i = 1}^{\infty}\newhat{O}a_if_i d^3r\nonumber\\ &= \int_{\mathbb{R}^3}\psi^\star\newhat{O}\sum_{i = 1}^{\infty}a_if_id^3r = \int_{\mathbb{R}^3}^{}\psi^\star\newhat{O}\psi d^3r. \end{align} \]

In general, the expectation value $\langle\newhat{O}\rangle$ of the operator $\newhat{O}$ in the state $|\psi\rangle$ is given by

\[ \begin{align} \langle\newhat{O}\rangle = \int_{\mathbb{R}^3}\psi^\star\newhat{O}\psi d^3r = \left\langle\psi|\newhat{O}|\psi\right\rangle. \end{align} \]

Now expand $\newhat{O}$ as a matrix with respect to the basis $\left(f_i\right)$ and write $O_{i, j}$ for the entry in the $j-$th column of the $i-$th row. As the initial state, choose $|\psi\rangle = f_j$, so the system is certainly in the $j-$th state. One writes

\[ \begin{align} |\chi\rangle \coloneqq \newhat{O}|\psi\rangle = \sum_{i = 1}^{\infty}b_if_i\left(\mathbf{r}\right) \end{align} \]

for the state $|\chi\rangle$ of the system under the action of the operator $\newhat{O}$. Let $k\in \mathbb {N}$ with $k\geq 1$; then

\[ \begin{align} b_k = \sum_{l = 1}^{\infty}O_{k, l}\langle f_l|\psi\rangle = \sum_{l = 1}^{\infty}O_{k, l}\delta_{l, j} = O_{k, j}. \end{align} \]

$\left|b_k\right|^2$ is the probability of finding the system in state $f_k$ after the operator acts, $\left|O_{k, j}\right|^2$ is therefore the probability that the system changes from state $j$ to state $k$ under the effect of the operator $\newhat{O}$. If $O_{k, j} = 0$, then the system cannot transition from state $j$ to state $k$ under the action of the operator $\newhat{O}$, and one obtains a forbidden transition. One can therefore write

\[ \begin{align} O_{i, j} = \left\langle f_i\left|\newhat{O}\right|f_j\right\rangle. \end{align} \]

This is a computationally evaluable formula for determining the matrix elements $O_{i, j}$ of the operator $\newhat{O}$ with respect to the basis $\left(f_k\right)$.

Now it is shown that every linear operator $\newhat{O}$ has at least one eigenvalue $\lambda\in \mathbb{C}$. With $x\in \mathbb{C}$ and $|\psi\rangle\in H$ the eigenvalue problem is

\[ \begin{align} \newhat{O}|\psi\rangle = x|\psi\rangle. \end{align} \]

Now choose an orthonormal basis $\left(|\psi_1\rangle, \dotsc\right)$ of $H$ and expand $\newhat{O}$ and $|\psi\rangle$ with respect to this basis, then the problem becomes

\[ \begin{align} O\mathbf{a} = x\mathbf{a} \end{align} \]

with $\mathbf{a} = \left(a_1, \dotsc\right)^T$ and $|\psi\rangle = \sum_{i = 1}^{\infty}a_i|\psi_i\rangle$. The existence of an eigenvalue $\lambda\in \mathbb{C}$ is equivalent to the characteristic polynomial

\[ \begin{align} p\left(x\right) \coloneqq \det\left(O - x\right) \end{align} \]

has at least one complex zero. Since $p\left(x\right)$ decomposes into linear factors over $\mathbb{C}$ according to the fundamental theorem of algebra , this is the case.

4.6.2 Hermitian operators

An operator $\newhat{O}$ is called Hermitian if

\[ \begin{align} \langle\newhat{O}f|g\rangle = \langle f|\newhat{O}g\rangle, \end{align} \]

holds, i.e. written out

\[ \begin{align} \int_{\mathbb{R}^3}\left(\newhat{O} f\right)^\star gd^3r = \int_{\mathbb{R}^3}f^\star\left(\newhat{O}g\right)d^3r. \end{align} \]

Hermitian operators are therefore self-adjoint,

\[ \begin{align} \newhat{O}^+= \newhat{O}. \end{align} \]

Now it is shown that the eigenvalues of Hermitian operators are real. So let a Hermitian operator $\newhat{O}$ be given and $|\psi\rangle\in H, \alpha\in \mathbb{C}$ with $\newhat{O}|\psi\rangle = \alpha|\psi\rangle$. Then one has

\[ \begin{align} \alpha^\star\langle\psi|\psi\rangle &= \langle\newhat{O}\psi|\psi\rangle = \langle\psi|\newhat{O}\psi\rangle = \alpha\langle\psi|\psi\rangle, \tag{4.111}\label{eq:hermit_ew_reell_herl} \end{align} \]

therefore $\alpha^\star = \alpha$ and therefore $\alpha\in \mathbb{R}$. Furthermore, the expectation values of Hermitian operators are also real:

\[ \begin{align} & \left\langle\psi\left|\newhat{O}\right|\psi\right\rangle^\star = \left(\int_{\mathbb{R}^3}\psi^\star\newhat{O}\psi d^3r\right)^\star = \int_{\mathbb{R}^3}\psi\left(\newhat{O}\psi\right)^\star d^3r\nonumber\\ &= \int_{\mathbb{R}^3}\left(\newhat{O}\psi\right)^\star\psi d^3r = \left\langle\newhat{O}\psi|\psi\right\rangle = \left\langle\psi\left|\newhat{O}\right|\psi\right\rangle \end{align} \]

Measurable quantities are always real. Hermitian operators are therefore suitable for defining measurable quantities.

Now it is shown that the eigenvectors of Hermitian operators belonging to different eigenvalues are orthogonal. So let a Hermitian operator $\newhat{O}$ be given and $|\psi\rangle, |\chi\rangle\in H, \alpha, \beta\in \mathbb{R}$ with $\alpha\not = \beta$ and

\[ \begin{align} \newhat{O}|\psi\rangle = \alpha|\psi\rangle, & {} & \newhat{O}|\chi\rangle = \beta|\chi\rangle. \end{align} \]

Then one has

\[ \begin{align} \beta\langle\psi|\chi\rangle = \langle\psi|\newhat{O}\chi\rangle &= \langle\newhat{O}\psi|\chi\rangle = \alpha\langle\psi|\chi\rangle, \end{align} \]

hence

\[ \begin{align} \langle\psi|\chi\rangle = 0. \end{align} \]

Of course, eigenvectors belonging to degenerate eigenvalues need not be orthonormal per se. However, the Gram-Schmidt orthonormalization procedure can be applied to them.

Now some operators are examined for Hermiticity. One has

\[ \begin{align} \int\left(-\frac{d}{dx}\varphi\right)^\star\psi dx = \int - \frac{d}{dx}\varphi^\star\psi dx = \int\varphi^\star\frac{d}{dx}\psi dx, \end{align} \]

hence

\[ \begin{align} \left(\frac{d}{dx}\right)^+= -\frac{d}{dx}. \end{align} \]

Partial derivatives are therefore not Hermitian. However, the x-component of the momentum operator $\newhat{p}_x = -i\hbar\frac{\partial}{\partial x}$ is Hermitian and so are the other components. One has

\[ \begin{align} \int\left(\left(-1\right)^n\frac{d^n}{dx^n}\varphi\right)^\star\psi dx = \left(-1\right)^n\int\frac{d^n}{dx^n}\varphi^\star\psi dx = \left(-1\right)^n\left(-1\right)^n\int\varphi^\star\frac{d^n\psi}{dx^n}dx, \end{align} \]

hence

\[ \begin{align} \left(\frac{d^n}{dx^n}\right)^+= \left(-1\right)^n\frac{d^n}{dx^n}. \end{align} \]

Because of

\[ \begin{align} & \int\left(\Delta\varphi\right)^\star\psi d^3r = \int\left(\sum_{i = 1}^{3}\frac{d^2}{dx_i^2}\varphi^\star\right)\psi d^3r = \sum_{i = 1}^{3}\int \left(\frac{d^2\varphi^\star}{dx_i^2}\right)\psi d^3r\nonumber\\ &= \sum_{i = 1}^{3}\int \varphi^\star\left(\frac{d^2\psi}{dx_i^2}\right) d^3r = \int\varphi^\star\Delta\psi d^3r \end{align} \]

is $\Delta^+= \Delta$. The Laplace operator is therefore Hermitian. Because of

\[ \begin{align} \int\left(\left(-1 - x\frac{d}{dx}\right)\varphi\right)^\star\psi dx &= -\int\varphi^\star\psi dx - \int\left(x\frac{d}{dx}\varphi^\star\right)\psi dx\nonumber\\ = -\int\varphi^\star\psi dx& + \int\varphi^\star\left(\psi + x\frac{d}{dx}\psi\right)dx = \int\varphi^\star x\frac{d}{dx}\psi dx \end{align} \]

one has

\[ \begin{align} \left(x\frac{d}{dx}\right)^+= -1 - x\frac{d}{dx}. \end{align} \]

The annihilation operator $\newhat{a}$ is defined by

\[ \begin{align} \newhat{a} \coloneqq\frac{1}{\sqrt{2}}\left(x + \frac{\partial}{\partial x}\right). \end{align} \]

One has

\[ \begin{align} \int\left(\frac{1}{\sqrt{2}}\left(x - \frac{\partial}{\partial x}\right)\varphi\right)^\star\psi dx &= \int\frac{1}{\sqrt{2}}\varphi^\star x\psi + \frac{1}{\sqrt{2}}\varphi^\star\frac{\partial\psi}{\partial x}dx = \int\varphi^\star\frac{1}{\sqrt{2}}\left(x + \frac{\partial}{\partial x}\right)\psi dx, \end{align} \]

that is,

\[ \begin{align} \newhat{a}^+ = \frac{1}{\sqrt{2}}\left(x - \frac{\partial}{\partial x}\right), \end{align} \]

this is called the creation operator. For the particle number operator $\newhat{N}$ one has

\[ \begin{align} \newhat{N} \coloneqq\frac{1}{2}\left(x^2 - \frac{\partial^2}{\partial x^2}\right). \end{align} \]

With this, one has

\[ \begin{align} & \int\left(\frac{1}{2}\left(x^2 - \frac{\partial^2}{\partial x^2}\right)\varphi\right)^\star\psi dx = \int\frac{1}{2}x^2\varphi^\star\psi - \frac{1}{2}\frac{\partial^2}{\partial x^2}\varphi^\star\psi dx\nonumber\\ &= \int\varphi^\star\frac{1}{2}\left(x^2 - \frac{\partial^2}{\partial x^2}\right)\psi dx = \int\varphi^\star\newhat{N}\psi dx, \end{align} \]

hence

\[ \begin{align} \newhat{N}^+= \newhat{N}. \end{align} \]

The particle number operator is therefore Hermitian.

At this point, some general statements about Hermitian operators are recorded. Let $\newhat{A}$, $\newhat{B}$ be two linear Hermitian operators. Then $\alpha\newhat{A} + \beta\newhat{B}$ with $\alpha, \beta\in \mathbb{R}$ is also Hermitian. Indeed, let $f, g\in H$. Then one has

\[ \begin{align} \left\langle f\newvline\left(\alpha\newhat{A} + \beta\newhat{B}\right)g\right\rangle &= \left\langle f\newvline\alpha\newhat{A}g\right\rangle + \left\langle f\newvline\beta\newhat{B}g\right\rangle = \left\langle\alpha\newhat{A} f\newvline g\right\rangle + \left\langle\beta\newhat{B}f\newvline g\right\rangle = \left\langle \left(\alpha\newhat{A} + \beta\newhat{B}\right)f\newvline g\right\rangle. \end{align} \]

Furthermore, in this case one has

\[ \begin{align} \left\langle f\newvline\newhat{A}\newhat{B}g\right\rangle = \left\langle\newhat{A}f\newvline\newhat{B}g\right\rangle = \left\langle \newhat{B}\newhat{A}f\newvline g\right\rangle, \end{align} \]

hence

\[ \begin{align} \left(\newhat{A}\newhat{B}\right)^+= \newhat{B}\newhat{A}. \end{align} \]

If $\newhat{A}$ and $\newhat{B}$ commute, then their successive application $\newhat{A}\newhat{B}$ is also Hermitian. Furthermore, with $n\in \mathbb{N}$, one has

\[ \begin{align} \left(\newhat{A}^n\right)^+= \newhat{A}^n. \end{align} \]

Thus, the momentum operator $\newhat{\mathbf{p}} = \sum_{j = 1}^{3}\newhat{p}_j\mathbf{e}_j$ as a linear combination of Hermitian operators is Hermitian. Furthermore, all potentials $\newhat{v}_h$ are Hermitian since they are real and contain no differential operators. Furthermore, the Hamilton operator is

\[ \begin{align} \newhat{H} = \frac{\newhat{\mathbf{p}}^2}{2m} + \newhat{v}_h \end{align} \]

Hermitian

4.6.3 Commuting operators

The commutator $\left[\newhat{A}, \newhat{B}\right]$ of two operators $\newhat{A}, \newhat{B}$ is defined by

\[ \begin{align} \left[\newhat{A}, \newhat{B}\right] \coloneqq\newhat{A}\newhat{B} - \newhat{B}\newhat{A}. \end{align} \]

Let $\left|\psi\right\rangle$ be a state. If the commutator of a constant operator $\newhat{A}$ with the Hamiltonian $\newhat{H}$ vanishes,

\[ \begin{align} \left[\newhat{H}, \newhat{A}\right] = 0, \end{align} \]

then $\left\langle\newhat{A}\right\rangle = \left\langle\psi\left|\newhat{A}\right|\psi\right\rangle$ is a conserved quantity:

\[ \begin{align} \frac{\partial}{\partial t}\left\langle\newhat{A}\right\rangle &= \frac{\partial}{\partial t}\left\langle\psi\left|\newhat{A}\right|\psi\right\rangle = \left\langle\frac{\partial}{\partial t}\psi\left|\newhat{A}\right|\psi\right\rangle + \left\langle\psi\left|\frac{\partial}{\partial t}\newhat{A}\right|\psi\right\rangle + \left\langle\psi\left|\newhat{A}\right|\frac{\partial}{\partial t}\psi\right\rangle \end{align} \]

Since $\newhat{A}$ is assumed to be constant, the SE gives

\[ \begin{align} \frac{\partial}{\partial t}\left\langle\newhat{A}\right\rangle &= -\frac{1}{i\hbar}\left\langle\newhat{H}\psi\left|\newhat{A}\right|\psi\right\rangle + \frac{1}{i\hbar}\left\langle\psi\left|\newhat{A}\right|\newhat{H}\psi\right\rangle = -\frac{1}{i\hbar}\left\langle\newhat{H}\psi\left|\newhat{A}\right|\psi\right\rangle + \frac{1}{i\hbar}\left\langle\psi\newvline\newhat{H}\newhat{A}\psi\right\rangle = 0, \end{align} \]

since $\newhat{H}$ is Hermitian. If $\left[\newhat{A}, \newhat{B}\right] = 0$ for Hermitian operators $\newhat{A}, \newhat{B}$, simultaneous eigenfunctions of the two operators can be found. This can be made clear as follows: First of all:

\[ \begin{align} \newhat{B}\newhat{A}\left|\psi\right\rangle = \newhat{B}a\left|\psi\right\rangle \Leftrightarrow \newhat{A}\left(\newhat{B}\left|\psi\right\rangle\right) = a\left(\newhat{B}\left|\psi\right\rangle\right), \tag{4.138}\label{eq:bew_kom_op_ef} \end{align} \]

so $\newhat{B}\left|\psi\right\rangle$ is an eigenvector of $\newhat{A}$ for the eigenvalue $a$. If $a$ is not degenerate, then the corresponding eigenspace is one-dimensional and therefore there exists a $b\in \mathbb{C}$ with $\newhat{B}\left|\psi\right\rangle = b\left|\psi\right\rangle$.

However, if $a$ as an eigenvalue of $\newhat{A}$ is degenerate $n-$fold, with $n\in\mathbb{N}$, $n\geq 2$, this means that the associated eigenspace is spanned by $n$ basis vectors $\left|\psi_1\right\rangle, \dotsc, \left|\psi_n\right\rangle$, which can be assumed to be orthonormal, i.e.

\[ \begin{align} \left\langle\psi_i\newvline\psi_j\right\rangle = \delta_{i, j}. \end{align} \]

Because of Eq. (4.138), there exist $k_i\in \mathbb{C}$ for $1\leq i\leq n$ with

\[ \begin{align} \newhat{B}\left|\psi\right\rangle = \sum_{i = 1}^{n}k_i|\psi_i\rangle. \end{align} \]

Since $\newhat{B}$ is Hermitian, the matrix $\left(B_{i, j}\right) \coloneqq\left\langle\psi_i\left|\newhat{B}\right|\psi_j\right\rangle$ can be converted into diagonal form by a unitary transformation $U$

\[ \begin{align} B' = U^TBU = \left(\begin{array}{cccc} \lambda_1& & & 0\\ &\lambda_2& &\\ & &\ddots&\\ 0 & & &\lambda_n \end{array} \right) \end{align} \]

with not necessarily distinct eigenvalues $\lambda_1, \dotsc, \lambda_n\in \mathbb{C}$. The transformed basis elements $\left|\psi_i'\right\rangle \coloneqq U\left|e_i\right\rangle$ are also eigenfunctions of $\newhat{A}$ as superpositions of the $\left|\psi_i\right\rangle$. Thus, $n$ simultaneous eigenfunctions of the two operators $\newhat{A}, \newhat{B}$ have been found.

4.6.4 Ladder operators

The ladder operators (also: creation and annihilation operators) $\newhat{a}^+$ and $\newhat{a}$ are defined by

\[ \begin{align} \newhat{a}^+ \coloneqq \frac{1}{\sqrt{2}}\left(u - \frac{\partial}{\partial u}\right). \end{align} \]

and

\[ \begin{align} \newhat{a} \coloneqq \frac{1}{\sqrt{2}}\left(u + \frac{\partial}{\partial u}\right) \end{align} \]

For the states $\psi_n\left(u\right)$ of the harmonic oscillator, Eqs. (4.82) and (4.83) give

\[ \begin{align} \newhat{a}\psi_n\left(u\right) = \sqrt{n}\psi_{n - 1}\left(u\right) \end{align} \]

and

\[ \begin{align} \newhat{a}^+\psi_n\left(u\right) = \sqrt{n + 1}\psi_{n + 1}\left(u\right). \end{align} \]

This is where the ladder operators get their name. One can, furthermore, generate any state $\psi_n\left(u\right)$ from the ground state $\psi_0\left(u\right)$:

\[ \begin{align} \left(\newhat{a}^+\right)^n\psi_0\left(u\right) = \sqrt{n!}\psi_n\left(u\right)\Leftrightarrow\psi_n\left(u\right) = \frac{1}{\sqrt{n!}}\left(\newhat{a}^+\right)^n\psi_0\left(u\right) \end{align} \]

The following should hold

\[ \begin{align} \newhat{a}\psi_0\left(u\right)\hastobe0. \end{align} \]

This means

\[ \begin{align} \left(u + \frac{\partial}{\partial u}\right)\psi_0\left(u\right) &\propto&\left(u + \frac{\partial }{\partial u}\right)\exp\left(-\frac{u^2}{2}\right) = \left(u - u\right)\exp\left(-\frac{u^2}{2}\right) = 0, \end{align} \]

So the required statement holds. For continuously differentiable $f, g:\mathbb{R}\to\mathbb{R}$, one has

\[ \begin{align} \left\langle f|\newhat{a}g\right\rangle &= \int_{ - \infty}^{\infty}f^\star\newhat{a}gdu = \int_{ - \infty}^{\infty}f^\star\frac{1}{\sqrt{2}}\left(u + \frac{\partial}{\partial u}\right)gdu\nonumber\\ &= \int_{ - \infty}^{\infty}g\frac{1}{\sqrt{2}}\left(u - \frac{\partial}{\partial u}\right)f^\star du = \left\langle \newhat{a}^+f|g\right\rangle \end{align} \]

The ladder operators are therefore adjoint in pairs, as was already taken into account in the notation $\newhat{a}, \newhat{a}^+$. For the operator

\[ \begin{align} \newhat{N} \coloneqq \newhat{a}^+\newhat{a} \end{align} \]

one has

\[ \begin{align} \newhat{N}\psi_n\left(u\right) = \newhat{a}^+\sqrt{n}\psi_{n - 1}\left(u\right) = n\psi_n\left(u\right). \end{align} \]

$\newhat{N}$ is called the particle number operator. For the commutator $\left[\newhat{a}, \newhat{a}^+\right]$ of the two operators, one has

\[ \begin{align} \left[\newhat{a}, \newhat{a}^+\right]\psi_n\left(u\right) &= \left(\newhat{a}\newhat{a}^+ - \newhat{a}^+\newhat{a}\right)\psi_n\left(u\right) = \newhat{a}\sqrt{n + 1}\psi_{n + 1}\left(u\right) - \newhat{a}^+\sqrt{n}\psi_{n - 1}\left(u\right)\nonumber\\ &= \left(n + 1\right)\psi_n\left(u\right) - n\psi_n\left(u\right) = \psi_n\left(u\right). \end{align} \]

Thus, in general,

\[ \begin{align} \left[\newhat{a}, \newhat{a}^+\right] = \newhat{1}. \end{align} \]

For the x-component $\newhat{x}$ of the position operator $\newhat{\mathbf{r}}$, one has

\[ \begin{align} \newhat{a} + \newhat{a}^+ &= \sqrt{2}u = \frac{\sqrt{2}x}{b}\Leftrightarrow \newhat{x} = \frac{b}{\sqrt{2}}\left(\newhat{a} + \newhat{a}^+\right). \end{align} \]

For the x-component $\newhat{p}_x = -i\hbar\frac{\partial}{\partial x}$ of the momentum operator $\newhat{\mathbf{p}}$, one has

\[ \begin{align} \newhat{a} - \newhat{a}^+ &= \sqrt{2}\frac{\partial}{\partial u} = \sqrt{2}b\frac{\partial}{\partial x} = ib\frac{\sqrt{2}}{\hbar}\newhat{p}_x\Leftrightarrow \newhat{p}_x = -i\frac{\hbar}{b\sqrt{2}}\left(\newhat{a} - \newhat{a}^+\right). \end{align} \]

For the Hamiltonian $\newhat{H}$ of the harmonic oscillator, one has

\[ \begin{align} \newhat{H} &= \frac{\newhat{p}_x^2}{2m} + \frac{1}{2}m\omega^2x^2 = - \frac{1}{2m}\frac{\hbar^2}{2b^2}\left(\newhat{a}^2 + \left(\newhat{a}^+\right)^2 - \newhat{a}\newhat{a}^+ - \newhat{a}^+\newhat{a}\right) + \frac{1}{4}m\omega^2b^2\left(\newhat{a}^2 + \left(\newhat{a}^+\right)^2 + \newhat{a}\newhat{a}^+ + \newhat{a}^+\newhat{a}\right)\nonumber\\ &= \frac{\hbar\omega}{4}\left(2\newhat{a}\newhat{a}^+ + 2\newhat{a}^+\newhat{a}\right) = \frac{\hbar\omega}{2}\left(2\newhat{N} + 1\right). \end{align} \]

4.7 Uncertainty principle

Let $\newhat{A}, \newhat{B}$ be two operators, and let a particle be in the state $\left|\psi\right\rangle$. Define the variances $\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle$, $\left\langle\left(\Delta\newhat{B}\right)^2\right\rangle$ of the operators $\newhat{A}, \newhat{B}$ in the state $\left|\psi\right\rangle$ by

\[ \begin{align} \left\langle\left(\Delta\newhat{A}\right)^2\right\rangle&\coloneqq \left\langle\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right)^2\right\rangle \end{align} \]

and analogously for $\left\langle\left(\Delta\newhat{B}\right)^2\right\rangle$. Furthermore, one has

\[ \begin{align} \left\langle\left(\Delta\newhat{A}\right)^2\right\rangle &= \left\langle\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right)^2\right\rangle = \left\langle\newhat{A}^2 + \left\langle\newhat{A}\right\rangle^2 - 2\newhat{A}\left\langle\newhat{A}\right\rangle\right\rangle\nonumber\\ &= \left\langle\newhat{A}^2\right\rangle - \left\langle\newhat{A}\right\rangle^2. \end{align} \]

Now assume that $\newhat{A}$ and $\newhat{B}$ are Hermitian. The expectation values $\left\langle\newhat{A}\right\rangle, \left\langle\newhat{B}\right\rangle$ of $\newhat{A}$ and $\newhat{B}$ are then real. Define the function $F:\mathbb{R}\to\mathbb{R}$ by

\[ \begin{align} F\left(x\right)&\coloneqq\int_{\mathbb{R}^3}\left|\left[x\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) - i\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\psi\right|^2d^3r\nonumber\\ &= \int_{\mathbb{R}^3}\left(\left[x\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) - i\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\psi\right)^\star\left[x\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) - i\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\psi d^3r\nonumber\\ &= \int_{\mathbb{R}^3}\psi^\star\left[x\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) + i\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\left[x\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) - i\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\psi d^3r. \end{align} \]

With

\[ \begin{align} & xi\left[\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right) - \left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right)\left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)\right]\nonumber\\ &= xi\left[\newhat{B}\newhat{A} - \newhat{B}\left\langle\newhat{A}\right\rangle - \left\langle\newhat{B}\right\rangle\newhat{A} + \left\langle\newhat{B}\right\rangle\left\langle\newhat{A}\right\rangle - \newhat{A}\newhat{B} + \newhat{A}\left\langle\newhat{B}\right\rangle + \left\langle\newhat{A}\right\rangle\newhat{B} - \left\langle\newhat{A}\right\rangle\left\langle\newhat{B}\right\rangle\right]\nonumber\\ &= -xi\left[\newhat{A}, \newhat{B}\right] \end{align} \]

follows

\[ \begin{align} F\left(x\right) &= \int_{\mathbb{R}^3}\psi^\star\left[x^2\left(\newhat{A} - \left\langle\newhat{A}\right\rangle\right)^2 + \left(\newhat{B} - \left\langle\newhat{B}\right\rangle\right)^2 - xi\left[\newhat{A}, \newhat{B}\right]\right]\psi d^3r\nonumber\\ &= x^2\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle + \left\langle\left(\Delta\newhat{B}\right)^2\right\rangle - x\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle\geq 0.\tag{4.161}\label{eq:heis_unsch_deriv_1} \end{align} \]

Let $f, g\in H$. Then one has

\[ \begin{align} & \left\langle i\left[A, B\right]f|g\right\rangle = \left\langle i\left(\newhat{A}\newhat{B} - \newhat{B}\newhat{A}\right)f|g\right\rangle = \left\langle i\newhat{A}\newhat{B}f - i\newhat{B}\newhat{A}f|g\right\rangle\nonumber\\ &= \left\langle i\newhat{A}\newhat{B}f|g\right\rangle - \left\langle i\newhat{B}\newhat{A}f|g\right\rangle = -\left\langle\newhat{B}f|i\newhat{A}g\right\rangle + \left\langle\newhat{A}f|i\newhat{B}g\right\rangle = \left\langle f|i\newhat{A}\newhat{B}g\right\rangle - \left\langle f|i\newhat{B}\newhat{A}g\right\rangle \nonumber\\ &= \left\langle f|i\left(\newhat{A}\newhat{B} - \newhat{B}\newhat{A}\right)g\right\rangle = \left\langle f|i\left[\newhat{A}, \newhat{B}\right]g\right\rangle. \end{align} \]

So $i\left[\newhat{A}, \newhat{B}\right]$ is Hermitian and $\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle$ is real. $F\left(x\right)$ is an upwardly opened parabola with exactly one minimum $x_0$. This is obtained by setting the derivative to zero

\[ \begin{align} 0 &= 2x_0\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle - \left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle\Leftrightarrow x_0 = \frac{\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle}{2\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle}. \end{align} \]

Substituting this into Eq. (4.161), one obtains

\[ \begin{align} & \frac{1}{4}\frac{\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle^2}{\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle} + \left\langle\left(\Delta\newhat{B}\right)^2\right\rangle - \frac{\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle^2}{2\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle} \geq 0\Leftrightarrow\left\langle\left(\Delta\newhat{A}\right)^2\right\rangle\left\langle\left(\Delta\newhat{B}\right)^2\right\rangle\geq\frac{1}{4}\left\langle i\left[\newhat{A}, \newhat{B}\right]\right\rangle^2 . \end{align} \]

This can also be written as

\[ \begin{align} \left\langle\left(\Delta\newhat{A}\right)^2\right\rangle\left\langle\left(\Delta\newhat{B}\right)^2\right\rangle\geq\frac{1}{4}\left|\left\langle\left[\newhat{A}, \newhat{B}\right]\right\rangle\right|^2\tag{4.165}\label{eq:heis_unschaerfe}. \end{align} \]

From this one can conclude that, in general, the expectation values of non-commuting operators cannot be determined simultaneously to arbitrary precision. The eigenvalues of Hermitian operators are the measurable quantities of quantum mechanics. Let a bound state be given that can be specified by $N \geq 1$ quantum numbers. A set of Hermitian operators $\newhat{A}_1, \dotsc, \newhat{A}_N$ is called a set of complete observables if the $\newhat{A}_i$ with $1\leq i\leq N$ commute pairwise. With this requirement, the quantum numbers can be measured simultaneously.

4.8 H atom

The matter on Earth is structured into atoms, which consist of a positively charged nucleus made of protons and neutrons with the atomic number $Z\in \mathbb {N}$ with $Z\geq 1$, as well as an electron shell. $Z$ is also called the atomic number, all atoms with the same atomic number are combined into a chemical element. The mass number $N\in \mathbb{N}$, $N\geq Z\geq 1$ corresponds to the number of protons and neutrons in the nucleus. Protons are simply positively charged (in units of elementary charge $e$), neutrons are neutral. Atoms with the same atomic number but different mass numbers are called isotopes. Since the electrons are also simply negatively charged, a neutral atom consists of as many protons as electrons. If electrons are missing or additional ones are present, this is an ion, the atom has been ionized. Each element receives a symbol, e.g. $A$, one writes $_Z^NA$ if $A$ has the atomic number $Z$ and isotopes of mass number $N$ are meant. Atomic numbers up to 92 occur in nature; elements with higher atomic numbers can be produced in nuclear reactors. For atmospheric physics, the exact structure of the nucleus is irrelevant; it is assumed to be a positively charged point mass.

The hydrogen atom (element symbol H) is the simplest atom; it consists of a proton as a nucleus and an electron in the shell. It is examined here. The concept of reduced mass is known from Sect. 2.1. This can also be applied quantum mechanically. To this end, one starts from the classical case and sets up the Lagrange function of the two-particle system:

\[ \begin{align} L\left(\newdot{\mathbf{r}}_1, \newdot{\mathbf{r}}_2, \mathbf{r}_1, \mathbf{r}_2\right) = \frac{m_1}{2}\newdot{\mathbf{r}}_1^2 + \frac{m_2}{2}\newdot{\mathbf{r}}_2^2 - V\left(\mathbf{r}_2 - \mathbf{r}_1\right) \end{align} \]

It makes sense to assume that the potential only depends on the relative vector. Introducing the definitions

\[ \begin{align} \mathbf{r} \coloneqq\mathbf{r}_2 - \mathbf{r}_1, & {} & \mathbf{R} \coloneqq\frac{m_1\mathbf{r}_1 + m_2\mathbf{r}_2}{m_1 + m_2},\\ M \coloneqq m_1 + m_2, & {} & \mu \coloneqq\frac{m_1m_2}{m_1 + m_2} \end{align} \]

one can compute

\[ \begin{align} \frac{M}{2}\newdot{\mathbf{R}}^2 + \frac{\mu}{2}\newdot{\mathbf{r}}^2 &= \frac{1}{2M}\left(m_1^2\newdot{\mathbf{r}}_1^2 + m_2^2\newdot{\mathbf{r}}_2^2 + 2m_1m_2\newdot{\mathbf{r}}_1\cdot\newdot{\mathbf{r}}_2\right) + \frac{m_1m_2}{2\left(m_1 + m_2\right)}\left(\newdot{\mathbf{r}}_1^2 + \newdot{\mathbf{r}}_2^2 - 2\newdot{\mathbf{r}}_1\cdot\newdot{\mathbf{r}}_2\right)\nonumber\\ &= \frac{1}{2}m_1\newdot{\mathbf{r}}_1^2 + \frac{1}{2}m_2\newdot{\mathbf{r}}_2^2. \end{align} \]

Thus, an alternative Lagrange function reads

\[ \begin{align} L\left(\newdot{\mathbf{R}}, \newdot{\mathbf{r}}, \mathbf{r}\right) = \frac{M}{2}\newdot{\mathbf{R}}^2 + \frac{\mu}{2}\newdot{\mathbf{r}}^2 - V\left(\mathbf{r}\right). \end{align} \]

For the canonical momenta, one obtains

\[ \begin{align} \mathbf{p} = \frac{\partial L}{\partial\newdot{\mathbf{r}}} = \mu\newdot{\mathbf{r}}, & {} & \mathbf{P} = \frac{\partial L}{\partial\newdot{\mathbf{R}}} = M\newdot{\mathbf{R}}. \end{align} \]

For the Hamilton function, one obtains

\[ \begin{align} H\left(\mathbf{P}, \mathbf{p}, \mathbf{r}\right) = \frac{\mathbf{P}^2}{2M} + \frac{\mathbf{p}^2}{2\mu} + V\left(\mathbf{r}\right). \end{align} \]

Using the replacement rules for operators, one writes

\[ \begin{align} \mathbf{P}&\to \newhat{\mathbf{P}} = -i\hbar\frac{\partial}{\partial\mathbf{R}},\\ \mathbf{p}&\to \newhat{\mathbf{p}} = -i\hbar\frac{\partial}{\partial\mathbf{r}}. \end{align} \]

This yields the Hamilton operator

\[ \begin{align} \newhat{H}' = -\frac{\hbar^2}{2m}\Delta_\mathbf{R} - \frac{\hbar^2}{2\mu}\Delta_\mathbf{r} + V\left(\mathbf{r}\right). \end{align} \]

It can be applied to a wave function $\psi = \psi\left(\mathbf{R}, \mathbf{r}\right)$:

\[ \begin{align} \newhat{H}'\psi\left(\mathbf{R}, \mathbf{r}\right) = E\psi\left(\mathbf{R}, \mathbf{r}\right). \end{align} \]

$\newhat{H}'$ does not explicitly depend on $\mathbf{R}$, so

\[ \begin{align} \left[\newhat{H}', \newhat{\mathbf{P}}\right] = 0. \end{align} \]

According to Sect. 4.6.3 $\newhat{H}'$ and $\newhat{\mathbf{P}}$ therefore have the same eigenfunctions. One therefore looks for solutions that are eigenfunctions of $\newhat{\mathbf{P}}$:

\[ \begin{align} \newhat{\mathbf{P}}\psi\left(\mathbf{R}, \mathbf{r}\right) = \hbar\mathbf{K}\psi\left(\mathbf{R}, \mathbf{r}\right). \end{align} \]

From this it follows

\[ \begin{align} \psi\left(\mathbf{R}, \mathbf{r}\right) = \exp\left(i\mathbf{K}\cdot\mathbf{R}\right)\varphi\left(\mathbf{r}\right). \end{align} \]

One substitutes this into the stationary SE:

\[ \begin{align} \left(-\frac{\hbar^2}{2M}\Delta_\mathbf{R} - \frac{\hbar^2}{2\mu}\Delta_\mathbf{r} + V\left(\mathbf{r}\right) - E\right)\exp\left(i\mathbf{K}\cdot\mathbf{R}\right)\varphi\left(\mathbf{r}\right) = 0 \end{align} \]

With the definitions

\[ \begin{align} \Delta \coloneqq \Delta_\mathbf{r}, & {} & \epsilon \coloneqq E - \frac{\hbar^2K^2}{2M} \end{align} \]

this becomes

\[ \begin{align} \left(-\frac{\hbar^2}{2\mu}\Delta + V\left(\mathbf{r}\right) - \epsilon\right)\varphi\left(\mathbf{r}\right) = 0. \end{align} \]

With the Hamilton operator

\[ \begin{align} \newhat{H} \coloneqq - \frac{\hbar^2}{2\mu}\Delta + V\left(\mathbf{r}\right) \end{align} \]

one obtains

\[ \begin{align} \newhat{H}\varphi\left(\mathbf{r}\right) = \epsilon\varphi\left(\mathbf{r}\right). \end{align} \]

The center of mass movement leads to a modulation of the wave function with $e^{i\mathbf{K}\cdot\mathbf{R}}$ and a contribution to the kinetic energy of $\hbar^2K^2/2m$. This has been decoupled here. Thus, through the transition

\[ \begin{align} m_e\to\mu \end{align} \]

one can achieve higher accuracy for the hydrogen eigenstates. Now these eigenfunctions are to be determined. The Schrödinger equation

\[ \begin{align} \newhat{H}\left|\psi\right\rangle = E\left|\psi\right\rangle \end{align} \]

must be solved for a potential $V = V\left(r\right)$ that only depends on the distance:

\[ \begin{align} - \frac{\hbar^2}{2m}\Delta\psi + V\left(r\right)\psi = E\psi, \end{align} \]

here $m$ is the mass of the particle under consideration. To solve it, one makes a product ansatz of the form

\[ \begin{align} \psi\left(r, \theta, \phi\right) = R\left(r\right)G\left(\theta, \phi\right).\tag{4.188}\label{eq:ansatz_radial} \end{align} \]

For the Laplace operator $\Delta$, one can write

\[ \begin{align} \Delta = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\Delta_{\theta, \phi} \end{align} \]

with a part $\Delta_{\theta, \phi}$ that contains only partial angular derivatives (see Eq. (B.94)). Substituting this together with Eq. (4.188) into the Schrödinger equation, one first obtains

\[ \begin{align} & -G\left(\theta, \phi\right)\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)R\left(r\right) - R\left(r\right)\frac{\hbar^2}{2m}\frac{1}{r^2}\Delta_{\theta, \phi}G\left(\theta, \phi\right)\nonumber\\ & + V\left(r\right)R\left(r\right)G\left(\theta, \phi\right) = ER\left(r\right)G\left(\theta, \phi\right). \end{align} \]

One knows that $\psi$ is nonzero outside null sets, so one can separate

\[ \begin{align} - \frac{1}{R\left(r\right)}\frac{\hbar^2}{2m}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)R\left(r\right) + r^2V\left(r\right) - Er^2 = \frac{1}{G\left(\theta, \phi\right)}\frac{\hbar^2}{2m}\Delta_{\theta, \phi}G\left(\theta, \phi\right). \end{align} \]

Both sides are therefore equal to a separation constant, which should be denoted by $-\frac{\hbar^2l\left(l + 1\right)}{2m}$ with $l\in \mathbb{N}$. The solutions therefore get an index $l$. First, one solves the angular part. The ODE for this reads:

\[ \begin{align} \Delta_{\theta, \phi}G_l\left(\theta, \phi\right) = -l\left(l + 1\right)G_l\left(\theta, \phi\right). \end{align} \]

This is satisfied by the spherical harmonics $Y_{l, m}\left(\theta, \phi\right)$, Eq. (C.155), see Eq. (C.166); here $\left|m\right|\leq l$ holds. The normalization condition reads

\[ \begin{align} \int_{\phi = 0}^{2\pi}\int_{\theta = 0}^{\pi}\left|G_l\right|^2\sin\left(\theta\right) d\theta d\phi \hastobe1. \end{align} \]

This too is satisfied by the spherical harmonics $Y_{l, m}\left(\theta, \phi\right)$. For the radial part one obtains

\[ \begin{align} - \frac{\hbar^2}{2m}\left[r^2R_l''\left(r\right) + 2rR_l'\left(r\right)\right] + r^2V\left(r\right)R_l\left(r\right) - Er^2R_l\left(r\right) &= -\frac{\hbar^2l\left(l + 1\right)}{2m}R_l\left(r\right)\nonumber\\ \Leftrightarrow \left[-\frac{\hbar^2}{2m}\left(\frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr}\right) + \frac{\hbar^2l\left(l + 1\right)}{2mr^2} + V\left(r\right) - E\right]R_l\left(r\right) &= 0. \end{align} \]

Define $U_l\left(r\right) \coloneqq R_l\left(r\right)r$, then $R_l\left(r\right) = \frac{U_l\left(r\right)}{r}$ and thus

\[ \begin{align} & \left(\frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr}\right)\frac{U_l\left(r\right)}{r} = \frac{d}{dr}\left(-\frac{U_l\left(r\right)}{r^2} + \frac{\frac{dU_l\left(r\right)}{dr}}{r}\right) + \frac{2}{r}\left(\frac{\frac{dU_l\left(r\right)}{dr}}{r} - \frac{1}{r^2}U_l\left(r\right)\right)\nonumber\\ &= 2\frac{U_l\left(r\right)}{r^3} - \frac{\frac{dU_l\left(r\right)}{dr}}{r^2} + \frac{\frac{d^2U_l\left(r\right)}{dr^2}}{r} - \frac{\frac{dU_l\left(r\right)}{dr}}{r^2} + \frac{2}{r^2}\frac{dU_l\left(r\right)}{dr} - \frac{2}{r^3}U_l\left(r\right), \end{align} \]

hence

\[ \begin{align} \left(\frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr}\right)R_l\left(r\right) = \frac{1}{r}\frac{d^2}{dr^2}U_l\left(r\right). \end{align} \]

The $U_l\left(r\right)$ therefore satisfy the differential equation

\[ \begin{align} \left[-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} + \frac{\hbar^2l\left(l + 1\right)}{2mr^2} + V\left(r\right) - E\right]U_l\left(r\right) &= 0\nonumber\\ \Leftrightarrow \left[-\frac{d^2}{dr^2} + \frac{l\left(l + 1\right)}{r^2} + \frac{2mV\left(r\right)}{\hbar^2} - \frac{2mE}{\hbar^2}\right]U_l\left(r\right) &= 0.\tag{4.197}\label{eq:radial_sg} \end{align} \]

Eq. (4.197) is the radial Schrödinger equation. Now the Coulomb potential is used for $V\left(r\right)$

\[ \begin{align} V\left(r\right) = -\frac{Ze^2}{r}. \end{align} \]

This generally assumes that there are $Z$ protons in the nucleus and only one electron in the shell. Eq. (4.197) becomes

\[ \begin{align} \left[-\frac{d^2}{dr^2} + \frac{l\left(l + 1\right)}{r^2} - \frac{2mZe^2}{\hbar^2r} - \frac{2mE}{\hbar^2}\right]U_l\left(r\right) &= 0.\tag{4.199}\label{eq:radial_sgh} \end{align} \]

At this point the Bohr radius $a_B$ is defined by

\[ \begin{align} a_B \coloneqq \frac{\hbar^2}{e^2m_e}, \tag{4.200}\label{eq:def_bohr} \end{align} \]

and further define $u \coloneqq \frac{r}{a_B}$. Then one has

\[ \begin{align} \frac{d^2}{dr^2} = \frac{d}{dr}\left(\frac{du}{dr}\frac{d}{du}\right) = \frac{d}{dr}\left(\frac{1}{a_B}\frac{d}{du}\right) = \frac{1}{a_B^2}\frac{d^2}{du^2}. \end{align} \]

With the replacement

\[ \begin{align} U_l\left(r\right)\to U_l\left(u\right) \end{align} \]

and the definitions

\[ \begin{align} E_{\text{at}} \coloneqq \frac{\hbar^2}{m_ea_B^2}, & {} & \epsilon \coloneqq \frac{E}{E_{\text{at}}} \end{align} \]

Eq. (4.199) becomes

\[ \begin{align} \left[-\frac{d^2}{du^2} + \frac{l\left(l + 1\right)}{u^2} - \frac{2Z}{u} - 2\epsilon\right]U_l\left(u\right) &= 0.\tag{4.204}\label{eq:radial_seq_rescaled} \end{align} \]

For large $u$, one has

\[ \begin{align} - \frac{d^2}{du^2}U_l\left(u\right) - 2\epsilon U_l\left(u\right)\approx 0. \end{align} \]

One makes an ansatz

\[ \begin{align} U_l\left(u\right) = \exp\left(-\lambda u\right) \end{align} \]

with $\lambda\in \mathbb{R}$, $\lambda >0$.

This gives

\[ \begin{align} \lambda^2 = -2\epsilon.\tag{4.207}\label{eq:energy_radial_sg} \end{align} \]

For small $r$, one has

\[ \begin{align} \frac{d^2U_l\left(u\right)}{du^2}&\approx l\left(l + 1\right)\frac{U_l\left(u\right)}{u^2}. \end{align} \]

Here an ansatz suggests itself

\[ \begin{align} U_l\left(u\right) = u^k \end{align} \]

with $k\in \mathbb {N}$, $k\geq 2$. It follows

\[ \begin{align} k\left(k - 1\right) = l\left(l + 1\right), \end{align} \]

so

\[ \begin{align} k = l + 1. \end{align} \]

Now make an ansatz for a function $P_l\left(u\right)$ with

\[ \begin{align} U_l\left(u\right) = P_l\left(u\right)u^{l + 1}\exp\left(-\lambda u\right) \end{align} \]

Then one has

\[ \begin{align} U_l' &= P_l'u^{l + 1}\exp\left(-\lambda u\right) + \left(l + 1\right)\frac{U_l}{u} - \lambda U_l = P_l'u^{l + 1}\exp\left(-\lambda u\right) + U_l\left(\frac{l + 1}{u} - \lambda\right),\\ U_l'' &= P_l''u^{l + 1}\exp\left(-\lambda u\right) + \left(l + 1\right)P_l'u^l\exp\left(-\lambda u\right)\nonumber\\ & - \lambda P_l'u^{l + 1}\exp\left(-\lambda u\right) + U_l'\left(\frac{l + 1}{u} - \lambda\right) - \frac{l + 1}{u^2}U_l = P_l''u^{l + 1}\exp\left(-\lambda u\right)\nonumber\\ & + \left(l + 1\right)P_l'u^l\exp\left(-\lambda u\right) - \lambda P_l'u^{l + 1}\exp\left(-\lambda u\right)\nonumber\\ & + \left(\frac{l + 1}{u} - \lambda\right)\left[P_l'u^{l + 1}\exp\left(-\lambda u\right) + U_l\left(\frac{l + 1}{u} - \lambda\right)\right] - \frac{l + 1}{u^2}U_l\nonumber\\ &= P_l''u^{l + 1}\exp\left(-\lambda u\right) + P_l'u^l\exp\left(-\lambda u\right)2\left(l + 1\right) - 2\lambda P_l'u^{l + 1}\exp\left(-\lambda u\right)\nonumber\\ & + \lambda^2P_lu^{l + 1}\exp\left(-\lambda u\right) - 2\lambda\left(l + 1\right)P_lu^l\exp\left(-\lambda u\right) + l\left(l + 1\right)P_lu^{l - 1}\exp\left(-\lambda u\right)\nonumber\\ &= \exp\left(-\lambda u\right)u^{l + 1}\left[P_l'' + \frac{2\left(l + 1\right)P_l'}{u} - 2\lambda P_l' + \lambda^2P_l - 2\lambda\frac{l + 1}{u}P_l + l\frac{l + 1}{u^2}P_l\right]. \end{align} \]

Substituting this into Eq. (4.204), one obtains

\[ \begin{align} & - \exp\left(-\lambda u\right)u^{l + 1}\left[P_l'' + \frac{2\left(l + 1\right)P_l'}{u} - 2\lambda P_l' + \lambda^2P_l - 2\lambda\frac{l + 1}{u}P_l + l\frac{l + 1}{u^2}P_l\right]\nonumber\\ & + l\frac{l + 1}{u^2}P_l\exp\left(-\lambda u\right)u^{l + 1} - 2Z\exp\left(-\lambda u\right)u^lP_l - 2\epsilon\exp\left(-\lambda u\right)u^{l + 1}P_l = 0\nonumber\\ &\Leftrightarrow -u^{l + 1}P_l'' - u^l2\left(l + 1\right)P_l' + 2\lambda P_l'u^{l + 1} - \lambda^2P_lu^{l + 1} + 2\lambda\left(l + 1\right)u^lP_l - 2Zu^lP_l - 2\epsilon u^{l + 1}P_l = 0\nonumber\\ &\Leftrightarrow uP_l'' + 2\left(l + 1\right)P_l' - 2\lambda P_l'u + \lambda^2P_lu - 2\lambda\left(l + 1\right)P_l + 2ZP_l + 2\epsilon uP_l = 0\nonumber\\ &\Leftrightarrow uP_l'' + \left(2l + 2 - 2\lambda u\right)P_l' + \left(\lambda^2u - 2\lambda\left(l + 1\right) + 2Z + 2\epsilon u\right)P_l = 0\nonumber\\ &\Leftrightarrow uP_l'' + \left(2l + 2 - 2\lambda u\right)P_l' + \left(-2\lambda\left(l + 1\right) + 2Z\right)P_l = 0, \end{align} \]

the last step follows because $\lambda^2 = -2\epsilon$, Eq. (4.207). This is divided by $2\lambda$:

\[ \begin{align} & \frac{1}{2\lambda}uP_l'' + \left(\frac{l + 1}{\lambda} - u\right)P_l' + \left(\frac{Z}{\lambda} - l - 1\right)P_l = 0 \end{align} \]

Define

\[ \begin{align} x \coloneqq 2\lambda u, \end{align} \]

then

\[ \begin{align} u = \frac{x}{2\lambda}, & {} & \frac{d}{du} = 2\lambda\frac{d}{dx}, & {} & \frac{d^2}{du^2} = 4\lambda^2\frac{d^2}{dx^2}. \end{align} \]

Substituting

\[ \begin{align} P_l\left(u\right)\to P_l\left(x\right), \end{align} \]

one obtains

\[ \begin{align} xP_l'' + \left(\frac{l + 1}{\lambda} - \frac{x}{2\lambda}\right)2\lambda P_l' + \left(\frac{Z}{\lambda} - l - 1\right)P_l &= 0\nonumber\\ \Leftrightarrow xP_l'' + \left(2l + 2 - x\right)P_l' + \left(\frac{Z}{\lambda} - l - 1\right)P_l &= 0. \end{align} \]

This is the Laguerre differential equation. In Sect. C.4 it is shown that the Laguerre polynomials $L_{n_r, 2l + 1}\left(x\right)$ defined there in Eq. (C.134) satisfy this equation; here $n_r\in \mathbb {N}$. Define $n \coloneqq \frac{Z}{\lambda}\in \mathbb{N}$, $n$ is the principal quantum number. For it, one has

\[ \begin{align} n = n_r + l + 1, \end{align} \]

from this it follows that $l<n$. Thus, for the energy eigenvalues, one has with Eq. (4.207)

\[ \begin{align} \epsilon = -\frac{\lambda^2}{2} = -\frac{Z^2}{2n^2}, \end{align} \]

so $n\geq1$. In the original units, one obtains

Thus, for the radial functions, one has

\[ \begin{align} R_{n_r, l}\left(r\right) &= C_{n_r, l}\frac{U_{n_r, l}\left(r\right)}{r} = C_{n_r, l}\frac{1}{r}L_{n_r, 2l + 1}^{}\left(\frac{2Zr}{na_B}\right)\left(\frac{2Zr}{na_B}\right)^{l + 1}\exp\left(-\frac{Zr}{na_B}\right)\nonumber\\ &= C_{n_r, l}\frac{2Z}{na_B}\left(\frac{2Zr}{na_B}\right)^lL_{n_r, 2l + 1}^{}\left(\frac{2Zr}{na_B}\right)\exp\left(-\frac{Zr}{na_B}\right) \end{align} \]

with a normalization constant $C_{n_r, l}\in \mathbb {R}$ that now needs to be determined. The probability $P\left(r\right)$ of encountering the particle at a distance of at most $r$ from the origin is given by

\[ \begin{align} P\left(r\right) &= \int_{0}^{r}\int_{0}^{2\pi}\int_{0}^{\pi}\left|\psi_{n, l, m}\left(r', \theta, \phi\right)\right|^2r'^2\sin\left(\theta\right) d\theta d\phi dr' = \int_{0}^{r}\left|R_{n_r, l}\left(r'\right)\right|^2r'^2dr', \end{align} \]

so the radial probability density $\rho_{n_r, l}\left(r\right)$ is given by

\[ \begin{align} \rho_{n_r, l}\left(r\right) = \frac{dP}{dr} = R_{n_r, l}\left(r\right)^2r^2. \end{align} \]

The normalization condition therefore reads

\[ \begin{align} \int_{0}^\infty R_{n_r, l}\left(r\right)^2r^2dr\hastobe1. \end{align} \] One sets

\[ \begin{align} \int_{0}^{\infty}U_{n_r, l}\left(r\right)^2dr\hastobe\frac{1}{C_{n_r, l}^2} \end{align} \]

and rescales the integral with $y = \frac{2Zr}{na_B}$:

\[ \begin{align} \int_{0}^{\infty}U_{n_r, l}\left(r\right)^2dr &= \int_{0}^{\infty}y^{2l + 2}L_{n_r, 2l + 1}\left(y\right)^2\exp\left(-y\right)\frac{na_B}{2Z}dy\nonumber\\ &= \frac{na_B}{2Z}\int_{0}^{\infty}y^{2l + 2}L_{n_r, 2l + 1}\left(y\right)^2\exp\left(-y\right)dy. \end{align} \]

For the remaining integral, with Eq. (C.143), one has

\[ \begin{align} & \int_{0}^{\infty}y^{2l + 2}L_{n_r, 2l + 1}\left(y\right)^2\exp\left(-y\right)dy = \frac{\left(n_r + 2l + 1\right)!}{n_r!}\left(2n_r + 2l + 2\right) = 2n\frac{\left(n + l\right)!}{n_r!}. \end{align} \]

It follows

\[ \begin{align} C_{n_r, l} = \sqrt{\frac{Z}{a_B}}\frac{1}{n}\sqrt{\frac{n_r!}{\left(n + l\right)!}}. \end{align} \]

The normalized radial functions therefore read

\[ \begin{align} R_{n, l}\left(r\right) &= \sqrt{\frac{Z^3}{a_B^3}}\sqrt{\frac{\left(n - l - 1\right)!}{\left(n + l\right)!}}\frac{2}{n^2}\left(\frac{2Zr}{na_B}\right)^lL_{n - l - 1, 2l + 1}\left(\frac{2Zr}{na_B}\right)\exp\left(-\frac{Zr}{na_B}\right). \end{align} \]

The hydrogen eigenfunctions $\psi_{n, l, m}\left(r, \theta, \phi\right)$ finally read

Designation of the hydrogen eigenfunctions depending on the angular momentum quantum number $l$. For higher $l$ it continues alphabetically.

$l$	Designation
$0$	s
$1$	p
$2$	d
$3$	f

The probability densities $\left|\psi_{n, l, m}\right|^2$ are also referred to as orbitals. According to the so-called angular momentum quantum number $l$, they are named as given in Tab. 4.1. Different eigenfunctions with the same eigenvalue are called degenerate. From the conditions $n\in\mathbb{N}$, $l<n$ and $\left|m\right|\leq l$ one can determine the multiplicity $N$ of the degeneracy for a given principal quantum number $n$ (the energy depends only on the principal quantum number $n$):

\[ \begin{align} N &= \sum_{l = 0}^{n - 1}\sum_{m = -l}^{l}1 = \sum_{l = 0}^{n - 1}\left(2l + 1\right) = \sum_{l = 1}^{n}\left(2l - 1\right) = n\left(n + 1\right) - n = n^2. \end{align} \]

The Gaussian summation formula, Eq. (A.6), was used. The hydrogen eigenfunctions for the energy eigenvalue $E_n$ are therefore $n^2-$fold degenerate. If the spin is taken into account and relativistic corrections are not taken into account, the degree of degeneracy doubles.

4.9 Angular momentum

The angular momentum $\mathbf{L}$ of a point mass with the momentum $\mathbf{p} = \left(p_1, p_2, p_3\right)^T$ at the position $\mathbf{r} = \left(x_1, x_2, x_3\right)^T$ is defined in classical mechanics by

\[ \begin{align} \mathbf{L} \coloneqq \mathbf{r}\times\mathbf{p} = \sum_{j, k, l = 1}^{3}\epsilon_{j, k, l}x_jp_k\mathbf{e}_l \end{align} \]

Replacing the Cartesian position components in this equation with the corresponding operators

\[ \begin{align} x_j\to\newhat{x}_j \end{align} \]

as well as the Cartesian momentum components through the momentum operators

\[ \begin{align} p_k\to\newhat{p}_k, \end{align} \]

one obtains the angular momentum operator

\[ \begin{align} \newhat{\mathbf{L}} \coloneqq \hbar\sum_{j, k, l = 1}^{3} - i\epsilon_{j, k, l}\newhat{x}_j\frac{\partial}{\partial x_k}\mathbf{e}_l. \end{align} \]

Written out in the position representation in Cartesian coordinates, one obtains

\[ \begin{align} \newhat{L}_x &= \frac{\hbar}{i}\left(y\frac{\partial }{\partial z} - z\frac{\partial}{\partial y}\right),\\ \newhat{L}_y &= \frac{\hbar}{i}\left(z\frac{\partial }{\partial x} - x\frac{\partial}{\partial z}\right),\\ \newhat{L}_z &= \frac{\hbar}{i}\left(x\frac{\partial }{\partial y} - y\frac{\partial}{\partial x}\right). \end{align} \]

The momentum operator $\newhat{\mathbf{p}}$ is Hermitian, just like the position operator $\newhat{\mathbf{r}}$. Since $\newhat{p}_j$ commutes with $\newhat{x}_k$ for $x\not = k$, the successive application $\newhat{p}_j\newhat{x}_k = \newhat{x}_k\newhat{p}_j$ is, according to Sect. 4.6.2, also Hermitian. Thus the angular momentum operators are Hermitian. One defines

\[ \begin{align} \newhat{L}_\pm \coloneqq \newhat{L}_x\pm i\newhat{L}_y. \end{align} \]

As a shorthand, one introduces a generalized angular momentum operator

\[ \begin{align} \newhat{\mathbf{J}} \coloneqq - i\newhat{\mathbf{r}}\times\nabla\tag{4.243}\label{eq:drehmomentum_op_allg} \end{align} \]

Now some commutator relations of the angular momentum operators are to be derived. In the position representation, one has

\[ \begin{align} \newhat{J}_l = -i\sum_{j, k = 1}^{3}\epsilon_{j, k, l}x_j\frac{\partial}{\partial x_k}. \end{align} \]

Thus, for the commutator of two components of the angular momentum operator, one obtains

\[ \begin{align} \left[\newhat{J}_j, \newhat{J}_k\right] &= \newhat{J}_j\newhat{J}_k - \newhat{J}_k\newhat{J}_j = \left(-i\sum_{l, m = 1}^{3}\epsilon_{l, m, j}x_l\frac{\partial}{\partial x_m}\right)\left(-i\sum_{l, m = 1}^{3}\epsilon_{l, m, k}x_l\frac{\partial}{\partial x_m}\right)\nonumber\\ & - \left(-i\sum_{l, m = 1}^{3}\epsilon_{l, m, k}x_l\frac{\partial}{\partial x_m}\right)\left(-i\sum_{l, m = 1}^{3}\epsilon_{l, m, j}x_l\frac{\partial}{\partial x_m}\right)\nonumber\\ &= -\left(\sum_{l, m = 1}^{3}\epsilon_{l, m, j}x_l\frac{\partial}{\partial x_m}\right)\left(\sum_{l, m = 1}^{3}\epsilon_{l, m, k}x_l\frac{\partial}{\partial x_m}\right) + \left(\sum_{l, m = 1}^{3}\epsilon_{l, m, k}x_l\frac{\partial}{\partial x_m}\right)\left(\sum_{l, m = 1}^{3}\epsilon_{l, m, j}x_l\frac{\partial}{\partial x_m}\right)\nonumber\\ &= -\sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, j}\epsilon_{n, o, k}x_l\frac{\partial}{\partial x_m}x_n\frac{\partial}{\partial x_o} + \sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, k}\epsilon_{n, o, j}x_l\frac{\partial }{\partial x_m}x_n\frac{\partial}{\partial x_o}\nonumber\\ &= -\sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, j}\epsilon_{n, o, k}x_lx_n\frac{\partial^2}{\partial x_o\partial x_m} + \sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, k}\epsilon_{n, o, j}x_lx_n\frac{\partial^2}{\partial x_o\partial x_m}\nonumber\\ & - \sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, j}\epsilon_{n, o, k}x_l\delta_{n, m}\frac{\partial}{\partial x_o} + \sum_{l, m, n, o = 1}^{3}\epsilon_{l, m, k}\epsilon_{n, o, j}x_l\delta_{mn}\frac{\partial}{\partial x_o}\nonumber\\ &= \sum_{l, m, n, o = 1}^{3}x_lx_n\left(\epsilon_{l, m, k}\epsilon_{n, o, j} - \epsilon_{l, m, j}\epsilon_{n, o, k}\right)\frac{\partial^2}{\partial x_o\partial x_m} - \sum_{l, n, o = 1}^{3}\epsilon_{l, n, j}\epsilon_{n, o, k}x_l\frac{\partial}{\partial x_o} + \sum_{l, n, o = 1}^{3}\epsilon_{l, n, k}\epsilon_{n, o, j}x_l\frac{\partial}{\partial x_o}\nonumber\\ &= \sum_{l, n, o = 1}^{3}x_l\frac{\partial}{\partial x_o}\left(\epsilon_{l, n, k}\epsilon_{n, o, j} - \epsilon_{l, n, j}\epsilon_{n, o, k}\right) = \sum_{n = 1}^{3}\sum_{l, o = 1}^{3}\left(\epsilon_{l, n, k}\epsilon_{n, o, j} - \epsilon_{l, n, j}\epsilon_{n, o, k}\right)x_l\frac{\partial}{\partial x_o}\nonumber\\ &= \sum_{n = 1}^{3}\sum_{l, o = 1}^{3}\epsilon_{j, k, n}\epsilon_{l, o, n}x_l\frac{\partial}{\partial x_o} = \sum_{n = 1}^{3}\epsilon_{j, k, n}\sum_{l, o = 1}^3\epsilon_{l, o, n}x_l\frac{\partial}{\partial x_o} = \sum_{n = 1}^{3}\epsilon_{j, k, n}\frac{1}{ - i}\newhat{J}_n = i\sum_{n = 1}^{3}\epsilon_{j, k, n}\newhat{J}_n.\tag{4.245}\label{eq:kommutator_angular_momentum_op} \end{align} \]

Here Eq. (A.24) was used. This means, in particular,

\[ \begin{align} \left[\newhat{J}_1, \newhat{J}_2\right] = i\newhat{J}_3, & {} & \left[\newhat{J}_2, \newhat{J}_3\right] = i\newhat{J}_1, & {} & \left[\newhat{J}_3, \newhat{J}_1\right] = i\newhat{J}_2. \end{align} \]

Thus,

\[ \begin{align} \left[\newhat{J}_3, \newhat{J}_\pm\right] &= \left[\newhat{J}_3, \newhat{J}_1\pm i\newhat{J}_2\right] = \left[\newhat{J}_3, \newhat{J}_1\right]\pm\left[\newhat{J}_3, i\newhat{J}_2\right] = i\newhat{J}_2\pm i\left(-i\newhat{J}_1\right) = \pm \newhat{J}_1 + i\newhat{J}_2 = \pm\newhat{J}_\pm.\tag{4.247}\label{eq:komm_aufsteige_1} \end{align} \]

Furthermore, one has

\[ \begin{align} \newhat{J}_\pm\newhat{J}_\mp &= \left(\newhat{J}_1\pm i\newhat{J}_2\right)\left(\newhat{J}_1\mp i\newhat{J}_2\right) = \newhat{J}_1^2 + \newhat{J}_2^2 \mp i\newhat{J}_1\newhat{J}_2\pm i\newhat{J}_2\newhat{J}_1\nonumber\\ &= \newhat{\mathbf{J}}^2 - \newhat{J}_3^2\mp i\left[\newhat{J}_1, \newhat{J}_2\right] = \newhat{\mathbf{J}}^2 - \newhat{J}_3\left(\newhat{J}_3\mp 1\right)\tag{4.248}\label{eq:komm_aufsteige_2} \end{align} \]

and

\[ \begin{align} \left[\newhat{\mathbf{J}}^2, \newhat{J}_1\right] &= \left[\newhat{J}_2^2 + \newhat{J}_3^2, \newhat{J}_1\right] = \newhat{J}_2\newhat{J}_2\newhat{J}_1 - \newhat{J}_1\newhat{J}_2\newhat{J}_2 + \newhat{J}_3\newhat{J}_3\newhat{J}_1 - \newhat{J}_1\newhat{J}_3\newhat{J}_3\nonumber\\ &= \newhat{J}_2\newhat{J}_2\newhat{J}_1 - \newhat{J}_1\newhat{J}_2\newhat{J}_2 + \newhat{J}_2\newhat{J}_1\newhat{J}_2 - \newhat{J}_2\newhat{J}_1\newhat{J}_2 + \newhat{J}_3\newhat{J}_3\newhat{J}_1 - \newhat{J}_1\newhat{J}_3\newhat{J}_3 + \newhat{J}_3\newhat{J}_1\newhat{J}_3 - \newhat{J}_3\newhat{J}_1\newhat{J}_3\nonumber\\ &= \newhat{J}_2\left[\newhat{J}_2, \newhat{J}_1\right] + \left[\newhat{J}_2, \newhat{J}_1\right]\newhat{J}_2 + \newhat{J}_3\left[\newhat{J}_3, \newhat{J}_1\right] + \left[\newhat{J}_3, \newhat{J}_1\right]\newhat{J}_3\nonumber\\ &= -i\newhat{J}_2\newhat{J}_3 - i\newhat{J}_3\newhat{J}_2 + i\newhat{J}_3\newhat{J}_2 + i\newhat{J}_2\newhat{J}_3 = 0, \tag{4.249}\label{eq:komm_tot_angular_momentum} \end{align} \]

analogously for $\newhat{J}_2$ and $\newhat{J}_3$ and thus also

\[ \begin{align} \left[\newhat{\mathbf{J}}^2, \newhat{J}_\pm\right] = 0. \end{align} \]

To deal with the angular momentum in the H atom, the angular momentum operators are now to be transformed into spherical coordinates. For this purpose, in addition to Eq. (B.91) the representation

\[ \begin{align} \nabla = \mathbf{e}_r\frac{\partial}{\partial r} + \mathbf{e}_\theta\frac{1}{r}\frac{\partial}{\partial\theta} + \mathbf{e}_\phi\frac{1}{r\sin\left(\theta\right)}\frac{\partial}{\partial\phi}. \end{align} \]

of the gradient. Furthermore, the well-known relations hold

\[ \begin{align} \mathbf{e}_r\times\mathbf{e}_\theta = \mathbf{e}_\phi \end{align} \]

as well as

\[ \begin{align} \mathbf{e}_r\times\mathbf{e}_\phi = -\mathbf{e}_\theta. \end{align} \]

Thus, in the position representation, one obtains

\[ \begin{align} \newhat{\mathbf{J}} &= -i\mathbf{r}\times\nabla = -ir\mathbf{e}_r\times\left( \mathbf{e}_r\frac{\partial}{\partial r} + \mathbf{e}_\theta\frac{1}{r}\frac{\partial}{\partial\theta} + \mathbf{e}_\phi\frac{1}{r\sin\left(\theta\right)}\frac{\partial}{\partial\phi}\right) = -i\mathbf{e}_\phi\frac{\partial}{\partial\theta} + i\mathbf{e}_\theta\frac{1}{\sin\left(\theta\right)}\frac{\partial}{\partial\phi}. \end{align} \]

By virtue of Eqs. (B.74) - (B.76), one has

\[ \begin{align} \newhat{J}_x &= -i\left(-\sin\left(\phi\right)\frac{\partial}{\partial\theta} - \cos\left(\phi\right)\cot\left(\theta\right)\frac{\partial}{\partial\phi}\right)\tag{4.255}\label{eq:drehmomentum_op_x},\\ \newhat{J}_y &= -i\left(\cos\left(\phi\right)\frac{\partial}{\partial\theta} - \sin\left(\phi\right)\cot\left(\theta\right)\frac{\partial}{\partial\phi}\right), \tag{4.256}\label{eq:eq:drehmomentum_op_y}\\ \newhat{J}_z &= -i\frac{\partial}{\partial \phi}\tag{4.257}\label{eq:eq:drehmomentum_op_z}. \end{align} \]

Furthermore, one has

\[ \begin{align} \newhat{\mathbf{J}}^2 &= \newhat{J}_x^2 + \newhat{J}_y^2 + \newhat{J}_z^2\nonumber\\ &= -\sin^2\left(\phi\right)\frac{\partial^2}{\partial\theta^2} - \cos^2\left(\phi\right)\cot^2\left(\theta\right)\frac{\partial^2}{\partial\phi^2} - \cos^2\left(\phi\right)\frac{\partial^2}{\partial\theta^2} - \sin^2\left(\phi\right)\cot^2\left(\theta\right)\frac{\partial^2}{\partial\phi^2} - \frac{\partial^2}{\partial\phi^2} - \cot\left(\theta\right)\frac{\partial}{\partial\theta}\nonumber\\ &= -\left(\frac{\partial^2}{\partial\theta^2} + \cot^2\left(\theta\right)\frac{\partial^2}{\partial\phi^2} + \frac{\partial^2}{\partial\phi^2} + \cot\left(\theta\right)\frac{\partial}{\partial\theta}\right) = -\left(\frac{1}{\sin\left(\theta\right)}\frac{\partial}{\partial\theta}\left(\sin\left(\theta\right)\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\left(\theta\right)}\frac{\partial^2}{\partial\phi^2}\right). \end{align} \]

By Eq. (B.96), this corresponds to the statement

\[ \begin{align} \newhat{\mathbf{J}}^2 = -\Delta_{\theta, \phi}.\tag{4.259}\label{eq:drehmomentum_op_quadrat} \end{align} \]

Let $|n, l, m\rangle$ be the state of an electron in the hydrogen atom with quantum numbers $n$, $l$ and $m$. By virtue of the position representation of $|n, l, m\rangle$, Eq. (4.233), the definition of the spherical harmonics, Eq. (C.155), the representation of the z-component of the angular momentum operator, Eq. (4.257), the just-derived Eq. (4.259), and the property Eq. (C.166) of the spherical harmonics, the following two statements hold:

\[ \begin{align} \newhat{\mathbf{L}}^2|n, l, m\rangle &= \hbar^2l\left(l + 1\right)|n, l, m\rangle,\\ \newhat{L}_z|n, l, m\rangle &= m\hbar|n, l, m\rangle \end{align} \]

From this, the following follow:

• The total angular momentum $L$ in state $|n, l, m\rangle$ is $\hbar\sqrt{l\left(l + 1\right)}$. Therefore $l$ is called the angular momentum quantum number.

• The z-component of the angular momentum $L_z$ in the state $|n, l, m\rangle$ is $L_z = m\hbar$. $m$ is called the magnetic quantum number.

• $\newhat{\mathbf{L}}^2$, $\newhat{L_z}$ and $\newhat{H}$ commute pairwise. Therefore, $L^2$ and $L_z$ are conserved quantities and since the hydrogen eigenstates depend on three quantum numbers, these three operators in the hydrogen atom are complete observables regardless of spin.

Furthermore, one finds

\[ \begin{align} \newhat{J}_\pm &= \newhat{J}_x\pm i\newhat{J}_y = -i\left(-\sin\left(\phi\right)\frac{\partial}{\partial\theta} - \cos\left(\phi\right)\cot\left(\theta\right)\frac{\partial}{\partial\phi}\right)\pm\left(\cos\left(\phi\right)\frac{\partial}{\partial\theta} - \sin\left(\phi\right)\cot\left(\theta\right)\frac{\partial}{\partial\phi}\right)\nonumber\\ &= \left(i\cos\left(\phi\right) - \pm\sin\left(\phi\right)\right)\cot\left(\theta\right)\frac{\partial}{\partial\phi} + \left(\pm\cos\left(\phi\right) + i\sin\left(\phi\right)\right)\frac{\partial}{\partial\theta}\nonumber\\ &= ie^{\pm i\phi}\cot\left(\theta\right)\frac{\partial}{\partial\phi} \pm e^{\pm i\phi}\frac{\partial}{\partial\theta} = \exp\left(\pm i\phi\right)\left(i\cot\left(\theta\right)\frac{\partial}{\partial\phi}\pm\frac{\partial}{\partial\theta}\right). \end{align} \]

Now the generalized angular momentum operator $\newhat{\mathbf{J}}$, Eq. (4.243), is treated a little further. It and its components were identified as Hermitian, so $\newhat{\mathbf{J}}^2$, as a successive application of Hermitian commuting operators, is also Hermitian according to Sect. 4.6.2. Furthermore, one has

\[ \begin{align} \newhat{J}_+^+ = \newhat{J}_-, & {} & \newhat{J}_-^+ = \newhat{J}_+. \end{align} \]

Among the four operators $\newhat{\mathbf{J}}^2, \newhat{J}_x, \newhat{J}_y, \newhat{J}_z$, by Eqs. (4.245) and (4.249), one can find two that commute, namely $\newhat{\mathbf{J}}^2$ and one other, for which $\newhat{J}_z$ is chosen. According to Sect. 4.6.3, these have the same eigenfunctions, so one can write

\[ \begin{align} \newhat{\mathbf{J}}^2\left|\lambda, m\right\rangle = \lambda\left|\lambda, m\right\rangle, & {} & \newhat{J}_z\left|\lambda, m\right\rangle = m\left|\lambda, m\right\rangle \end{align} \]

with real eigenvalues $\lambda, m\in \mathbb{R}$. Now one examines the states

\[ \begin{align} \newhat{J}_+\left|\lambda, m\right\rangle \end{align} \]

and

\[ \begin{align} \newhat{J}_-\left|\lambda, m\right\rangle \end{align} \]

Since $\newhat{\mathbf{J}}^2$ commutes with $\newhat{J}_\pm$, one has

\[ \begin{align} \newhat{\mathbf{J}}^2\newhat{J}_\pm\left|\lambda, m\right\rangle = \newhat{J}_\pm\newhat{\mathbf{J}}^2\left|\lambda, m\right\rangle = \lambda\newhat{J}_\pm \left|\lambda, m\right\rangle. \end{align} \]

With Eq. (4.247) follows

\[ \begin{align} \newhat{J}_z\newhat{J}_\pm - \newhat{J}_\pm\newhat{J}_z = \pm\newhat{J}_\pm\Leftrightarrow \newhat{J}_z\newhat{J}_\pm = \newhat{J}_\pm\newhat{J}_z\pm\newhat{J}_\pm \end{align} \]

and thus

\[ \begin{align} \newhat{J}_z\newhat{J}_\pm\left|\lambda, m\right\rangle = \left(\newhat{J}_\pm\newhat{J}_z\pm\newhat{J}_\pm\right)\left|\lambda, m\right\rangle = \newhat{J}_\pm\left(\newhat{J}_z\pm 1\right)\left|\lambda, m\right\rangle = \left(m\pm 1\right)\newhat{J}_\pm\left|\lambda, m\right\rangle. \end{align} \]

Thus one has

\[ \begin{align} \newhat{J}_\pm\left|\lambda, m\right\rangle = c_\pm\left|\lambda, m\pm 1\right\rangle.\tag{4.270}\label{eq:auf_ab_op_allg} \end{align} \]

with $c_\pm\in \mathbb{R}$. Therefore, $\newhat{J}_+$ is called the raising operator and $\newhat{J}_-$ the lowering operator. One now has to calculate the normalization constants $c_\pm$. To this end, one starts from the normalization

\[ \begin{align} \left\langle\lambda, m|\lambda, m\right\rangle \hastobe 1 \end{align} \]

of the states, which one can simply require. One thus obtains

\[ \begin{align} c_\pm^2 = \left\langle \lambda, m\pm 1\left|c_\pm c_\pm\right|\lambda, m\pm 1\right\rangle = \left\langle\newhat{J}_\pm\lambda, m\left|\newhat{J}_\pm\right|\lambda, m\right\rangle = \left\langle\lambda, m\left|\newhat{J}_\mp\newhat{J}_\pm\right|\lambda, m\right\rangle. \end{align} \]

Here one can insert Eq. (4.248) and obtains, with the requirement $c_\pm > 0$,

\[ \begin{align} c_\pm^2 &= \left\langle\lambda, m\left|\newhat{\mathbf{J}}^2 - \newhat{J}_z^2\mp\newhat{J}_z\right|\lambda, m\right\rangle = \lambda - m^2\mp m\nonumber\\ \Leftrightarrow c_\pm &= \sqrt{\lambda - m^2\mp m} = \sqrt{\lambda - m\left(m \pm 1\right)}.\tag{4.273}\label{eq:dreh_allg_deriv_norm} \end{align} \]

Now one still has to derive a condition for $\lambda$ and $m$. For a Hermitian operator $\newhat{A}$, one has

\[ \begin{align} \left\langle f\big|\newhat{A}^2f\right\rangle = \left\langle\newhat{A}f\big|\newhat{A}f\right\rangle \geq 0. \end{align} \]

From this it follows

\[ \begin{align} 0\leq \left\langle\lambda, m\left|\newhat{J}_x^2\right|\lambda, m\right\rangle + \left\langle\lambda, m\left|\newhat{J}_y^2\right|\lambda, m\right\rangle = \left\langle\lambda, m\left|\newhat{\mathbf{J}}^2 - \newhat{J}_z^2\right|\lambda, m\right\rangle = \lambda - m^2. \end{align} \]

Consequently, one has

\[ \begin{align} \lambda \geq m^2 \geq 0.\tag{4.276}\label{eq:deriv_allg_angular__1} \end{align} \]

From a state $|\lambda, m\rangle$ one obtains the states $|\lambda, m\pm 1\rangle$, $|\lambda, m\pm 2\rangle$ and so on by applying the ascending and descending operators. However, this must terminate somewhere because of (4.276), so the normalization, Eq. (4.273), must vanish at a maximum value $m_{{\mathrm{max}}}$ and at a minimum value $m_{{\mathrm{min}}}$:

\[ \begin{align} \newhat{J}_ + \left|\lambda, m_{\text{max}}\right\rangle = 0\Rightarrow c_ + = 0\Rightarrow \lambda &= m_{\text{max}}\left(m_{\text{max}} + 1\right),\\ \newhat{J}_ - \left|\lambda, m_{\text{min}}\right\rangle = 0\Rightarrow c_ - = 0\Rightarrow\lambda &= m_{\text{min}}\left(m_{\text{min}} - 1\right)\\ \Rightarrow\left(m_{\text{max}} + m_{\text{min}}\right)\left(m_{\text{max}} - m_{\text{min}} + 1\right) &= 0 \end{align} \]

Because $m_{\mathrm{max}}\geq m_{\mathrm{min}}$ the second bracket is not equal to zero, so the first bracket is equal to zero:

\[ \begin{align} m_{\text{max}} = -m_{\text{min}}\eqqcolon j \end{align} \]

One has

\[ \begin{align} \lambda = j\left(j + 1\right). \end{align} \]

The raising operator thus leads from $-j$ to $j$ in integer steps. Hence $2j$ is an integer, so $j$ is an integer or half-integer. One now introduces a change of notation

\[ \begin{align} \left|\lambda, m\right\rangle = \left|j\left(j + 1\right), m\right\rangle\to \left|j, m\right\rangle \end{align} \]

These states are normalized; furthermore, as eigenstates of Hermitian operators belonging to different eigenvalues, they are orthogonal:

\[ \begin{align} \left\langle j', m'\big|j, m\right\rangle &= \delta_{j, j'}\delta_{m, m'}. \end{align} \]

If $j$ is an integer, the position representation of the solution is given by the spherical harmonics, Eq. (C.155). This describes, for example, the orbital angular momentum of particles in a central force field, see Sect. 4.8. The case that $j$ is half-integer is important for particles with half-integer spin. Particles with half-integer spin are called Fermions, while particles with integer spin are called Bosons. Examples of fermions are electrons, protons and neutrons, photons are bosons.

From now on, half-integer spin is assumed, namely $j = \frac{1}{2}$. In this case, the states $\left|j, m\right\rangle$ are denoted by $\left|ss_z\right\rangle = \left|\frac{1}{2}, s_z\right\rangle$. One defines

\[ \begin{align} \left(\begin{array}{c} 1\\ 0 \end{array}\right) \coloneqq \left|\frac{1}{2}, \frac{1}{2}\right\rangle, \tag{4.284}\label{eq:spin_up}\\ \left(\begin{array}{c} 0\\ 1 \end{array}\right) \coloneqq \left|\frac{1}{2}, - \frac{1}{2}\right\rangle\tag{4.285}\label{eq:spin_down}. \end{align} \]

Eq. (4.284) is called the spin-up state and Eq. (4.285) the spin-down state. One has $\newhat{J}_z\left|\frac{1}{2}, m\right\rangle = m\left|\frac{1}{2}, m\right\rangle$, so in matrix notation it follows

\[ \begin{align} J_z\left(\begin{array}{c} 1\\ 0 \end{array}\right) = \left(\begin{array}{c} \frac{1}{2}\\ 0 \end{array}\right), & {} & J_z\left(\begin{array}{c} 0\\ 1 \end{array}\right) = \left(\begin{array}{c} 0\\ - \frac{1}{2} \end{array}\right), \end{align} \]

which is satisfied by

\[ \begin{align} J_z = \frac{1}{2}\left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right). \end{align} \]

For the normalization constant $c_\pm$ from Eq. (4.273), with $\lambda = j\left(j + 1\right) = \frac{1}{2}\frac{3}{2} = \frac{3}{4}$, one has

\[ \begin{align} c_ + &= \sqrt{\frac{3}{4} + \frac{1}{2}\left(-\frac{1}{2} + 1\right)} = 1,\\ c_ - &= \sqrt{\frac{3}{4} - \frac{1}{2}\left(\frac{1}{2} - 1\right)} = 1. \end{align} \]

Therefore, because of $\newhat{J}_ + \left|\frac{1}{2}, \frac{1}{2}\right\rangle = 0$ and $\newhat{J}_ + \left|\frac{1}{2}, - \frac{1}{2}\right\rangle = \left|\frac{1}{2}, \frac{1}{2}\right\rangle$, one also has

\[ \begin{align} J_ + = \left(\begin{array}{cc} 0&1\\ 0&0 \end{array}\right). \end{align} \]

Because $\newhat{J}_ - \left|\frac{1}{2}, \frac{1}{2}\right\rangle = \left|\frac{1}{2}, - \frac{1}{2}\right\rangle$ and $\newhat{J}_ - \left|\frac{1}{2}, - \frac{1}{2}\right\rangle = 0$, one has

\[ \begin{align} J_ - = \left(\begin{array}{cc} 0&0\\ 1&0 \end{array}\right). \end{align} \]

Thus, one has

\[ \begin{align} J_x &= \frac{1}{2}\left(J_ + + J_ -\right) = \frac{1}{2}\left(\begin{array}{cc} 0&1\\ 1&0 \end{array}\right),\\ J_y &= -i\frac{1}{2}\left(J_ + - J_ -\right) = \frac{1}{2}\left(\begin{array}{cc} 0& -i\\ i&0 \end{array}\right). \end{align} \]

The matrices appearing here

\[ \begin{align} \sigma_x \coloneqq \left(\begin{array}{cc} 0&1\\ 1&0 \end{array}\right), & {} & \sigma_y \coloneqq \left(\begin{array}{cc} 0& -i\\ i&0 \end{array}\right), & {} & \sigma_z \coloneqq \left(\begin{array}{cc} 1&0\\ 0& -1 \end{array}\right) \end{align} \]

are also called Pauli matrices or spin matrices.

4.10 Spin

The question arises as to whether the half-integer states have any physical relevance at all or are just purely mathematical phenomena. In fact, these states describe the intrinsic angular momenta of the elementary particles, namely the spin. Experimentally, one can only change the direction, but not the magnitude, of the spin, which is a particle property, just like charge and mass.

Now the Schrödinger equation is to be generalized to a particle with mass $m$, charge $q$ and spin $1/2$. If one is now given a particle with charge $q$ and spin $1/2$, a simple wave function $\mathbb{R}^3\to\mathbb{C}$ is no longer sufficient to describe the state of this particle. The probability density no longer depends only on the position, but also on the spin state. Using the definitions, Eqs. (4.284) and (4.285), one can, however, write for the state of such a particle

\[ \begin{align} \left|\psi\left(\mathbf{r}, t\right)\right\rangle = \varphi_ + \left(\mathbf{r}, t\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right) + \varphi_ - \left(\mathbf{r}, t\right)\left(\begin{array}{c} 0\\ 1 \end{array}\right)\tag{4.295}\label{eq:spinor} \end{align} \]

with position- and time-dependent components $\varphi_ + \left(\mathbf{r}, t\right)$ and $\varphi_ - \left(\mathbf{r}, t\right)$. Such a wave function is called a spinor. This state is of course normalized:

\[ \begin{align} \langle\psi|\psi\rangle = \int_{\mathbb{R}^3}\left(\left|\varphi_ + \left(\mathbf{r}\right)\right|^2 + \left|\varphi_ - \left(\mathbf{r}\right)\right|^2\right)d^3r \end{align} \]

Using Eq. (3.23), one substitutes

\[ \begin{align} \newhat{\mathbf{p}}\to \newhat{\mathbf{p}} - \frac{q}{c}\newhat{\mathbf{A}}, \end{align} \]

to take into account the energy of a particle in the electromagnetic field. Furthermore, the potential energy term must be expanded. A magnetic moment $\mubi$ has potential energy in a B field $\mathbf{B}$

\[ \begin{align} E_{\text{pot}} = -\mubi\cdot\mathbf{B}. \end{align} \]

Since an elementary particle knows only one distinguished direction in its rest frame, namely that of the spin, the magnetic moment must be parallel to the spin. The spin is replaced by the spin operator

\[ \begin{align} \mathbf{s}\to\newhat{\mathbf{s}}. \end{align} \]

For particles with spin $1/2$ this can be written using the operators $\newhat{\sigma}_x, \newhat{\sigma}_y, \newhat{\sigma}_z$ defined by the Pauli matrices as

\[ \begin{align} \newhat{\mathbf{s}} = \frac{\hbar}{2}\newhat{\sigmabi} = \frac{\hbar}{2}\sigmabi. \end{align} \]

Here, a vector

\[ \begin{align} \sigmabi = \sigma_x\mathbf{e}_x + \sigma_y\mathbf{e}_y + \sigma_z\mathbf{e}_z\in\mathbb{C}^{2\times 2\times 3} \end{align} \]

was introduced. If $\phi$ is the scalar potential of the electromagnetic field, then the desired Hamiltonian is

\[ \begin{align} \newhat{H} = \frac{1}{2m}\left(\newhat{\mathbf{p}} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi - \newhat{\mubi}\cdot\mathbf{B}. \end{align} \]

For the magnetic moment, one has

\[ \begin{align} \mubi = g\frac{q}{2mc}\mathbf{s}\tag{4.303}\label{eq:def_mag_mom_e} \end{align} \]

with a factor $g$ that cannot be derived classically, called the Landé factor. For the Pauli-Hamilton operator, one obtains

\[ \begin{align} \newhat{H}_P = \frac{1}{2m}\left(\newhat{\mathbf{p}} - \frac{q}{c}\mathbf{A}\right)^2 + q\phi - g\frac{q\hbar}{4mc}\sigmabi\cdot\mathbf{B}.\tag{4.304}\label{eq:pauli_hamiltonian} \end{align} \]

The Pauli equation therefore reads

\[ \begin{align} \newhat{H}_P\left|\psi\left(\mathbf{r}, t\right)\right\rangle = i\hbar\frac{\partial}{\partial t}\left|\psi\left(\mathbf{r}, t\right)\right\rangle. \end{align} \]

for a spinor $\left|\psi\left(\mathbf{r}, t\right)\right\rangle$, see Eq. (4.295).

4.11 Perturbation theory

4.11.1 Time-independent case

Let an eigenvalue problem

\[ \begin{align} \newhat{H}_0\left|\psi^{(0)}\right\rangle = E^{(0)}\left|\psi^{(0)}\right\rangle \end{align} \]

be given, with a Hamiltonian $\newhat{H}_0$, orthonormal and complete eigenfunctions $\left|\psi_n^{(0)}\right\rangle$, and associated eigenvalues $E_n^{(0)}$. Now one replaces the Hamiltonian $\newhat{H}_0$ by a new Hamiltonian $\newhat{H}$, given by

\[ \begin{align} \newhat{H} \coloneqq\newhat{H}_0 + \newhat{v}_h \end{align} \]

with a perturbation potential $\newhat{v}_h$. One seeks solutions $\left|\psi\right\rangle$ of the eigenvalue equation

\[ \begin{align} \newhat{H}\left|\psi\right\rangle = E\left|\psi\right\rangle\tag{4.308}\label{eq:ew_gestoert} \end{align} \]

To this end, one first makes the ansatz

\[ \begin{align} \newhat{H}\left(\lambda\right) \coloneqq \newhat{H}_0 + \lambda\newhat{v}_h \end{align} \]

with $\lambda\in\left[0, 1\right]$. Now one expands the eigenstates $\left|\psi\right\rangle$ and eigenvalues $E$ in powers of $\lambda$:

\[ \begin{align} E\left(\lambda\right) = \sum_{i = 0}^{\infty}\lambda^{(i)}E^{(i)}, & {} & \left|\psi\left(\lambda\right)\right\rangle = \sum_{i = 0}^{\infty}\lambda^{(i)}\left|\psi^{(i)}\right\rangle \end{align} \]

This is substituted into Eq. (4.308):

\[ \begin{align} \newhat{H}\left(\lambda\right)\left|\psi\left(\lambda\right)\right\rangle &= E\left(\lambda\right)\left|\psi\left(\lambda\right)\right\rangle\tag{4.311}\label{eq:sg_gestoert}\\ \Rightarrow\left(\newhat{H}_0 + \lambda\newhat{v}_h\right)\sum_{i = 0}^{\infty}\lambda^{(i)}\left|\psi^{(i)}\right\rangle &= \left(\sum_{i = 0}^{\infty}\lambda^{(i)}E^{(i)}\right)\left(\sum_{i = 0}^{\infty}\lambda^{(i)}\left|\psi^{(i)}\right\rangle\right)\tag{4.312}\label{eq:zeitunabh_stoerung_allg} \end{align} \]

One now sorts Eq. (4.312) by powers of $\lambda$; for the $m-$th order, one then has

\[ \begin{align} \newhat{H}_0\left|\psi^{(m)}\right\rangle + \newhat{v}_h\left|\psi^{(m - 1)}\right\rangle = \sum_{k = 0}^{m}E^{(m - k)}\left|\psi^{(k)}\right\rangle.\tag{4.313}\label{eq:zeitunabh_stoer_nte} \end{align} \]

There exist $C_{n, k}^{(1)}\in \mathbb {C}$ with

\[ \begin{align} \left|\psi_n^{(1)}\right\rangle = \sum_{k = 1}^{\infty}C_{n, k}^{(1)}\left|\psi_k^{(0)}\right\rangle. \end{align} \]

Applying the unperturbed Hamiltonian $\newhat{H}_0$ to this, one obtains

\[ \begin{align} \newhat{H}_0\left|\psi_n^{(1)}\right\rangle = \sum_{k = 1}^{\infty}C_{n, k}^{(1)}E_k^{(0)}\left|\psi_k^{(0)}\right\rangle. \end{align} \]

Substituting $m = 1$ and $\left|\psi\right\rangle = \left|\psi_n\right\rangle$ into Eq. (4.313) and multiplying this from the left by $\left\langle\psi_n^{(0)}\right|$, one obtains

\[ \begin{align} \left\langle\psi_n^{(0)}\left|\newhat{H}_0\right|\psi_n^{(1)}\right\rangle + \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle &= E_n^{(1)} + \left\langle\psi_n^{(0)}\left|E_n^{(0)}\right|\psi_n^{(1)}\right\rangle. \end{align} \]

Moreover, one has

\[ \begin{align} \left\langle\psi_n^{(0)}\left|\newhat{H}_0\right|\psi_n^{(1)}\right\rangle = C_{n, n}^{(1)}E_n^{(0)} = \left\langle\psi_n^{(0)}\left|E_n^{(0)}\right|\psi_n^{(1)}\right\rangle. \end{align} \]

Thus, for the first-order energy correction, one has

\[ \begin{align} E_n^{(1)} = \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle. \end{align} \]

Now the $\left|\psi_n^{(1)}\right\rangle$ are sought, i.e. the $C_{n, k}^{(1)}, k\in \mathbb{N}$ with $k\not = n$. Multiply Eq. (4.313) with $m = 2$ from the left by $\left\langle\psi_k^{(0)}\right|$. This yields

\[ \begin{align} \left\langle\psi_k^{(0)}\left|\newhat{H}_0\right|\psi_n^{(1)}\right\rangle + \left\langle\psi_k^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle &= \left\langle\psi_k^{(0)}\left|E_n^{(0)}\right|\psi_n^{(1)}\right\rangle\nonumber\\ \Leftrightarrow C_{n, k}^{(1)}E_k^{(0)} + \left\langle\psi_k^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle &= E_n^{(0)}C_{n, k}^{(1)}\nonumber\\ \Leftrightarrow C_{n, k}^{(1)} &= \frac{\left\langle\psi_k^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle}{E_n^{(0)} - E_k^{(0)}}. \end{align} \]

This only works in the case of non-degenerate energy levels $E_n^{(0)}, E_k^{(0)}$. For the state $\left|\psi_n\left(\lambda\right)\right\rangle$, one now obtains

\[ \begin{align} \left|\psi_n\left(\lambda\right)\right\rangle = \left|\psi_n^{(0)}\right\rangle + \lambda\sum_{\substack{m = 1,\\m\not = n}}^{\infty}\frac{\left\langle\psi_m^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle}{E_n^{(0)} - E_m^{(0)}}\left|\psi_m^{(0)}\right\rangle + C_{n, n}^{(1)}\left|\psi_n^{(0)}\right\rangle + \mathcal{O}\left(\lambda^2\right). \end{align} \]

The norm of this is

\[ \begin{align} \left\langle\psi_n\left(\lambda\right)|\psi_n\left(\lambda\right)\right\rangle &= 1 + C_{n, n}^{(1)\star} + C_{n, n}^{(1)} + \mathcal{O}\left(\lambda^2\right) \end{align} \]

Hence $C_{n, n}^{(1)} = ir$ with $r\in \mathbb{R}$. Therefore, the amplitude of $\left|\psi_n^{(0)}\right\rangle$ in $\left|\psi_n\left(\lambda\right)\right\rangle$ is given by

\[ \begin{align} 1 + i\lambda r = \exp\left(i\lambda r\right) + \mathcal{O}\left(\lambda ^2\right). \end{align} \]

The coefficient $C_{n, n}^{(1)}$ therefore leads, to first order, to a phase shift of $\left|\psi_n^{(0)}\right\rangle$; this phase shift can be absorbed into $\left|\psi_n^{(0)}\right\rangle$, and one may set

\[ \begin{align} C_{n, n}^{(1)} = 0 \end{align} \]

Thus, to first order, in the case of non-degenerate energy levels, one obtains

\[ \begin{align} E_n &= E_n^{(0)} + E_n^{(1)} = E_n^{(0)} + \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle,\\ \left|\psi_n\right\rangle &= \left|\psi_n^{(0)}\right\rangle + \left|\psi_n^{(1)}\right\rangle = \left|\psi_n^{(0)}\right\rangle + \sum_{\substack{m = 1,\\m\not = n}}^{\infty}\frac{\left\langle\psi_m^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle}{E_n^{(0)} - E_m^{(0)}}\left|\psi_{m}^{(0)}\right\rangle.\tag{4.325}\label{eq:stoerung_zeitun_erste_ordnung_zustand} \end{align} \]

A necessary condition for meaningful applicability of first-order perturbation theory is that the admixtures of states are small.

You can already see the advantage of perturbation theory: while stationary problems in quantum mechanics are eigenvalue problems in which one has to determine the eigenvalue and eigenvector simultaneously, in perturbation theory one can compute the energy corrections and the state corrections successively, one after another.

To compute the second order, one first substitutes $\left|\psi\right\rangle = \left|\psi_n\right\rangle$ with $m = 2$ into Eq. (4.313) and multiplies this from the left by $\left\langle \psi_n^{(0)}\right|$:

\[ \begin{align} \left\langle\psi_n^{(0)}\left|\newhat{H}_0\right|\psi_n^{(2)}\right\rangle + \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_n^{(1)}\right\rangle = E_n^{(2)} + \left\langle\psi_n^{(0)}\left|E_n^{(1)}\right|\psi_n^{(1)}\right\rangle + \left\langle\psi_n^{(0)}\left|E_n^{(0)}\right|\psi_n^{(2)}\right\rangle \end{align} \]

For the $\left|\psi_n^{(2)}\right\rangle$, one now makes the ansatz

\[ \begin{align} \left|\psi_n^{(2)}\right\rangle = \sum_{m = 1}^{\infty}C_{n, m}^{(2)}\left|\psi_m^{(0)}\right\rangle. \end{align} \]

From this it follows

\[ \begin{align} \left\langle\psi_n^{(0)}\newvline\newhat{H}_0\sum_{m = 1}^{\infty}C_{n, m}^{(2)}\left|\psi_m^{(0)}\right\rangle\right\rangle = \left\langle\psi_n^{(0)}\newvline\sum_{m = 1}^{\infty}C_{n, m}^{(2)}E_m^{(0)}\left|\psi_m^{(0)}\right\rangle\right\rangle = C_{n, n}^{(2)}E_n^{(0)}. \end{align} \]

Furthermore, one has

\[ \begin{align} \left\langle\psi_n^{(0)}\left|E_n^{(1)}\right|\psi_n^{(1)}\right\rangle = \left\langle\psi_n^{(0)}\left|E_n^{(1)}\right|\sum_{m = 1}^{\infty}C_{n, m}^{(1)}|\psi_m^{(0)}\rangle\right\rangle = E_n^{(1)}C_{n, n}^{(1)} = 0. \end{align} \]

Thus, one obtains

\[ \begin{align} E_n^{(2)} &= \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_n^{(1)}\right\rangle = \left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\sum_{m = 1}^{\infty}C_{n, m}^{(1)}|\psi_m^{(0)}\rangle\right\rangle = \sum_{m = 1}^{\infty}C_{n, m}^{(1)}\left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_m^{(0)}\right\rangle\nonumber\\ &= \sum_{\substack{m = 1,\\m\not = n}}^\infty\frac{\left\langle\psi_m^{(0)}\left|\newhat{v}_h\right|\psi_n^{(0)}\right\rangle\left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_m^{(0)}\right\rangle}{E_n^{(0)} - E_m^{(0)}} = \sum_{\substack{m = 1,\\m\not = n}}^\infty\frac{\left|\left\langle\psi_n^{(0)}\left|\newhat{v}_h\right|\psi_m^{(0)}\right\rangle\right|^2}{E_n^{(0)} - E_m^{(0)}}. \end{align} \]

The coefficients $C_{m, n}^{(2)}$ are no longer calculated here, since one usually only goes up to the first order in the state and up to the second order in the energy. In this way, one can successively compute further orders.

Now the case of degenerate energy eigenvalues is considered. So let $N\in \mathbb{N}$ with $N\geq 2$, and let the first $N$ unperturbed states $\left|\psi_1^{(0)}\right\rangle, \dotsc, \left|\psi_N^{(0)}\right\rangle$ be degenerate, i.e.

\[ \begin{align} \newhat{H}_0\left|\psi_m^{(0)}\right\rangle = E_0\left|\psi_m^{(0)}\right\rangle \end{align} \]

for $1\leq m\leq N$. Formula (4.325) is not applicable. However, one can see that with small energy differences $E_n^{(0)} - E_m^{(0)}$ the admixture of the states is very strong. Thus, even for a very weak perturbation $\lambda\to 0$, one must assume the admixture of degenerate states, so one makes the ansatz

\[ \begin{align} \left|\psi\left(\lambda\right)\right\rangle &= \sum_{l = 1}^{N}C_l\left|\psi_l^{(0)}\right\rangle + \lambda\sum_{m = N + 1}^{\infty}C_{m}^{(1)}\left|\psi_m^{(0)}\right\rangle + \mathcal{O}\left(\lambda^2\right),\\ E\left(\lambda\right) &= E_0 + \lambda E^{(1)} + \mathcal{O}\left(\lambda^2\right). \end{align} \]

This is substituted into the Schrödinger equation, Eq. (4.311):

\[ \begin{align} & \sum_{l = 1}^{N}C_l\newhat{H}_0\left|\psi_l^{(0)}\right\rangle + \sum_{l = 1}^{N}C_l\lambda\newhat{v}_h\left|\psi_l^{(0)}\right\rangle + \lambda\sum_{m = N + 1}^{\infty}C_m^{(1)}\newhat{H}_0\left|\psi_m^{(0)}\right\rangle + \mathcal{O}\left(\lambda^2\right)\nonumber\\ &= E_0\sum_{l = 1}^{N}C_l\left|\psi_l^{(0)}\right\rangle + \lambda E^{(1)}\sum_{l = 1}^{N}C_l\left|\psi_l^{(0)}\right\rangle + \lambda E_0\sum_{m = N + 1}^{\infty}C_m^{(1)}\left|\psi_m^{(0)}\right\rangle + \mathcal{O}\left(\lambda^2\right)\tag{4.334}\label{eq:sg_gestoert_entartet} \end{align} \]

This is satisfied to zeroth order. In the following, only the first order is considered. Let the $\left|\psi_l^{(0)}\right\rangle, 1\leq l\leq N$, be orthonormalized. Let $j\in \mathbb{N}$ with $1\leq j\leq N$ be given. Multiplying the first-order terms in Eq. (4.334) from the left by $\left\langle\psi_j^{(0)}\right|$, one obtains

\[ \begin{align} \sum_{l = 1}^{N}C_l\left\langle\psi_j^{(0)}\left|\newhat{v}_h\right|\psi_l^{(0)}\right\rangle &= E^{(1)}C_j. \end{align} \]

The solution $\left(C, \dotsc, C_N, E^{(1)}\right)$ gives the zeroth-order correction in the state and the first-order correction in the energy. The equation holds for all $1\leq l\leq N$, so one can also write this as a matrix

\[ \begin{align} v_h\mathbf{C} = E^{(1)}\mathbf{C} \end{align} \]

with the matrix elements

\[ \begin{align} V_{j, l} \coloneqq \left\langle\psi_j^{(0)}\left|\newhat{v}_h\right|\psi_l^{(0)}\right\rangle \end{align} \]

and the vector $\mathbf{C} \coloneqq \left(C, \dotsc, C_N\right)^T$. Since $\newhat{v}_h$ is Hermitian, the matrix $v_h$ is also Hermitian. As a solution, one therefore obtains $N$ orthonormal eigenvectors $\mathbf{C}^{k}$ for the eigenvalues $E_k^{(1)}$. The associated $N$ states are

\[ \begin{align} \left|\psi_k\right\rangle = \sum_{l = 1}^{N}C_l^{(k)}\left|\psi_l^{(0)}\right\rangle \end{align} \]

with the respective energy

\[ \begin{align} E_k = E_0 + E_k^{(1)}. \end{align} \]

These states are orthonormal. Indeed, let $k, m\in \mathbb {N}$ with $1\leq k, m\leq N$ be given. Then one has

\[ \begin{align} \left\langle\psi_k\newvline\psi_m\right\rangle &= \sum_{l = 1}^{\infty}C_l^{(k)\star}C_l^{(m)} = \left\langle\mathbf{C}^{(k)}\newvline\mathbf{C}^{(m)}\right\rangle = \delta_{m, n}. \end{align} \]

4.11.2 Spin-orbit coupling in the H atom

The spin-orbit coupling describes the effect of the interaction of the magnetic moment of the electron with the magnetic field of the nucleus; It is a relativistic effect because the nuclear motion arises from a coordinate transformation into the rest frame of the electron. This effect is treated here in the H atom in first order perturbation theory.

First, one recalls here the definition of the Bohr radius $a_B$ according to Eq. (4.200):

\[ \begin{align} a_B = \frac{\hbar^2}{m_ee^2} \end{align} \]

For the energy eigenvalues in the H atom, one has with Eq. (4.223)

\[ \begin{align} E_n = -\frac{Z^2}{2n^2}E_{\text{at}} \end{align} \]

For $E_{\mathrm{at}}$ one can now compute further

\[ \begin{align} E_{\text{at}} &= \frac{\hbar^2}{m_ea_B^2} = \frac{1}{a_B}\frac{\hbar^2}{m_ea_B} = \frac{1}{a_B}\frac{\hbar^2}{m_e}\frac{m_ee^2}{\hbar^2} = \frac{e^2}{a_B}. \end{align} \]

For a $\left(Z - 1\right)-$fold ionized atom, the atomic energy scale is

\[ \begin{align} \epsilon_{\text{at}} &= Z^2E_{\text{at}} = Z^2\frac{e^2}{a_B} = Z^2e^2\frac{m_ee^2}{\hbar^2} = Z^2e^4\frac{m_e}{\hbar^2}\frac{c^2}{c^2} = m_ec^2\left(\frac{Ze^2}{\hbar c}\right)^2 = m_ec^2\left(Z\alpha\right)^2. \end{align} \]

$\alpha$ here is the fine structure constant; for it, one has

\[ \begin{align} \alpha \coloneqq\frac{e^2}{\hbar c}\stackrel{\text{SI}}{=}\frac{e^2}{4\pi\epsilon_0\hbar c}\approx\frac{1}{137}. \end{align} \]

The spin-orbit coupling does not arise from a possible magnetic nuclear dipole. In a semiclassical treatment, the electron moves around the nucleus with an instantaneous velocity $\mathbf{v}$. In the electron's instantaneous rest frame, by Eq. (3.105), there exists a magnetic field

\[ \begin{align} \mathbf{B'} = -\gamma\frac{\mathbf{v}}{c}\times\mathbf{E}, \end{align} \]

which the electron experiences. The function

\[ \begin{align} \mathbf{u}\left(\mathbf{v'}\right) \coloneqq\frac{\mathbf{v'}}{\sqrt{1 - v'^2}} \end{align} \]

with

\[ \begin{align} \mathbf{v'} \coloneqq\frac{\mathbf{v}}{c} \end{align} \]

describes the strength of the relativistic effect. The linear Taylor expansion of this function is

\[ \begin{align} \mathbf{u} = \mathbf{v'} + \mathcal{O}\left(\mathbf{v'}^2\right), \end{align} \]

so that, for the relativistic B field, one makes the ansatz

\[ \begin{align} \mathbf{B'} = -\frac{\mathbf{v}}{c}\times\mathbf{E} + \mathcal{O}\left(\left(\frac{\mathbf{v}}{c}\right)^2\right). \end{align} \]

The higher order terms are no longer noted. With $g\approx2$ the magnetic moment of the electron is given by Eq. (4.303)

\[ \begin{align} \mubi &= -\frac{e}{m_ec}\mathbf{s}. \end{align} \]

This gives the energy

\[ \begin{align} - \mubi\cdot\mathbf{B'} = \frac{e}{m_ec}\mathbf{s}\cdot\mathbf{B'} = -\frac{e}{m_ec^2}\mathbf{s}\cdot\left(\mathbf{v}\times\mathbf{E}\right). \end{align} \]

For the E field, $\mathbf{E} = \frac{Ze\mathbf{r}}{r^3}$, with the definition $\mathbf{l} = \mathbf{r}\times\mathbf{p}$ one obtains the spin-orbit interaction

\[ \begin{align} V &= -\mubi\cdot\mathbf{B'} = -\frac{e}{m_ec^2}\mathbf{s}\cdot\left(\mathbf{v}\times\mathbf{E}\right) = \frac{Ze^2}{m_e^2c^2}\frac{\mathbf{l}\cdot\mathbf{s}}{r^3}. \end{align} \]

A relativistic quantum-mechanical treatment of the accelerated electron using the Dirac equation yields an additional factor $1/2$ at this point. The perturbation operator sought is:

\[ \begin{align} \newhat{v}_h = \frac{Ze^2}{2m_e^2c^2}\frac{\newhat{\mathbf{l}}\cdot\newhat{\mathbf{s}}}{\newhat{r}^3}. \end{align} \]

Using the total angular momentum operator

\[ \begin{align} \newhat{\mathbf{j}} \coloneqq\newhat{\mathbf{l}} + \newhat{\mathbf{s}} \end{align} \]

one can rewrite this as

\[ \begin{align} \newhat{v}_h &= \frac{Ze^2}{4m_e^2c^2}\frac{\newhat{2\mathbf{l}}\cdot\newhat{\mathbf{s}}}{\newhat{r}^3} = \frac{Ze^2}{4m_e^2c^2}\frac{\newhat{2\mathbf{l}}\cdot\newhat{\mathbf{s}} + \newhat{\mathbf{s}}^2 - \newhat{\mathbf{s}}^2}{\newhat{r}^3} = \frac{Ze^2}{4m_e^2c^2}\frac{\left(2\newhat{\mathbf{l}} + \newhat{\mathbf{s}}\right)\cdot\newhat{\mathbf{s}} - \newhat{\mathbf{s}}^2}{\newhat{r}^3} = \frac{Ze^2}{4m_e^2c^2}\frac{\left(\newhat{\mathbf{j}} + \newhat{\mathbf{l}}\right)\cdot\left(\newhat{\mathbf{j}} - \newhat{\mathbf{l}}\right) - \newhat{\mathbf{s}}^2}{\newhat{r}^3}\nonumber\\ &= \frac{Ze^2}{4m_e^2c^2}\frac{\newhat{\mathbf{j}}^2 - \newhat{\mathbf{l}}^2 - \newhat{\mathbf{s}}^2}{\newhat{r}^3} \end{align} \]

which can be rewritten as follows. To work with this further, one couples the angular momentum eigenstates of the H atom to

\[ \begin{align} \left|n, l, m, s, s_z\right\rangle &= \left|n, l\right\rangle\left|l, m, s, s_z\right\rangle = \left|n, l\right\rangle\left|j, l, s, m_j\right\rangle = \left|n, j, l, s, m_j\right\rangle. \end{align} \]

Since the angular momentum eigenstates are orthogonal and $\newhat{v}_h$ only acts on the radial coordinates, for the energy splitting one has

\[ \begin{align} \left\langle n, j, l, s, m_j\left|\newhat{v}_h\right|n, j', l', s, m_j'\right\rangle &= \Delta E\delta_{j, j'}\delta_{l, l'}\delta_{m_j, m_j'} \end{align} \]

with

\[ \begin{align} \Delta E &= \frac{Ze^2\hbar^2}{4m_e^2c^2}\left\langle n, l\left|\frac{1}{\newhat{r}^3}\right|n, l\right\rangle\left[j\left(j + 1\right) - l\left(l + 1\right) - s\left(s + 1\right)\right]. \end{align} \]

For $l\not = 0$, one has

\[ \begin{align} \left\langle\frac{1}{\newhat{r}^3}\right\rangle &= \int_{\mathbb{R}^3}\frac{1}{r^3}\left|\psi\right|^2d^3r = \int_{r = 0}^\infty\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{r^3}\left|\psi\right|^2r^2\sin\left(\vartheta\right)d\vartheta d\varphi dr\nonumber\\ &= \int_{r = 0}^\infty\int_{\varphi = 0}^{2\pi}\int_{\vartheta = 0}^{\pi}\frac{1}{r}R_{n, l}\left(r\right)^2\left|Y_{l, m}\left(\vartheta, \varphi\right)\right|^2\sin\left(\vartheta\right)d\vartheta d\varphi dr\nonumber\\ &= \int_{0}^\infty\frac{1}{r}R_{n, l}\left(r\right)^2dr = \frac{4Z^3}{a_B^3n^4}\frac{\left(n - l - 1\right)!}{\left(n + l\right)!}\int_{0}^{r}\frac{1}{r}\left(\frac{2Zr}{na_B}\right)^{2l}\left[L_{n - l - 1, 2l + 1}\left(\frac{2Zr}{na_B}\right)\right]^2\exp\left(-\frac{2Zr}{na_B}\right)dr\nonumber\\ &= \frac{4Z^3}{a_B^3n^4}\frac{\left(n - l - 1\right)!}{\left(n + l\right)!}\int_{0}^{\infty}r^{2l - 1}\left[L_{n - l - 1, 2l + 1}\left(r\right)\right]^2\exp\left(-r\right)dr.\nonumber \end{align} \]

One identifies

\[ \begin{align} m& \coloneqq n - l - 1\geq0\Rightarrow n = m + l + 1,\\ k& \coloneqq 2l + 1\geq3. \end{align} \]

From this, the transformations follow

\[ \begin{align} n& \coloneqq m + 1 + \frac{1}{2}\left(k - 1\right) = m + \frac{1}{2}\left(k + 1\right),\\ l& \coloneqq\frac{1}{2}\left(k - 1\right). \end{align} \]

With Eq. (C.147), it follows

\[ \begin{align} & \int_{0}^{\infty}r^{2l - 1}\left[L_{n - l - 1, 2l + 1}\left(r\right)\right]^2\exp\left(-r\right)dr = \int_{0}^{\infty}r^{k - 2}\left[L_{m, k}\left(r\right)\right]^2\exp\left(-r\right)dr\nonumber\\ &= \frac{\left(m + k\right)!}{m!}\frac{2m + 1 + k}{\left(k - 1\right)k\left(k + 1\right)} = \frac{\left(m + k\right)!}{m!}\frac{2n}{\left(k - 1\right)k\left(k + 1\right)} = \frac{\left(n + l\right)!}{\left(n - l - 1\right)!}\frac{n}{2l\left(2l + 1\right)\left(l + 1\right)}\nonumber\\ &= \frac{\left(n + l\right)!}{\left(n - l - 1\right)!}\frac{n}{4l\left(l + \frac{1}{2}\right)\left(l + 1\right)}. \end{align} \]

Thus, one has

\[ \begin{align} \left\langle\frac{1}{\newhat{r}^3}\right\rangle &= \frac{Z^3}{a_B^3n^3l\left(l + \frac{1}{2}\right)\left(l + 1\right)} = \frac{Z^3m_e^3e^6}{\hbar^6n^3l\left(l + \frac{1}{2}\right)\left(l + 1\right)}. \end{align} \]

For $l = 0$

\[ \begin{align} \left\langle\frac{1}{\newhat{r}^3}\right\rangle\not\in&\mathbb{R}, \end{align} \]

However, in this case the spin-orbit coupling is zero anyway because of $j = l = \frac{1}{2}$. Thus, with

\[ \begin{align} \frac{Ze^2\hbar^2}{4m_e^2c^2}\frac{Z^3m_e^3e^6}{\hbar^6n^3} = \frac{m_eZ^4e^8}{4\hbar^4n^3c^2} = \frac{m_ec^2}{4n^3}\frac{Z^4e^8}{\hbar^4c^4} \end{align} \]

as a summary

\[ \begin{align} \Delta E = \begin{cases} 0, \text{ }l = 0,\\ \frac{m_ec^2\left(Z\alpha\right)^4}{4n^3}\frac{j\left(j + 1\right) - l\left(l + 1\right) - s\left(s + 1\right)}{l\left(l + \frac{1}{2}\right)\left(l + 1\right)}, \text{ }l\not = 0. \end{cases} \end{align} \]

Relativistic effects therefore influence the eigenenergies in the H atom and thus also the spectrum.

4.11.3 Time-dependent case

Previously only time-constant perturbations $\newhat{V}$ were considered, now a time dependence $\newhat{V}\left(t\right)$ is allowed. Into the time-dependent Schrödinger equation

\[ \begin{align} \newhat{H}\left(t\right)\left|\psi\left(t\right)\right\rangle = i\hbar\frac{\partial}{\partial t}\left|\psi\left(t\right)\right\rangle\tag{4.369}\label{eq:sg_zeitabh_stoer} \end{align} \]

a perturbation operator

\[ \begin{align} \newhat{H}\left(t\right) = \newhat{H}_0 + \newhat{V}\left(t\right) \end{align} \]

is now to be substituted. The solutions $\left(\left|\psi_n^{(0)}\right\rangle, E_n\right)$ of the stationary problem

\[ \begin{align} \newhat{H}_0\left|\psi_n^{(0)}\right\rangle = E_n\left|\psi_n^{(0)}\right\rangle \end{align} \]

are known again. The $\left|\psi_n^{(0)}\right\rangle$ can be provided with a time-dependent part; with an abuse of notation, one can write

\[ \begin{align} \left|\psi_n^{(0)}\right\rangle = \left|\psi_n^{(0)}\right\rangle\exp\left(-i\frac{E_n}{\hbar}t\right). \end{align} \]

This section does not go further than the first order. One now sets, with $\lambda\in\left[0, 1\right]$,

\[ \begin{align} \newhat{H}\left(\lambda, t\right) &= \newhat{H}_0 + \lambda\newhat{v}_h\left(t\right),\\ \left|\psi\left(\lambda, t\right)\right\rangle &= \sum_{m = 1}^{\infty}C_m^{(0)}\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \lambda\sum_{m = 1}^{\infty}C_m^{(1)}\left(t\right)\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \mathcal{O}\left(\lambda^2\right). \end{align} \]

This gives

\[ \begin{align} i\hbar\frac{\partial}{\partial t}\left|\psi\left(\lambda, t\right)\right\rangle &= \sum_{m = 1}^{\infty}E_mC_m^{(0)}\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right)\nonumber\\ & + \lambda\sum_{m = 1}^{\infty}\left(i\hbar\frac{\partial C_m^{(1)}}{\partial t} + C_m^{(1)}E_m\right)\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \mathcal{O}\left(\lambda^2\right). \end{align} \]

Substituting this into Eq. (4.369), one obtains

\[ \begin{align} & \newhat{H}_0\sum_{m = 1}^{\infty}C_m^{(0)}\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \lambda\newhat{H}_0\sum_{m = 1}^{\infty}C_m^{(1)}\left(t\right)\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right)\nonumber\\ & + \lambda\newhat{v}_h\left(t\right)\sum_{m = 1}^{\infty}C_m^{(0)}\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \mathcal{O}\left(\lambda^2\right) = \sum_{m = 1}^{\infty}C_m^{(0)}E_m\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right)\nonumber\\ & + \lambda\sum_{m = 1}^{\infty}\left(i\hbar\frac{\partial C_m^{(1)}}{\partial t} + C_m^{(1)}E_m\right)\left|\psi_m^{(0)}\right\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) + \mathcal{O}\left(\lambda^2\right). \end{align} \]

The zeroth order is trivially satisfied. Now let $k\in \mathbb {N}$ with $k\geq 1$ be given, and multiply the first-order terms from the left by $\langle\psi_k^{(0)}|$. One obtains

\[ \begin{align} & C_k^{(1)}\left(t\right)E_k\exp\left(-i\frac{E_k}{\hbar}t\right) + \sum_{m = 1}^{\infty}C_m^{(0)}\langle\psi_k^{(0)}\left|\newhat{v}_h\left(t\right)\right|\psi_m^{(0)}\rangle\exp\left(-i\frac{E_m}{\hbar}t\right) = \left(i\hbar\frac{\partial C_k^{(1)}}{\partial t} + C_k^{(1)}E_k\right)\exp\left(-i\frac{E_k}{\hbar}t\right)\nonumber\\ &\Leftrightarrow \frac{\partial C_k^{(1)}}{\partial t} = -\frac{i}{\hbar}\sum_{m = 1}^{\infty}C_m^{(0)}\langle\psi_k^{(0)}\left|\newhat{v}_h\left(t\right)\right|\psi_m^{(0)}\rangle\exp\left(i\frac{E_k - E_m}{\hbar}t\right). \end{align} \]

Thus, for the admixture of the states, one obtains

\[ \begin{align} C_k^{(1)}\left(t'\right) = C_k^{(1)}\left(t_0\right) - \frac{i}{\hbar}\int_{t_0}^{t'}\sum_{m = 1}^{\infty}C_m^{(0)}\left\langle\psi_k^{(0)}\left|\newhat{v}_h\left(t\right)\right|\psi_m^{(0)}\right\rangle\exp\left(i\frac{E_k - E_m}{\hbar}t\right)dt. \end{align} \]

If the system is in state $n$ at time $t_0$, then one obtains

\[ \begin{align} C_n^{(0)} &= 1, \nonumber\\ C_{m\not = n}^{(0)} &= 0 \end{align} \]

and thus

\[ \begin{align} C_k^{(1)}\left(t'\right) = -\frac{i}{\hbar}\int_{t_0}^{t'}\left\langle\psi_k^{(0)}\left|\newhat{v}_h\left(t\right)\right|\psi_n^{(0)}\right\rangle\exp\left(i\frac{E_k - E_n}{\hbar}t\right)dt.\tag{4.380}\label{eq:zust_beimischung} \end{align} \]

From this, one can construct the time-dependent solution

\[ \begin{align} \left|\psi\left(t\right)\right\rangle\approx\left|\psi_n^{(0)}\right\rangle\exp\left(-i\frac{E_n}{\hbar}t\right) + \sum_{k = 1}^{\infty}C_k^{(1)}\left(t\right)\left|\psi_k^{(0)}\right\rangle\exp\left(-i\frac{E_k}{\hbar}t\right) \end{align} \]

The probability of encountering the state $\left|\psi_m^{(0)}\right\rangle$ is given by

\[ \begin{align} p_{m}\left(t\right) = \left|\delta_{mn} + C_m^{(1)}\left(t\right)\right|^2 \end{align} \]

To understand the absorption and emission induced by electromagnetic waves, one needs knowledge of the effect of a periodic perturbation operator

\[ \begin{align} \newhat{V}\left(t\right) = \newhat{V}_0\exp\left(-i\omega t\right) + \newhat{V}_0^\star\exp\left(i\omega t\right), \end{align} \]

the addition of the adjoint term guarantees Hermiticity and thus real measurable quantities. Substituting this into Eq. (4.380), it follows

\[ \begin{align} i\hbar C_k^{(1)}\left(t\right) &= \left\langle\psi_k^{(0)}\left|\newhat{V}_0\right|\psi_n^{(0)}\right\rangle\int_{0}^\exp\left(i\left(\omega_{k, n} - \omega\right)t'\right)dt'\nonumber\\ & + \left\langle\psi_k^{(0)}\left|\newhat{V}_0^\star\right|\psi_n^{(0)}\right\rangle\int_{0}^\exp\left(i\left(\omega_{k, n} + \omega\right)t'\right)dt'.\tag{4.384}\label{eq:period_stoer_deriv_1} \end{align} \]

here, $t_0 = 0$ was set and

\[ \begin{align} \omega_{k, n} \coloneqq\omega_k - \omega_n \end{align} \]

defined. Define further

\[ \begin{align} \Omega_{\pm} \coloneqq\omega_{k, n}\pm\omega, \end{align} \]

then

\[ \begin{align} \int_{0}^\exp\left(i\Omega_{\pm}t'\right)dt' &= \frac{\exp\left(i\Omega_\pm t\right) - 1}{i\Omega_\pm}\nonumber\\ \Rightarrow \left|\int_{0}^\exp\left(i\Omega_{\pm}t'\right)dt'\right|^2 &= \left|\frac{\exp\left(i\Omega_\pm t\right) - 1}{i\Omega_\pm}\right|^2 = \frac{1}{\Omega_\pm^2}\left[\left(\cos\left(\Omega_\pm t\right) - 1\right)^2 + \sin^2\left(\Omega_\pm t\right)\right]\nonumber\\ &= \frac{1}{\Omega_\pm^2}\left[\cos^2\left(\Omega_\pm t\right) + 1 - 2\cos\left(\Omega_\pm t\right) + \sin^2\left(\Omega_\pm t\right)\right] = \frac{2}{\Omega_\pm^2}\left[1 - \cos\left(\Omega_\pm t\right)\right]\nonumber\\ &= \frac{2}{\Omega_\pm^2}\left[1 - \cos^2\left(\frac{\Omega_\pm t}{2}\right) + \sin^2\left(\frac{\Omega_\pm t}{2}\right)\right] = \frac{4\sin^2\left(\frac{\Omega_\pm t}{2}\right)}{\Omega_\pm^2}. \end{align} \]

This only makes a significant contribution when $\Omega_\pm$ is small; because of $\Omega_ + = \Omega_ - + 2\omega$, therefore, usually only one of the two integrals in Eq. (4.384) contributes to the result. With $\newhat{V} - \coloneqq\newhat{V}_0$ and $\newhat{V} + \coloneqq\newhat{V}_0^\star$ one obtains

\[ \begin{align} R_{n\to k} = \frac{\left|C_k^{(1)}\left(t\right)\right|^2} = \frac{\left|\left\langle\psi_k^{(0)}\left|\newhat{V}_\pm\right|\psi_n^{(0)}\right\rangle\right|^2}{\hbar^2}\frac{4\sin^2\left(\frac{\Omega_\pm t}{2}\right)}{\Omega_\pm^2t}. \end{align} \]

Here, $R_{n\to k}$ is the probability that the system changes from state $n$ to state $k$ in the time interval $\left[0, t\right]$, divided by the time available for this. Define

\[ \begin{align} f\left(\Omega\right) \coloneqq\frac{4\sin^2\left(\frac{\Omega t}{2}\right)}{\Omega^2t}, \end{align} \]

then with Eq. (A.85)

\[ \begin{align} \lim_{t\to \infty}f\left(\Omega\right) = 2\pi\delta\left(\Omega\right). \end{align} \]

This may be used in the case $t\gg1/\Omega$; it follows

\[ \begin{align} R_{n\to k} = \frac{2\pi}{\hbar}\left|\left\langle\psi_k^{(0)}\left|\newhat{V}_\pm\right|\psi_n^{(0)}\right\rangle\right|^2\delta\left(E_k - E_n\pm\hbar\omega\right).\tag{4.391}\label{eq:ubergangsraten_period} \end{align} \]

The minus sign must be used for $E_k>E_n$. In the case of a time-independent perturbation $\omega = 0$, it follows

\[ \begin{align} R_{n\to k} = \frac{2\pi}{\hbar}\left|\left\langle\psi_k^{(0)}\left|\newhat{V}_\pm\right|\psi_n^{(0)}\right\rangle\right|^2\delta\left(E_k - E_n\right).\tag{4.392}\label{eq:ubergangsraten_konst} \end{align} \]

4.11.4 Transitions in the H atom

Take a vector potential of the form

\[ \begin{align} \mathbf{A}\left(\mathbf{r}, t\right) &= A_0\epsilonbi\cos\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right) \end{align} \]

with $\omega = ck$ and $\epsilonbi\cdot\mathbf{k} = 0$, where $\epsilonbi$ is the normalized polarization vector. Then it follows

\[ \begin{align} \mathbf{B} = \nabla\times\mathbf{A}&\stackrel{\text{Eq. }\href{ch-40-vector-analysis.html#eq:diff_op_rule_2}{(B.48)}}{=} -A_0\epsilonbi\times\mathbf{k}\cos\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right) = A_0\mathbf{k}\times\epsilonbi\cos\left(\mathbf{k}\cdot\mathbf{r} - \omega t\right). \end{align} \]

This describes an electromagnetic wave. For the Hamilton operator, one obtains

\[ \begin{align} \newhat{H} &= \frac{1}{2m_e}\left(-i\hbar\nabla + \frac{e}{c}\mathbf{A}\right)^2 - \frac{e^2}{r}\stackrel{\text{Eq. }\href{ch-40-vector-analysis.html#eq:diff_op_rule_3}{(B.49)}}{=}\frac{\newhat{\mathbf{p}}^2}{2m_e} - \frac{e^2}{r} + \frac{e}{m_ec}\newhat{\mathbf{A}}\cdot\newhat{\mathbf{p}} = \newhat{H}_0 + \newhat{v}_h\left(t\right) \end{align} \]

with the perturbation operator

\[ \begin{align} \newhat{v}_h\left(t\right) = \frac{e}{m_ec}\newhat{\mathbf{A}}\cdot\newhat{\mathbf{p}}. \end{align} \]

The quadratic terms in the vector product were neglected. With

\[ \begin{align} \cos = \frac{\exp\left( +\right) + \exp\left(-\right)}{2} \end{align} \]

one can write

\[ \begin{align} \newhat{V}\left(t\right) &= \newhat{V}_0\exp\left(-i\omega t\right) + \newhat{V}_0^+\exp\left(i\omega t\right), \end{align} \]

here,

\[ \begin{align} \newhat{V}_0 = \frac{eA_0}{2m_ec}\exp\left(i\mathbf{k}\cdot\mathbf{r}\right)\epsilonbi\cdot\newhat{\mathbf{p}} \end{align} \]

was inserted. With the long-wave approximation

\[ \begin{align} \exp\left(i\mathbf{k}\cdot\mathbf{r}\right) &= 1 + i\mathbf{k}\cdot\mathbf{r} + \dotsc = 1 + \mathcal{O}\left(\left\langle r\right\rangle/\lambda\right)\approx 1 \end{align} \]

one obtains

\[ \begin{align} \newhat{V}_0 &\approx \frac{eA_0}{2m_ec}\epsilonbi\cdot\newhat{\mathbf{p}}. \end{align} \]

This is justified here because one can calculate with $\left\langle r\right\rangle\sim a_B$

\[ \begin{align} \frac{\left\langle r\right\rangle}{\lambda} &\sim \frac{a_B}{\lambda} = \frac{a_B\omega}{2\pi c} = \frac{a_B\Delta E}{2\pi\hbar c}\sim\frac{a_BE_{\text{at}}}{20\pi\hbar c} = \frac{\hbar}{m_ea_B20\pi c} = \frac{e^2}{20\hbar\pi c}\to\frac{e^2}{4\pi\epsilon_020\hbar\pi c}\sim 10^{-4}. \end{align} \]

Here, $\Delta E\sim\frac{E_{\text{at}}}{10}$ was inserted. With Eq. (4.392), one obtains

\[ \begin{align} R_{a\to b} &= \frac{\pi e^2A_0^2}{2m_e^2c^2\hbar}\left|\left\langle b\left|\epsilonbi\cdot\newhat{\mathbf{p}}\right|a\right\rangle\right|^2\left[\delta\left(E_b - E_a + \hbar\omega\right) + \delta\left(E_b - E_a - \hbar\omega\right)\right]. \end{align} \]

So the transition rates are proportional to $A_0^2$, which in turn is proportional to the energy density of the electromagnetic field. The transition rates are therefore proportional to the magnitude of the matrix element

\[ \begin{align} M_{b, a} = \left\langle b\left|\epsilonbi\cdot\newhat{\mathbf{p}}\right|a\right\rangle. \end{align} \]

This needs to be rephrased a bit. First, one computes

\[ \begin{align} \left[\newhat{H}_0, \newhat{\mathbf{r}}\right] &= \newhat{H}_0\newhat{\mathbf{r}} - \newhat{\mathbf{r}}\newhat{H}_0 = -\frac{\hbar^2}{2m_e}\nabla\nabla\cdot\newhat{\mathbf{r}} + \newhat{\mathbf{r}}\frac{\hbar^2}{2m_e}\Delta. \end{align} \]

One has

\[ \begin{align} \nabla\cdot\left(\mathbf{r}\psi\right) &= 3\psi + \mathbf{r}\cdot\nabla\psi\nonumber\\ \Rightarrow\nabla^2\left(\mathbf{r}\psi\right) &= 6\nabla\psi + \mathbf{r}\Delta\psi. \end{align} \]

Thus it follows

\[ \begin{align} \left[\newhat{H}_0, \newhat{\mathbf{r}}\right] &= -\frac{\hbar^2}{2m_e}6\nabla\psi = \frac{3\hbar}{im_e}\newhat{\mathbf{p}}\Rightarrow\newhat{\mathbf{p}} = \frac{im_e}{3\hbar}\left[\newhat{H}_0, \newhat{\mathbf{r}}\right]. \end{align} \]

Thus one can write

\[ \begin{align} M_{b, a} &= im_e\frac{E_b - E_a}{3\hbar}\left\langle b\left|\epsilonbi\cdot\newhat{\mathbf{r}}\right|a\right\rangle. \end{align} \]

The eigenstates in the H atom are known from Eq. (4.233), so one has

\[ \begin{align} M_{b, a} &\propto \left\langle n, l, m\left|\epsilonbi\cdot\newhat{\mathbf{r}}\right|n, l, m\right\rangle. \end{align} \]

Spin is omitted here. The radial part does not provide any selection rules for $\Delta n$, so here only

\[ \begin{align} M_{b, a} &\propto \int Y_{l_b, m_b}^\star\epsilonbi\cdot\newhat{\mathbf{r}}Y_{l_a, m_a}d\Omega. \end{align} \]

is relevant. One writes

\[ \begin{align} \epsilonbi\cdot\mathbf{r} &= r\left(\epsilon_x\sin\left(\theta\right)\cos\left(\phi\right) + \epsilon_y\sin\left(\theta\right)\sin\left(\phi\right) + \epsilon_z\cos\left(\theta\right)\right). \end{align} \]

From Eqs. (C.171) - (C.172), it follows

\[ \begin{align} \cos\left(\theta\right)Y_{l, m} = \alpha Y_{l - 1, m} + \beta Y_{l + 1, m}, & {} & \sin\left(\theta\right)Y_{l, m} = \gamma e^{-i\phi}Y_{l - 1, m + 1} + \delta e^{-i\phi}Y_{l + 1, m + 1} \end{align} \]

with coefficients $\alpha, \beta, \gamma, \delta$ that are not relevant here. Thus one has

\[ \begin{align} \Delta l &= l_b - l_a = \pm 1,\\ \Delta m &= 0, \pm 1. \end{align} \]

4.11.5 Relativistic corrections in the H atom

4.12 Multi-particle systems

Let $N\in \mathbb{N}$ with $N\geq1$, and let $N$ particles with spin be given. The state of such a system is described by a wave function

\[ \begin{align} \psi = \psi\left(x_i\right) \end{align} \]

with $1\leq i\leq N$ and $x_i = \left(\mathbf{r}_i, \sigma_i\right)$, where $\mathbf{r}_i$ denotes the position of the $i-$th particle and $\sigma_i$ its spin. In the case of spin 1/2 particles, one can imagine such a function as $2^N$ functions $\mathbb {R}^{3N}\to \mathbb {C}$.It is easy to see that it is not possible to tabulate such a function for a realistic system. Now define for $i, j\in \mathbb {N}$ with $1\leq i, j\leq N$ the permutation operator $\newhat{P}_{i, j}$ by

\[ \begin{align} \newhat{P}_{i, j}\psi\left(\dotsc, x_i, \dotsc, x_j, \dotsc\right) \coloneqq\psi\left(\dotsc, x_j, \dotsc, x_i, \dotsc\right). \end{align} \]

This operator swaps the arguments belonging to two particles. If there are $N$ particles of the same type, the Hamiltonian $\newhat{H}$ is invariant under permutation:

\[ \begin{align} \newhat{P}_{i, j}\newhat{H} = \newhat{H}. \end{align} \]

From this it follows

\[ \begin{align} \left[\newhat{P}_{i, j}, \newhat{H}\right] = 0, \end{align} \]

and that means the existence of a $\lambda\in \mathbb{C}$ with

\[ \begin{align} \newhat{P}_{i, j}\psi\left(x_k\right) = \lambda\psi\left(x_k\right) \end{align} \]

for eigenfunctions $\psi\left(x_k\right)$ of the Hamiltonian; this may depend on $i, j$, which has been neglected in the notation here. In any case, applying the same operator once more, it follows

\[ \begin{align} \lambda^2 = 1, \end{align} \]

therefore $\lambda = \pm1$. In the case $\lambda = -1$ one speaks of antisymmetry, in the case $\lambda = 1$ of symmetry. Particles with half-integer spin (fermions) have antisymmetric wave functions, while particles with integer spin (bosons) have symmetric wave functions. This has far-reaching implications. Assume that an N-fermion system is in a state in which particles $i$ and $j$ occupy the same orbitals, i.e.

\[ \begin{align} \psi\left(\dotsc, x_i, \dotsc, x_j, \dotsc\right) = \psi\left(\dotsc, x_j, \dotsc, x_i, \dotsc\right) \end{align} \]

holds. Applying $\newhat{P}_{i, j}$ to this now, it follows

\[ \begin{align} - \psi\left(\dotsc, x_i, \dotsc, x_j, \dotsc\right) = \psi\left(\dotsc, x_i, \dotsc, x_j, \dotsc\right), \end{align} \]

so $\psi = 0$, which is a contradiction to the normalization condition. Two fermions can only have the same wave function (occupy the same orbital) if they have a different spin. This is not the case with bosons. This repulsion of fermions is not due to a force, but rather to symmetry properties of their wave function, which is known as the Pauli principle.

4.12.1 Molecular spectra

To calculate the interaction of radiation with matter present in the atmosphere, Planck's radiation law is not sufficient. One also needs the spectra, i.e. the energy levels and transition probabilities of molecules under excitation by dipole radiation. An H$_2$O molecule consists of ten electrons; in order to store the state of the electron shell when each axis is discretized into ten intervals (which is hardly sufficient), one would already need

\[ \begin{align} N_{\text{points}} = \left(10^{3}\cdot 2\right)^N = 2^N\cdot 10^{3N}\approx10^{33} \end{align} \]

data points. These are two complex numbers each, so one needs approximately

\[ \begin{align} S = 16\cdot 10^{33}\approx10^{22}\text{ Terrabyte} \end{align} \]

of storage space, which is currently an unrealistically large amount. If one wants to derive spectra theoretically, one has to devise approximation methods (this is almost always the case in QM). The most common methods are:

• Perturbation theory

• Monte Carlo simulation (a statistical procedure)

• Density functional theory

4.12.2 Chemical reactions

Chemical reactions are material transformations. Let a mixture of $N\in\mathbb{N}$ components be given with $N\geq1$ and let the particle densities be given by $n_i$ with $1\leq i\leq N$. For $1\leq j, k\leq N$, define the number $U_{j, k}$ by the rate at which the substance $k$ transforms into the substance $j$ (dimension: particles per time and volume). Then, for $1\leq i\leq N$, one has

\[ \begin{align} \frac{dn_i}{dt} = \sum_{j = 1}^{N}U_{i, j} - U_{j, i}. \end{align} \]

If this takes place in the atmosphere, the composition of the air changes and thus also the gas constant $R_d=\frac{R}{M_d}$. The matrix $v$ depends on the thermodynamic quantities, the radiation field, and the material densities present.