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 i. A. 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,

The Schrödinger equation (SG) applies

with the Hamilton operator

Here $\hbar \coloneqq \frac{h}{2\pi}$ with the Planck's constant $h$. An operator turns a function into a new function and is identified by an umbrella 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 pass through

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

be noted. Here it is found that

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

applies, so the right side of the SG pulls the energy forward. The same applies to the left side

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

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

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

you define

as the momentum operator. The term of the partial time derivative $i\hbar\frac{\partial}{\partial t}$ is used for wave functions $\psi\left(\mathbf{r}, t\right)$, whose time dependence is trivial and can be separated from the position dependence, i.e

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

to

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

The Schrödinger equation then becomes 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

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

For $\psi\left(\mathbf{r}, t\right)$ we again use the approach of a plane wave

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

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 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 $c_{\mathrm{gr}}$ applies to the group velocity

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

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

4.2 potential well

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

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

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

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

You make the approach

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

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 approach applies

\[ \begin{align} \psi\left(x\right) = C\left(\exp\left(ik_1x\right) - \exp\left(ik_2x\right)\right). \end{align} \] 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 $\exp\left(ik_2x\right)$ are allowed to be linearly independent, therefore the following applies

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

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, otherwise the wave function disappears. The normalization condition applies

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

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

For the energy eigenvalues ​​$E_{n_1, n_2}$ follows

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 level

The potential is given

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

with $V_0, L>0$. If a classical particle with an energy $0L$. In quantum mechanics this is different; this is called the tunnel effect because, to put it bluntly, 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 $V_0>0$. Two different stationary Schrödinger equations apply in the two areas $1$ and $2$

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

for the two wave functions $\psi_1\left(x\right)$, $\psi_2\left(x\right)$, $E

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

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 approach

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

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

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

This 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 area $2$:

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

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
• If the motion of the particle is limited, 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 expand the problem to three dimensions, so

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

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 do a product approach 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 applies

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

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

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

So 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 needs to be solved. You can write this 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 are

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

The stationary SG will start with the replacement

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

to

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

In infinity this becomes

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

This is the approach

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

made with $\sigma>0$. This results in twice derived

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

for large $u$. If you put this in, you get

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

One does the approach for $\psi$

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

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 $P\left(u\right)$ a power series approach is used

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

made. This results in

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

and

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

Putting this into Eq. (4.69), you get

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

For this to apply to all $y\in \mathbb{R}$, the individual coefficients must disappear:

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

So you get the recursion formula for $a_i$

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

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

These are the energy eigenvalues ​​of the oscillator. $n$ defines a state. The states are therefore quantized, their energies are equidistant with the distance $\hbar\omega$ and the ground state energy $E_0>0$ is positive.

If you rewrite the recursion formula, you get

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

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 section C.3.

The normalization condition applies

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

so with Eq. (C.124)

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

So 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 relationships are noted:

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

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 you choose 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 which $a_i$ holds, let $k\in \mathbb {N}$ with $k\geq 1$,

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

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 broken off somewhere depending on the desired precision. The vector $\left(a_i\right)$ determines the state $\psi\left(\mathbf{r}\right)$ of a particle. You 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 you have solved the Schrödinger equation for a system and found a quantization of the states, you can define a state $\psi\left(\mathbf{r}\right)$ by $N\geq 1$ quantum numbers $n_1, \dotsc, n_N$, $\psi\left(\mathbf{r}, t\right) = \psi\left(n_1, \dotsc, n_N,\mathbf{r},t\right)$. So there are infinitely many ways to determine a state of a particle. If you don't want to commit to such a possibility, just write

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

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 you need two quantum numbers $n, m$ to define a state, you write $|\psi\rangle = |n, m\rangle$. You define

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

This means: If $|\psi\rangle = \psi\left(\mathbf{r}, t\right)$ applies to the state $|\psi\rangle$ in spatial space, then $\langle\psi| applies = \psi^\star\left(\mathbf{r}, t\right)$. Derived from the English word bracket (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 section A.8), so it is a mapping in Hilbert space. You write

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

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 you use only a finite number of base 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 should hold for all $\psi, \chi\in H$. This sets $\newhat{O}^+$. In linear algebra, a matrix $A$ over $\mathbb{C}$ is called unitary, if

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

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

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

and $A^+$ is the adjoint matrix

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

You write for a state

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

so applies

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

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 expected 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 $\newhat{O}|f_i\rangle = o_i|f_i\rangle$ applies to an operator $\newhat{O}$, then the expected value $\langle o\rangle$ of size $o$ 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 expected 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 develop $\newhat{O}$ as a matrix with respect to the base $\left(f_i\right)$ and write $O_{i, j}$ for the entry in the $j-$th column of the $i-$th row. As the initial state you choose $|\psi\rangle = f_j$, so the system is definitely in the $j-$th state. You write

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

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 is

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

$\left|b_k\right|^2$ 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 $\newhat{O}$, you get a forbidden transition. So you can 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 to determine the matrix elements $O_{i, j}$ of the operator $\newhat{O}$ with respect to the base $\left(f_k\right)$.

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 that the characteristic polynomial

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

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

4.6.2 Hermitian operators

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

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

so advertised

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

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 applies

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

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

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

Measured variables are always real. Hermite operators are therefore suitable for determining measured variables.

Now it is shown that the eigenvectors of Hermitian operators are orthogonal to different eigenvalues. 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 applies

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

so is

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

Of course, eigenvectors to degenerate eigenvalues ​​do not have to be orthonormal per se. However, the Gram-Schmidt orthonormalization method can be applied to them.

Now some operators will be examined for hermitism. It applies

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

so is

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

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. It applies

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

so is

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

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

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

is

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

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

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

It applies

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

That means

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

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

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

This applies

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

so is

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

The particle number operator is therefore Hermitian.

At this point some general statements about Hermite operators are made. 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 Let $f, g\in H$. Then applies

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

This also applies in this case

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

so

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

If $\newhat{A}$ and $\newhat{B}$ commute, then the consecutive execution $\newhat{A}\newhat{B}$ is also Hermitian. Furthermore, the following applies with $n\in \mathbb{N}$

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

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 Hermitean. 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 $\newhat{A}$ disappears with the Hamiltonian $\newhat{H}$

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

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 SG follows

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

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 from $\newhat{A}$ to 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) 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 different 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 destruction 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, the equations (4.82) and (4.83) apply

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

and

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

This is where ladder operators get their name. You can still create any state $\psi_n\left(u\right)$ from the ground state $\psi_0\left(u\right)$:

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

It should apply

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

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 is correct. It holds for $f, g:\mathbb{R}\to\mathbb{R}$ continuously-differentiable

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

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

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

applies

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

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

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

So it's general

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

For the x-component $\newhat{x}$ of the location operator $\newhat{\mathbf{r}}$ applies

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

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

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

For the Hamiltonian $\newhat{H}$ of the harmonic oscillator applies

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

4.7 Uncertainty principle

Let $\newhat{A}, \newhat{B}$ be two operators and be a particle 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 $\left\langle\left(\Delta\newhat{B}\right)^2\right\rangle$. It still applies

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

Now assume that $\newhat{A}$ and $\newhat{B}$ are Hermitian. The expected 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 applies

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

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

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

Putting this into Eq. (4.161), you get

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

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 it can be concluded that in general the expected values ​​of non-commuting operators cannot be determined with any precision at the same time. The eigenvalues ​​of Hermitian operators are the measured variables 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 complete observables if the $\newhat{A}_i$ commute in pairs with $1\leq i\leq N$. 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. B. $A$, 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 do this, we start from the classic case and set 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. If you do 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, you can calculate

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

Thus, an alternative Lagrange function is:

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

For the canonical impulses follows

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

For the Hamilton function follows

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

The replacement rules for operators are used to write

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

This is how we get 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 section 4.6.3 $\newhat{H}'$ and $\newhat{\mathbf{P}}$ therefore have the same eigenfunctions. One therefore looks for solutions that are eigenfunctions of $\newhat{\mathbf{P}}$:

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

It follows

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

This is what you put into the stationary SG:

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

With the definitions

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

this will be too

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

With the Hamilton operator

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

you get

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

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. So you can go through the transition

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

achieve greater accuracy in the hydrogen eigenstates. Now these eigenfunctions should 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. The solution is to make a product batch of the form

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

One can write $\Delta$ for the Laplace operator

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

with a portion $\Delta_{\theta, \phi}$, which only contains partial angle derivatives (see equation (B.94)). If you put this with Eq. (4.188) into the Schrödinger equation, you first get

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

One knows that $\psi$ is not zero outside of zero 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 you solve the angle component. The DGL for this is:

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

This is determined by the spherical surface functions $Y_{l, m}\left(\theta, \phi\right)$ Eq. (C.155) is fulfilled, see Eq. (C.166), where $\left|m\right|\leq l$ applies. The normalization condition is:

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

This is also fulfilled by the spherical surface functions $Y_{l, m}\left(\theta, \phi\right)$. 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

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

so is

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

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 drilling 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 continue to define $u \coloneqq \frac{r}{a_B}$. Then applies

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

With the replacement

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

and the definitions

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

becomes Eq. (4.199) to

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

For large $u$ applies

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

You make an approach

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

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

This results

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

For small $r$ holds

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

There is an approach here

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

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 set a function $P_l\left(u\right)$ with

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

to. Then apply

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

Putting this into Eq. (4.204), you get

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

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 are

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

You replace it

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

you get

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

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

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

from this it follows $l(4.207)

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

so $n\geq1$. It follows in original units

This therefore applies to the radial functions

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

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 is therefore:

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

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

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, Eq. (C.143)

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

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 are:

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

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

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

$l$	Bezeichnung
$0$	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 $l$, it is named as given in Tab. 4.1. Different eigenfunctions with the same eigenvalue are called degenerate. From the conditions $n\in\mathbb{N}$, $l

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

The Gaussian molecular formula Eq. (A.6) is 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 location $\mathbf{r} = \left(x_1, x_2, x_3\right)^T$ is given in classical mechanics by

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

defined. Replace 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

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

The location representation is written out in Cartesian coordinates

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

The momentum operator $\newhat{\mathbf{p}}$ is Hermitian just like the position operator $\newhat{\mathbf{r}}$ Since $\newhat{p}_j$ is swapped with $\newhat{x}_k$ for $x\not = k$, according to Sect. 4.6.2 is also the sequential execution $\newhat{p}_j\newhat{x}_k = \newhat{x}_k\newhat{p}_j$ Hermitian Thus the angular momentum operators are Hermitian. You define

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

As a shorthand notation, one uses a generalized angular momentum operator

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

a. Now some commutator relations of the angular momentum operators will be derived. It applies in the representation of the location

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

This means that the commutator has two components of the angular momentum operator

\[ \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} \]

Eq. (A.22) is 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 is

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

Still apply

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

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 should now 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. The well-known relations still apply

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

as well as

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

This follows in the location representation

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

Due to the equations (B.74) - (B.76) apply

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

Still applies

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

This corresponds to Eq. (B.96) of 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 $m$. Due to the location representation of $|n, l, m\rangle$ Eq. (4.233), the definition of the spherical area functions Eq. (C.155), the representation of the z-components of the angular momentum operator Eq. (4.257), the equation just derived. (4.259) and the property Eq. (C.166) of the spherical harmonic functions, the following two statements apply:

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

From this follow:

• The total torque $L$ in state $|n, l, m\rangle$ is $\hbar\sqrt{l\left(l + 1\right)}$. Therefore $l$ is called 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 magnetic quantum number.
• $\newhat{\mathbf{L}}^2$, $\newhat{L_z}$ and $\newhat{H}$ commute in pairs. 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.

You can still find

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

Now the generalized angular momentum operator $\newhat{\mathbf{J}}$ becomes Eq. (4.243) is treated a little further. It and its components were identified as Hermitian, so $\newhat{\mathbf{J}}^2$ is also Hermitian as a successive execution of Hermitian commutative operators according to section 4.6.2. It still applies

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

Among the four operators $\newhat{\mathbf{J}}^2, \newhat{J}_x, \newhat{J}_y, \newhat{J}_z$ one can use Eq. (4.245) and Eq. (4.249) find two that commute, namely $\newhat{\mathbf{J}}^2$ and another, for this $\newhat{J}_z$ is chosen. According to section 4.6.3, these have the same eigenfunctions, so you 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 look at the conditions

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

and

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

to. Since $\newhat{\mathbf{J}}^2$ commutes with $\newhat{J}_\pm$, we have

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

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

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

So it applies

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

with $c_\pm\in \mathbb{R}$. Therefore, $\newhat{J}_+$ is called ascend operator and $\newhat{J}_-$ descend operator. You now have to calculate the normalization constants $c_\pm$. This is done by normalization

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

of the conditions, you can simply demand this. So you get

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

Here you can see Eq. Insert (4.248) and get $c_\pm > 0$ with the requirement

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

Now we have to derive a condition for $\lambda$ and $m$. For a Hermitian operator $\newhat{A}$ holds

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

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 applies

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

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 has to stop somewhere because of (4.276), so the normalization Eq. (4.273) at a maximum $m_{{\mathrm{max}}}$ and at a minimum $m_{{\mathrm{min}}}$ disappear:

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

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

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

It applies

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

The ascending operator therefore leads from $-j$ to $j$ in integer steps. So $2j$ is an integer, therefore $j$ is an integer or half integer. Now there is a name change

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

a. These states are normalized and, as eigenstates of Hermitian operators, they are orthogonal to different eigenvalues:

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

If $j$ is an integer, the spatial representation of the solution is given by the spherical area functions Eq. (C.155). This describes, for example, the orbital torque of particles in the central force field, see section 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 we assume half-integer spin, namely $j = \frac{1}{2}$. In this case, the states $\left|j, m\right\rangle$ are denoted by $\left|ss_z\right\rangle = \left|\frac{1}{2}, s_z\right\rangle$. You define

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

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

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

what is fulfilled for

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

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

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

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

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

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

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

Therefore apply

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

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 torque of the elementary particles, namely the spin. Experimentally, you can only change the direction, but not the amount, of spin, which is a particle property, just like charge and mass. It is not linked to a rotating mass distribution, so no angular momentum in the pure sense.

Now the Schrödinger equation should be generalized to a particle with mass $m$, charge $q$ and spin 1/2. If you have now been 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 only depends on the location, but also on the spin state. However, one can write for the state of such a particle using the definitions Equations (4.284) and (4.285)

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

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

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

You can use Eq. (3.23)

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

replace 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 only knows one distinct direction in its rest system, 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

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

This was a vector

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

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 applies

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

with a classically inferable factor $g$, which is called Landé factor. For the Pauli-Hamilton operator follows

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

The Pauli equation is therefore:

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

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

Be an intrinsic value problem

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

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

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

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

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

find. To do this, you first set

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

with $\lambda\in\left[0, 1\right]$. Now one develops the eigenstates $\left|\psi\right\rangle$ and eigenvalues ​​$E$ according to 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 put into Eq. (4.308):

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

You sort Eq. (4.312) now to powers of $\lambda$, for the $m-$th order then applies

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

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

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

If you apply the undisturbed Hamiltonian $\newhat{H}_0$ to this, you get

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

If you put in Eq. (4.313) $m = 1$ and $\left|\psi\right\rangle = \left|\psi_n\right\rangle$ and multiply this from the left by $\left\langle\psi_n^{(0)}\right|$, you get

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

Also applies

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

The following applies to the first-order energy correction:

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

Now the $\left|\psi_n^{(1)}\right\rangle$ are searched for, 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 with $\left\langle\psi_k^{(0)}\right|$. It emerges

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

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

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

The norm of this is

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

So $C_{n, n}^{(1)} = ir$ 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 $C_{n, n}^{(1)}$ leads to a first order phase shift of $\left|\psi_n^{(0)}\right\rangle$, this phase shift can be absorbed in $\left|\psi_n^{(0)}\right\rangle$ and

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

set. So you get first order in the case of non-degenerate energy levels

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

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 you have to determine the eigenvalue and eigenvector simultaneously, in perturbation theory you can calculate the energy corrections and the state corrections successively one after the other.

To calculate the second order, one inserts into Eq. (4.313) with $m = 2$ first enter $\left|\psi\right\rangle = \left|\psi_n\right\rangle$ and multiply 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 $\left|\psi_n^{(2)}\right\rangle$ you now do the approach

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

This follows

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

Still applies

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

Thus follows

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

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. You can successively calculate further orders.

Now the case of degenerate energy eigenvalues ​​is considered. So let $N\in \mathbb{N}$ with $N\geq 2$ and the first $N$ undisturbed states $\left|\psi_1^{(0)}\right\rangle, \dotsc, \left|\psi_N^{(0)}\right\rangle$ are 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. Even with a very weak perturbation $\lambda\to 0$ you have to assume that degenerate states are mixed in, so you do this approach

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

This is put into the Schrödinger equation Eq. (4.311) a:

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

This is true to zero order. In the following only the first order is considered. The $\left|\psi_l^{(0)}\right\rangle, 1\leq l\leq N$, are orthonormalized. Let $j\in \mathbb{N}$ with $1\leq j\leq N$ given. By multiplying the first-order terms in Eq. (4.334) from the left with $\left\langle\psi_j^{(0)}\right|$ is obtained

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

The solution $\left(C, \dotsc, C_N, E^{(1)}\right)$ gives the zero-order correction in the state and first-order correction in the energy. The equation applies to all $1\leq l\leq N$, so you 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 obtains $N$ orthonormal eigenvectors $\mathbf{C}^{k}$ to 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 conditions are orthonormal. Namely, let $k, m\in \mathbb {N}$ with $1\leq k, m\leq N$ be given. Then applies

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

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 of all, let us consider the definition of Bohr's radius $a_B$ according to Eq. (4.200) reminds:

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

For the energy eigenvalues ​​in the H atom, Eq. (4.223)

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

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

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

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

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

$\alpha$ is the fine structure constant for which applies

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

The spin-orbit coupling does not arise from a possible magnetic core dipole. In a semi-classical view, the electron moves around the nucleus with an instantaneous speed $\mathbf{v}$. In the current rest frame of the electron, according to Eq. (3.105) a magnetic field

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

which the electron feels. 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 one approaches here for the relativistic B field

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

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 you 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 observation of the accelerated electron using the Dirac equation results in an additional factor $1/2$ at this point. The perturbation operator we are looking for 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 torque operator

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

you can do this

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

rewrite. In order to be able to continue working with this, the angular momentum eigenstates of the H atom are coupled

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

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

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

with

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

It holds for $l\not = 0$

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

You identify

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

This is where 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) follows

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

Therefore applies

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

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 $j = l = \frac{1}{2}$. So you get 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 have an influence on the own energies in the H atom and thus also on 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

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

is now a perturbation operator

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

to use. 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; one can write with a misuse of the term

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

This section does not go further than the first order. Now you set 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 you

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

Putting this into Eq. (4.369), you get

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

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

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

So you get for the admixture of the states

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

If the system is in state $n$ at time $t_0$, then we get:

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

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 you can get the time-dependent solution

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

construct. 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 measurements. Putting this into Eq. (4.380), follows

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

$t_0 = 0$ were 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 is

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

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

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

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.83)

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

This can 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 disturbance $\omega = 0$ 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

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

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

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

This describes an electromagnetic wave. For the Hamilton operator follows

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

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

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

you can write

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

this was done

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

used. With the long-wave approximation

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

you get

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

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

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

$\Delta E\sim\frac{E_{\text{at}}}{10}$ was used. With Eq. (4.392) is obtained

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

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 amount 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 you do the math

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

It applies

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

Thus follows

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

So you 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 given by Eq. (4.233) is known, so it applies

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

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

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

relevant. You write

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

From the equations (C.171) - (C.172) 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. So you have

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

4.11.5 Relativistic corrections in the H atom

4.12 Multi-particle systems

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

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

described with $1\leq i\leq N$ and $x_i = \left(\mathbf{r}_i, \sigma_i\right)$, where $\mathbf{r}_i$ denotes the location 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 $\newhat{P}_{i, j}$ through

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

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, which can depend on $i, j$, which has been neglected in the notation here. It follows anyway by using the same operator again

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

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 symmetrical 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

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

applies. If you now apply $\newhat{P}_{i, j}$ to this, 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 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. You also need 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, so you would need to store the state of the electron shell when each axis is discretized into ten intervals (which is unlikely to be enough).

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

Data points. These are two complex numbers each, so you need approx.

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

of storage space, which is currently an unrealistic amount. If you want to derive spectra theoretically, you have to think about approximation methods (this is almost always the case in QM). The most common procedures 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 $1\leq i\leq N$

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

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 variables, the radiation field and the existing material densities.