These online lecture notes are currently growing out of a 13 weeks course with 3 academic hours per week on Advanced Methods of Theoretical Physics within an MSc program on Quantum Physics and Technology at the Technical University Braunschweig. These notes are used as-is and one-to-one in the lectures.
The course is aimed at our students with a 3-year BSc in physics and an elementary knowledge of quantum mechanics and quantum statistical mechanics from the introductory theory courses. Contrary to its title, this course is anything but 'advanced' and also does not refer to theoretical physics in general. Rather, it tries to incoherently fill in some of the most embarrassing gaps in quantum and statistical mechanics, which are a consequence of curtailing our past Diploma education down to BSc standards.
These notes are a moving target - regarding everything! So, you don't want to print or store them! Instead, you want to visit
http://www.fkt.tu-bs.de/vl/
Wolfram Brenig

# Contents

Dirac's Perturbation Theory
1.1  The time evolution operator
1.2  Transition propabilities
2.1  Quantizing Electromagnetic Fields in Vacuum
2.2  Classical States of the Radiation Field
Quantum Theory of Scattering
4.1  Differential Cross Section
4.2  Born Approximation
4.3  T-Matrix and Higher-Order Born Approximation
4.4  The Optical Theorem
4.5  Partial Wave Expansion
4.6  How to Calculate Phase Shifts
Index

## 1Dirac's Perturbation Theory

### 1.1The time evolution operator

Dirac's perturbation theory, is also frequently called 'time dependent' perturbation theory. Consider the time evolution operator $U\left(t,{t}_{0}\right)$
 $|\psi ,t⟩=U\left(t,{t}_{0}\right)|\psi ,{t}_{0}⟩$ $\left(1\right)$
of a state $|\psi ,t⟩$, which is set by Schrödinger's equation for a general, time dependent Hamiltonian $H\left(t\right)$
 $i\hslash {\partial }_{t}|\psi ,t⟩=H\left(t\right)|\psi ,t⟩$ $\left(2\right)$
or
 $i\hslash {\partial }_{t}U\left(t,{t}_{0}\right)=H\left(t\right)U\left(t,{t}_{0}\right)$ $\left(3\right)$
i.e. $U\left(t,{t}_{0}\right)$ is valid for any state. The time evolution operator should have the following properties
 $\begin{array}{cccc}\multicolumn{1}{c}{U\left({t}_{0},{t}_{0}\right)}& =\hfill & 1\hfill & \hfill \left(4\right)\\ \multicolumn{1}{c}{U\left({t}_{2},{t}_{1}\right)U\left({t}_{1},{t}_{0}\right)}& =\hfill & U\left({t}_{2},{t}_{0}\right)\hfill & \hfill \left(5\right)\end{array}$

Eqn.(3) is formally solved by
 $U\left(t,{t}_{0}\right)=1-\frac{i}{\hslash }{\int }_{{t}_{0}}^{t}d\tau H\left(\tau \right)U\left(\tau ,{t}_{0}\right)$ $\left(6\right)$
From eqns.(6,4) and from $t={t}_{0}+\mathrm{dt}$
 $U\left(t+\mathrm{dt},t\right)=1-\frac{i}{\hslash }H\left(t\right)\mathrm{dt}$ $\left(7\right)$
I.e. an infinitesimal time-translation is realized a
unitary transformation, since
 $U\left(t+\mathrm{dt},t{\right)}^{+}U\left(t+\mathrm{dt},t\right)=1+\frac{i}{\hslash }H\left(t{\right)}^{+}\mathrm{dt}-\frac{i}{\hslash }H\left(t\right)\mathrm{dt}+O\left({\mathrm{dt}}^{2}\right)=1+O\left({\mathrm{dt}}^{2}\right)$ $\left(8\right)$
The latter is true since the Hamiltonian is hermitean, even for an explicitly time dependent situation. Since any time-translation can be viewed as an infinite succession of infinitesimal time-translations, we find that $U\left(t,{t}_{0}\right)$ is unitary
 $\begin{array}{cccc}\multicolumn{1}{c}{U\left(t,{t}_{0}{\right)}^{+}U\left(t,{t}_{0}\right)}& =\hfill & 1\hfill & \hfill \left(9\right)\\ \multicolumn{1}{c}{⇒ U\left(t,{t}_{0}{\right)}^{+}}& =\hfill & U\left({t}_{0},t\right)\hfill & \hfill \left(10\right)\end{array}$

In the
time-independent case the solution of eqn.(3) is simply
 $U\left(t,{t}_{0}\right)={e}^{-\mathrm{iH}\left(t-{t}_{0}\right)/\hslash }$ $\left(11\right)$
Consider now the case of
 $H\left(t\right)={H}_{0}+V\left(t\right)$ $\left(12\right)$
where we assume that we are able to diagonalize $H$. This resembles a time-depen
t perturbation $V\left(t\right)$ to an unperturbed Hamiltonian $H$. Note, that implicitely this also includes the case of $V$ being time-independent.
The solution of eqn.(3) proceeds via an Ansatz
 $U\left(t,{t}_{0}\right)={e}^{-{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }{U}_{D}\left(t,{t}_{0}\right)$ $\left(13\right)$
Note that any state $|\psi ,t{⟩}_{D}$, which evolves in time via ${U}_{D}\left(t,{t}_{0}\right)$, i.e. by $|\psi ,t{⟩}_{D}={U}_{D}\left(t,{t}_{0}\right)|\psi ,{t}_{0}{⟩}_{D}$ is obtained from $|\psi ,t⟩$ by
 $|\psi ,t{⟩}_{D}={e}^{{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }|\psi ,t⟩$ $\left(14\right)$
I.e. the states $|\psi ,t{⟩}_{D}$ are simply the representations of $|\psi ,t⟩$ in the
Dirac (or interaction) picture of $H\left(t\right)$ or of the Heisenberg picture of $H$. Therefore ${U}_{D}\left(t,{t}_{0}\right)$ is the time evolution operator in the Dirac picture. If $‖V‖/‖H‖\to 0$, we expect ${U}_{D}\left(t,{t}_{0}\right)\to 1$, therefore, we expect that ${U}_{D}\left(t,{t}_{0}\right)-1$ will allow for some kind of perturbation expansion in terms of $V$. Inserting eqn.(13) into eqn.(3) we get

In the last line of this equation we have used the usual tranformation of operators bewteen quantum mechanical pictures, i.e. we have defined the
interaction in the Dirac picture
 ${V}_{D}\left(t,{t}_{0}\right)={e}^{{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }V\left(t\right){e}^{-{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }$ $\left(16\right)$
Eqn.(15) can be integrated formally
 ${U}_{D}\left(t,{t}_{0}\right)=1-\frac{i}{\hslash }{\int }_{{t}_{0}}^{t}d\tau {V}_{D}\left(\tau ,{t}_{0}\right){U}_{D}\left(\tau ,{t}_{0}\right)$ $\left(17\right)$
Now we are in a position to apply the main idea of Dirac's perturbation theory, which lies in the fact that for $‖V‖/‖H‖\to 0$ an expansion of ${U}_{D}\left(t,{t}_{0}\right)-1$ in powers of ${V}_{D}$ can be obtained by solving eqn.(15)
iteratively
 $\begin{array}{ccc}\multicolumn{1}{c}{{U}_{D}\left(t,{t}_{0}\right)}& =\hfill & 1+\left(-\frac{i}{\hslash }\right){\int }_{{t}_{0}}^{t}d{\tau }_{1}{V}_{D}\left({\tau }_{1},{t}_{0}\right)+\left(-\frac{i}{\hslash }{\right)}^{2}{\int }_{{t}_{0}}^{t}{\int }_{{t}_{0}}^{{\tau }_{1}}d{\tau }_{1}d{\tau }_{2}{V}_{D}\left({\tau }_{1},{t}_{0}\right){V}_{D}\left({\tau }_{2},{t}_{0}\right)\hfill \\ \multicolumn{1}{c}{}& \hfill & +\dots \hfill & \hfill \left(18\right)\end{array}$

Formally this can be exponentiated by introducing the so called time ordering of operators
 ${T}_{\tau }\left[A\left({t}_{1}\right)B\left({t}_{2}\right)\right]=\left\{\begin{array}{cc}\hfill A\left({t}_{1}\right)B\left({t}_{2}\right),\hfill & \hfill {t}_{1}\geqslant{t}_{2}\hfill \\ \hfill B\left({t}_{2}\right)A\left({t}_{1}\right),\hfill & \hfill {t}_{2}>{t}_{1}\hfill \end{array}$ $\left(19\right)$
Since there are $n!$ ways to order $n$ times ${t}_{1\dots n}$ we may write
 $\begin{array}{ccc}\multicolumn{1}{c}{{U}_{D}\left(t,{t}_{0}\right)}& =\hfill & {T}_{\tau }\left[\sum _{n=0}^{\infty }\frac{1}{n!}\left(-\frac{i}{\hslash }{\right)}^{n}{\int }_{{t}_{0}}^{t}d{\tau }_{1}\dots {\int }_{{t}_{0}}^{t}d{\tau }_{n} {V}_{D}\left({\tau }_{1},{t}_{0}\right)\dots {V}_{D}\left({\tau }_{n},{t}_{0}\right)\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & {T}_{\tau }\mathrm{exp}\left[-\frac{i}{\hslash }{\int }_{{t}_{0}}^{t}d\tau {V}_{D}\left(\tau ,{t}_{0}\right)\right]\hfill & \hfill \left(20\right)\end{array}$

Note that in contrast to the 2nd line in eqn.(18), in the 1st line of eqn.(20)
all upper intergration bounds extend to $t$.
In the trivial case, that $V$ is time independent and $\left[{H}_{0},V\right]=0$ a perturbation expansion in terms of $V$ would not have been necessary right from the start. In fact, in that case
 $\begin{array}{ccc}\multicolumn{3}{c}{{V}_{D}\left(\tau ,{t}_{0}\right)={e}^{{\mathrm{iH}}_{0}\left(\tau -{t}_{0}\right)/\hslash }{\mathrm{Ve}}^{-{\mathrm{iH}}_{0}\left(\tau -{t}_{0}\right)/\hslash }=V}\\ \multicolumn{1}{c}{}& ⇒\hfill & {T}_{\tau }\mathrm{exp}\left[-\frac{i}{\hslash }{\int }_{{t}_{0}}^{t}d\tau {V}_{D}\left(\tau ,{t}_{0}\right)\right]={T}_{\tau }\mathrm{exp}\left[-\frac{i}{\hslash }{\int }_{{t}_{0}}^{t}d\tau V\right]={e}^{-\mathrm{iV}\left(t-{t}_{0}\right)/\hslash }\hfill \\ \multicolumn{1}{c}{}& ⇒\hfill & U\left(t,{t}_{0}\right)={e}^{-{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }{e}^{-\mathrm{iV}\left(t-{t}_{0}\right)/\hslash }={e}^{-i\left({H}_{0}+V\right)\left(t-{t}_{0}\right)/\hslash }\hfill & \hfill \left(21\right)\end{array}$

which is exactly what we expect.
Each addends from the 2nd line in eqn. (20) to the total time evolution operator $U\left(t,{t}_{0}\right)={e}^{-{\mathrm{iH}}_{0}\left(t-{t}_{0}\right)/\hslash }{U}_{D}\left(t,{t}_{0}\right)$ can be written somewhat differently using eqn. (16) and defining the time evolution operator ${U}_{0}\left({t}_{1},{t}_{2}\right)={e}^{-{\mathrm{iH}}_{0}\left({t}_{1}-{t}_{2}\right)/\hslash }$ of the unperturbed sytem, namely
 $\begin{array}{ccc}\multicolumn{1}{c}{{U\left(t,{t}_{0}\right)|}_{O\left({V}^{n}\right)}}& =\hfill & \left(-\frac{i}{\hslash }{\right)}^{n}{\int }_{{t}_{0}}^{t}\dots {\int }_{{t}_{0}}^{{\tau }_{n-2}}{\int }_{{t}_{0}}^{{\tau }_{n-1}}d{\tau }_{1}\dots d{\tau }_{n-2}d{\tau }_{n}\left[{U}_{0}\left(t,{\tau }_{1}\right)V\left({\tau }_{1}\right){U}_{0}\left({\tau }_{1},{\tau }_{2}\right)\hfill \\ \multicolumn{1}{c}{}& \hfill & \dots {U}_{0}\left({\tau }_{n-2},{\tau }_{n-1}\right)V\left({\tau }_{n-1}\right){U}_{0}\left({\tau }_{n-1},{\tau }_{n}\right)V\left({\tau }_{n}\right){U}_{0}\left({\tau }_{n},{t}_{0}\right)\right]\hfill & \hfill \left(22\right)\end{array}$

with this we may express the $O\left({V}^{n}\right)$ contribution to the time evolution ${U\left(t,{t}_{0}\right)|}_{O\left({V}^{n}\right)}|\psi ,{t}_{0}⟩$ of state in a diagrammatic way
Figure 1: Diagrammatic expression for Dirac's state propagator at order ${V}^{n}$

### 1.2Transition propabilities

Consider a quantum system described by $H={H}_{0}+V$, where
 ${H}_{0}|n⟩={e}_{n}|n⟩ ⟨m|n⟩={\delta }_{\mathrm{mn}}$ $\left(23\right)$
we know how to diagonalize ${H}_{0}$. Now, assume that at time ${t}_{0}=0$ the system is in an
initial state $|n⟩$. We seek to obtain the propability ${P}_{\mathrm{mn}}\left(t\right)$ by which we will find the system in the final state $|m⟩$ at time $t$
 ${P}_{\mathrm{mn}}\left(t\right)={|⟨m|{e}^{-{\mathrm{iH}}_{0}t/\hslash }{U}_{D}\left(t,0\right)|n⟩|}^{2}$ $\left(24\right)$
The corresponding amplitude is

For the remainder of this section we confine ourselves to the case $m\ne n$ and to only the leading order contribution in $V$ to ${P}_{\mathrm{mn}}\left(t\right)$, which is $O\left({V}^{2}\right)$. Because of the ${\delta }_{\mathrm{mn}}$-function in the 1st term of eqn.(25) the order $O\left({V}^{2}\right)$ term in eqn.(25) does not contribute to ${P}_{m\ne n}\left(t\right)$. We get
 ${P}_{m\ne n}\left(t\right)=|⟨m|V|n⟩{|}^{2}\frac{4{\mathrm{sin}}^{2}\left[\left({e}_{m}-{e}_{n}\right)t/\left(2\hslash \right)\right]}{\left({e}_{m}-{e}_{n}{\right)}^{2}}$ $\left(26\right)$
where we have used $1-\mathrm{cos}\left(x\right)=2{\mathrm{sin}}^{2}\left(x/2\right)$.
Exercise 1 Consider the quantum system eqn.(23) in state $|n⟩$ at time ${t}_{0}=0$. Show that the expression corresponding to eqn.(26) for $m=n$ is
 ${P}_{\mathrm{nn}}\left(t\right)=1-\sum _{m\ne n}|⟨m|V|n⟩{|}^{2}\frac{4{\mathrm{sin}}^{2}\left[\left({e}_{m}-{e}_{n}\right)t/\left(2\hslash \right)\right]}{\left({e}_{m}-{e}_{n}{\right)}^{2}}$ $\left(27\right)$
Interpret this result in terms of eqn.(26) and probability conservation.

#### Fermis Golden Rule

If we ask for the rate ${R}_{\mathrm{mn}}\left(t\right)$ of transition from state $|n⟩$ into state $|m⟩$ we have to consider
 ${R}_{\mathrm{mn}}\left(t\right)=\frac{{\mathrm{dP}}_{\mathrm{mn}}\left(t\right)}{\mathrm{dt}}=|⟨m|V|n⟩{|}^{2}\frac{2\pi }{\hslash {}^{2}} \frac{\mathrm{sin}\left[\left({e}_{m}-{e}_{n}\right)t/\hslash \right]}{\pi \left({e}_{m}-{e}_{n}\right)/\hslash }$ $\left(28\right)$
Using that
 $\underset{t\to \infty }{lim}\frac{\mathrm{sin}\left[\left({e}_{m}-{e}_{n}\right)t/\hslash \right]}{\pi \left({e}_{m}-{e}_{n}\right)/\hslash }=\hslash \delta \left({e}_{m}-{e}_{n}\right)$ $\left(29\right)$
we find that for very large times the transition rate approaches a time independent value
 ${R}_{\mathrm{mn}}=\frac{2\pi }{\hslash }|⟨m|V|n⟩{|}^{2}\delta \left({e}_{m}-{e}_{n}\right)$ $\left(30\right)$
This result is rather useless: either ${e}_{m}={e}_{n}$, then ${R}_{\mathrm{mn}}=\infty$ i.e. perturbaion theory is invalid, or ${e}_{m}\ne {e}_{n}$ then ${R}_{\mathrm{mn}}=0$ which is not too interesting. In fact, eqn.(30) is used differently usually, namely, it is integrated over a
dense spectrum either of the initial or final states. Eg., let the states $|m⟩$ depend on a set of quantum numbers, i.e. $|m\left(a,b\dots \right)⟩$ and $\epsilon \left(a,b\dots \right)$ is the energy. Consider the sum of all transition rates into states in the vicinity of ${e}_{n}$
 $\begin{array}{ccc}\multicolumn{1}{c}{{r}_{\mathrm{mna}\dots }}& =\hfill & \sum _{\epsilon ={e}_{n}-\Delta }^{\epsilon ={e}_{n}+\Delta }{R}_{\mathrm{mn}}{|}_{a,\dots }={\int }_{{e}_{n}-\Delta }^{{e}_{n}+\Delta }\mathrm{de}\sum _{\epsilon }\delta \left(e-\epsilon \right){|}_{a,\dots }{R}_{\mathrm{mn}}={\int }_{{e}_{n}-\Delta }^{{e}_{n}+\Delta }\mathrm{de}{\rho }_{a\dots }\left(e\right){R}_{\mathrm{mn}}\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{2\pi }{\hslash }|⟨m|V|n⟩{|}^{2}{\rho }_{a\dots }\left({e}_{n}\right)\hfill & \hfill \left(31\right)\end{array}$

This is Fermi's famous golden rule. The constraint ${|}_{a,\dots }$ symbolizes, that in performing the sum over energies certain quantum number may be kept fixed and ${\rho }_{a\dots }\left(e\right)=\sum _{\epsilon }\delta \left(e-\epsilon \right){|}_{a,\dots }$ is the density of states. Obviously the size $2\Delta$ of the energy interval is not relevant - as long as it contains the initial state energy ${e}_{n}$. Note that ${r}_{\mathrm{mna}\dots }~\left[{\mathrm{erg}}^{2}/\mathrm{erg}/\hslash \right]$ has dimension $\left[1/\mathrm{time}\right]$, as has ${R}_{\mathrm{mn}}$.
Exercise 2 Given the two-level system
 $H={H}_{0}+W, {H}_{0}=\left[\begin{array}{cc}\hfill E\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -E\hfill \end{array}\right], W=\left[\begin{array}{cc}\hfill 0\hfill & \hfill V\hfill \\ \hfill V\hfill & \hfill 0\hfill \end{array}\right]$ $\left(32\right)$
The eigenstates of ${H}_{0}$ are $|0⟩$ and $|1⟩$, with energies $E$ and $-E$. Show that the time evolution operator in Schrödinger's picture is given by
 $U\left(t,0\right)=\left[\begin{array}{cc}\hfill \mathrm{cos}\left(a\tau \right)-\frac{\mathrm{iE}}{a}\mathrm{sin}\left(a\tau \right)\hfill & \hfill -\frac{\mathrm{iV}}{a}\mathrm{sin}\left(a\tau \right)\hfill \\ \hfill -\frac{\mathrm{iV}}{a}\mathrm{sin}\left(a\tau \right)\hfill & \hfill \mathrm{cos}\left(a\tau \right)+\frac{\mathrm{iE}}{a}\mathrm{sin}\left(a\tau \right)\hfill \end{array}\right]$ $\left(33\right)$
with $a=\sqrt{{E}^{2}+{V}^{2}}$ and $\tau =t/\hslash$. Compare this result with eqn.(26,27). What are the differences? Assuming that the system is in state $|0⟩$ at $t=0$, plot ${P}_{\mathrm{mn}}\left(t\right)$ from eqn.(24) for $m,n=1,0$ and $m,n=0,0$ setting $\hslash =1$, $E=1$, and $V=0.1$

### 2.1Quantizing Electromagnetic Fields in Vacuum

In cases where there are no charges or currents the electromagnetic fields $E$ and $B$ can most conveniently be derived from the vector potential $A$ only using the Coulomb gauge
 $\phi =0 , \nabla ·A=0$ $\left(34\right)$
 $E=-\frac{1}{c}{\partial }_{t}A , B=\nabla ×A$ $\left(35\right)$
In this situation the vector potential satisfies the
homogeneous wave equation
 $▵A-\frac{1}{{c}^{2}}{\partial }_{t}^{2}A=0$ $\left(36\right)$
Quantization of the electromagnetic fields is based on a "quantization" of the vector potential. This needs some clarification: as for the coordinate and momentum of a single particle in basic quantum mechanics, we seek to express physical observables of Maxwell's theory, in terms of linear operators. The 'observable' we chose is the vector potential. Strictly speaking $A$ is not an observable, however, simple differential operations on it, i.e. eqn. (35) lead to such. Now, in contrast to $x$ and $p$ of a single particle $A\left(r\right)$ is a vector field, which takes on values for
all $r\in {R}^{3}$. It is customary to avoid the mathematical nuances which may arise from such a distribution of observables by discretizing the set of solutions to eqn. (36). To this end we confine the vector potential to some suitably chosen periodicity volume $V$ which allows to expand $A$ in terms of an othonormal set of eigenfunctions of eqn. (36) subject to the boundary conditions set by the periodicity volume. This procedure has nothing to do with quantization. Without loss of generality we choose $V$ to be a cube of Volume $V$ and linear dimension $L$. The periodicity conditions are
 $A\left(x+L,y,z\right)=A\left(x,y,z\right) & \mathrm{cyclic}$ $\left(37\right)$
This allows to expand $A\left(r,t\right)$ in a Fourier series
 $A\left(r,t\right)=\sum _{k,\sigma =1,2}{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{k}}\right)}^{1/2}{u}_{k,\sigma }{a}_{k,\sigma }\left(t\right) {e}^{ik·r}$ $\left(38\right)$
where the wave vector $k=\left({l}_{x},{l}_{y},{l}_{z}\right)2\pi /L$ and ${l}_{i}\in Z$. The normalization prefactor in the $\left(\right)$-brackets has been set for later convenience. Remember, that because of $\nabla ·A=0$, eqns. (34-38) allow for two wave polarizations transverse to $k$. These are accounted for by the sum over the two unit vectors ${u}_{k,\sigma }\perp k$. In view of our goal to develop a quantum theory for the vector potential, the time dependence of anticipated
operator $A\left(r,t\right)$ is analogous to a time-dependence of to operators $x\left(t\right)$ or $p\left(t\right)$ of a single particle, i.e. this time dependence should be interpreted as to result from the Heisenberg picture. Because $A$ is real $A={A}^{☆}$ eqn. (38) is an overcomplete representation of $A$ which has to satisfy
 ${a}_{k,\sigma }\left(t\right)={a}_{-k,\sigma }^{☆}\left(t\right)$ $\left(39\right)$
where we set ${u}_{k,\sigma }={u}_{-k,\sigma }$ and ${u}_{k,\sigma }={u}_{k,\sigma }^{☆}$. Therefore
 $A\left(r,t\right)=\sum _{k,\sigma =1,2}\frac{1}{2}{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{k}}\right)}^{1/2}{u}_{k,\sigma }\left({a}_{k,\sigma }\left(t\right) {e}^{ik·r}+{a}_{k,\sigma }^{☆}\left(t\right) {e}^{-ik·r}\right)$ $\left(40\right)$
The Fourier coefficients ${a}_{k,\sigma }\left(t\right)$ satisfy the
2nd order linear DEQN
 $\left({d}_{t}^{2}-{\omega }_{k}^{2}\right){a}_{k,\sigma }\left(t\right)=0 ⇒ {a}_{k,\sigma }\left(t\right)={b}_{k,\sigma }{e}^{-i{\omega }_{k}t}+{b}_{-k,\sigma }^{☆}{e}^{i{\omega }_{k}t}$ $\left(41\right)$
with the dispersion relation ${\omega }_{k}=\mathrm{kc}$ and $k=|k|$. The particular setting of the amplitudes ${b}_{k,\sigma }, {b}_{-k,\sigma }^{☆}$ for the two linearly independent solutions has been chosen to fulfill eqn. (39). Because of the latter, the amplitudes ${a}_{k,\sigma }\left(t\right)$ cannot be viewed as independent from each other. Inserting into eqn. (40) and using ${\omega }_{k}={\omega }_{-k}$ we get
 $A\left(r,t\right)=\sum _{k,\sigma =1,2}{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{k}}\right)}^{1/2}{u}_{k,\sigma }\left({b}_{k,\sigma }\left(t\right) {e}^{ik·r}+{b}_{k,\sigma }^{☆}\left(t\right) {e}^{-ik·r}\right)$ $\left(42\right)$
where ${b}_{k,\sigma }\left(t\right)$ has to satisfy only the
1st order linear DEQN
 $\left({d}_{t}+i{\omega }_{k}\right){b}_{k,\sigma }\left(t\right)=0 ⇒ {b}_{k,\sigma }\left(t\right)={b}_{k,\sigma }{e}^{-i{\omega }_{k}t}$ $\left(43\right)$
In contrast to ${a}_{k,\sigma }\left(t\right)$, the ${b}_{k,\sigma }\left(t\right)$ can now be viewed as independent amplitudes for each $k$ and $\sigma$ and eqn. (43) can be considered as their equation of motion. Eqn. (43) is the point of entry into quantizing $A$. In fact it can be interpreted in terms of Heisenberg's equation of motion for ${b}_{k,\sigma }\left(t\right)$, which we will show now.
The Hamiltonian function of the electromagnetic field can be expressed in terms of ${b}_{k,\sigma }\left(t\right)$
 $\begin{array}{ccc}\multicolumn{1}{c}{H}& =\hfill & \frac{1}{8\pi }\int {d}^{3}r\left({E}^{2}+{B}^{2}\right)=\frac{1}{8\pi }\int {d}^{3}r\left(\frac{1}{{c}^{2}}|{\partial }_{t}A{|}^{2}+|\nabla ×A{|}^{2}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \sum _{k,\mu =1,2} \sum _{q,\nu =1,2}\left[{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{k}}\right)}^{1/2}{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{q}}\right)}^{1/2}\hfill \\ \multicolumn{1}{c}{}& \hfill & \left(-\right){u}_{k,\mu }·{u}_{q,\nu }\frac{{\omega }_{k}{\omega }_{q}}{{c}^{2}}\int \left({b}_{k,\mu }\left(t\right) {e}^{ik·r}-{b}_{k,\mu }^{☆}\left(t\right) {e}^{-ik·r}\right)×\hfill \\ \multicolumn{1}{c}{}& \hfill & \left({b}_{q,\nu }\left(t\right) {e}^{iq·r}-{b}_{q,\nu }^{☆}\left(t\right) {e}^{-iq·r}\right){d}^{3}r\hfill \\ \multicolumn{1}{c}{}& \hfill & -\left(k×{u}_{k,\mu }\right)·\left(q×{u}_{q,\nu }\right)\int \left({b}_{k,\mu }\left(t\right) {e}^{ik·r}-{b}_{k,\mu }^{☆}\left(t\right) {e}^{-ik·r}\right)×\hfill \\ \multicolumn{1}{c}{}& \hfill & \left({b}_{q,\nu }\left(t\right) {e}^{iq·r}-{b}_{q,\nu }^{☆}\left(t\right) {e}^{-iq·r}\right){d}^{3}r\right]\hfill & \hfill \left(44\right)\end{array}$

Using
 $\begin{array}{ccc}\multicolumn{1}{c}{\int {e}^{ik·r}{e}^{iq·r}{d}^{3}r}& =\hfill & V{\delta }_{k,q}\hfill \\ \multicolumn{1}{c}{{u}_{k,\mu }·{u}_{q,\nu }}& =\hfill & {\delta }_{\mu ,\nu }\hfill & \hfill \left(45\right)\end{array}$

we get

where we have discarded the time dependence of the ${b}_{k,\sigma }$ for the moment. Eqns. (46) and (47) are surely identical for classical C-number variables ${b}_{k,\sigma }$.
The main point of the preceeding is, that eqns. (43) and (46, 47) allows for a consistent application of the correspondence principle. Namely eqns. (46) and (47) can be viewed as a sum of Hamiltonians of independent harmonic oscillators with energy ${E}_{k}=\hslash {\omega }_{k}$ for each wave vector and polarization. In this context, the C-number variables ${b}_{k,\sigma }, {b}_{k,\sigma }^{☆}$ are interpreted as boson operators ${b}_{k,\sigma }, {b}_{k,\sigma }^{+}$ with commutation relations
 $\left[{b}_{k,\mu },{b}_{q,\nu }^{+}\right]={\delta }_{k,q}{\delta }_{\mu ,\nu }, \left[{b}_{k,\mu },{b}_{q,\nu }\right]=0, \left[{b}_{k,\mu }^{+},{b}_{q,\nu }^{+}\right]=0$ $\left(48\right)$
Heisenberg's equation of motion yields
 ${d}_{t}{b}_{k,\mu }\left(t\right)=\frac{i}{\hslash }\left[H,{b}_{k,\mu }\right]\left(t\right)=\frac{i}{\hslash }\sum _{q,\sigma =1,2}\hslash {\omega }_{q}\left[{b}_{q,\sigma }^{☆}{b}_{q,\sigma },{b}_{k,\mu }\right]\left(t\right)=-i{\omega }_{k}{b}_{k,\mu }\left(t\right)$ $\left(49\right)$
which is exactly eqn. (43). A priori, and because of the commutation relations eqn. (48), the final from of the Hamiltonian is fixed only up to a constant - the socalled
zero-point energy of the vacuum. As we will see later, and because of actual verification via experiment (eg. the Casimir effect) eqn. (46) is already of the proper form. Inserting eqn. (48) we get
 $H=\sum _{k,\sigma =1,2}\hslash {\omega }_{k}\left({b}_{k,\sigma }^{+}{b}_{k,\sigma }+\frac{1}{2}\right)=\sum _{k,\sigma =1,2}\hslash {\omega }_{k}\left({N}_{k,\sigma }+\frac{1}{2}\right)$ $\left(50\right)$
where ${N}_{k,\sigma }={b}_{k,\sigma }^{+}{b}_{k,\sigma }$ is the boson occupation number operator for each mode at $\left(k,\sigma \right)$. Therefore the
zero-point energy of the vacuum is
 ${E}_{0}=\sum _{k,\sigma =1,2}\hslash {\omega }_{k}/2$ $\left(51\right)$
which is present even if the boson occupation in each harmonic oscillator is identical to zero. Simple insepction of ${E}_{0}$ shows, that the sum on the right hand side diverges. I.e. the zero-point energy of the vacuum is
Finally note, that having no explicit time-dependence displayed on eqn.(50) is exactly what we should expect since ${d}_{t}H=i\left[H,H\right]/\hslash =0$. I.e. as usual, the Hamiltonian is time independent in the Schrödinger and in the Heisenberg picture.
The eigenstates of the radiation field are direct products of eigenstates of each harmonic oscillator
 $|\left\{{n}_{k,\sigma }\right\}⟩\equiv |{n}_{{k}_{1},{\sigma }_{1}}⟩\otimes \dots \otimes |{n}_{{k}_{s},{\sigma }_{s}}⟩\otimes \dots \equiv |{n}_{{k}_{1},{\sigma }_{1}},\dots ,{n}_{{k}_{s},{\sigma }_{s}},\dots ⟩$ $\left(52\right)$
Remember the usual things
 $\begin{array}{ccc}\multicolumn{1}{c}{{N}_{k,\sigma }|\left\{{n}_{q,\mu }\right\}⟩}& =\hfill & {n}_{k,\sigma }|\left\{{n}_{q,\mu }\right\}⟩\hfill \\ \multicolumn{1}{c}{{b}_{k,\sigma }^{+}|\left\{{n}_{q,\mu }\right\}⟩}& =\hfill & \sqrt{{n}_{k,\sigma }+1}|\dots \left({n}_{k,\sigma }+1\right)\dots ⟩\hfill \\ \multicolumn{1}{c}{{b}_{k,\sigma }|\left\{{n}_{q,\mu }\right\}⟩}& =\hfill & \sqrt{{n}_{k,\sigma }}|\dots \left({n}_{k,\sigma }-1\right)\dots ⟩\hfill & \hfill \left(53\right)\end{array}$

For interpretational purposes it is interesting to evaluate the
Poynting vector $P$ of the electromagnetic field
 $\begin{array}{ccc}\multicolumn{3}{c}{P\left(t\right)=\frac{1}{4\pi c}\int {d}^{3}r E×B=-\frac{1}{4\pi {c}^{2}}\int {d}^{3}r {\partial }_{t}A×\left(\nabla ×A\right)=}\\ \multicolumn{1}{c}{}& \hfill & -\frac{1}{4\pi {c}^{2}}\sum _{k,q,\mu ,\nu =1,2}\left[\frac{2\pi \hslash {c}^{2}}{V\sqrt{{\omega }_{k}{\omega }_{q}}} {u}_{k,\mu }×\left(iq×{u}_{q,\nu }\right)\left(-i{\omega }_{k}\right)\hfill & \hfill \left(54\right)\\ \multicolumn{1}{c}{}\\ \multicolumn{1}{c}{}& \hfill & \int {d}^{3}r\left({b}_{k,\sigma }\left(t\right) {e}^{ik·r}-{b}_{k,\sigma }^{+}\left(t\right) {e}^{-ik·r}\right)\left({b}_{q,\nu }\left(t\right) {e}^{iq·r}-{b}_{q,\nu }^{+}\left(t\right) {e}^{-iq·r}\right)\right]\hfill \end{array}$

There are two cases $k=q$ and $k=-q$. However, since we have set ${u}_{q\mu }={u}_{-q\mu }$ the cross product ${u}_{k,\mu }×\left(q×{u}_{q,\nu }\right)$ yields
 ${u}_{q,\mu }×\left(q×{u}_{q,\nu }\right)={\delta }_{\mu \nu }{u}_{\mu }×\left(q×{u}_{\mu }\right)=\left\{\begin{array}{c}\hfill \mu =1: {u}_{1}×\left(q×{u}_{1}\right)={u}_{1}×\left(-q{u}_{2}\right)=q\hfill \\ \hfill \mu =2: {u}_{2}×\left(q×{u}_{2}\right)={u}_{2}×\left(q{u}_{1}\right)=q\hfill \end{array}$ $\left(55\right)$
in both cases. I.e.
 $P=\frac{-1}{4\pi {c}^{2}}\sum _{q,\mu =1,2}\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{q}}\right){\omega }_{q}\left[q{b}_{-q,\mu }{b}_{q,\mu }-q{b}_{q,\mu }{b}_{q,\mu }^{+}-q{b}_{q,\mu }^{+}{b}_{q,\mu }+q{b}_{-q,\mu }^{+}{b}_{q,\mu }^{+}\right]$ $\left(56\right)$
where we have switched to the Schrödinger picture, i.e. we have droped the time-dependence on the ${b}_{q,\mu }^{\left(+\right)}$s. Now
 $\sum _{q}q {b}_{-q,\mu }{b}_{q,\mu }=-\sum _{q}q {b}_{q,\mu }{b}_{-q,\mu }=-\sum _{q}q {b}_{-q,\mu }{b}_{q,\mu }=0$ $\left(57\right)$
because $\left[{b}_{k,\mu },{b}_{q,\nu }\right]=\left[{b}_{k,\mu }^{+},{b}_{q,\nu }^{+}\right]=0$ and
 $\sum _{q}q {b}_{q,\mu }{b}_{q,\mu }^{+}=\sum _{q}q {b}_{q,\mu }^{+}{b}_{q,\mu }+\sum _{q}q=\sum _{q}q {b}_{q,\mu }^{+}{b}_{q,\mu }$ $\left(58\right)$
because $\left[{b}_{k,\mu },{b}_{k,\mu }^{+}\right]=1$. Therefore
 $P=\sum _{k,\mu =1,2}\hslash k{b}_{k,\mu }^{+}{b}_{k,\mu }$ $\left(59\right)$
If not done in lecture, then do:
Exercise 3 Using the 1st equality in eqn. (54), and eqns. (35,42), with eqn. (48), derive eqn. (59).
Eqns.(50) and (59) are at the heart of interpreting the electromagnetic field as a gas of particles, the so called photons. Not only does each of these particles carry an energy quantum ${\epsilon }_{k\mu }=\hslash {\omega }_{k\mu }$ as is obvious from eqn.(50), but even more so the momentum of the electromagnetic field can be understood as being carried the sum of these photons: one unit of momentum $\hslash k=p$ per photon. So the photons are particles. Moreover since they satisfy ${\epsilon }_{k\mu }=\mathrm{pc}$, where $c$ is the speed of light, these particles move at the speed of light.

### 2.2Classical States of the Radiation Field

Let us simplify the situation for this section by assuming, that there exists only one state of the radiation field with one particular $k,\sigma$, which is occupied with $n$ photons $|\left\{{n}_{k,\sigma }\right\}⟩\equiv |{n}_{k,\sigma }⟩\otimes |0⟩\otimes \dots \equiv |n⟩$. The expectation - and also eigen - value of the energy (discarding the zero point energy) is
 $⟨H⟩=n \hslash {\omega }_{k}$ $\left(60\right)$
where
 $n=⟨n|{b}^{+}b|n⟩$ $\left(61\right)$
is the expectation - and also eigen - value of the photon number operator in state $|n⟩$. The expectation value of the electric field is
 $⟨E⟩=-i{\left(\frac{2\pi \hslash {\omega }_{k}}{V}\right)}^{1/2}u\left(⟨n|b|n⟩ {e}^{ik·r}-⟨n|{b}^{+}|n⟩ {e}^{-ik·r}\right)=0$ $\left(62\right)$
(at any time) since ${b}^{\left(+\right)}$ (in)decrease the number of photons by one.
Classically speaking therefore the $n$-photon state at $k,\sigma$ must correspond to a state of the electric field which has a certain well defined absolute value of its amplitude - otherwise it could not have a non-zero energy - but the classical phase must be completely uncertain such as to yield a vanishing expectation value of the signed value of the field amplitude. This implies, that $n$-photon are very non-classical.
We are now going to construct a new type of state, the so-called Glauber or coherent state, which has somewhat more 'classical behavior'. To this end consider the following superposition
 $|g⟩=\sum _{n=0}^{\infty }\frac{{g}^{n}{e}^{-|g{|}^{2}/2}}{\sqrt{n!}}|n⟩\equiv \sum _{n=0}^{\infty }{\beta }_{n}|n⟩$ $\left(63\right)$
The coefficients ${\beta }_{n}$ have been chosen such as to have $|g⟩$ normalized
 ${e}^{-|g{|}^{2}}\sum _{n=0}^{\infty }\frac{|g{|}^{2n}}{n!}={e}^{-|g{|}^{2}}{e}^{|g{|}^{2}}=1$ $\left(64\right)$
The physical meaning of $|{\beta }_{n}{|}^{2}$ is the propability to find $n$ photons in state $|g⟩$ when measuring the photon number operator ${b}^{+}b$. Note that $|{\beta }_{n}{|}^{2}$ is the Poisson distribution $\Pi \left({g}^{2},n\right)$, which is sharply peeked at ${g}^{2}$ if ${g}^{2}>>1$. Since $|{\beta }_{n}{|}^{2}\ne 0 \forall n\ge 0$ this implies, that $|g⟩$ is a state of no well-defined photon number. Such a state may seem rather strange at first: in classical physics we are used to 'know' the number of particles in a state of a physical system - at least in principle
${}^{1}$.
Now, let us look at various expectation values using $|g⟩$
 $\begin{array}{ccc}\multicolumn{1}{c}{⟨g|b|g⟩}& =\hfill & \sum _{l=0}^{\infty }\sum _{m=0}^{\infty }{\beta }_{l}^{☆}{\beta }_{m}⟨l|b|m⟩=\sum _{l=0}^{\infty }\sum _{m=0}^{\infty }{\beta }_{l}^{☆}{\beta }_{m}\sqrt{m}⟨l|m-1⟩=\sum _{l=0}^{\infty }{\beta }_{l}^{☆}{\beta }_{l+1}\sqrt{l+1}\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{{g}^{☆l}}{\sqrt{l!}}\frac{{g}^{l+1}}{\sqrt{\left(l+1\right)!}}\sqrt{l+1}=g {e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{|g{|}^{2l}}{l!}=g\hfill & \hfill \left(65\right)\end{array}$

From hermiticity it follows $⟨g|{b}^{+}|g⟩={g}^{☆}$. Finally
 $\begin{array}{ccc}\multicolumn{1}{c}{⟨n⟩\equiv ⟨g|{b}^{+}b|g⟩}& =\hfill & \sum _{l=0}^{\infty }\sum _{m=0}^{\infty }{\beta }_{l}^{☆}{\beta }_{m}m⟨l|m⟩=\sum _{l=0}^{\infty }|{\beta }_{l}{|}^{2}l={e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{|g{|}^{2l}}{l!}l\hfill \\ \multicolumn{1}{c}{}& =\hfill & |g{|}^{2}{e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{|g{|}^{2l}}{l!}=|g{|}^{2}\hfill & \hfill \left(66\right)\\ \multicolumn{1}{c}{⟨{n}^{2}⟩\equiv ⟨g|{b}^{+}{\mathrm{bb}}^{+}b|g⟩}& =\hfill & \sum _{l=0}^{\infty }\sum _{m=0}^{\infty }{\beta }_{l}^{☆}{\beta }_{m}{m}^{2}⟨l|m⟩=\sum _{l=0}^{\infty }|{\beta }_{l}{|}^{2}{l}^{2}\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{|g{|}^{2l}}{l!}{l}^{2}={e}^{-|g{|}^{2}}\sum _{l=0}^{\infty }\frac{|g{|}^{2l}}{l!}\left(l\left(l-1\right)+l\right)=\hfill \\ \multicolumn{1}{c}{}& =\hfill & |g{|}^{4}+|g{|}^{2}=⟨n{⟩}^{2}+⟨n⟩\hfill & \hfill \left(67\right)\end{array}$

From eqns. (65-66) we see, that in the Glauber state, $g$ plays the role of a complex, non-zero amplitude. In particular
 $⟨g|E|g⟩=-i{\left(\frac{2\pi \hslash {\omega }_{k}}{V}\right)}^{1/2}u\left(g {e}^{ik·r}-{g}^{☆} {e}^{-ik·r}\right)$ $\left(68\right)$
Exercise 4 Show that the relative uncertainty, i.e. the relative variance, of the photon number in the Glauber state is
 $\Delta n/⟨n⟩=⟨g|\left({b}^{+}b-⟨n⟩{\right)}^{2}|g{⟩}^{1/2}/⟨n⟩=⟨n{⟩}^{-1/2}$ $\left(69\right)$
Which mathematical theorem is this related to? Show that the squared deviation of $E$ from its average in a Glauber state is independent of $g$ and vanishes only in the classical limit
 $⟨g|\left(E-⟨g|E|g⟩{\right)}^{2}|g⟩=\frac{2\pi \hslash \omega }{V}$ $\left(70\right)$

Consider non-relativistic matter, i.e. 'some' $n$ particles, described by
 ${H}_{m}=\sum _{i=1}^{n}\frac{{p}_{i}^{2}}{2{m}_{i}}+U\left({r}_{1},\dots {r}_{n}\right)$ $\left(71\right)$
where $\left({m}_{i},{r}_{i},{p}_{i}\right)$ are masses, coordinates and momenta, and $U$ is some interaction potential. Including a finite electromagnetic field, the Hamiltonian changes into
 $H=\sum _{i=1}^{n}\frac{|{p}_{i}-{q}_{i}A\left({r}_{i}\right)/c{|}^{2}}{2{m}_{i}}+U+\frac{1}{8\pi }\int {d}^{3}r\left({E}^{2}+{B}^{2}\right)={H}_{m}+{H}_{\mathrm{em}}+V$ $\left(72\right)$
The contribution $V$ is
 $V=\sum _{i}\left[-\frac{{q}_{i}}{{m}_{i}c}{p}_{i}A\left({r}_{i}\right)+\frac{{q}_{i}^{2}}{2{m}_{i}{c}^{2}}A\left({r}_{i}{\right)}^{2}\right]$ $\left(73\right)$
The 1st addend ${V}_{1}$ describes interaction between matter and radiation, the 2nd ${V}_{2}$ is a self-interaction of radiation in the presence of charged matter. To simplify: only
one particle

${V}_{1,2}$ are treated as perturbations to ${H}_{m,\mathrm{em}}$. We will justify this later.

Assume the particle moves in a central potential, say an electron moving in the Coulomb potential of a nucleus, which we keep fixed at the origin, i.e. an atom. We seek for the transition rate $|I⟩\to |F⟩$
 $\begin{array}{ccc}\multicolumn{1}{c}{|I⟩}& =\hfill & |i⟩|\dots {n}_{k\sigma }\dots ⟩\hfill \\ \multicolumn{1}{c}{|F⟩}& =\hfill & |f⟩|\dots {n}_{k\sigma }+1\dots ⟩\hfill & \hfill \left(76\right)\end{array}$

where $|i\left(f\right)⟩$ and $|\dots {n}_{k\sigma }\left({n}_{k\sigma }+1\right)\dots ⟩$ correspond to the inital(final) states of the atom and the radiation field. This transition describes the emission of one photon. It is generated by ${V}_{1}$ only. Using eqns.(30,31) consider 1st
 $⟨F|V|I⟩=-\frac{q}{\mathrm{mc}}{\left(\frac{2\pi \hslash {c}^{2}}{V{\omega }_{k}}\right)}^{1/2}⟨f|p{e}^{-ik·r}|i⟩·{u}_{k,\sigma }\sqrt{{n}_{k\sigma }+1}$ $\left(77\right)$
This is
non-zero even if ${n}_{k\sigma }=0$, i.e. we may have spontaneous transitions of the atom state - without any radiation field present - this is called spontaneous emission of photons. Classically, an electron moving arround the nucleus will also emit radiation, i.e. Bremsstrahlung (see eqn. (83)) 'spontaneously' because of its finite acceleration, however, with the wrong outcome, that the atom would collapse completely. Transitions due ${n}_{k\sigma }\ne 0$ are called stimulated emission. Let's focus on spontaneous emission 1st.
Total rate summing all possible photons emitted (see eqn.(31)):
 ${r}_{\mathrm{fi}}=\frac{4{\pi }^{2}{q}^{2}}{{\mathrm{Vm}}^{2}}\sum _{k\sigma }\frac{1}{{\omega }_{k}}|⟨f|p{e}^{-ik·r}|i⟩{|}^{2}{\mathrm{sin}}^{2}\Theta \delta \left({e}_{f}+\hslash {\omega }_{k}-{e}_{i}\right)$ $\left(78\right)$
where $\Theta$ is the angle between $p$ and $k$ in the plane of
one polarization, say ${u}_{k,1}$ and $k$. Now we use the wave-zone approximation $|k|<<|r|$, i.e. the wave-lenght of the em radiation ${\lambda }_{\mathrm{em}}$ is large compared to the atomic size $r$, which remains valid even up to hard X-Ray transitions
 ${e}^{-ik·r}=1-ik·r-\frac{1}{2}\left(k·r{\right)}^{2}+\dots \approx 1$ $\left(79\right)$
With $\sum _{k}=V/\left(2\pi {\right)}^{3}\int {d}^{3}k$, putting ${k}_{z}$ along $p$, and polar coordinates, see fig. 2, we get
Figure 2: Geometry for dipol matrix element

 $\begin{array}{ccc}\multicolumn{1}{c}{{r}_{\mathrm{fi}}}& =\hfill & \frac{{q}^{2}}{2\pi {m}^{2}}\int {d}^{3}k|⟨f|p|i⟩{|}^{2}\frac{1}{\mathrm{kc}}{\mathrm{sin}}^{2}\Theta \delta \left({e}_{f}-{e}_{i}+\hslash \mathrm{ck}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{{q}^{2}}{2\pi {m}^{2}c}|⟨f|p|i⟩{|}^{2}\int \mathrm{dk}{\int }_{0}^{\pi }d\Theta {\int }_{0}^{2\pi }d\phi k{\mathrm{sin}}^{3}\Theta \delta \left({e}_{f}-{e}_{i}+\hslash \mathrm{ck}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{{q}^{2}{\omega }_{\mathrm{if}}}{{m}^{2}\hslash {c}^{3}}|⟨f|p|i⟩{|}^{2}{\int }_{0}^{\pi }d\Theta {\mathrm{sin}}^{3}\Theta =\underset{‾}{\underset{‾}{\frac{4{q}^{2}{\omega }_{\mathrm{if}}}{3{m}^{2}\hslash {c}^{3}}|⟨f|p|i⟩{|}^{2}}}\hfill & \hfill \left(80\right)\end{array}$

where $\hslash {\omega }_{\mathrm{if}}={e}_{i}-{e}_{f}$ is the energy difference between the initial and final state of the atom. From eqn.(71)
 $\begin{array}{ccc}\multicolumn{1}{c}{\left[r,{H}_{m}\right]}& =\hfill & \left[r,\frac{{p}^{2}}{2m}\right]=\frac{i\hslash }{m}p ⇒\hfill \\ \multicolumn{1}{c}{}\\ \multicolumn{1}{c}{{r}_{\mathrm{fi}}}& =\hfill & \frac{4{q}^{2}{\omega }_{\mathrm{if}}^{3}}{3\hslash {c}^{3}}|⟨f|r|i⟩{|}^{2}=\alpha \frac{4{\omega }_{\mathrm{if}}^{3}}{3{c}^{2}}|⟨f|r|i⟩{|}^{2}\hfill & \hfill \left(81\right)\end{array}$

The presence of $⟨f|r|i⟩$ demonstrates the meaning of
dipole radiation. ${r}_{\mathrm{fi}}$ is 'small' in two ways. To check this we have to look at ${r}_{\mathrm{fi}}/{\omega }_{\mathrm{if}}$, i.e. the decay per one oscillation period. Now, (i) the fine-structure constant $\alpha ={q}^{2}/\left(\hslash c\right)\approx 1/137$ (for $q$=electron charge) is small and (ii) ${\omega }_{\mathrm{if}}^{2}/{c}^{2}|⟨f|r|i⟩{|}^{2}~{a}_{0}^{2}/{\lambda }_{\mathrm{if}}^{2}$ is small up to hard X-Ray transitions.
Using Heisenbergs equation of motion we may rewrite
 $|⟨f|\frac{{d}^{2}r}{{\mathrm{dt}}^{2}}|i⟩{|}^{2}=\frac{1}{\hslash {}^{4}}|⟨f|\left[H,\left[H,r\right]\right]|i⟩{|}^{2}=\frac{1}{\hslash {}^{4}}|⟨f|{H}^{2}r-2HrH-r{H}^{2}|i⟩{|}^{2}={\omega }_{\mathrm{if}}^{4}|⟨f|r|i⟩{|}^{2}$ $\left(82\right)$
i.e.
 $\mathrm{emitted} \mathrm{power}=\hslash {\omega }_{\mathrm{if}}{r}_{\mathrm{fi}}=\frac{4{q}^{2}}{3{c}^{3}}|⟨f|\frac{{d}^{2}r}{{\mathrm{dt}}^{2}}|i⟩{|}^{2}$ $\left(83\right)$
This shows that the result for classical
Bremstrahlung is corrected by a factor of two by Quantum mechanics.

Absorption of a quantum of radiation corresponds to $|I⟩\to |F⟩$
 $\begin{array}{ccc}\multicolumn{1}{c}{|I⟩}& =\hfill & |i⟩|\dots {n}_{k\sigma }\dots ⟩\hfill \\ \multicolumn{1}{c}{|F⟩}& =\hfill & |f⟩|\dots {n}_{k\sigma }-1\dots ⟩\hfill & \hfill \left(84\right)\end{array}$

Analogous to eqns.(78,77) we ${R}_{\mathrm{fi}}$ from eqn.(30) is obtained as
 ${R}_{\mathrm{fi}}=\frac{4{\pi }^{2}{q}^{2}}{{\mathrm{Vm}}^{2}{\omega }_{k}}|⟨f|p{e}^{ik·r}|i⟩{|}^{2}{n}_{k\sigma }{\mathrm{sin}}^{2}\Theta \delta \left({e}_{f}-\hslash {\omega }_{k}-{e}_{i}\right)$ $\left(85\right)$
Inspecting eqn.(77) and because of $⟨f|p{e}^{ik·r}|i⟩=⟨i|p{e}^{-ik·r}|f⟩$ it clear, that the transition rate for stimulated emission $|f⟩\to |i⟩$ is identical to that of absorption $|i⟩\to |f⟩$. Dividing by the
photon current
 ${j}_{k\sigma }=\frac{{n}_{k\sigma }}{V}c$ $\left(86\right)$
the absorption cross section ${\Sigma }_{\mathrm{fi},k\sigma }={R}_{\mathrm{fi}}/{j}_{k\sigma }$ (dimension $\left[\Sigma \right]=\left(1/\mathrm{sec}\right)/\left(1/\left({\mathrm{cm}}^{2}\mathrm{sec}\right)={\mathrm{cm}}^{2}$ = area) is
 ${\Sigma }_{\mathrm{fi},k\sigma }=\frac{4{\pi }^{2}{q}^{2}}{{\mathrm{cm}}^{2}{\omega }_{k}}|⟨f|p{e}^{ik·r}|i⟩{|}^{2}{\mathrm{sin}}^{2}\Theta \delta \left({e}_{f}-\hslash {\omega }_{k}-{e}_{i}\right)$ $\left(87\right)$
As for eqn.(31), eqns.(85,87) are meaningful only, if integrated over a certain energy interval of an incomming flux of photons - which depends on the particular situation.
Figure 3: Coordinate layout for the photoelectric effect
Exercise 5 Consider photons with $k=k{e}_{z}$ and polarization ${u}_{k,\sigma }={e}_{x}$ absorbed by an atom, such that the atom emitts an electron from one of its bound states $|i⟩$ to scattering state $|f⟩$, i.e. a state in the continuum. This is the photoelectric effect .
a) Write down the overall initial and final states of this process $|I⟩$ and $|F⟩$
b) Assume that the atom is placed in a cubical box of volume $V={L}^{3}$, and that the emitted electron is approximately in a plane-wave state $⟨r|q⟩={e}^{i\mathrm{rq}}/\sqrt{V}$ where $\hslash q$ is the electron momentum. The latter approximation is the so-called neglect of final state interactions and is ok. if the emitted electron has not to low an energy. Let the wave-vector $k=\left({n}_{x},{n}_{y},{n}_{z}\right)2\pi /L$ with ${n}_{x}\in Z$ be quantized w.r.t. the box. The differential cross section for photo-electrons into a solid angle $d\Omega$ is
 $\frac{d\sigma }{d\Omega }d\Omega =\frac{\mathrm{number} \mathrm{of} \mathrm{electrons} \mathrm{into} d\Omega /\mathrm{dt}}{\mathrm{number} \mathrm{of} \mathrm{photon} \mathrm{per} \mathrm{unit} \mathrm{area}/\mathrm{dt}}$

Using eqn. (85-87), show that
 $\frac{d\sigma }{d\Omega }=\frac{\alpha {q}_{f}V}{2\pi m\hslash {\omega }_{k}}|⟨{q}_{f}|{e}_{x}·p{e}^{ik·r}|i⟩{|}^{2}$

where $\hslash {}^{2}|{q}_{f}{|}^{2}/\left(2m\right)={e}_{f}={e}_{i}+\hslash {\omega }_{k}$ is the photo-electrons energy and ${q}_{f}/|{q}_{f}|$ is the direction of its emission. $\alpha$ is the
fine-structure constant. Verify dimensions, i.e. $\left[d\sigma /d\Omega \right]={\mathrm{length}}^{2}$.
c) Chose the 1s state of a hydrogen-like atom with atomic number $Z$ for $|i⟩$
 $⟨r|i⟩={\left(\frac{Z}{{a}_{0}}\right)}^{3}{e}^{-\mathrm{Zr}/{a}_{0}}$

Show that
 $\frac{d\sigma }{d\Omega }=32{q}^{2}{q}_{f}\frac{\left({e}_{x}·{q}_{f}{\right)}^{2}}{\mathrm{mc}{\omega }_{k}}\frac{{Z}^{5}}{{a}_{0}^{5}}\frac{1}{{\left[{Z}^{2}/{a}_{0}^{2}+{s}^{2}\right]}^{4}}$

where $s={q}_{f}-{\omega }_{k}{e}_{z}/c$. [Hint: Use partial integration w.r.t. $p=-i\hslash \nabla$ and ${e}_{x}k=0$. With that $⟨{q}_{f}|{e}_{x}·p{e}^{ik·r}|i⟩$ is directly related to the Fourier transform of the 1s state.]

TODO .................
 $\begin{array}{ccc}\multicolumn{1}{c}{\frac{{\mathrm{dN}}_{i}}{\mathrm{dt}}}& =\hfill & {N}_{f}{t}_{\mathrm{fi}}-{N}_{i}{t}_{\mathrm{if}}\hfill \\ \multicolumn{1}{c}{\frac{{\mathrm{dN}}_{f}}{\mathrm{dt}}}& =\hfill & -{N}_{f}{t}_{\mathrm{fi}}+{N}_{i}{t}_{\mathrm{if}}\hfill \end{array}$

 $\frac{{\mathrm{dN}}_{i}}{\mathrm{dt}}=\frac{{\mathrm{dN}}_{f}}{\mathrm{dt}}\mathrm{ }\mathrm{and}\mathrm{ }\frac{{N}_{i}}{{N}_{f}}={e}^{\beta \left({e}_{f}-{e}_{I}\right)}$

where $\beta ={k}_{B}T$
TODO .....................
Figure 4: Scattering process

## 4Quantum Theory of Scattering

Consider
 $H={H}_{0}+V$ $\left(88\right)$
For a typical scattering problem ${H}_{0}={p}^{2}/2m$ could be the Hamiltonian of - simplification for the rest of this section: - one particle coming in on some potential well $V$. Let $|\phi ⟩$ ( $|\psi ⟩$) be solutions of the Schrödinger equation for ${H}_{0}$ ( $H$) for
one and the same energy $E$ as in fig. 4
 $\begin{array}{ccc}\multicolumn{1}{c}{\left(E-{H}_{0}\right)|\psi ⟩}& =\hfill & V|\psi ⟩\hfill \\ \multicolumn{1}{c}{0}& =\hfill & \left(E-{H}_{0}\right)|\phi ⟩\hfill \\ \multicolumn{1}{c}{\oplus ⇒\left(E-{H}_{0}\right)|\psi ⟩}& =\hfill & V|\psi ⟩+\left(E-{H}_{0}\right)|\phi ⟩\hfill & \hfill \left(89\right)\end{array}$

This is trivial yet. The problem is, that $E$ is real and the operator $E-{H}_{0}$
cannot be inverted. Now consider two possible solutions $|{\psi }^{±}⟩$ of eqn. (89), meant to represent an outgoing(incoming) scattered wave for the superscript $+\left(-\right)$ for times $t\to +\left(-\right)\infty$. Wile the ' $+$'-sign represents the physically conceivable, causal or retarded case where the scattered wave develops in the future, the ' $-$'-sign, the advanced case corresponds to a situation where the scattered wave occurs in past. Mathematically, both are necessary the construct the most general solution to eqn. (89). For both cases we will be interested only in times long after(before) the scattering has occured for $+\left(-\right)$. This allows to replace
 ${H}_{0}\to {H}_{0}\mp i\eta$ $\left(90\right)$
with $\eta \to {0}^{+}$ in eqn. (89) since within the time evolution $\mathrm{exp}\left(-{\mathrm{iH}}_{0}t\mp \eta t\right)$ the replacement eqn.(90) is irrelevant for $t\to ±\infty$. With eqn.(90) the
operator $\left(E-{H}_{0}±i\eta \right)$ is invertable for real $E$ leading to

 $|{\psi }^{±}⟩=|\phi ⟩+\frac{1}{E-{H}_{0}±i\eta }V|{\psi }^{±}⟩$ $\left(91\right)$
This is the
Lippman-Schwinger equation. It is frequently stated in its real space representation
 $⟨r|{\psi }^{±}⟩=⟨r|\phi ⟩+\int {d}^{3}r\text{'}⟨r|\frac{1}{E-{H}_{0}±i\eta }|r\text{'}⟩⟨r\text{'}|V|{\psi }^{±}⟩$ $\left(92\right)$
Exercise 6 Show that
 ${G}_{±}\left(r,r\text{'}\right)=\frac{\hslash {}^{2}}{2m}⟨r|\frac{1}{E-{H}_{0}±i\eta }|r\text{'}⟩=-\frac{1}{4\pi }\frac{{e}^{±\mathrm{ik}|r-r\text{'}|}}{|r-r\text{'}|}$ $\left(93\right)$
where $\hslash {}^{2}{k}^{2}/2m=E$. To that end:

a) Use the real space representation of the momentum eigenstates $⟨r|p⟩={e}^{i\mathrm{rp}/\hslash }/\left(2\pi \hslash {\right)}^{3/2}$, with $⟨p|p\text{'}⟩=\delta \left(p-p\text{'}\right)$ to insert two suitable $1$-operators into the central equation of eqn. (93).
b) Use the method of residues to evaluate the 3D integral resulting from a)
In many cases $⟨r|V|r\text{'}⟩=V\left(r\right)\delta \left(r-r\text{'}\right)$ is diagonal in real space. Inserting $\int {d}^{3}r"|r"⟩⟨r"|=1$ into the right side of the last matrix element in eqn. (92) we therefore get
 $⟨r|{\psi }^{±}⟩=⟨r|\phi ⟩-\frac{m}{2\pi \hslash {}^{2}}\int {d}^{3}r\text{'}\frac{{e}^{±\mathrm{ik}|r-r\text{'}|}}{|r-r\text{'}|}V\left(r\text{'}\right)⟨r\text{'}|{\psi }^{±}⟩$ $\left(94\right)$
Usually scattering is investiated far away from the scattering potential, i.e. in the
far-field approximation, $r\text{'}/r<<1$. There
 $|r-r\text{'}|=\sqrt{\left(r-r\text{'}{\right)}^{2}}=r\sqrt{1-2\mathrm{rr}\text{'}/{r}^{2}+r{\text{'}}^{2}/{r}^{2}}=r-r\text{'}{e}_{r}+O\left(\left(r\text{'}/r{\right)}^{2}\right)$ $\left(95\right)$
Using this in the $\mathrm{exp}$-function and replacing $1/|r-r\text{'}|\approx 1/r$ we get
 $\begin{array}{ccc}\multicolumn{1}{c}{⟨r|{\psi }^{±}⟩{|}_{r/r\text{'}>>1}}& =\hfill & ⟨r|\phi ⟩-\frac{m}{2\pi \hslash {}^{2}}\int {d}^{3}r\text{'}{e}^{±\mathrm{ik}\left(r-r\text{'}{e}_{r}\right)}\frac{1}{r}V\left(r\text{'}\right)⟨r\text{'}|{\psi }^{±}⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & ⟨r|\phi ⟩-\frac{m}{2\pi \hslash {}^{2}}\frac{{e}^{±\mathrm{ikr}}}{r}\int {d}^{3}r\text{'}{e}^{\mp ik\text{'}r\text{'}}V\left(r\text{'}\right)⟨r\text{'}|{\psi }^{±}⟩\hfill & \hfill \left(96\right)\end{array}$

where $k\text{'}=k{e}_{r}$ is the wave vector corresponding to the energy of the incomming(outgoing) particle for the sign +(-) but into the direction of the outgoing(incomming) scattered wave.
Watch out for the prime! It is customary to chose $⟨r|\phi ⟩$ to be a plane-wave state normalized in a wave-vector, i.e. $k=p/\hslash$ (and not in a momentum $p$), representation: $⟨r|k⟩={e}^{i\mathrm{rk}}/\left(2\pi {\right)}^{3/2}$
 $\begin{array}{cccc}\multicolumn{1}{c}{⟨r|{\psi }^{±}⟩{|}_{r/r\text{'}>>1}}& =\hfill & \frac{1}{\left(2\pi {\right)}^{3/2}}\left[{e}^{i\mathrm{kr}}+\frac{{e}^{±\mathrm{ikr}}}{r}f\left(k\text{'},k\right)\right]\hfill & \hfill \left(97\right)\\ \multicolumn{1}{c}{}\\ \multicolumn{1}{c}{f\left(k\text{'},k\right)}& =\hfill & -\frac{m}{2\pi \hslash {}^{2}}\left(2\pi {\right)}^{3/2}\int {d}^{3}r\text{'}{e}^{\mp ik\text{'}r\text{'}}V\left(r\text{'}\right)⟨r\text{'}|{\psi }^{±}⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{\left(2\pi {\right)}^{2}m}{\hslash {}^{2}}⟨±k\text{'}|V|{\psi }^{±}⟩\hfill & \hfill \left(98\right)\end{array}$

Obviously eqn. (98) decomposes the solution of the scattering problem $⟨r|{\psi }^{±}⟩{|}_{r/r\text{'}>>1}$ into the particle with wave-vector $k$ which is to be scattered and an outgoing(incomming) sperical wave with a
scattering amplitude $f\left(k\text{'},k\right)$. While the variable $k\text{'}$ of $f\left(k\text{'},k\right)$ is explicit in eqn. (98) the dependence on $k$ is implicit through the plane-wave state $⟨r|\phi ⟩$. Note, that eqns. (97,98) don't 'solve' the scattering problem. They are still an integral equation for $⟨r|{\psi }^{±}⟩$.

### 4.1Differential Cross Section

Consider the 'upper' sign situation in eqn. (97). Far away from the scattering center, $r/r\text{'}>>$1, the absolute value of the current density of the incomming particle is ${j}_{i}={v}_{i}|⟨r|\phi ⟩{|}^{2}$. For the outgoing spherical wave ${j}_{s}={v}_{s}|f\left(k\text{'},k\right){|}^{2}/{r}^{2}$, where the velocities ${v}_{i}={v}_{s}=\hslash k/m$ are the same. The differential cross section $d\sigma /d\Omega$ is defined by
 $\begin{array}{ccc}\multicolumn{1}{c}{d\sigma }& =\hfill & \frac{\mathrm{num}. \mathrm{particles} \mathrm{scattered} \mathrm{into} d\Omega /\mathrm{dt}}{\mathrm{num}. \mathrm{particles} \mathrm{incomming} \mathrm{per} \mathrm{unit} \mathrm{area}/\mathrm{dt}}=\frac{{j}_{s}{r}^{2}d\Omega }{{j}_{i}}=|f\left(k\text{'},k\right){|}^{2}d\Omega \hfill \\ \multicolumn{1}{c}{\frac{d\sigma }{d\Omega }}& =\hfill & |f\left(k\text{'},k\right){|}^{2}\hfill & \hfill \left(99\right)\end{array}$

I.e. absolute value of scattering amplitude = differential cross section.

### 4.2Born Approximation

Consider the 'upper' sign situation in eqn. (97). Similar to the solution of the integral equation for the time-evoution operator eqn. (18) in section 1.1, for 'small' $V$, we may attempt to solve eqns. (96, 97, 98) by iteration. I.e., we set $⟨r\text{'}|{\psi }^{+}⟩=⟨r\text{'}|\phi ⟩={e}^{ir\text{'}k}/\left(2\pi {\right)}^{3/2}$
 $f\left(k\text{'},k\right)=-\frac{m}{2\pi \hslash {}^{2}}\int {d}^{3}{\mathrm{re}}^{i\left(k-k\text{'}\right)r}V\left(r\right)+O\left({V}^{2}\right)$ $\left(100\right)$
Confining to a sperically symmetric potential $V\left(r\right)=V\left(r\right)$ and to 1st order in $V$ only, we get
 $\begin{array}{ccc}\multicolumn{1}{c}{f\left(k\text{'},k\right)}& \approx \hfill & -\frac{m}{2\pi \hslash {}^{2}}{\int }_{0}^{\infty }\mathrm{dr}{\int }_{\Omega }d\alpha d\beta {r}^{2}\mathrm{sin}\beta {e}^{-\mathrm{iqr}\mathrm{cos}\beta }V\left(r\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{m}{2\pi \hslash {}^{2}}2\pi \left(-\frac{1}{\mathrm{iq}}\right){\int }_{0}^{\infty }\mathrm{dr} r\left({e}^{-\mathrm{iqr}}-{e}^{\mathrm{iqr}}\right)V\left(r\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{2m}{q\hslash {}^{2}}{\int }_{0}^{\infty }\mathrm{dr} r\mathrm{sin}\left(\mathrm{qr}\right)V\left(r\right)=f\left(q\right)\hfill & \hfill \left(101\right)\end{array}$

where we have used fig. 5 and $q=k-k\text{'}$ and $|k\text{'}|=|k|=k$ and $q=2k\mathrm{sin}\left(\Theta /2\right)$
Figure 5: Geometry for integral in Born approximation
Let us look at the differential cross section of a 3D potential well
 $V\left(r\right)=\left\{\begin{array}{cc}\hfill u,\hfill & \hfill r\le R\hfill \\ \hfill 0,\hfill & \hfill r>R\hfill \end{array}$ $\left(102\right)$
for that
 $f\left(q\right)=-\frac{2\mathrm{mu}}{q\hslash {}^{2}}\frac{\mathrm{sin}\left(\mathrm{qR}\right)-\mathrm{qR}\mathrm{cos}\left(\mathrm{qR}\right)}{{q}^{2}}=-\frac{2\mathrm{mu}}{3\hslash {}^{2}}{R}^{3}+O\left({q}^{2}\right)\approx -\frac{u}{e\left(R\right)}R$ $\left(103\right)$
where, in the 2nd equation we have made a
low-energy expansion $\mathrm{qR}<<1$. In the last equation we have introduced the energy $e\left(R\right)=3\hslash {}^{2}/\left(2{\mathrm{mR}}^{2}\right)$ of a particle with a wave-length of the size of the potential well. The ratio $u/e$ is a dimensionless measure for the 'strength' of the scattering potential. Note that $f\left(q\right)$ has dimension [length] and is independent of $q$ for $\mathrm{qR}<<1$. Finally
 $\frac{d\sigma }{d\Omega }={\left(\frac{u}{e\left(R\right)}\right)}^{2}{R}^{2}\mathrm{ }\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{ }{\sigma }_{\mathrm{tot}}=\int d\Omega \frac{d\sigma }{d\Omega }=4\pi {\left(\frac{u}{e\left(R\right)}\right)}^{2}{R}^{2}$ $\left(104\right)$
This is 'kind of' what we expect: the cross section has dimension [area] and is proportional to the square of the scattering potential - however, the size of ${\sigma }_{\mathrm{tot}}$ is different from $\pi {R}^{2}$ even for $k\to 0$. A commonly used unit for $d\sigma /d\Omega$ and ${\sigma }_{\mathrm{tot}}$ is
1 barn = 1b = ${10}^{-24}{\mathrm{cm}}^{2}$ (c.f. am. slang: as big as a barn's door).

### 4.3T-Matrix and Higher-Order Born Approximation

From eqn. (107) we may expect a one-to-one correspondence between $|{\psi }^{±}⟩$ and $|\phi ⟩$. Therefore we may define the transition operator
 $T|\phi ⟩=V|{\psi }^{+}⟩$ $\left(105\right)$
using the retarded solution $|{\psi }^{+}⟩$. With $V·$eqn. (107)

 $T|\phi ⟩=V|\phi ⟩+V\frac{1}{E-{H}_{0}+i\eta }T|\phi ⟩$ $\left(106\right)$
from which we may remove $|\phi ⟩$, since the set of all plane-wave states is complete.
 $T=V+V\frac{1}{E-{H}_{0}+i\eta }T$ $\left(107\right)$
This is the so-called
T-matrix equation. The scattering amplitude can be obtained from plane-wave matrix elements of the T-matrix
 $f\left(k\text{'},k\right)=-\frac{\left(2\pi {\right)}^{2}m}{\hslash {}^{2}}⟨k\text{'}|T|k⟩$ $\left(108\right)$
Using this iteration on eqn. (107) leads to a perturbative expansion of the scattering amplitude
 $\begin{array}{ccc}\multicolumn{1}{c}{T}& =\hfill & V+V\frac{1}{E-{H}_{0}+i\eta }V+V\frac{1}{E-{H}_{0}+i\eta }V\frac{1}{E-{H}_{0}+i\eta }V+\dots \hfill \\ \multicolumn{1}{c}{}& =\hfill & V\frac{1}{1-G\left(E\right)V}=\frac{1}{1-\mathrm{VG}\left(E\right)}V\hfill \\ \multicolumn{1}{c}{G\left(E\right)}& =\hfill & \frac{1}{E-{H}_{0}+i\eta }\hfill & \hfill \left(109\right)\end{array}$

Eqn. (109) plays a crucial role in evaluating $f\left(k\text{'},k\right)$ in practice. $G\left(E\right)$ is the so-called
retarded bare resolvent or retarded bare Greens function. The word 'bare' refers the unperturbed Hamiltonian in the denominator.
Exercise 7 Approximate(!) the T-matrix for particle scattered by a delta-potential
 $V\left(r\right)=\left(2\pi {\right)}^{3}v\delta \left(r\right)$

which is, eg., a model for scattering an electrons off a localized impurity in a solid.

a) Show that
 $⟨k\text{'}|T|k⟩=v+v\int {d}^{3}q\frac{1}{E-{ϵ}_{q}+i\eta }⟨q|T|k⟩$

where $⟨r|k⟩={e}^{ir·k}/\left(2\pi {\right)}^{3/2}$ with $H|k⟩={ϵ}_{k}|k⟩$. Show that the solution of this integral equation is
 $⟨k\text{'}|T|k⟩=T\left(E\right)=\frac{v}{1-v\int {d}^{3}q\frac{1}{E-{ϵ}_{q}+i\eta }}$

b) Now, lets assume that the density of states $\rho \left(ϵ\right)=\int {d}^{3}q\delta \left(ϵ-{ϵ}_{q}\right)$ is constant in energy interval $\rho \left(ϵ\right)={\rho }_{0}$ for $ϵ\in \left[-W,W\right]$ and zero elsewhere. $W$ sets the 'bandwidth' of the solid. This is a very rough approximation, the applicability of which may vary. Show that
 $T\left(E\right)=\frac{v}{1-v{\rho }_{0}\mathrm{ln}\left(\frac{E-W+i\eta }{E+W+i\eta }\right)}$

c) Make a sketch of $\mathrm{Re}\left(T\left(E\right)\right)$ and $\mathrm{Im}\left(T\left(E\right)\right)$ for $\left(v,W,{\rho }_{0}\right)=\left(±0.4\mathrm{eV},2\mathrm{eV},1/\mathrm{eV}\right)$. Under which conditions are there energies for which $T\left(E\right)$ is purely imaginary. At which energies does this happen. [Hint: search for zeros of the real part denominator] What is the physical significance of these energies.

### 4.4The Optical Theorem

The optical theorem states the remarkable fact, that knowing the imaginary part of the forward scattering amplitude is sufficient to know the total cross section:
 $\frac{4\pi }{k}\mathrm{Im} f\left(\Theta =0\right)={\sigma }_{\mathrm{tot}}$ $\left(110\right)$
with $f\left(\Theta =0\right)=f\left(k,k\right)$. To prove eqn. (110), remember eqn. (105) with $|k⟩=|\phi ⟩$ and use eqn. (91
 $\begin{array}{ccc}\multicolumn{1}{c}{\mathrm{Im}⟨k|T|k⟩}& =\hfill & \mathrm{Im}⟨k|V|{\psi }^{+}⟩=\mathrm{Im}\left[⟨{\psi }^{+}|-⟨{\psi }^{+}|V\frac{1}{E-{H}_{0}-i\eta }\right]V|{\psi }^{+}⟩=\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\pi ⟨{\psi }^{+}|V\delta \left(E-{H}_{0}\right)V|{\psi }^{+}⟩\hfill & \hfill \left(111\right)\end{array}$

where we have used that $⟨{\psi }^{+}|V|{\psi }^{+}⟩\in R$ and $⟨{\psi }^{+}|\mathrm{VP}\left(E-{H}_{0}{\right)}^{-1}V|{\psi }^{+}⟩\in R$, because ${V}^{+}=V$ and ${H}_{0}^{+}={H}_{0}$. Now we may use eqn. (105) again
 $\begin{array}{ccc}\multicolumn{1}{c}{\mathrm{Im}⟨k|T|k⟩}& =\hfill & -\pi ⟨k|{T}^{+}\delta \left(E-{H}_{0}\right)T|k⟩=-\pi \int {d}^{3}k\text{'}⟨k|{T}^{+}|k\text{'}⟩\delta \left(E-\frac{\hslash {}^{2}k{\text{'}}^{2}}{2m}\right)⟨k\text{'}|T|k⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{\pi \mathrm{mk}}{\hslash {}^{2}}\int d\Omega \text{'}|⟨k\text{'}|T|k⟩{|}^{2}\hfill & \hfill \left(112\right)\end{array}$

Where $k\text{'}=k$ has been used. Finally, using eqn. (108) and then eqn. (99)
 $\frac{4\pi }{k}\mathrm{Im} f\left(\Theta =0\right)=\frac{4\pi }{k} \frac{\pi \mathrm{mk}}{\hslash {}^{2}} \frac{\hslash {}^{2}}{\left(2\pi {\right)}^{2}m} \int d\Omega \text{'}|f\left(k\text{'},k\right){|}^{2}=\int d\Omega \text{'}\frac{d\sigma }{d\Omega \text{'}}={\sigma }_{\mathrm{tot}}$ $\left(113\right)$
Note, that ${\sigma }_{\mathrm{tot}}~{f}^{2}$, i.e. the Born approximation cannot satisfy the optical theorem. The physical significance of the optical theorem is rooted in unitarity: the scattered particles must be removed from the incoming wave of particles.

### 4.5Partial Wave Expansion

Many scattering potentials $V$ are spherically symmetric. I.e. $\left[V,{L}^{z}\right]=\left[V,{L}^{2}\right]=0$. Since $\left[{H}_{0},{L}^{z}\right]=\left[{H}_{0},{L}^{2}\right]=0$ anyway, we may look at the scattering problem in terms of the partial-wave states $|E,l,m⟩$ which are eigenstates of the energy, as well as of the total and the z-component of the angular momentum
 ${H}_{0} / {L}^{2} / {L}^{z}|E,l,m⟩=E / \hslash {}^{2}l\left(l+1\right) / \hslash m|E,l,m⟩$ $\left(114\right)$
Due to spherical symmetry the $T$-matrix will be diagonal in $l$ and $m$
 $⟨E\text{'},l\text{'},m\text{'}|T|E,l,m⟩=T\left(E,l,m\right){\delta }_{l\text{'}l}{\delta }_{m\text{'}m}=T\left(E,l\right){\delta }_{l\text{'}l}{\delta }_{m\text{'}m}$ $\left(115\right)$
where the last equality, i.e. diagonality in $E$, is physically conceivable, but a rigorous proof would require the Wigner-Eckart theorem - which we skip. Inserting into eqn. (108) we get
 $\begin{array}{ccc}\multicolumn{1}{c}{f\left(k\text{'},k\right)}& =\hfill & -\frac{\left(2\pi {\right)}^{2}m}{\hslash {}^{2}}⟨k\text{'}|T|k⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{\left(2\pi {\right)}^{2}m}{\hslash {}^{2}}\sum _{\mathrm{ll}\text{'}\mathrm{mm}\text{'}}\int \mathrm{dEdE}\text{'}⟨k\text{'}|E\text{'},l\text{'},m\text{'}⟩⟨E\text{'},l\text{'},m\text{'}|T|E,l,m⟩⟨E,l,m|k⟩\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{\left(2\pi {\right)}^{2}m}{\hslash {}^{2}}\sum _{\mathrm{lm}}\int \mathrm{dEdE}\text{'}⟨k\text{'}|E\text{'},l,m⟩⟨E,l,m|k⟩T\left(E,l\right)\hfill & \hfill \left(116\right)\end{array}$

To proceed we need the partial wave representation $⟨k|E,l,m⟩$ of the plane wave states (or vice versa).
In principle, the following does not need Schrödinger's equation, as in many textbooks. In fact, all we need is rotational invarince and unitarity. Yet, we do the following
Reminder: In spherical coordinates, the eigenstates ${\phi }_{\mathrm{Elm}}\left(r\right)\equiv {A}_{l}\left(r\right){Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)$ of the Schrödinger equation are obtained from solving the 'equivalent' 1D or radial Schrödinger equation
 $\left\{\frac{{d}^{2}}{{\mathrm{dr}}^{2}}+\left[{k}^{2}-\frac{2m}{\hslash {}^{2}}V\left(r\right)-\frac{l\left(l+1\right)}{{r}^{2}}\right]\right\}\left({\mathrm{rA}}_{l}\left(r\right)\right)=0$ $\left(117\right)$
where $E=\hslash {}^{2}{k}^{2}/2m$ has been used. ${Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)$ are the spherical harmonics, which are the eigenfunctions of the orbital angular momentum operators ${L}^{2}$ and ${L}^{z}$ represented in spherical coordinates. They are orthonormal in these coordinates, i.e. the satisfy
 $\begin{array}{ccc}\multicolumn{1}{c}{{\int }_{\Omega }{Y}_{l\text{'}m\text{'}}^{☆}\left(\Theta ,\varphi \right){Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)d\Omega }& =\hfill & {\delta }_{l\text{'}l}{\delta }_{m\text{'}m}\hfill \\ \multicolumn{1}{c}{\sum _{l=0}^{\infty }\sum _{m=-l}^{l}{Y}_{\mathrm{lm}}^{☆}\left(\Theta \text{'},\varphi \text{'}\right){Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)}& =\hfill & \delta \left(\Omega \text{'}-\Omega \right)=\frac{\delta \left(\Theta \text{'}-\Theta \right)\delta \left(\varphi \text{'}-\varphi \right)}{\mathrm{sin}\left(\Theta \right)}\hfill & \hfill \left(118\right)\end{array}$

For a free particle, i.e. $V\left(r\right)=0$ and fixed $k$ eqn. (117) has two particular solutions, the spherical Bessel functions ${A}_{l}\left(r\right)={j}_{l}\left(\mathrm{kr}\right)$ (of the first kind) and ${A}_{l}\left(r\right)={n}_{l}\left(\mathrm{kr}\right)$ (of the second kind). Their small $r$-behavior is ${j}_{l}\left(0\le x<<1\right)~{x}^{l}$ and ${n}_{l}\left(0\le x<<1\right)~{x}^{-l-1}$.

TODO: give small/large x expressions for sp.Bess.
 $\mathrm{TODO}$ $\left(119\right)$

Exercise 9 a
a) Look up the analytic expressions for ${j}_{l}\left(x\right), {n}_{l}\left(x\right)$ for $l=0,1$. Make a plot of them. Try to sove eqn. (117) by a power series ansatz.
b) Show that
 $⟨k|E,l,m⟩=\frac{\hslash }{\sqrt{\mathrm{mk}}}\delta \left(E-\frac{\hslash {}^{2}{k}^{2}}{2m}\right){Y}_{\mathrm{lm}}\left(\stackrel{^}{k}\right)$ $\left(120\right)$
where $\stackrel{^}{k}=k/k$ and ${Y}_{\mathrm{lm}}\left(\stackrel{^}{k}\right)$ is a spherical harmonic.
Inserting eqn. (120) into eqn. (116) and using $|k\text{'}|=|k{e}_{r}|=k=|k|$ we get
 $f\left(k\text{'},k\right)=-\frac{\left(2\pi {\right)}^{2}}{k}\sum _{\mathrm{lm}}T\left(\frac{\hslash {}^{2}{k}^{2}}{2m},l\right){Y}_{\mathrm{lm}}^{☆}\left(\stackrel{^}{k\text{'}}\right){Y}_{\mathrm{lm}}\left(\stackrel{^}{k}\right)$ $\left(121\right)$
Now we specify the scattering geometry by choosing $k=k{e}_{z}$ and without loss of generality $k\text{'}=k\left({e}_{z}\mathrm{cos}\Theta +{e}_{x}\mathrm{sin}\Theta \right)$, i.e.
 $\begin{array}{ccc}\multicolumn{1}{c}{{Y}_{\mathrm{lm}}\left(\stackrel{^}{k}\right)}& =\hfill & {Y}_{\mathrm{lm}}\left(0,0\right)={\delta }_{m0}\sqrt{\frac{2l+1}{4\pi }}\hfill \\ \multicolumn{1}{c}{{Y}_{l0}\left(\stackrel{^}{k\text{'}}\right)}& =\hfill & {Y}_{l0}\left(\Theta ,0\right)=\sqrt{\frac{2l+1}{4\pi }}{P}_{l}\left(\mathrm{cos}\Theta \right)\hfill & \hfill \left(122\right)\end{array}$

where ${P}_{l}\left(\mathrm{cos}\Theta \right)$ are the Legendre polynomials. Defining the
partial-wave amplitude
 ${f}_{l}\left(E\right)=-\frac{\pi }{k}T\left(\frac{\hslash {}^{2}{k}^{2}}{2m},l\right)$ $\left(123\right)$
we finally get
 $f\left(k\text{'},k\right)=f\left(\Theta \right)=\sum _{l}\left(2l+1\right){f}_{l}\left(E\right){P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(124\right)$
This expression for $f\left(k\text{'},k\right)$ allows for a very intuitive interpretation of the scattering process. For that we need the real-space representation of the decomposition of plane waves into partial waves. From the 'reminder' eqn. (117-118) we know: its a linear combination of ${j}_{l}\left(\mathrm{kr}\right){Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)$ and ${n}_{l}\left(\mathrm{kr}\right){Y}_{\mathrm{lm}}\left(\Theta ,\varphi \right)$. The latter drops out because ${n}_{l}\left(0\le x<<1\right)~{x}^{-l-1}$. From the former only $m=0$ survives, because ${e}^{ik·r}$ is rotanionally symmetric around ${e}_{z}$. What remains is
 ${e}^{ik·r}=\sum _{l}\left(2l+1\right){i}^{l}{j}_{l}\left(\mathrm{kr}\right){P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(125\right)$
where ${j}_{l}\left(x\right)$ is the spherical Bessel function of the first kind of order $l$ and ${P}_{l}\left(x\right)$ is the Legendre polynomial.
Exercise 10 Prove eqn. (125).
Using
 ${j}_{l}\left(x\right){|}_{x\to \infty }=\frac{1}{x}\mathrm{sin}\left(x-\frac{l\pi }{2}\right)=\frac{1}{2\mathrm{ix}}\left({e}^{i\left(x-l\pi /2\right)}-{e}^{-i\left(x-l\pi /2\right)}\right)$ $\left(126\right)$
and eqn. (97) we get
 $⟨r|{\psi }^{+}⟩{|}_{\frac{r}{r\text{'}}>>1}=\frac{1}{\left(2\pi {\right)}^{3/2}}\sum _{l}\frac{\left(2l+1\right)}{2\mathrm{ik}}\left[\left(1+2\mathrm{ik} {f}_{l}\left(E\right)\right)\frac{{e}^{\mathrm{ikr}}}{r}-\frac{{e}^{-i\left(\mathrm{kr}-l\pi \right)}}{r}\right]{P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(127\right)$
Keeping in mind the time-evolution of $|{\psi }^{+}⟩~{e}^{-\mathrm{iEt}/\hslash }$ with $E=\hslash {}^{2}{k}^{2}/2m>0$, the 1st (2nd) exponential in the $\left[\right]$-brackets describes an
out-going (in-comming) spherical wave. Comparing eqns. (125,126) and eqn. (127), the sole action of the scattering potential is to modify the amplitude and phase of the out-going wave - the in-comming waves remain as is.
Due to $\left[H,L\right]=0$, propability conservation, i.e. unitarity of the time evolution, applies to each $l$-channel separately. I.e. in-comming and out-outgoing waves have equal magnitude amplitudes per channel
 $|1+2\mathrm{ik} {f}_{l}\left(E\right)|=1$ $\left(128\right)$
I.e. the out-going wave aquires a
scattering phase-shift ${\delta }_{l}\left(E\right)\in \mathbbR$
 ${f}_{l}\left(E\right)=\frac{{e}^{2i{\delta }_{l}\left(E\right)}-1}{2\mathrm{ik}}$ $\left(129\right)$
which allows to rewrite eqn. (124) as
 $f\left(\Theta \right)=\frac{1}{k}\sum _{l}\left(2l+1\right){e}^{i{\delta }_{l}\left(E\right)}\mathrm{sin}\left({\delta }_{l}\left(E\right)\right){P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(130\right)$
The total cross-section is
 ${\sigma }_{\mathrm{tot}}={\int }_{4\pi }|f\left(\Theta \right){|}^{2}d\Omega$ $\left(131\right)$
using ${\int }_{0}^{\pi }d\Theta \mathrm{sin}\left(\Theta \right){P}_{l}\left(\mathrm{cos}\Theta \right){P}_{m}\left(\mathrm{cos}\Theta \right)={\int }_{-1}^{1}{\mathrm{dxP}}_{l}\left(x\right){P}_{m}\left(x\right)=2/\left(2l+1\right){\delta }_{\mathrm{lm}}$ we get
 ${\sigma }_{\mathrm{tot}}=\frac{4\pi }{{k}^{2}}\sum _{l}\left(2l+1\right){\mathrm{sin}}^{2}\left({\delta }_{l}\left(E\right)\right)$ $\left(132\right)$
It is instructive to see, that the optical theorem is fulfilled, i.e. plugging $\Theta =0$ into eqn. (130)
 $\mathrm{Im} f\left(\Theta \right)=\frac{1}{k}\sum _{l}\left(2l+1\right){\mathrm{sin}}^{2}\left({\delta }_{l}\left(E\right)\right)=\frac{k}{4\pi }{\sigma }_{\mathrm{tot}}$ $\left(133\right)$
which is exactly eqn. (110).

### 4.6How to Calculate Phase Shifts

Let us assume a spherically symmteric scattering potential $V\left(r\right)$ which fulfills $V\left(r>R\right)=0$. Since the incident plane wave is rotationally invariant about $k$, any solution of the scattering problem is also is rotationally invariant about $k$ and can be expanded as

 $⟨r|{\psi }^{+}⟩{|}_{r>R}=\frac{1}{\left(2\pi {\right)}^{3/2}}\sum _{l}\left(2l+1\right){i}^{l}{A}_{l}\left(r\right){P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(134\right)$
where ${\mathrm{rA}}_{l}\left(r\right)$ is the solution of the radial Schrödinger equation in the angular momentum channel $l$
 $\left\{\frac{{d}^{2}}{{\mathrm{dr}}^{2}}+\left[{k}^{2}-\frac{2m}{\hslash {}^{2}}V\left(r\right)-\frac{l\left(l+1\right)}{{r}^{2}}\right]\right\}\left({\mathrm{rA}}_{l}\left(r\right)\right)=0$ $\left(135\right)$
For $r>R$, ${A}_{l}\left(r\right)$ can be expressed in terms spherical Bessel functions
 $2{A}_{l}\left(r\right)={a}_{l}{j}_{l}\left(\mathrm{kr}\right)+{b}_{l}{n}_{l}\left(\mathrm{kr}\right)$ $\left(136\right)$
where ${n}_{l}\left(x\right)$ is the spherical Bessel function of the second kind of order $l$. In contrast to ${j}_{l}\left(x\right)$, ${n}_{l}\left(x\right)$ diverges $~1/{x}^{l+1}$ as $x\to 0$. This is why it plays no role in the expansion of eqn. (125). However, since eqn. (134) is considered only for $r>R$ we must allow for ${b}_{l}\ne 0$. The asymptotic behavior of eqn. (134) follows from
 $\begin{array}{ccc}\multicolumn{1}{c}{{i}^{l}{A}_{l}\left(r\right){|}_{\mathrm{kr}>>1}}& =\hfill & {e}^{\mathrm{il}\pi /2}\left({a}_{l}\frac{\mathrm{sin}\left(\mathrm{kr}-\frac{l\pi }{2}\right)}{2\mathrm{kr}}-{b}_{l}\frac{\mathrm{cos}\left(\mathrm{kr}-\frac{l\pi }{2}\right)}{2\mathrm{kr}}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\frac{{\mathrm{ia}}_{l}+{b}_{l}}{2\mathrm{kr}}{e}^{\mathrm{ikr}}+\frac{{\mathrm{ia}}_{l}-{b}_{l}}{2\mathrm{kr}}{e}^{-i\left(\mathrm{kr}-l\pi \right)}\hfill & \hfill \left(137\right)\end{array}$

i.e.
 $⟨r|{\psi }^{+}⟩{|}_{r>>R}=\frac{1}{\left(2\pi {\right)}^{3/2}}\sum _{l}\frac{\left(2l+1\right)}{2\mathrm{ik}}\left[\left({a}_{l}-{\mathrm{ib}}_{l}\right)\frac{{e}^{\mathrm{ikr}}}{r}-\left({a}_{l}+{\mathrm{ib}}_{l}\right)\frac{{e}^{-i\left(\mathrm{kr}-l\pi \right)}}{r}\right]{P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(138\right)$
from eqn. (127,129) we have
 $⟨r|{\psi }^{+}⟩{|}_{r>>R}=\frac{1}{\left(2\pi {\right)}^{3/2}}\sum _{l}\frac{\left(2l+1\right)}{2\mathrm{ik}}\left[{e}^{2i{\delta }_{l}\left(E\right)}\frac{{e}^{\mathrm{ikr}}}{r}-\frac{{e}^{-i\left(\mathrm{kr}-l\pi \right)}}{r}\right]{P}_{l}\left(\mathrm{cos}\Theta \right)$ $\left(139\right)$
comparison of the last two eqns. leads to
 ${a}_{l}-{\mathrm{ib}}_{l}={e}^{2i{\delta }_{l}\left(E\right)} ,\mathrm{ }{a}_{l}+{\mathrm{ib}}_{l}=1$ $\left(140\right)$
or
 ${a}_{l}={e}^{i{\delta }_{l}\left(E\right)}\mathrm{cos}\left({\delta }_{l}\left(E\right)\right) ,\mathrm{ }{b}_{l}=-{e}^{i{\delta }_{l}\left(E\right)}\mathrm{sin}\left({\delta }_{l}\left(E\right)\right)$ $\left(141\right)$
i.e.
 ${A}_{l}\left(r\right)={e}^{i{\delta }_{l}\left(E\right)}\left[\mathrm{cos}\left({\delta }_{l}\left(E\right)\right){j}_{l}\left(\mathrm{kr}\right)-\mathrm{sin}\left({\delta }_{l}\left(E\right)\right){n}_{l}\left(\mathrm{kr}\right)\right]$ $\left(142\right)$
Now, once we have determined ${a}_{l}$ and ${b}_{l}$ by solving Schrödinger's equation for $r>R$. We can calculate the
logarithimic derivative
 ${\alpha }_{l}={r\frac{d\mathrm{ln}\left({A}_{l}\left(r\right)\right)}{\mathrm{dr}}|}_{r=R}=\mathrm{kR}\frac{\mathrm{cos}\left({\delta }_{l}\left(E\right)\right){j}_{l}\text{'}\left(\mathrm{kR}\right)-\mathrm{sin}\left({\delta }_{l}\left(E\right)\right){n}_{l}\text{'}\left(\mathrm{kR}\right)}{\mathrm{cos}\left({\delta }_{l}\left(E\right)\right){j}_{l}\left(\mathrm{kR}\right)-\mathrm{sin}\left({\delta }_{l}\left(E\right)\right){n}_{l}\left(\mathrm{kR}\right)}$ $\left(143\right)$
which can be inverted to yield
 $\mathrm{tan}\left({\delta }_{l}\left(E\right)\right)=\frac{{\alpha }_{l}{j}_{l}\left(\mathrm{kR}\right)-{\mathrm{kRj}}_{l}\text{'}\left(\mathrm{kR}\right)}{{\alpha }_{l}{n}_{l}\left(\mathrm{kR}\right)-{\mathrm{kRn}}_{l}\text{'}\left(\mathrm{kR}\right)}$ $\left(144\right)$
To summarize:
the complete solution of the scattering problem, i.e. the phase shifts are obtained from the logarithmic derivative of the solution of the radial Schrödinger equation at the potential's boundary.
Example: Scattering from a hard sphere
 $v\left(r\right)=\left\{\begin{array}{cc}\infty \hfill & , r\leqslantR\hfill \\ 0\hfill & , \mathrm{else}\hfill \end{array}$ $\left(145\right)$
In this case ${\delta }_{l}$ can be inferred directly from eqn. (142) and $A\left(R\right)=0$, i.e.
 $\begin{array}{ccc}\multicolumn{1}{c}{\mathrm{cos}\left({\delta }_{l}\right){j}_{l}\left(\mathrm{kR}\right)-\mathrm{sin}\left({\delta }_{l}\right){n}_{l}\left(\mathrm{kR}\right)}& =\hfill & 0\hfill \\ \multicolumn{1}{c}{\mathrm{tan}\left({\delta }_{l}\right)}& =\hfill & \frac{{j}_{l}\left(\mathrm{kR}\right)}{{n}_{l}\left(\mathrm{kR}\right)}\hfill & \hfill \left(146\right)\end{array}$

In the
s-wave channel, $l=0$, this implies
 $\mathrm{tan}\left({\delta }_{l}\right)=-\mathrm{tan}\left(\mathrm{kR}\right) ⇒ {\delta }_{l}=-\mathrm{kR}$ $\left(147\right)$

 $\begin{array}{ccc}\multicolumn{1}{c}{{A}_{l}\left(r\right)}& =\hfill & {e}^{\mathrm{ikR}}\left[\mathrm{cos}\left(\mathrm{kR}\right){j}_{l}\left(\mathrm{kr}\right)+\mathrm{sin}\left(\mathrm{kR}\right){n}_{l}\left(\mathrm{kr}\right)\right]\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{\mathrm{ikR}}\frac{1}{\mathrm{kr}}\mathrm{sin}\left(k\left(r-R\right)\right)\hfill & \hfill \left(148\right)\end{array}$

I.e. the wave function is a ' $\mathrm{sin}$' phase-shifted by $R$. This clearly highlights the significance of the wording 'scattering phase-shift'. Based on the asymptotic behavior of the spherical Bessel functions we can discuss the low and high energy behavior of the phase shifts, i.e. $\mathrm{kR}<<1$ and $\mathrm{kR}>>1$. For $\mathrm{kR}<<1$ we use
 ${j}_{l}\left(x<<1\right)=\frac{{x}^{l}}{\left(2l+1\right)!!}\mathrm{ }{n}_{l}\left(x<<1\right)=-\frac{\left(2l-1\right)!!}{{x}^{l+1}}$ $\left(149\right)$

 $\mathrm{tan}\left({\delta }_{l}\right)=-\frac{{\mathrm{kR}}^{2l+1}}{\left(2l+1\right)\left[\left(2l-1\right)!!{\right]}^{2}}$ $\left(150\right)$
This implies ${\delta }_{0}/{\delta }_{l>0}<<1$, i.e. only s-wave scattering is relevant. From eqn. (132) we get
 ${\sigma }_{\mathrm{tot}}=\frac{4\pi }{{k}^{2}}{\mathrm{sin}}^{2}\left({\delta }_{0}=\mathrm{kR}<<1\right)=4\pi {R}^{2}\ne \pi {R}^{2}$ $\left(151\right)$
This $4×$the classical area of the scattering target. At $k<<{R}^{-1}$ the particle's wave-length is large compared to the scatteres linear dimension. Therefore we may not necessarily expect a classical result. For $\mathrm{kR}>>1$ we have
 ${\mathrm{sin}}^{2}\left({\delta }_{l}\right)=\frac{{\mathrm{tan}}^{2}\left({\delta }_{l}\right)}{1+{\mathrm{tan}}^{2}\left({\delta }_{l}\right)}=\frac{\left(\frac{\mathrm{sin}\left(\mathrm{kR}-l\pi /2\right)}{\mathrm{cos}\left(\mathrm{kR}-l\pi /2\right)}{\right)}^{2}}{1+\left(\frac{\mathrm{sin}\left(\mathrm{kR}-l\pi /2\right)}{\mathrm{cos}\left(\mathrm{kR}-l\pi /2\right)}{\right)}^{2}}={\mathrm{sin}}^{2}\left(\mathrm{kR}-l\pi /2\right)$ $\left(152\right)$
where the asymptotic forms eqn. (119) of $j\left(n{\right)}_{l}\left(x\right)$ have been used. The meaning of $x\to \infty$ in the latter, is that $x-l\pi /2>>1$. I.e. eqn. (152) can be used only for $l\leqslant{l}_{\mathrm{max}}<~\mathrm{kR}$. Since ${l}_{\mathrm{max}}>>1$, many $l$-values contribute and we approximate
 $\begin{array}{ccc}\multicolumn{1}{c}{{\mathrm{sin}}^{2}\left(\mathrm{kR}-l\pi /2\right)}& \approx \hfill & \frac{1}{2}\hfill \\ \multicolumn{1}{c}{\sum _{l}^{\mathrm{kR}}\left(2l+1\right)}& \approx \hfill & \left(\mathrm{kR}{\right)}^{2}\hfill & \hfill \left(153\right)\end{array}$

and therefore
 ${\sigma }_{\mathrm{tot}}=\frac{4\pi }{{k}^{2}}\sum _{l}\left(2l+1\right){\mathrm{sin}}^{2}\left(\mathrm{kR}-l\pi /2\right)\approx \frac{4\pi }{{k}^{2}}\left(\mathrm{kR}{\right)}^{2}\frac{1}{2}=2\pi {R}^{2}\ne \pi {R}^{2}$ $\left(154\right)$
Up to eqn. (151) everything has been exact. Eqn. (154) is only a
rough lower bound. Therefore, at high energies ${\sigma }_{\mathrm{tot}}$ is twice the geometrical cross section.
This can be understood in words as follows: a) At short wave lengths ('ray'-limit), the incoming particles (wave) with impact paramter $>R$ 'just pass'. I.e. $f\left(\Theta \right)$ is zero for all angles between the backward and forward direction. b) All incoming particles with impact paramter $ are exactly reflected. I.e. $f\left(\Theta \right)$ has a peak into the backward direction. The differential cross section integrated over this peak (partial cross section) must be $\pi {R}^{2}$ for all particles to be reflected. c) Into the forward direction there are no particles at impact paramter $. I.e. $f\left(\Theta \right)$ must peak into the forward direction to interfere destructively with the incoming wave. d) Because of unitarity the forward and backward scattered currents must have identcal magnitude. I.e. the backward and forward partial cross sections are identical. In summary: ${\sigma }_{\mathrm{tot}}=\pi {R}^{2}+\pi {R}^{2}$
Looking at this mathematically we consider the following decomposition
 $\begin{array}{ccc}\multicolumn{1}{c}{f\left(\Theta \right)}& =\hfill & {f}_{r}\left(\Theta \right)+{f}_{s}\left(\Theta \right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & R\frac{1}{2\mathrm{iRk}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right){e}^{2i{\delta }_{l}}{P}_{l}\left(\mathrm{cos}\Theta \right)+R\frac{i}{2\mathrm{Rk}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right){P}_{l}\left(\mathrm{cos}\Theta \right)\hfill & \hfill \left(155\right)\end{array}$

a) ${f}_{r}$ is complex, while ${f}_{s}$ is purely imaginary
 ${f}_{r}\left(\Theta \right)\in \mathbbC ,\mathrm{ }{f}_{s}\left(\Theta \right)\in i\mathbbR$

Therefore all scattered wave contributions from ${f}_{s}\left(\Theta \right)$
into the forward direction will interfere destructively with the incoming wave (see eqn.(97))
b)
 $\begin{array}{ccc}\multicolumn{1}{c}{{\sigma }_{\mathrm{tot}}^{r\left(s\right)}\equiv {\int }_{4\pi }d\Omega |{f}_{r\left(s\right)}\left(\Theta \right){|}^{2}}& =\hfill & \frac{\pi }{2{k}^{2}}\sum _{l}^{\mathrm{kR}}\left(2l+1{\right)}^{2}{\int }_{-1}^{1}{P}_{l}\left(x{\right)}^{2}\mathrm{dx}=\frac{\pi }{{k}^{2}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right)\approx \pi {R}^{2}\hfill \\ \multicolumn{1}{c}{{\int }_{4\pi }d\Omega {f}_{r}^{☆}\left(\Theta \right){f}_{s}\left(\Theta \right)}& =\hfill & \frac{\pi }{{k}^{2}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right){e}^{-2i{\delta }_{l}}\approx 0\hfill \\ \multicolumn{1}{c}{⇒{\sigma }_{\mathrm{tot}}}& =\hfill & {\sigma }_{\mathrm{tot}}^{r}+{\sigma }_{\mathrm{tot}}^{s}\hfill \end{array}$

For the last eqn. we have used eqn. (152) from wich ${\delta }_{l}=±\left(\mathrm{kR}-l\pi /2\right)$, where it is sufficient to use the upper sign. I.e. the phase shift oscillated by $l\pi$. This makes the sum negelegible compared to $\left(\mathrm{kR}{\right)}^{2}$.
c) As $\mathrm{kR}>>1$, ${f}_{r\left(s\right)}\left(\Theta \right)$ is strongly peaked into the backward(forward) direction and $|{f}_{r\left(s\right)}\left(\Theta \right){|}^{2}$ shows a diffraction pattern
Figure 6: Red(Black) = $|{f}_{r\left(s\right)}\left(\Theta \right){|}^{2}/{R}^{2}$ for $\Theta \in \left[0,2\pi \right]$ and for $\mathrm{kR}=10$
Therefore ${f}_{r\left(s\right)}\left(\Theta \right)$ is called the reflection(shadow) amplitude.
Relation to the optical theorem:
 $\begin{array}{ccc}\multicolumn{1}{c}{\frac{4\pi }{k}\mathrm{Im}\left(f\left(0\right)\right)}& =\hfill & \frac{4\pi }{k}\mathrm{Im}\left({f}_{r}\left(0\right)\right)+\frac{4\pi }{k}\mathrm{Im}\left({f}_{s}\left(0\right)\right)\hfill \\ \multicolumn{1}{c}{}& \hfill & \frac{4\pi }{k}\mathrm{Im}\left[R\frac{1}{2\mathrm{iRk}}{e}^{2\mathrm{ikR}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right)\left(-{\right)}^{l+1}\right]+\frac{4\pi }{k}\mathrm{Im}\left[R\frac{i}{2\mathrm{Rk}}\sum _{l}^{\mathrm{kR}}\left(2l+1\right)\right]\hfill \\ \multicolumn{1}{c}{}& \approx \hfill & 0+\frac{4\pi }{k}\mathrm{Im}\left({f}_{s}\left(0\right)\right)=\frac{4\pi }{k}\frac{1}{2k}\left(\mathrm{kR}{\right)}^{2}=2\pi {R}^{2}={\sigma }_{\mathrm{tot}}\hfill \end{array}$

where we have used eqn. (152) with ${\delta }_{l}=±\left(\mathrm{kR}-l\pi /2\right)$ and have chosen the positive sign.
Exercise 11 Low energy scattering and bound states and scattering length.
a) Consider a spherically symmetric potential
 $v\left(r\right)=\left\{\begin{array}{cc}{v}_{0}\hfill & , r\leqslantR\hfill \\ 0\hfill & , \mathrm{else}\hfill \end{array}$

In the low-energy limit $k\to 0$ only $l=0$ scattering is relevant. The solution of the corresponding radial Schrödinger equation is ${\mathrm{rA}}_{0}\left(r\right)=\phi \left(r\right)$. Show that for $r>R$ and $k\to 0$
 $\phi \left(r\right)=\mathrm{const}×\left(r-a\right)$

Use ${A}_{0}\left(r\right)$'s representation in terms of spherical Bessel functions to show that also
 $\phi \left(r\right)={e}^{i{\delta }_{0}}\frac{1}{k}\mathrm{sin}\left(\mathrm{kr}+{\delta }_{0}\right)$

Use the logarithmic derivative of $\phi \left(r\right)$ to conclude that
 $\underset{k\to 0}{lim}k\mathrm{cot}\left(\mathrm{kr}+{\delta }_{0}\right)=\frac{1}{r-a}$

and from that
 ${\sigma }_{\mathrm{tot}}{|}_{k\to 0}=4\pi {a}^{2}$

b) Roughly sketch the potential and the radial wave function and discuss the scattering length $a$ qualitatively for the following three different cases:
i) ${v}_{0}>0$ and such that ${v}_{0}>>\hslash {}^{2}{k}^{2}/2m$.
ii) ${v}_{0}<0$ and such that the energy ${\epsilon }_{b}$ of the highest of any potentially present bound state of the potential fulfills $|{\epsilon }_{b}|>>\hslash {}^{2}{k}^{2}/2m$.
iii) ${v}_{0}<0$ and such that the energy ${\epsilon }_{b}$ of the highest of any potentially present bound state of the potential fulfills $|{\epsilon }_{b}|~\hslash {}^{2}{k}^{2}/2m$.

## Index (showing section)

 absorption cross section, 3.2 advanced, 4.0 Bremstrahlungs, 3.1 Casimir effect, 2.1 coherent state, 2.2 correspondence principle, 2.1 Coulomb gauge, 2.1 differential cross section, 4.1 dipole radiation, 3.1 Dirac picture, 1.1 Dirac's perturbations theory, 1.1 far-field approximation, 4.0 Fermi's golden rule, 1.2 final state, 1.2 Glauber state, 2.2 Greens function, 4.3 Heisenberg picture, 1.1 homogeneous wave equation, 2.1 initial state, 1.2 Lippman-Schwinger, 4.0 optical theorem, 4.4 partial-wave amplitude, 4.5 partial-wave states, 4.5 photoelectric effect, 3.2 photon current, 3.2 photons, 2.1 Poynting vector, 2.1 radial Schrödinger equation, 4.5 resolvent, 4.3 retarded, 4.0 s-wave channel, 4.6 scattering amplitude, 4.0 scattering phase-shift, 4.5 spontaneous emission, 3.1 stimulated emission, 3.1 T-matrix equation, 4.3 time dependent perturbation theory, 1.1 time evolution operator, 1.1 time ordering, 1.1 transition operator, 4.3 two-level system, 1.2 unitarity, 4.5 wave zone approximation, 3.1 zero-point energy of the vacuum, 2.1

### Footnotes:

${}^{1}$Don't confuse this with the notions of canonical vs. grand canonical in statistical mechanics. We are not discussing ensembles, but only one microstate of a physical system here.

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On 18 Nov 2009, 10:21.