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Dirac's Perturbation Theory
    1.1  The time evolution operator
    1.2  Transition propabilities
Quantum Theory of Radiation
    2.1  Quantizing Electromagnetic Fields in Vacuum
    2.2  Classical States of the Radiation Field
Radiation Interacting with Matter
    3.1  Emission of Radiation
    3.2  Absorption of Radiation
    3.3  Black-Body Radiation
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

1  Dirac's Perturbation Theory

1.1  The time evolution operator

Dirac's perturbation theory, is also frequently called 'time dependent' perturbation theory. Consider the time evolution operator U(t, t0 )
|ψ,t=U(t, t0 )|ψ, t0 (1)
of a state |ψ,t, which is set by Schrödinger's equation for a general, time dependent Hamiltonian H(t)
i t |ψ,t=H(t)|ψ,t (2)
i t U(t, t0 )=H(t)U(t, t0 ) (3)
i.e. U(t, t0 ) is valid for any state. The time evolution operator should have the following properties
U( t0 , t0 ) = 1       (4) U( t2 , t1 )U( t1 , t0 ) = U( t2 , t0 )       (5)

Eqn.(3) is formally solved by
U(t, t0 )=1- i t0 t dτH(τ)U(τ, t0 ) (6)
From eqns.(6,4) and from t= t0 +dt
U(t+dt,t)=1- i H(t)dt (7)
I.e. an infinitesimal time-translation is realized a
unitary transformation, since
U(t+dt,t )+ U(t+dt,t)=1+ i H(t )+ dt- i H(t)dt+O( dt2 )=1+O( dt2 ) (8)
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(t, t0 ) is unitary
U(t, t0 )+ U(t, t0 ) = 1       (9) U(t, t0 )+ = U( t0 ,t)       (10)

In the
time-independent case the solution of eqn.(3) is simply
U(t, t0 )= e-iH(t- t0 )/ (11)
Consider now the case of
H(t)= H0 +V(t) (12)
where we assume that we are able to diagonalize H. This resembles a time-depen
t perturbation V(t) 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(t, t0 )= e- iH0 (t- t0 )/ UD (t, t0 ) (13)
Note that any state |ψ,t D , which evolves in time via UD (t, t0 ), i.e. by |ψ,t D = UD (t, t0 )|ψ, t0 D is obtained from |ψ,t by
|ψ,t D = e iH0 (t- t0 )/ |ψ,t (14)
I.e. the states |ψ,t D are simply the representations of |ψ,t in the
Dirac (or interaction) picture of H(t) or of the Heisenberg picture of H. Therefore UD (t, t0 ) is the time evolution operator in the Dirac picture. If V/H0, we expect UD (t, t0 )1, therefore, we expect that UD (t, t0 )-1 will allow for some kind of perturbation expansion in terms of V. Inserting eqn.(13) into eqn.(3) we get
e- iH0 (t- t0 )/ H0 UD (t, t0 )+ e- iH0 (t- t0 )/ i t UD (t, t0 )=H(t) e- iH0 (t- t0 )/ UD (t, t0 ) i t UD (t, t0 )=- H0 UD (t, t0 )+ e iH0 (t- t0 )/ H(t) e- iH0 (t- t0 )/ UD (t, t0 )     i t UD (t, t0 )= e iH0 (t- t0 )/ V(t) e- iH0 (t- t0 )/ UD (t, t0 ) i t UD (t, t0 )= VD (t, t0 ) UD (t, t0 )       (15)

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
VD (t, t0 )= e iH0 (t- t0 )/ V(t) e- iH0 (t- t0 )/ (16)
Eqn.(15) can be integrated formally
UD (t, t0 )=1- i t0 t dτ VD (τ, t0 ) UD (τ, t0 ) (17)
Now we are in a position to apply the main idea of Dirac's perturbation theory, which lies in the fact that for V/H0 an expansion of UD (t, t0 )-1 in powers of VD can be obtained by solving eqn.(15)
UD (t, t0 ) = 1+(- i ) t0 t d τ1 VD ( τ1 , t0 )+(- i )2 t0 t t0 τ1 d τ1 d τ2 VD ( τ1 , t0 ) VD ( τ2 , t0 ) +       (18)

Formally this can be exponentiated by introducing the so called time ordering of operators
Tτ [A( t1 )B( t2 )]={ A( t1 )B( t2 ), t1 \geqslant t2 B( t2 )A( t1 ), t2 > t1 (19)
Since there are n! ways to order n times t1n we may write
UD (t, t0 ) = Tτ [ n=0 1 n! (- i )n t0 t d τ1 t0 t d τn VD ( τ1 , t0 ) VD ( τn , t0 )] = Tτ exp[- i t0 t dτ VD (τ, t0 )]       (20)

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 [ H0 ,V]=0 a perturbation expansion in terms of V would not have been necessary right from the start. In fact, in that case
VD (τ, t0 )= e iH0 (τ- t0 )/ Ve- iH0 (τ- t0 )/ =V Tτ exp[- i t0 t dτ VD (τ, t0 )]= Tτ exp[- i t0 t dτV]= e-iV(t- t0 )/ U(t, t0 )= e- iH0 (t- t0 )/ e-iV(t- t0 )/ = e-i( H0 +V)(t- t0 )/       (21)

which is exactly what we expect.
Each addends from the 2nd line in eqn. (20) to the total time evolution operator U(t, t0 )= e- iH0 (t- t0 )/ UD (t, t0 ) can be written somewhat differently using eqn. (16) and defining the time evolution operator U0 ( t1 , t2 )= e- iH0 ( t1 - t2 )/ of the unperturbed sytem, namely
U(t, t0 )|O( Vn ) = (- i )n t0 t t0 τn-2 t0 τn-1 d τ1 d τn-2 d τn [ U0 (t, τ1 )V( τ1 ) U0 ( τ1 , τ2 ) U0 ( τn-2 , τn-1 )V( τn-1 ) U0 ( τn-1 , τn )V( τn ) U0 ( τn , t0 )]       (22)

with this we may express the O( Vn ) contribution to the time evolution U(t, t0 )|O( Vn ) |ψ, t0 of state in a diagrammatic way
Figure 1: Diagrammatic expression for Dirac's state propagator at order Vn

1.2  Transition propabilities

Consider a quantum system described by H= H0 +V, where
H0 |n= en |nm|n= δmn (23)
we know how to diagonalize H0 . Now, assume that at time t0 =0 the system is in an
initial state |n. We seek to obtain the propability Pmn (t) by which we will find the system in the final state |m at time t
Pmn (t)= |m| e- iH0 t/ UD (t,0)|n|2 (24)
The corresponding amplitude is
m| e- iH0 t/ UD (t,0)|n=m| e- iH0 t/ +(- i ) 0 t dτ e iH0 (τ-t)/ Ve- iH0 τ/ |n+O( V2 ) e- iem t/ [ δmn +(- i ) 0 t dτ ei( em - en )τ/ m|V|n]+O( V2 )                                       = e- iem t/ δmn +[ e- iem t/ em - en + e- ien t/ en - em ]m|V|n+O( V2 )       (25)

For the remainder of this section we confine ourselves to the case mn and to only the leading order contribution in V to Pmn (t), which is O( V2 ). Because of the δmn -function in the 1st term of eqn.(25) the order O( V2 ) term in eqn.(25) does not contribute to Pmn (t). We get
Pmn (t)=|m|V|n |2 4 sin2 [( em - en )t/(2)] ( em - en )2 (26)
where we have used 1-cos(x)=2 sin2 (x/2).
Exercise 1 Consider the quantum system eqn.(23) in state |n at time t0 =0. Show that the expression corresponding to eqn.(26) for m=n is
Pnn (t)=1- mn |m|V|n |2 4 sin2 [( em - en )t/(2)] ( em - en )2 (27)
Interpret this result in terms of eqn.(26) and probability conservation.

Fermis Golden Rule

If we ask for the rate Rmn (t) of transition from state |n into state |m we have to consider
Rmn (t)= dPmn (t) dt =|m|V|n |2 2π 2 sin[( em - en )t/] π( em - en )/ (28)
Using that
limt sin[( em - en )t/] π( em - en )/ =δ( em - en ) (29)
we find that for very large times the transition rate approaches a time independent value
Rmn = 2π |m|V|n |2 δ( em - en ) (30)
This result is rather useless: either em = en , then Rmn = i.e. perturbaion theory is invalid, or em en then Rmn =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(a,b) and ε(a,b) is the energy. Consider the sum of all transition rates into states in the vicinity of en
rmna = ε= en -Δ ε= en +Δ Rmn |a, = en -Δ en +Δ de ε δ(e-ε) |a, Rmn = en -Δ en +Δ de ρa (e) Rmn = 2π |m|V|n |2 ρa ( en )       (31)

This is Fermi's famous golden rule. The constraint |a, symbolizes, that in performing the sum over energies certain quantum number may be kept fixed and ρa (e)= ε δ(e-ε) |a, is the density of states. Obviously the size 2Δ of the energy interval is not relevant - as long as it contains the initial state energy en . Note that rmna ~[ erg2 /erg/] has dimension [1/time], as has Rmn .
Exercise 2 Given the two-level system
H= H0 +W, H0 =[ E0 0-E ],W=[ 0V V0 ] (32)
The eigenstates of H0 are |0 and |1, with energies E and -E. Show that the time evolution operator in Schrödinger's picture is given by
U(t,0)=[ cos(aτ)- iE a sin(aτ)- iV a sin(aτ) - iV a sin(aτ)cos(aτ)+ iE a sin(aτ) ] (33)
with a= E2 + V2 and τ=t/. Compare this result with eqn.(26,27). What are the differences? Assuming that the system is in state |0 at t=0, plot Pmn (t) from eqn.(24) for m,n=1,0 and m,n=0,0 setting =1, E=1, and V=0.1

2  Quantum Theory of Radiation

2.1  Quantizing 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
φ=0,·A=0 (34)
which leads to
E=- 1 c t A,B=×A (35)
In this situation the vector potential satisfies the
homogeneous wave equation
A- 1 c2 t 2 A=0 (36)
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(r) is a vector field, which takes on values for
all r R3 . 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(x+L,y,z)=A(x,y,z)&cyclic (37)
This allows to expand A(r,t) in a Fourier series
A(r,t)= k,σ=1,2 ( 2π c2 V ωk )1/2 uk,σ ak,σ (t) eik·r (38)
where the wave vector k=( lx , ly , lz )2π/L and li Z. The normalization prefactor in the ()-brackets has been set for later convenience. Remember, that because of ·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 uk,σ k. In view of our goal to develop a quantum theory for the vector potential, the time dependence of anticipated
operator A(r,t) is analogous to a time-dependence of to operators x(t) or p(t) 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
ak,σ (t)= a-k,σ (t) (39)
where we set uk,σ = u-k,σ and uk,σ = uk,σ . Therefore
A(r,t)= k,σ=1,2 1 2 ( 2π c2 V ωk )1/2 uk,σ ( ak,σ (t) eik·r + ak,σ (t) e-ik·r ) (40)
The Fourier coefficients ak,σ (t) satisfy the
2nd order linear DEQN
( dt 2 - ωk 2 ) ak,σ (t)=0 ak,σ (t)= bk,σ e-i ωk t + b-k,σ ei ωk t (41)
with the dispersion relation ωk =kc and k=|k|. The particular setting of the amplitudes bk,σ , b-k,σ for the two linearly independent solutions has been chosen to fulfill eqn. (39). Because of the latter, the amplitudes ak,σ (t) cannot be viewed as independent from each other. Inserting into eqn. (40) and using ωk = ω-k we get
A(r,t)= k,σ=1,2 ( 2π c2 V ωk )1/2 uk,σ ( bk,σ (t) eik·r + bk,σ (t) e-ik·r ) (42)
where bk,σ (t) has to satisfy only the
1st order linear DEQN
( dt +i ωk ) bk,σ (t)=0 bk,σ (t)= bk,σ e-i ωk t (43)
In contrast to ak,σ (t), the bk,σ (t) can now be viewed as independent amplitudes for each k and σ 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 bk,σ (t), which we will show now.
The Hamiltonian function of the electromagnetic field can be expressed in terms of bk,σ (t)
H = 1 8π d3 r( E2 + B2 )= 1 8π d3 r( 1 c2 | t A |2 +|×A |2 ) = k,μ=1,2 q,ν=1,2 [ ( 2π c2 V ωk )1/2 ( 2π c2 V ωq )1/2 (-) uk,μ · uq,ν ωk ωq c2 ( bk,μ (t) eik·r - bk,μ (t) e-ik·r )× ( bq,ν (t) eiq·r - bq,ν (t) e-iq·r ) d3 r -(k× uk,μ )·(q× uq,ν )( bk,μ (t) eik·r - bk,μ (t) e-ik·r )× ( bq,ν (t) eiq·r - bq,ν (t) e-iq·r ) d3 r]       (44)

eik·r eiq·r d3 r = V δk,q uk,μ · uq,ν = δμ,ν       (45)

we get
H = k,σ=1,2 1 4 ωk {[( bk,σ bk,σ + bk,σ bk,σ )-( bk,σ b-k,σ + bk,σ b-k,σ )]                     +[( bk,σ bk,σ + bk,σ bk,σ )+( bk,σ b-k,σ + bk,σ b-k,σ )]} = k,σ=1,2 1 2 ωk ( bk,σ bk,σ + bk,σ bk,σ )       (46) = k,σ=1,2 ωk bk,σ bk,σ       (47)

where we have discarded the time dependence of the bk,σ for the moment. Eqns. (46) and (47) are surely identical for classical C-number variables bk,σ .
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 Ek = ωk for each wave vector and polarization. In this context, the C-number variables bk,σ , bk,σ are interpreted as boson operators bk,σ , bk,σ + with commutation relations
[ bk,μ , bq,ν + ]= δk,q δμ,ν ,[ bk,μ , bq,ν ]=0,[ bk,μ + , bq,ν + ]=0 (48)
Heisenberg's equation of motion yields
dt bk,μ (t)= i [H, bk,μ ](t)= i q,σ=1,2 ωq [ bq,σ bq,σ , bk,μ ](t)=-i ωk bk,μ (t) (49)
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= k,σ=1,2 ωk ( bk,σ + bk,σ + 1 2 )= k,σ=1,2 ωk ( Nk,σ + 1 2 ) (50)
where Nk,σ = bk,σ + bk,σ is the boson occupation number operator for each mode at (k,σ). Therefore the
zero-point energy of the vacuum is
E0 = k,σ=1,2 ωk /2 (51)
which is present even if the boson occupation in each harmonic oscillator is identical to zero. Simple insepction of E0 shows, that the sum on the right hand side diverges. I.e. the zero-point energy of the vacuum is
infinite. We will return to this issue later.
Finally note, that having no explicit time-dependence displayed on eqn.(50) is exactly what we should expect since dt H=i[H,H]/=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
|{ nk,σ }| n k1 , σ1 | n ks , σs | n k1 , σ1 ,, n ks , σs , (52)
Remember the usual things
Nk,σ |{ nq,μ } = nk,σ |{ nq,μ } bk,σ + |{ nq,μ } = nk,σ +1|( nk,σ +1) bk,σ |{ nq,μ } = nk,σ |( nk,σ -1)       (53)

For interpretational purposes it is interesting to evaluate the
Poynting vector P of the electromagnetic field
P(t)= 1 4πc d3 rE×B=- 1 4π c2 d3 r t A×(×A)= - 1 4π c2 k,q,μ,ν=1,2 [ 2π c2 V ωk ωq uk,μ ×(iq× uq,ν )(-i ωk )       (54) d3 r( bk,σ (t) eik·r - bk,σ + (t) e-ik·r )( bq,ν (t) eiq·r - bq,ν + (t) e-iq·r )]

There are two cases k=q and k=-q. However, since we have set uqμ = u-qμ the cross product uk,μ ×(q× uq,ν ) yields
uq,μ ×(q× uq,ν )= δμν uμ ×(q× uμ )={ μ=1: u1 ×(q× u1 )= u1 ×(-q u2 )=q μ=2: u2 ×(q× u2 )= u2 ×(q u1 )=q (55)
in both cases. I.e.
P= -1 4π c2 q,μ=1,2 ( 2π c2 V ωq ) ωq [q b-q,μ bq,μ -q bq,μ bq,μ + -q bq,μ + bq,μ +q b-q,μ + bq,μ + ] (56)
where we have switched to the Schrödinger picture, i.e. we have droped the time-dependence on the bq,μ (+) s. Now
q q b-q,μ bq,μ =- q q bq,μ b-q,μ =- q q b-q,μ bq,μ =0 (57)
because [ bk,μ , bq,ν ]=[ bk,μ + , bq,ν + ]=0 and
q q bq,μ bq,μ + = q q bq,μ + bq,μ + q q= q q bq,μ + bq,μ (58)
because [ bk,μ , bk,μ + ]=1. Therefore
P= k,μ=1,2 k bk,μ + bk,μ (59)
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 εkμ = ωkμ 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 k=p per photon. So the photons are particles. Moreover since they satisfy εkμ =pc, where c is the speed of light, these particles move at the speed of light.

2.2  Classical 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,σ, which is occupied with n photons |{ nk,σ }| nk,σ |0|n. The expectation - and also eigen - value of the energy (discarding the zero point energy) is
H=n ωk (60)
n=n| b+ b|n (61)
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 ( 2π ωk V )1/2 u(n|b|n eik·r -n| b+ |n e-ik·r )=0 (62)
(at any time) since b(+) (in)decrease the number of photons by one.
Classically speaking therefore the n-photon state at k,σ 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= n=0 gn e-|g |2 /2 n! |n n=0 βn |n (63)
The coefficients βn have been chosen such as to have |g normalized
e-|g |2 n=0 |g |2n n! = e-|g |2 e|g |2 =1 (64)
The physical meaning of | βn |2 is the propability to find n photons in state |g when measuring the photon number operator b+ b. Note that | βn |2 is the Poisson distribution Π( g2 ,n), which is sharply peeked at g2 if g2 >>1. Since | βn |2 0n0 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
g|b|g = l=0 m=0 βl βm l|b|m= l=0 m=0 βl βm ml|m-1= l=0 βl βl+1 l+1 = e-|g |2 l=0 gl l! gl+1 (l+1)! l+1=g e-|g |2 l=0 |g |2l l! =g       (65)

From hermiticity it follows g| b+ |g= g . Finally
ng| b+ b|g = l=0 m=0 βl βm ml|m= l=0 | βl |2 l= e-|g |2 l=0 |g |2l l! l = |g |2 e-|g |2 l=0 |g |2l l! =|g |2       (66) n2 g| b+ bb+ b|g = l=0 m=0 βl βm m2 l|m= l=0 | βl |2 l2 = e-|g |2 l=0 |g |2l l! l2 = e-|g |2 l=0 |g |2l l! (l(l-1)+l)= = |g |4 +|g |2 =n 2 +n       (67)

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 ( 2π ωk V )1/2 u(g eik·r - g e-ik·r ) (68)
Exercise 4 Show that the relative uncertainty, i.e. the relative variance, of the photon number in the Glauber state is
Δn/n=g|( b+ b-n )2 |g 1/2 /n=n -1/2 (69)
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|(E-g|E|g )2 |g= 2πω V (70)

3  Radiation Interacting with Matter

Consider non-relativistic matter, i.e. 'some' n particles, described by
Hm = i=1 n pi 2 2 mi +U( r1 , rn ) (71)
where ( mi , ri , pi ) are masses, coordinates and momenta, and U is some interaction potential. Including a finite electromagnetic field, the Hamiltonian changes into
H= i=1 n | pi - qi A( ri )/c |2 2 mi +U+ 1 8π d3 r( E2 + B2 )= Hm + Hem +V (72)
The contribution V is
V= i [- qi mi c pi A( ri )+ qi 2 2 mi c2 A( ri )2 ] (73)
The 1st addend V1 describes interaction between matter and radiation, the 2nd V2 is a self-interaction of radiation in the presence of charged matter. To simplify: only
one particle
V1 = - q mc k,σ=1,2 ( 2π c2 V ωk )1/2 pi · uk,σ ( bk,σ eik·r + bk,σ + e-ik·r )       (74) V2 = q2 2 mc2 k,q,σ,μ 2π c2 V ωk ωq uk,σ · uq,μ ( bk,σ bq,μ ei(q+k)·r + bk,σ + bq,μ + e-i(q+k)·r                      + bk,σ bq,μ + ei(k-q)·r + bk,σ + bq,μ ei(q-k)·r )       (75)

V1,2 are treated as perturbations to Hm,em . We will justify this later.

3.1  Emission of Radiation

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|F
|I = |i| nkσ |F = |f| nkσ +1       (76)

where |i(f) and | nkσ ( nkσ +1) 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 V1 only. Using eqns.(30,31) consider 1st
F|V|I=- q mc ( 2π c2 V ωk )1/2 f|p e-ik·r |i· uk,σ nkσ +1 (77)
This is
non-zero even if nkσ =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 nkσ 0 are called stimulated emission. Let's focus on spontaneous emission 1st.
Total rate summing all possible photons emitted (see eqn.(31)):
rfi = 4 π2 q2 Vm2 kσ 1 ωk |f|p e-ik·r |i |2 sin2 Θδ( ef + ωk - ei ) (78)
where Θ is the angle between p and k in the plane of
one polarization, say uk,1 and k. Now we use the wave-zone approximation |k| << |r|, i.e. the wave-lenght of the em radiation λem is large compared to the atomic size r, which remains valid even up to hard X-Ray transitions
e-ik·r =1-ik·r- 1 2 (k·r )2 +1 (79)
With k =V/(2π )3 d3 k, putting kz along p, and polar coordinates, see fig. 2, we get
Figure 2: Geometry for dipol matrix element

rfi = q2 2π m2 d3 k|f|p|i |2 1 kc sin2 Θδ( ef - ei +ck) = q2 2π m2 c |f|p|i |2 dk 0 π dΘ 0 2π dφk sin3 Θδ( ef - ei +ck) = q2 ωif m2 c3 |f|p|i |2 0 π dΘ sin3 Θ= 4 q2 ωif 3 m2 c3 |f|p|i |2       (80)

where ωif = ei - ef is the energy difference between the initial and final state of the atom. From eqn.(71)
[r, Hm ] = [r, p2 2m ]= i m p rfi = 4 q2 ωif 3 3 c3 |f|r|i |2 =α 4 ωif 3 3 c2 |f|r|i |2       (81)

The presence of f|r|i demonstrates the meaning of
dipole radiation. rfi is 'small' in two ways. To check this we have to look at rfi / ωif , i.e. the decay per one oscillation period. Now, (i) the fine-structure constant α= q2 /(c)1/137 (for q=electron charge) is small and (ii) ωif 2 / c2 |f|r|i |2 ~ a0 2 / λif 2 is small up to hard X-Ray transitions.
Using Heisenbergs equation of motion we may rewrite
|f| d2 r dt2 |i |2 = 1 4 |f|[H,[H,r]]|i |2 = 1 4 |f| H2 r-2HrH-r H2 |i |2 = ωif 4 |f|r|i |2 (82)
emittedpower = ωif rfi = 4 q2 3 c3 |f| d2 r dt2 |i |2 (83)
This shows that the result for classical
Bremstrahlung is corrected by a factor of two by Quantum mechanics.

3.2  Absorption of Radiation

Absorption of a quantum of radiation corresponds to |I|F
|I = |i| nkσ |F = |f| nkσ -1       (84)

Analogous to eqns.(78,77) we Rfi from eqn.(30) is obtained as
Rfi = 4 π2 q2 Vm2 ωk |f|p eik·r |i |2 nkσ sin2 Θδ( ef - ωk - ei ) (85)
Inspecting eqn.(77) and because of f|p eik·r |i=i|p e-ik·r |f it clear, that the transition rate for stimulated emission |f|i is identical to that of absorption |i|f. Dividing by the
photon current
jkσ = nkσ V c (86)
the absorption cross section Σfi,kσ = Rfi / jkσ (dimension [Σ]=(1/sec)/(1/( cm2 sec)= cm2 = area) is
Σfi,kσ = 4 π2 q2 cm2 ωk |f|p eik·r |i |2 sin2 Θδ( ef - ωk - ei ) (87)
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 ez and polarization uk,σ = ex 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= L3 , and that the emitted electron is approximately in a plane-wave state r|q= eirq /V where 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=( nx , ny , nz )2π/L with nx Z be quantized w.r.t. the box. The differential cross section for photo-electrons into a solid angle dΩ is
dσ dΩ dΩ= numberofelectronsinto dΩ/dt numberofphotonperunitarea /dt

Using eqn. (85-87), show that
dσ dΩ = α qf V 2πm ωk | qf | ex ·p eik·r |i |2

where 2 | qf |2 /(2m)= ef = ei + ωk is the photo-electrons energy and qf /| qf | is the direction of its emission. α is the
fine-structure constant. Verify dimensions, i.e. [dσ/dΩ]= length2 .
c) Chose the 1s state of a hydrogen-like atom with atomic number Z for |i
r|i= ( Z a0 )3 e-Zr/ a0

Show that
dσ dΩ =32 q2 qf ( ex · qf )2 mc ωk Z5 a0 5 1 [ Z2 / a0 2 + s2 ]4

where s= qf - ωk ez /c. [Hint: Use partial integration w.r.t. p=-i and ex k=0. With that qf | ex ·p eik·r |i is directly related to the Fourier transform of the 1s state.]

3.3  Black-Body Radiation

TODO .................
dNi dt = Nf tfi - Ni tif dNf dt = - Nf tfi + Ni tif

dNi dt = dNf dt       and        Ni Nf = eβ( ef - eI )

where β= kB T
TODO .....................
Figure 4: Scattering process

4  Quantum Theory of Scattering

H= H0 +V (88)
For a typical scattering problem H0 = p2 /2m could be the Hamiltonian of - simplification for the rest of this section: - one particle coming in on some potential well V. Let |φ ( |ψ) be solutions of the Schrödinger equation for H0 ( H) for
one and the same energy E as in fig. 4
(E- H0 )|ψ = V|ψ 0 = (E- H0 )|φ (E- H0 )|ψ = V|ψ+(E- H0 )|φ       (89)

This is trivial yet. The problem is, that E is real and the operator E- H0
cannot be inverted. Now consider two possible solutions | ψ± of eqn. (89), meant to represent an outgoing(incoming) scattered wave for the superscript +(-) for times t+(-). 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 +(-). This allows to replace
H0 H0 iη (90)
with η 0+ in eqn. (89) since within the time evolution exp(- iH0 tηt) the replacement eqn.(90) is irrelevant for t±. With eqn.(90) the
operator (E- H0 ±iη) is invertable for real E leading to

| ψ± =|φ+ 1 E- H0 ±iη V| ψ± (91)
This is the
Lippman-Schwinger equation. It is frequently stated in its real space representation
r| ψ± =r|φ+ d3 r'r| 1 E- H0 ±iη |r'r'|V| ψ± (92)
Exercise 6 Show that
G± (r,r')= 2 2m r| 1 E- H0 ±iη |r'=- 1 4π e±ik|r-r'| |r-r'| (93)
where 2 k2 /2m=E. To that end:

a) Use the real space representation of the momentum eigenstates r|p= eirp/ /(2π )3/2 , with p|p'=δ(p-p') 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'=V(r)δ(r-r') is diagonal in real space. Inserting d3 r"|r"r"|=1 into the right side of the last matrix element in eqn. (92) we therefore get
r| ψ± =r|φ- m 2π2 d3 r' e±ik|r-r'| |r-r'| V(r')r'| ψ± (94)
Usually scattering is investiated far away from the scattering potential, i.e. in the
far-field approximation, r'/r << 1. There
|r-r'|=(r-r' )2 =r1-2rr'/ r2 +r '2 / r2 =r-r' er +O((r'/r )2 ) (95)
Using this in the exp-function and replacing 1/|r-r'|1/r we get
r| ψ± |r/r'>>1 = r|φ- m 2π2 d3 r' e±ik(r-r' er ) 1 r V(r')r'| ψ± = r|φ- m 2π2 e±ikr r d3 r' eik'r' V(r')r'| ψ±       (96)

where k'=k er 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|φ to be a plane-wave state normalized in a wave-vector, i.e. k=p/ (and not in a momentum p), representation: r|k= eirk /(2π )3/2
r| ψ± |r/r'>>1 = 1 (2π )3/2 [ eikr + e±ikr r f(k',k)]       (97) f(k',k) = - m 2π2 (2π )3/2 d3 r' eik'r' V(r')r'| ψ± = - (2π )2 m 2 ±k'|V| ψ±       (98)

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

4.1  Differential Cross Section

Consider the 'upper' sign situation in eqn. (97). Far away from the scattering center, r/r'>>1, the absolute value of the current density of the incomming particle is ji = vi |r|φ |2 . For the outgoing spherical wave js = vs |f(k',k) |2 / r2 , where the velocities vi = vs =k/m are the same. The differential cross section dσ/dΩ is defined by
dσ = num.particlesscatteredinto dΩ/dt num.particlesincommingperunitarea /dt = js r2 dΩ ji =|f(k',k) |2 dΩ dσ dΩ = |f(k',k) |2       (99)

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

4.2  Born 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'| ψ+ =r'|φ= eir'k /(2π )3/2
f(k',k)=- m 2π2 d3 rei(k-k')r V(r)+O( V2 ) (100)
Confining to a sperically symmetric potential V(r)=V(r) and to 1st order in V only, we get
f(k',k) - m 2π2 0 dr Ω dαdβ r2 sinβ e-iqrcosβ V(r) = - m 2π2 2π(- 1 iq ) 0 drr( e-iqr - eiqr )V(r) = - 2m q2 0 drrsin(qr)V(r)=f(q)       (101)

where we have used fig. 5 and q=k-k' and |k'|=|k|=k and q=2ksin(Θ/2)
Figure 5: Geometry for integral in Born approximation
Let us look at the differential cross section of a 3D potential well
V(r)={ u,rR 0,r>R (102)
for that
f(q)=- 2mu q2 sin(qR)-qRcos(qR) q2 =- 2mu 32 R3 +O( q2 )- u e(R) R (103)
where, in the 2nd equation we have made a
low-energy expansion qR << 1. In the last equation we have introduced the energy e(R)=32 /(2 mR2 ) 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(q) has dimension [length] and is independent of q for qR << 1. Finally
dσ dΩ = ( u e(R) )2 R2         and         σtot =dΩ dσ dΩ =4π ( u e(R) )2 R2 (104)
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 σtot is different from π R2 even for k0. A commonly used unit for dσ/dΩ and σtot is
1 barn = 1b = 10-24 cm2 (c.f. am. slang: as big as a barn's door).

4.3  T-Matrix and Higher-Order Born Approximation

From eqn. (107) we may expect a one-to-one correspondence between | ψ± and |φ. Therefore we may define the transition operator
T|φ=V| ψ+ (105)
using the retarded solution | ψ+ . With V·eqn. (107)

T|φ=V|φ+V 1 E- H0 +iη T|φ (106)
from which we may remove |φ, since the set of all plane-wave states is complete.
T=V+V 1 E- H0 +iη T (107)
This is the so-called
T-matrix equation. The scattering amplitude can be obtained from plane-wave matrix elements of the T-matrix
f(k',k)=- (2π )2 m 2 k'|T|k (108)
Using this iteration on eqn. (107) leads to a perturbative expansion of the scattering amplitude
T = V+V 1 E- H0 +iη V+V 1 E- H0 +iη V 1 E- H0 +iη V+ = V 1 1-G(E)V = 1 1-VG(E) V G(E) = 1 E- H0 +iη       (109)

Eqn. (109) plays a crucial role in evaluating f(k',k) in practice. G(E) 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(r)=(2π )3 vδ(r)

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

a) Show that
k'|T|k=v+v d3 q 1 E- ϵq +iη q|T|k

where r|k= eir·k /(2π )3/2 with H|k= ϵk |k. Show that the solution of this integral equation is
k'|T|k=T(E)= v 1-v d3 q 1 E- ϵq +iη

b) Now, lets assume that the density of states ρ(ϵ)= d3 qδ(ϵ- ϵq ) is constant in energy interval ρ(ϵ)= ρ0 for ϵ[-W,W] 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(E)= v 1-v ρ0 ln( E-W+iη E+W+iη )

c) Make a sketch of Re(T(E)) and Im(T(E)) for (v,W, ρ0 )=(±0.4eV,2eV,1/eV). Under which conditions are there energies for which T(E) 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.

4.4  The 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:
4π k Imf(Θ=0)= σtot (110)
with f(Θ=0)=f(k,k). To prove eqn. (110), remember eqn. (105) with |k=|φ and use eqn. (91
Imk|T|k = Imk|V| ψ+ =Im[ ψ+ |- ψ+ |V 1 E- H0 -iη ]V| ψ+ = = -π ψ+ |Vδ(E- H0 )V| ψ+       (111)

where we have used that ψ+ |V| ψ+ R and ψ+ |VP(E- H0 )-1 V| ψ+ R, because V+ =V and H0 + = H0 . Now we may use eqn. (105) again
Imk|T|k = -πk| T+ δ(E- H0 )T|k=-π d3 k'k| T+ |k'δ(E- 2 k '2 2m )k'|T|k = - πmk 2 dΩ'|k'|T|k |2       (112)

Where k'=k has been used. Finally, using eqn. (108) and then eqn. (99)
4π k Imf(Θ=0)= 4π k πmk 2 2 (2π )2 m dΩ'|f(k',k) |2 =dΩ' dσ dΩ' = σtot (113)
Note, that σtot ~ f2 , 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.
Exercise 8 Shadow scattering. TODO

4.5  Partial Wave Expansion

Many scattering potentials V are spherically symmetric. I.e. [V, Lz ]=[V, L2 ]=0. Since [ H0 , Lz ]=[ H0 , L2 ]=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
H0 / L2 / Lz |E,l,m=E/2 l(l+1)/m|E,l,m (114)
Due to spherical symmetry the T-matrix will be diagonal in l and m
E',l',m'|T|E,l,m=T(E,l,m) δl'l δm'm =T(E,l) δl'l δm'm (115)
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
f(k',k) = - (2π )2 m 2 k'|T|k = - (2π )2 m 2 ll'mm' dEdE'k'|E',l',m'E',l',m'|T|E,l,mE,l,m|k = - (2π )2 m 2 lm dEdE'k'|E',l,mE,l,m|kT(E,l)       (116)

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 φElm (r) Al (r) Ylm (Θ,ϕ) of the Schrödinger equation are obtained from solving the 'equivalent' 1D or radial Schrödinger equation
{ d2 dr2 +[ k2 - 2m 2 V(r)- l(l+1) r2 ]}( rAl (r))=0 (117)
where E=2 k2 /2m has been used. Ylm (Θ,ϕ) are the spherical harmonics, which are the eigenfunctions of the orbital angular momentum operators L2 and Lz represented in spherical coordinates. They are orthonormal in these coordinates, i.e. the satisfy
Ω Yl'm' (Θ,ϕ) Ylm (Θ,ϕ)dΩ = δl'l δm'm l=0 m=-l l Ylm (Θ',ϕ') Ylm (Θ,ϕ) = δ(Ω'-Ω)= δ(Θ'-Θ)δ(ϕ'-ϕ) sin(Θ)       (118)

For a free particle, i.e. V(r)=0 and fixed k eqn. (117) has two particular solutions, the spherical Bessel functions Al (r)= jl (kr) (of the first kind) and Al (r)= nl (kr) (of the second kind). Their small r-behavior is jl (0x << 1)~ xl and nl (0x << 1)~ x-l-1 .

TODO: give small/large x expressions for sp.Bess.
TODO (119)

Exercise 9 a
a) Look up the analytic expressions for jl (x), nl (x) 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= mk δ(E- 2 k2 2m ) Ylm ( k ^ ) (120)
where k ^ =k/k and Ylm ( k ^ ) is a spherical harmonic.
Inserting eqn. (120) into eqn. (116) and using |k'|=|k er |=k=|k| we get
f(k',k)=- (2π )2 k lm T( 2 k2 2m ,l) Ylm ( k' ^ ) Ylm ( k ^ ) (121)
Now we specify the scattering geometry by choosing k=k ez and without loss of generality k'=k( ez cosΘ+ ex sinΘ), i.e.
Ylm ( k ^ ) = Ylm (0,0)= δm0 2l+1 4π Yl0 ( k' ^ ) = Yl0 (Θ,0)= 2l+1 4π Pl (cosΘ)       (122)

where Pl (cosΘ) are the Legendre polynomials. Defining the
partial-wave amplitude
fl (E)=- π k T( 2 k2 2m ,l) (123)
we finally get
f(k',k)=f(Θ)= l (2l+1) fl (E) Pl (cosΘ) (124)
This expression for f(k',k) 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 jl (kr) Ylm (Θ,ϕ) and nl (kr) Ylm (Θ,ϕ). The latter drops out because nl (0x << 1)~ x-l-1 . From the former only m=0 survives, because eik·r is rotanionally symmetric around ez . What remains is
eik·r = l (2l+1) il jl (kr) Pl (cosΘ) (125)
where jl (x) is the spherical Bessel function of the first kind of order l and Pl (x) is the Legendre polynomial.
Exercise 10 Prove eqn. (125).
jl (x) |x = 1 x sin(x- lπ 2 )= 1 2ix ( ei(x-lπ/2) - e-i(x-lπ/2) ) (126)
and eqn. (97) we get
r| ψ+ | r r' >>1 = 1 (2π )3/2 l (2l+1) 2ik [(1+2ik fl (E)) eikr r - e-i(kr-lπ) r ] Pl (cosΘ) (127)
Keeping in mind the time-evolution of | ψ+ ~ e-iEt/ with E=2 k2 /2m>0, the 1st (2nd) exponential in the []-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 [H,L]=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+2ik fl (E)|=1 (128)
I.e. the out-going wave aquires a
scattering phase-shift δl (E)\mathbbR
fl (E)= e2i δl (E) -1 2ik (129)
which allows to rewrite eqn. (124) as
f(Θ)= 1 k l (2l+1) ei δl (E) sin( δl (E)) Pl (cosΘ) (130)
The total cross-section is
σtot = 4π |f(Θ) |2 dΩ (131)
using 0 π dΘsin(Θ) Pl (cosΘ) Pm (cosΘ)= -1 1 dxPl (x) Pm (x)=2/(2l+1) δlm we get
σtot = 4π k2 l (2l+1) sin2 ( δl (E)) (132)
It is instructive to see, that the optical theorem is fulfilled, i.e. plugging Θ=0 into eqn. (130)
Imf(Θ)= 1 k l (2l+1) sin2 ( δl (E))= k 4π σtot (133)
which is exactly eqn. (110).

4.6  How to Calculate Phase Shifts

Let us assume a spherically symmteric scattering potential V(r) which fulfills V(r>R)=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| ψ+ |r>R = 1 (2π )3/2 l (2l+1) il Al (r) Pl (cosΘ) (134)
where rAl (r) is the solution of the radial Schrödinger equation in the angular momentum channel l
{ d2 dr2 +[ k2 - 2m 2 V(r)- l(l+1) r2 ]}( rAl (r))=0 (135)
For r>R, Al (r) can be expressed in terms spherical Bessel functions
2 Al (r)= al jl (kr)+ bl nl (kr) (136)
where nl (x) is the spherical Bessel function of the second kind of order l. In contrast to jl (x), nl (x) diverges ~1/ xl+1 as x0. 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 bl 0. The asymptotic behavior of eqn. (134) follows from
il Al (r) |kr>>1 = eilπ/2 ( al sin(kr- lπ 2 ) 2kr - bl cos(kr- lπ 2 ) 2kr ) = - ial + bl 2kr eikr + ial - bl 2kr e-i(kr-lπ)       (137)

r| ψ+ |r>>R = 1 (2π )3/2 l (2l+1) 2ik [( al - ibl ) eikr r -( al + ibl ) e-i(kr-lπ) r ] Pl (cosΘ) (138)
from eqn. (127,129) we have
r| ψ+ |r>>R = 1 (2π )3/2 l (2l+1) 2ik [ e2i δl (E) eikr r - e-i(kr-lπ) r ] Pl (cosΘ) (139)
comparison of the last two eqns. leads to
al - ibl = e2i δl (E) ,       al + ibl =1 (140)
al = ei δl (E) cos( δl (E)),       bl =- ei δl (E) sin( δl (E)) (141)
Al (r)= ei δl (E) [cos( δl (E)) jl (kr)-sin( δl (E)) nl (kr)] (142)
Now, once we have determined al and bl by solving Schrödinger's equation for r>R. We can calculate the
logarithimic derivative
αl = r dln( Al (r)) dr |r=R =kR cos( δl (E)) jl '(kR)-sin( δl (E)) nl '(kR) cos( δl (E)) jl (kR)-sin( δl (E)) nl (kR) (143)
which can be inverted to yield
tan( δl (E))= αl jl (kR)- kRjl '(kR) αl nl (kR)- kRnl '(kR) (144)
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(r)={ ,r\leqslantR 0,else (145)
In this case δl can be inferred directly from eqn. (142) and A(R)=0, i.e.
cos( δl ) jl (kR)-sin( δl ) nl (kR) = 0 tan( δl ) = jl (kR) nl (kR)       (146)

In the
s-wave channel, l=0, this implies
tan( δl )=-tan(kR) δl =-kR (147)

Al (r) = eikR [cos(kR) jl (kr)+sin(kR) nl (kr)] = eikR 1 kr sin(k(r-R))       (148)

I.e. the wave function is a ' 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. kR << 1 and kR>>1. For kR << 1 we use
jl (x << 1)= xl (2l+1)!!        nl (x << 1)=- (2l-1)!! xl+1 (149)

tan( δl )=- kR2l+1 (2l+1)[(2l-1)!! ]2 (150)
This implies δ0 / δl>0 << 1, i.e. only s-wave scattering is relevant. From eqn. (132) we get
σtot = 4π k2 sin2 ( δ0 =kR << 1)=4π R2 π R2 (151)
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 kR>>1 we have
sin2 ( δl )= tan2 ( δl ) 1+ tan2 ( δl ) = ( sin(kR-lπ/2) cos(kR-lπ/2) )2 1+( sin(kR-lπ/2) cos(kR-lπ/2) )2 = sin2 (kR-lπ/2) (152)
where the asymptotic forms eqn. (119) of j(n )l (x) have been used. The meaning of x in the latter, is that x-lπ/2>>1. I.e. eqn. (152) can be used only for l\leqslant lmax <~kR. Since lmax >>1, many l-values contribute and we approximate
sin2 (kR-lπ/2) 1 2 l kR(2l+1) (kR )2       (153)

and therefore
σtot = 4π k2 l (2l+1) sin2 (kR-lπ/2) 4π k2 (kR )2 1 2 =2π R2 π R2 (154)
Up to eqn. (151) everything has been exact. Eqn. (154) is only a
rough lower bound. Therefore, at high energies σtot 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(Θ) is zero for all angles between the backward and forward direction. b) All incoming particles with impact paramter <R are exactly reflected. I.e. f(Θ) has a peak into the backward direction. The differential cross section integrated over this peak (partial cross section) must be π R2 for all particles to be reflected. c) Into the forward direction there are no particles at impact paramter <R. I.e. f(Θ) 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: σtot =π R2 +π R2
Looking at this mathematically we consider the following decomposition
f(Θ) = fr (Θ)+ fs (Θ) = R 1 2iRk l kR(2l+1) e2i δl Pl (cosΘ)+R i 2Rk l kR(2l+1) Pl (cosΘ)       (155)

a) fr is complex, while fs is purely imaginary
fr (Θ)\mathbbC,       fs (Θ)i\mathbbR

Therefore all scattered wave contributions from fs (Θ)
into the forward direction will interfere destructively with the incoming wave (see eqn.(97))
σtot r(s) 4π dΩ| fr(s) (Θ) |2 = π 2 k2 l kR(2l+1 )2 -1 1 Pl (x )2 dx= π k2 l kR(2l+1)π R2 4π dΩ fr (Θ) fs (Θ) = π k2 l kR(2l+1) e-2i δl 0 σtot = σtot r + σtot s

For the last eqn. we have used eqn. (152) from wich δl =±(kR-lπ/2), where it is sufficient to use the upper sign. I.e. the phase shift oscillated by lπ. This makes the sum negelegible compared to (kR )2 .
c) As kR>>1, fr(s) (Θ) is strongly peaked into the backward(forward) direction and | fr(s) (Θ) |2 shows a diffraction pattern
Figure 6: Red(Black) = | fr(s) (Θ) |2 / R2 for Θ[0,2π] and for kR=10
Therefore fr(s) (Θ) is called the reflection(shadow) amplitude.
Relation to the optical theorem:
4π k Im(f(0)) = 4π k Im( fr (0))+ 4π k Im( fs (0)) 4π k Im[R 1 2iRk e2ikR l kR(2l+1)(- )l+1 ]+ 4π k Im[R i 2Rk l kR(2l+1)] 0+ 4π k Im( fs (0))= 4π k 1 2k (kR )2 =2π R2 = σtot

where we have used eqn. (152) with δl =±(kR-lπ/2) and have chosen the positive sign.
Exercise 11 Low energy scattering and bound states and scattering length.
a) Consider a spherically symmetric potential
v(r)={ v0 ,r\leqslantR 0,else

In the low-energy limit k0 only l=0 scattering is relevant. The solution of the corresponding radial Schrödinger equation is rA0 (r)=φ(r). Show that for r>R and k0
φ(r)=const ×(r-a)

Use A0 (r)'s representation in terms of spherical Bessel functions to show that also
φ(r)= ei δ0 1 k sin(kr+ δ0 )

Use the logarithmic derivative of φ(r) to conclude that
limk0 kcot(kr+ δ0 )= 1 r-a

and from that
σtot |k0 =4π a2

b) Roughly sketch the potential and the radial wave function and discuss the scattering length a qualitatively for the following three different cases:
i) v0 >0 and such that v0 >>2 k2 /2m.
ii) v0 <0 and such that the energy εb of the highest of any potentially present bound state of the potential fulfills | εb |>>2 k2 /2m.
iii) v0 <0 and such that the energy εb of the highest of any potentially present bound state of the potential fulfills | εb |~2 k2 /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


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.