Download Energy Levels for the Hydrogen Atom (from Ph234)

Ph235-08
F. Merritt
24-Nov-2008
Lecture 25 (November 24, 2008):
(version 1.0; 24-Nov-2008::00:00)
Electromagnetic perturbations: The Photoelectric Effect
We found last week that the E&M Hamiltonian is given by
2
1 
A
(25.1)
H
 P  q   e
2m 
C
which gives a perturbation Hamiltonian of
e
H 1 (t ) 
( P A  A P)
(25.2)
2mc
with A  A0 cos k r  t . Using the Coulomb gauge and neglected the negative-


frequency part of H 1  t  , this became
e ik r
e A0 P e it  H 1e it
(25.3)
2mc
This form is applicable to a variety of problems. For the photoelectric effect on
hydrogen, we require the final state to be a free particle, with a kinetic energy of a few
times the binding energy. Assuming that the initial state is the ground state, we have
H 1 (t ) 
H 1fi  f 0 H 1 i 0    *f  H 1  i  d 3r
1/ 2
 1 3/ 2
 r   3 (25.4)
 ip f r    e ik r 
   1 

  
e A0      3  exp     d r
 
 exp  
2

2
mc


 i     a0 


 
 a0  

To simplify the integral, we also restrict the problem to the k-range ka0 1 , so that the
atom sees an electromagnetic field that is spatially uniform over the region of integration.
In that case, we can approximate exp  ikr   1 . This is called the “dipole approximation”.
We showed in class that this matrix element, in spite of initial appearance, is just
what we would expect since it reduces to
H 1fi  f 0  eEr i 0
(25.5)
 i

 e  A0  f 0 R i 0
 2c

1
To evaluate H fi , it is easiest to return to equation (25.4) and integrate by parts. The
integrals are straightforward after the first integration by parts, and the result is
8
1
A0 p f
2
a0
1/ a 2   p / 2 
0
f


To obtain the transition rate, we plug this into Fermi’s Golden Rule, giving
H 1fi  N
(25.6)
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Ph235-08
F. Merritt
24-Nov-2008
Ri  f 
2
f H1 i
2
  E f  Ei   
(25.7)
But how are we to interpret this? The delta-function implies that  must be
exactly equal to E f  Ei , in which case the rate is infinite!
The important thing to realize is that equation (25.7) gives us the transition rate
into a particular final state f 0 , but in this problem there is a continuum of final states
available. Since we dropped the higher order terms from exp  ik r  , the direction of the
initial state k-vector is immaterial; the electron can be emitted in all directions. And
when we measure the final state momentum, we are necessarily measuring over a range
of momenta defined by the resolution of our detector. So we do not want the rate to a
particular discrete state f 0 , but rather the summed rate into all final states that are
consistent with the   function . This is (almost) always going to be the case when we
are applying FGR. The   function tells us that we need to integrate over all available
states.
To get the final total rate, we need to integrate (25.7) over the density of final
states:
2
2
R tot  
f H 1 i   E f  Ei    d 3 p f
(25.8)
where d 3 p f is the “density of final states”, or “phase-space factor”. We can write the
  function in terms of p f as
 p 2f

 Ei   

 2m

m

 p f   2m  Ei    


pf
  E f  Ei      


(25.9)
Then (25.8) becomes
R tot  


2
2
f H1 i
2
f H1 i
2


 m
 2
  p f   2m  Ei      p f d p f d 

 p f
mp f d 

4a03e 2 p f A0 p f

(25.10)
2
m c 1   p f a0 /

4 2

2


4
or equivalently
dR tot
d
d

4a03e 2 p f A0 p f

m
c 1   p f a0 /

4 2

2

2


4
(25.11)
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Ph235-08
F. Merritt
24-Nov-2008
Two things may be bothering you. First, how we even check the dimensionality
of an expression like this? That is a good question, and you can figure it out. Note that
in the numerator, p f a0 has dimensions of , and eA0 has dimensions of energy [see
equation (25.5)], which we will write as  . So the dimensionality of the numerator is
5 2
 . The dimensionality of the denominator is similarly 4   , so the whole
expression has dimensions of  , or t 1 , which is what we want for a rate.
Secondly, where did the phase space factor of d 3 p come from? And speaking of
dimensionality, this factor has a dimension of momentum cubed! Fermi’s Golden Rule
looks like it gives us a rate, so how can we multiply by p3f and still have a rate? The
answer to the first of these of course answers both questions, and it all comes from our
normalization of the final-state wavefunction. We wrote this as
3/ 2
 ip f r 
 1 
exp
(25.12)




 2 


which is the normalization factor we learned in Ph234. But if we think back to where this
comes from, it’s what we called “delta-function normalization”. We required
2
 1 
(25.13)

  exp i  p  p  x /  dx    p  p 
 2 
which of course has the dimensions of 1/ p . Whenever we use this normalization, we
must eventually integrate over dp to reach a physical result. So Fermi’s Golden Rule is
perhaps better stated as
2
2
R
f H i   E f  Ei      Phase-space
(25.14)
Electromagnetic perturbations: Magnetic transitions
We have already encountered perturbations due to a magnetic field, and we
treated this as a perturbation
H    B
(25.15)
in time-independent perturbation theory. (Recall that    e / mc  L ). But how do we
get this from equation (25.5), which seems to be purely electric?
The answer is, we don’t – for two reasons. In getting equation (25.5), we used the
approximation exp  ikr /   1 , corresponding to a long-wavelength perturbation. This
was equivalent to assuming that the field was spatially constant in the region of the atom.
But the magnetic field is proportional to the curl of A, so it requires that the field have a
spatial variation over the region of integration. In the “electric dipole approximation”,
there is no such variation and therefore no such transition.
If we had expanded the exponential to first order, then we would have an
additional term:
exp(ik r )  1  ik r
(25.16)
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Ph235-08
F. Merritt
24-Nov-2008
The second term does indeed have the potential to produce magnetic transitions. To see
this, go back to (25.15) and write  in terms of the angular momentum L :
 e

H B    B   
L  A
 2mc 
(25.17)
e
 
 R  P  ik  A
2mc
Compare this to the expression we derived for the photoelectric effect:
e
H 1 (t ) 
( P A  A P)
(25.18)
2mc
We have the same factors of P and A, but in (25.17) we have additional factors of R and
k. These are just the additional factors that we get from the second term in (25.16).
But even if we include this term, the magnetic force is


FB  e     A   e  ikA 
(25.19)
c
c
while the electric force is
 1 A 
 i 
FE  eE  e  
(25.20)
  e A
 c t 
 c 
So
FB  k


 
(25.21)
FE c  / c c
for the ground state of hydrogen. So the magnetic effects will be much smaller.




All of this of course is derived from the first-order electromagnetic Hamiltonian
of equation (25.18). It is worthwhile to show that we can derive H B    B directly
from that for a uniform B-field. In that case, recall that we found that the vector potential
producing a uniform solenoidal field B  Bzˆ was given by Stokes Law:
B   A   B da   A dl  B   r 2  A  2 r
(25.22)
B
B
1
 A  r  ˆ    yxˆ  xyˆ   B  r
2
2
2
Using this expression for A, and using the Coulomb gauge, we have
e
e
H B  
P A
P Br
mc
2mc
e
e

Pk  Bi rj  ijk   
Bi  rj Pk  ijk 
(25.23)
2mc
2mc
e
e

B r  P  
LB
2mc
2mc
  B
QED
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Ph235-08
F. Merritt
24-Nov-2008
Spontaneous Decay Rates
So far, we have calculated transition probabilities in terms of an external
perturbation to the system we are considering. For an isolated atom in an excited energy
state, there is no such external perturbation. From the formalism we have used up until
now, we would say that if the atom is in an eigenstate, then it is in a “stationary state”,
and the wavefunction will evolve only by a phase factor exp(eEt / ). This does not
seem reasonable, though. We know an atom in the ground state can be kicked into an
excited state by absorbing an incident photon, and it seems reasonable that the timereversed process should also occur. That is, an atom in an excited state should be able to
emit a photon as it transitions to the ground state, or some lower-energy state. And of
course this is what is seen experimentally.
To see how to incorporate this kind of phenomenon into our theoretical
formalism, I will start by following the treatment given by Gasciorowicz (p.346-9). The
energy density of an electromagnetic field is given by
2
1
1   1 A 
2
2
2
E

B

 


     A  
8
8   c t 

(25.24)
2
2   i 
2

  A0    k  A0  
8   c


where the factor of 2 comes from the 2 polarization states of the field. The second term
can be simplified:
 k  A  k  A   ki Aj ijk   k A  k 
(25.25)
  ijk  ki Aj ki Aj  ki Aj k j Ai 
Choosing the Coulomb gauge,  A  iki Ai  0 so the second term vanishes. Then
2
  2
 2
*
  2 A0 A   k 2 A0 A*0  
(25.26)
A0 A*
2
 c

2 c



If the field is confined to a box of volume V, then the total energy of the system, if there
are N photons, is
1
 2V
3
2
2
(25.27)
d
r
E

B

 2 c2 A0 2  N 
 8 
The direction of the A vector is specified by the unit polarization vector ˆ , which must
be perpendicular to the direction of propagation k̂ , so
(25.28)
ˆ ˆ  1 and ˆ kˆ  0
Then

1
1
E 2  B2  

8
4
1/ 2

 2 c 2 N 
A(r , t )  ˆ 
(25.29)
 exp i k r  t 

V


The quadratic nature of the energy, and the quantized energy levels, are both reminiscent
of the harmonic oscillator. Just as we have raising and lowering operators there, we
define creation and annihilation operators here, that either create or annihilate one
photon. For the transition from a state of N photons to one of N+1 photons,


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Ph235-08
F. Merritt
24-Nov-2008
 2 c 2  N  1 
A  r , t   ˆ 
 exp  i k r  t 

V


and for the transition from a state of N photons to N-1,
1/ 2


(25.30)
 2 c 2  N  
(25.31)
A  r , t   ˆ 
 exp  i k r  t 

V


Consequently, the transition matrix element for the emission of a single photon from a
state that has no photons,
1/ 2

1/ 2

e  2 c 2 
H (t ) 
(25.32)

 ˆ P exp  i k r  t 
mc  V 
Substituting this into Fermi’s Golden Rule gives
2
2  2 e2 
 ik r
0
0
Rk m  k m 
 m2V  m e  p k  Ek  Em     Phase space (25.33)


The phase space factor is the same the one we have developed for the final-state electron
in the photoelectric effect, but now it applies to the final-state photon wavefunction:
d3 p
(25.34)
[Phase space]=
3
 2 
for each final state particle.
Gasciorowicz goes through the steps leading up to this result (p.349-351), but this
is really just the same development that we have already seen in Ph234 for the freeparticle wavefunction, and also for the Free Electron Gas in Ph235. Note that the factor
of V (for volume) which Gasciorowicz includes always drops out – so I have omitted it
everywhere. It is useful, however, if one want to keep track of the dimensionality of each
term throughout the development.

1



6