To make sure we don't do something crazy like divide by zero we project out the. | n0> ket by introducing a projection operator Ï = â k=n. | k0>
Perturbation Theory D. Rubin December 2, 2010 Lecture 32-41 November 10- December 3, 2010
1 1.1
Stationary state perturbation theory Nondegenerate Formalism
We have a Hamiltonian H = H0 + V and we suppose that we have determined the complete set of solutions to H0 with ket | n0 i so that H0 | n0 i = En0 | n0 i. And we suppose that there is no degeneracy. The eigenkets of H satisfy H| ni = En | ni → (H0 + V )| ni = (En0 + ∆n )| ni
(1)
(En0 − H0 )| ni = (V − ∆n )| ni
(2)
where En = En0 + ∆n . We multiply the perturbative term by a real parameter λ and insist that ∆n is analytic in λ as λ goes from 0 to 1. Then Equation ?? becomes (En0 − H0 )| ni = (λV − ∆n )| ni
(3)
The perturbation is small if the energy shift is small compared to the spacing of the unperturbed levels that might be connected by the perturbation. That last equation is our Schrodinger equation and we could invert it like so | ni =
1 (λV − ∆n )| ni En0 − H0 1
To make sure we don’t do something crazy like divide by zero we project out the | n0 i ket by introducing a projection operator φ=
X
| k 0 ihk 0 |
k6=n
Then we make the equation safe from nan. | ni =
En0
φ (λV − ∆n )| ni − H0
We are not losing anything here since Equation ?? says that hn0 | λV − ∆n | ni = 0
(4)
Finally, so we have something sensible as λ → 0 we add the solution to the unperturbed part and write | ni = | n0 i +
En0
φ (λV − ∆n )| ni − H0
One last thing to note. Normalization. It seems that hn | n0 i = 1. We will use this fact. | ni is not normalized by itself. Now Equation ?? says that ∆n = λ hn0 | V | ni Now we suppose that | ni = | n0 i + λ| n1 i + λ2 | n2 i + . . . and ∆n = λ∆1n + λ2 ∆2n + . . . Then equation powers of λ. Our energy equation becomes
λ∆1n + λ2 ∆2n + . . . = λhn0 |V | n0 i + λ| n1 i + λ2 | n2 i + . . . So ∆1n = hn0 | V | n0 i ∆2n = hn0 | V | n1 i ∆3n = hn0 | V | n2 i
2
And for the wave function | n0 i+λ| n1 i+λ2 | n2 i+. . . = | n0 i+
En0
φ (λV −(λ∆1n −λ2 ∆2n +. . .))| n0 i+λ| n1 i+λ2 | n2 i+. . . − H0
Equating stuff linear in λ we get | n1 i =
En0
φ V | n0 i − H0
(5)
Next everything quadratic in λ gives | n2 i =
φ 1 1 1 V | n i − ∆ | n i n En0 − H0
φ φ φ V 0 |n i= 0 V | n0 i − hn0 | V | n0 i 0 V | n0 i En − H0 En − H0 En − H0
!
2
And we can use our solution for | n1 i to see that ∆2n = hn0 |V
X | hn0 | V | k0 i |2 φ V | n i = 0 En0 − H0 En0 − Ek0 k6=n
It is evident that degeneracy could get us in trouble.
1.2 1.2.1
Examples Helium
To first approximation, the energy of the ground state of helium is 2Z 2 E0 = −2Z 2
e2 2a
!
= 2(4)(−13.6ev) = −108.8ev
The next approximation would be to include the interaction of the electrons as a perturbation. e2 V = |r1 − r2 | The unperturbed wave function is taken to be ψ(r1 , r2 ) = ψ0 (r1 )ψ0 (r2 ) = 3
8 −2(r1 +r2 )/a e πa3
where a=
h ¯c αmc2
Set the polar axis along r1 and 1
2
E =e
8 πa3
e−4(r1 +r2 )/a
2 Z q
r12 + r22 − 2r1 r2 cos θ2
r12 dΩ1 r22 dΩ2 = e2
5 4a
Then the total energy is −e2 E = 2(4) 2a 1.2.2
!
+
5e2 11e2 11 =− = − (13.6ev) = −74.8ev 4a 4a 2
Stark effect for rigid rotator
The hamilonian for the rigid rotator is H=
L2 2I
where I is the moment of inertia and the vibrational levels are assumed infinitely far apart as compared to the spacing of the rotational levels. Then the energy eigenstates 2 and eigenvalues are | ψi = | l, mi and El = ¯h l(l+1) . In the presence of a uniform 2I electric field in the z direction we introduce V = −ezE = −erE cos θ We suppose that ehri = d is the electric dipole moment of the rotator. (All states have the same radial wave function). We can write s
T01
V =
= −dE
4π Y1,0 (θ) 3
We need to evaluate D
l2 , m2 | T01 | l1 , m1
E
The matrix element is non zero if m1 = m2 and (|l2 − l1 | ≤ 1, (Wigner Eckart), l1 6= l2 , (parity). Then the non zero matrix element can be written s D
E
l ± 1, m | T01 | l, m
= −dE
4π Z ∗ Yl±1,m Y1,0 Yl,m dΩ 3 4
s
= −dE =
v u 4π u t (2(1) + 1)(2l + 1) h1, 0, l, m
3
v u u −dE t s
= −dE
4π(2(l ± 1) + 1)
| l ± 1, mih1, 0, l, 0 | l ± 1, 0i
(2l + 1) h1, 0, l, m | l ± 1, mih1, 0, l, 0 | l ± 1, 0i (2(l ± 1) + 1) l2 − m2 4l2 − 1
The unperturbed levels are all degenerate. But since H0 and V both commute with Lz , there is no mixing, and non degenerate theory is OK. In each of the m supspaces, the spectrum is non degenerate. The first order correction is zero, by the rules above, (hl, m | T01 | l, mi = 0. The second order correction reduces to the two terms corresponding to l = ±1. Finally lm 2
=
dE El0
!2
l(l + 1) − 3m2 2(2l − 1)(2l + 3)
The degeneracy is only partly removed, since it depends on m2 . Some symmetry remains. We could compute the electric dipole moment. µ ∼ e hn | z | ni We have that
hk0 | T01 | n0 i | k0 i | ni ∼ | n0 i + En0 − Ek0 k6=n X
Then µ∼
1.2.3
hn0 | z | k0 i hk0 | T01 | n0 i 1 X hn0 | T01 | k0 i hk0 | T01 | n0 i = (En0 − Ek0 ) E k6=n (En0 − Ek0 ) k6=n X
Stark effect in hydrogen
As in the case of the rigid rotator, the perturbation commutes with Lz so there is no mixing of states with different m and we use non degenerate perturbation theory. Also, since all of the eigenstates with definite angular momentum have definite parity, there is no first order correction. Another way to see this is to note that although the perturbed Hamiltonian does not commute with L2 , it does commute with Lz . So there is no change to the z-component and therefore no mixing of states with
5
m 6= m0 . Parity eigenstates have zero electric dipole moment. The electric dipole moment is Z d = hα, l, m | x | α, l, mi = x|ψ(x)|2 d3 x = 0 since x has odd parity. An electric dipole emerges in second order since the first order correction to the wave function includes contributions of opposite parity. That also means that the electric dipole moment scales with the applied electric field. The energy shift is quadratic in E The polarizability is α where 1 ∆ = − α|E|2 2 The energy shift is second order ∆2 = e2 E 2
| hk 0 | z | 1, 0, 0i |2 (E00 − Ek0 ) k6=0 X
We approximate the sum by assuming that the denominator is constant, (this will give us an upper limit on the energy shift) and noting that X D
E
| k 0 | z | 1, 0, 0 |2 =
k6=0
X D
E
| k 0 | z | 1, 0, 0
all k
since h1, 0, 0 | z | 1, 0, 0i = 0. The sum over all k becomes E D E 1 | k 0 | z | 1, 0, 0 = 1, 0, 0 | z 2 | 1, 0, 0 = hr2 i = a20 3 k
X D all
Therefore, ∆2 ≤ e2 E 2
2
3 a20 a20 2 2 2 2 8a0 = −E = ∆ ≤ e E (E00 − Ek0 ) 3 (−e2 /2a0 − −1 e2 /2a0 ) 4
Degenerate perturbation theory
Our formalism falls apart if there are degeneracies. But actually, it is OK as long as the perturbation does not couple the degenerate states. If it does couple degenerate states we are in trouble since then we have finite numerator and an energy denominator that is zero in Equation ??. The strategy is to find a new basis for the degenerate states in which none of the states in that basis are coupled by the perturbation. 6
2.1
Formalism
So here’s what we do. Suppose that there is a degenerate set of states and we are interested in the effect on the perturbation of states at that energy. The states are | ii, i = 1, 2, 3 for example. Now if hi | V | ji = |V |δij then we do not have a problem. Business as usual. If there are off-diagonal terms, then we need to find a different linear combination of the degenerate states for which the perturbation matrix is diagonal. That is we diagonalize
h1 | V | 1i h1 | V | 2i h1 | V | 3i hI | V | Ii 0 0 V = h2 | V | 1i h2 | V | 2i h2 | V | 3i → 0 hII | V | IIi 0 h3 | V | 1i h3 | V | 2i h3 | V | 3i 0 0 hIII | V | IIIi The new basis is related to the old by a unitary transformation. | Ii = a| 1i + b| 2i + c| 3i The states | Ii, | IIi, | IIIi are the eigenkets of the perturbation matrix and the diagonal elements of the perturbation matrix in the new basis are the eigenvalues. In fact we see that the eigenvalues are the first order energy shift. The first order shift in the state vector is given as before by | n1 i =
hk 0 | V | n0 i 0 |k i 0 0 k6=D (En − Ek ) X
The states in the degenerate subspace do not contribute because all of the off diagonal matrix elements are zero. Likewise for the second order correction to the energy. 2.1.1
Projection operators
We could more formally use the projection operator approach. We go back to the Schrodinger equation 0 = (E − H0 − λV )| li = (E − ED − λV )P0 | li + (E − H0 − λV )P1 | li Projecting from the left by P0 and P1 we get P0 ((E − ED − λV )P0 | li + (E − H0 − λV )P1 | li) = (E − ED − λP0 V P0 | li − λP0 V P1 | li P1 ((E − ED − λV )P0 | li + (E − H0 − λV )P1 | li) = −λP1 V P0 | li + (E − H0 − λP1 V P1 )| li 7
(6) (7)
We solve Equation ?? for | li and then project out the states not part of the degenerate subspace and then the denominator is safe 1 λP1 V P0 | li E − H0 − λP1 V P1 X hk 0 | V | l0 i ∼ P1 | l 1 i ∼ | k0i 0 ED − Ek0 k6=D
P1 | li = P1
(8)
Then substitute Equation ?? into Equation ?? for P1 | li multiply by P0 and we have (E − ED − λP0 V P0 − λP0 V P1
1 λP1 V )P0 | li E − H0 − λP1 V P1
∼ (E − ED − λP0 V P0 )| li = 0 The energies are the eigenvalues of the perturbation matrix. In summary we divide the space into the subspace of degenerate states, and the all the rest. The exact state vector is a linear combination of all | l0 i kets. | li =
X
ai | li i
Then the Schrodinger equation including the perturbation is (El0 − H0 )| li = (λV − ∆n )| li
(9)
Substitute the expansion for ∆l and | li = | l0 i + λ| l1 i + λ2 | l2 i + . . . . Here | l0 i =
X
ai | ii
i=D
| l0 i is a normalized linear combination of the degenerate states in the basis in which the perturbation matrix is diagonalized. Multiply by hm0 | from the left D
E
m0 | En0 − H0 | l = λ hm0 | V | li − ∆l hm0 | li
and keep the first order in λ, when m0 = lD and we get that 0 = hlD | V | l0 i − ∆1l hlD | li → hlD | V ilD = ∆iD When m0 6= lD 0 (En0 − Em )hm0 | li = λhm0 | V | li − ∆l hm0 | li
8
To first order in λ we have D
E
0 (En0 − Em ) m0 | l1 = hm0 | V | l0 i
and D
E
m0 | l1 ∼
and
hm0 | V | l0 i 0 El0 − Em
hm0 | V | l0 i | m0 i 0 0 l6=D En − Em
| l1 i =
X
Note that we can always add states with eigenvalue El0 . The above gives no information about them. We define | l1 i and all higher order contributions to include no | l0 i or for that matter any of the | lD i degenerate states. Then the exact solution is | li =
X
ai | ii + | l1 i + | l2 i + . . .
i=D
and | l0 i =
X
ai | ii
i=D
hl0 | lj i = 0, for j 6= 0. We suppose that the ai are chosen so that | l0 i is normalized. We see that in this way | li is a solution to Equation ?? independent of the coefficients ai . And the second order energy shift D
E
∆2l = l0 | V | n1 = 2.1.2
Second order degenerate perturbation theory
Another strategy1 for getting second order correction of degenerate states is to write the true state X X | ai = cα | αi + dµ | µi α
µ
where the sum over α includes the degenerate states and the sum over µ is all the others. We know that (H − Ea )| ai = (H0 + λH1 − Ea )| ai = 0 1
Gottfried and Yan
9
Then 0 =
X
cα (H − Ea )| αi +
X
α
0 =
dµ (H − Ea )| µi
µ
X
cα (Eα − Ea + λH1 )| αi +
X
α
dµ (Eµ − Ea + λH1 )| µi
µ
Then multiply from the left first by hβ | which is an element of the degenerate space and then by hν | which is outside of the degenerate space. 0 = cβ (Eβ − Ea ) + λ
X
cα hβ | H1 | αi + λ
α
0 =
X
X
dµ hβ | H1 | µi
(10)
dµ hν | H1 | µi
(11)
µ
cα λ hν | H1 | αi + dν (Eν − Eα ) + λ
α
X µ
Drop the last term in Equation ?? as it is higher order in λ. Justification for this step is that the state | ai is mostly a linear combination of the states in the degenerate space. The coefficients dµ would all be zero in the λ → 0 limit so they are at most first order in λ. Then that last term is second order in λ. You might then argue that we should drop the last term in Equation ??, but probably hβ | H1 | µi is more important that hν | H1 | µi since it connects to the relevant states. Finally, we could keep the last term in Equation ??, solve for dν , substitute into Equation ?? and then drop the highest order term. That would be equivalent to solving for P
dν = −λ
hν | H1 | αi (Eν − ED )
α cα
(12)
Substitution into ?? gives 0 = cβ (Eβ − Eα ) +
X
cα λ hβ | H1 | αi + λ
2
X
α
µ
hµ | H1 | αi hβ | H1 | µi Eµ − ED
!
(13)
which is the eigenvalue problem in the subspace D for the effective Hamiltonian Hef f hβ | Hef f | αi = λ hβ | H1 | αi + λ2
X µ
hβ | H1 | µi hµ | H1 | αi . ED − Eµ
So we diagonalize hβ | H1 | αi. Define P to be the projection operator onto the degenerate subspace D, P = P α | αihα |. Then we can write Hef f = λP H1 P + λ2 P H1 10
1−P H1 P E − H0
2.1.3
Example
Suppose we have a system with 3 states. The first two are degenerate. The third is at energy ∆ above the two.
0 H1 = 0 λM The Hef f
0 0 λM
(λM )2 =− ∆
λM λM ∆ 1 1 1 1
The energies and eigenstates are ES = −2
(λM )2 , ∆
1 | ES i = √ (| 1i + | 2i) 2
and
1 | ES i = √ (| 1i − | 2i) 2 That’s what happens if ∆ λM . On the other hand, if ∆ ∼ λM , then we treat all three states as though they were degenerate. Assume the unperturbed energies are all the same. Then the shift in the energy of eachqof the three states is given by the eigenvalues of H1 which are, E0 = 0, E± = 12 (∆ ± ∆2 + 8(λM )2 ). The eigenvectors are E± | 3i) (| 1i + | 2i + λM 1 | 0i = √ (| 1i − | 2i), | ±i = q 2 2 + E±2 /λ2 M 2 EA = 0,
What if we went back to the weak perturbation case where M ∆, but we had started with states | Si = √12 (| 1i + | 2i) and | Ai = √12 (| 1i − | 2i) and | IIIi = | 3i. Then √ 0 0 2M V = 0 0 √0 2M 0 ∆ Now
Hef f =
−2M 2 /∆ 0 0 0
11
2.2 2.2.1
Fine Structure Spin orbit coupling
The magnetic field at the electron due to its motion through the E field of the nucleus is v×E B=− c |E| r r
The electric field is radial so E = B= Next, note that E = 1e ∇V =
1 ∂V e ∂r
and p = mv and then
r × p |E| L |E| = mc r mc r and we get that B=
L 1 ∂V mec r ∂r
The magnetic moment of the electron is µe = −
e S mc
Finally ∆H = −µ · B =
1 m2 c2 r
∂V L·S ∂r
and if V = e/r then e2 L · S e2 L · S → m2 c2 r3 2m2 c2 r3 1 Then a factor of 2 for the Thomas Precession. The best way to do this is to take the nonrelativistic limit of the Dirac equation with a vector potential. Or we could use the Biot Savart Law which states that ∆H =
B=
Z
(JdV )×r c|r|3
For a current loop, we find the magnetic field at the center is I Z er × p B = 3 eλdlv × r = − cr cmr3 The unperturbed states have orbital angular momentum and spin. We can use | l, ml i| s, ms i as base kets with eigenvalues l, m, s, and ms , or we can take the linear 12
combinations that would be eigenkets of j, jz , l, s. Since L · S = 12 (J 2 − L2 − S 2 ), it is clear that it will be more convenient to use the latter. Then e2 h ¯2 1 3 = h 3 i j(j + 1) − l(l + 1) − 2 2 4m c r ( 4 ) [j(j + 1) − l(l + 1) − 3/4] h ¯ 2 e2 1 = 4m2 c2 n3 a30 l(l + 21 )(l + 1)
En1 ESO
where h For j = l ±
1 1 i = 3 r3 n3 a0 l(l + 1/2)(l + 1)
1 2
ESO =
=
l −(l + 1)
h ¯ 2 e2 1 1 4m2 c2 a30 n3 l(l + 2 )(l + 1) mc2 α4
4
l −(l + 1) 1 3 n l(l + 2 )(l + 1)
We remember that 1 e2 e2 h ¯2 E0 = − α2 mc2 = − → a0 = 2 2 = 2 2a0 α mc me2 Substitution into the above gives En1 ∼
h ¯ 2 c2 (α2 mc2 )3 1 3 e2 h ¯ 2 (α2 mc2 )3 = = α2 mc2 (α2 ) j(j + 1) − l(l + 1) − 2 2 6 4 2 2 2m c e 2e (mc ) 2 4
No dependence on ml . No mixing of levels. No need for degenerate formalism. Each level is split into j = l ± 12 . 2.2.2
Relativistic correction
Really K=
q
(mc2 )2 + (pc)2 − mc2 ∼ 13
p2 p4 − 2m 8m3 c2
And *
ψ|
p4 2m
!
+
|ψ
*
=
p2 p2 ψ| ψ = h(E0 − V )ψ | (E0 − V )ψi 2m 2m +
D
ψ | (E02 − 2E0 V + V 2 | ψ
=
E
We use the fact that p2 is Hermitian in the first step. (2m)2 −e2 e4 4m2 2 ∼ − 3 2 h(En − V (r))2 i ∼ E − 2E h i + h i n n 8m c 8m3 c2 r r2 ! −e2 e4 4m2 2 ∼ − 3 2 En − 2En 2 + 8m c n a0 (l + 21 )n3 a30 !
∆E01
4m2 4n ∼ − 3 2 En2 1 − 4 + 8m c (l + 12 ) 1 4n ∼ − En2 −3 + 2 2mc (l + 12 )
!
!
1 mc2 α4 4n ∼ − −3 + 2 4n4 (l + 21 ) 5 1 2 2 5 ∼ − E ( α mc ) = E0 α 2 0 2mc2 2 4 !
E
D
D
E
(Note that 1, 0, 0 | 1r | 1, 0, 0 = a10 and 1, 0, 0 | r12 | 1, 0, 0 = a22 .) Depends on n, l, 0 not on j. Again there is no mixing of degenerate levels. In general 1 1 h i= 2 , r n a0 2.2.3
h
1 1 i= 1 2 r (l + 2 )n3 a20
Darwin term
The particle cannot be localized to better than its Compton wavelength h ¯ /mc. The potential that is relevant is not V (r) but some smeared average about the point r. V (r) = V (r) +
X i
∂V 1 X X ∂ 2V δri + δri δrj + O(δr3 ) ∂ri 2! i j ∂ri ∂rj
1 = V (r) + (δr)2 ∇2 V + O(δr3 ) 6 14
If δr ∼ h ¯ /mc then 2 e2 h ¯2 HD ∼ π 2 2 δ 3 (r) 3 mc Note that according to Merzbacher (p. 206 and 208) s
ψn,l,m (0) =
1 n! 1 =q 3 (na) 2n(n − 1)! 2(na0 )3
Then 2π 1 2 e2 h ¯2 ¯ 2 1 m3 e6 1 2π e2 h hn00 | HD | n00i = π 2 2 = mc2 α4 6 = 3 2 2 3 3 m c 2(na0 ) 3 m c 2n h 3 2n3 ¯ and it exactly compensates for the l = 0 term that we got from the spin orbit term. Or it would if we did it properly using the Dirac equation. 2.2.4
Fine Structure
Combining all order α2 corrections we have ESO
E2 = n2 mc
(
n[j(j + 1) − l(l + 1) − 3/4] l(l + 12 )(l + 1)
)
The relativistic correction is 4n En2 −3 =− 2 2mc l + 12
#
"
Er1 The total α2 correction is
E1 α2 − 2 1+ 2 n n "
2.3
n j+
1 2
3 − 4
!#
Stark effect
To determine the effect of an external uniform electric field on a set of degenerate levels we need to exercise a bit more care. Consider n = 2 in hydrogen. There are 2l+1 = 3 levels with l = 1 and 1 level with l = 0. Spin is not relevant. The candidate
15
set of base states are the orbital angular momentum kets. The perturbation matrix elements hi | V | ji, where V = −ez|E| are 0 h2, 0, 0 | V | 2, 1, 0i h2, 1, 0 | V | 2, 0, 0i 0 V = 0 0 0 0
0 0 0 0
0 0 0 0
Only the upper left is relevant. All of the other states are not coupled. The eigenvectors of the upper left 2X2 matrix are ∆1± = − ± e|E| h2, 0 | r | 2, 1i h2, 0, 0 | cos θ | 2, 1, 0i = ±3ea0 |E| The eigenvectors are 1 | ±i = √ (| 2, 0, 0i ± | 2, 1, 0i) 2 These two states are shifted positively and negatively with an electric field. The other 2 states are not effected by the E-field.
2.4
Zeeman effect H=
e e2 p2 + Vc (r) − |B|Lz + |B|2 (x2 + y 2 ) 2m 2mc 8mc2
We have HZ = −
e |B|(Lz + 2Sz ) 2mc
and then Hspin = −µ · B = −
e e S · B = − |B|Sz mc mc
and
1 1 dV (r) L·S 2m2 c2 r dr Assuming B is very weak so that the spin-orbit piece dominates, we need to evaluate hj, m, l, s | Lz + 2Sz | j, m, l, si = hj, m, l, s | Jz + Sz | j, m, l, si Since Hspin−orbit =
s
l ± m + 12 1 1 1 1 | l ± , mi = ± | l, m − i| , i + 2 2l + 1 2 2 2
s
l ∓ m + 12 1 1 1 | l, m + i| , − i 2l + 1 2 2 2
we can compute
1 1 m¯h l ± , m | Sz | l ± , m = ± 2 2 2l + 1
16
D
E
Meanwhile l ± 12 , m | Jz | l ± 21 , m = h ¯ m. Then ∆1B 2.4.1
e¯hB 1 =− m 1± 2mc 2l + 1
Paschen-Back limit
If the magnetic field is very strong, then total angular momentum is not conserved. (There is an external torque). But the Lz and Sz are fixed. So now it is better to use the basis | , ml i| s, ms i. It is easy to evaluate hLz + 2Sz i but not quite as simple to compute hL · Si for the spin orbit coupling. But if we write 1 hL · Si = hLz Sz + (L+ S− + L− S+ )i 2 and note that hL± i = hS± i = 0 then we get that hL · Si = h ¯ 2 ml m2 2.4.2
Intermediate field
Let’s examine the effect of an external magnetic field on the n = 2 states of hydrogen. And we suppose that hHspin−orbit i ≈ hHz i. Now we have to exercise some care in computing energy shifts. There are 8 degenerate states. We can choose as a basis eigenstates of j, jz , l, s, (l = 1, j = ±3/2, ±1/2,or l = 0, j = ± 21 or eigenstates of l, ml , s, ms , (l = 1, ml = ±1, 0, ms = ±1/2, l = 0, ms = ±1/2). We have to construct the perturbation matrix elements hi | V | ji in whatever basis we have chosen and V =−
e2 1 e B(Lz + 2Sz ) + L·S 2mc 2m2 c2 r3
Let’s pick the basis of eigenstates of j, jz , l, s. Then the L · S term only contributes along the diagonal. So let’s look at the effect of the external field. We had better label the states. 1 1 1 1 1 | 1i = | , − , 0, i = | 0, 0i| , − i 2 2 2 2 2 1 1 1 1 1 | 2i = | , , 0, i = | 0, 0i| , i 2 2 2 2 2 1 1 1 | 3i = | , − , 1, i = 2 2 2
s
1 1 1 | 1, 0i| , − i − 3 2 2 17
s
2 1 1 | 1, −1i| , i 3 2 2
s
1 1 1 1 1 1 | 1, 0i| , i + | 4i = | , , 1, i = − 2 2 2 3 2 2 3 1 1 | 5i = | , − , 1, i = 2 2 2
s
s
2 1 1 | 1, 0i| , − i + 3 2 2
s
3 1 2 1 1 1 | 6i = | , , 1, i = | 1, 0i| , i + 2 2 2 3 2 2 1 1 1 3 3 | 7i = | , − , 1, i = | 1, −1i| , − i 2 2 2 2 2 3 3 1 1 1 | 8i = | , , 1, i = | 1, 1i| , i 2 2 2 2 2
s
2 1 1 | 1, 1i| , − i 3 2 2
s
1 1 1 | 1, −1i| , i 3 2 2
1 1 1 | 1, 1i| , − i 3 2 2
Now we can compute the matrix elements
α0 − b
0
0
0
α0 + b
0
0
0
0
0
0
0
0
0
0
0
0
0
α1 − b 1/3
0
√ −b 2/3
0
0
0
0
0
α1 + b( 1/3)
0
√ −b 2/3
0
0
0
0
√ −b( 2/3)
0
α1 − b( 2/3)
0
0
0
0
0
0
√ −b( 2/3(1))
0
α1 + b( 2/3)
0
0
0
0
0
0
0
0
α1 + b(−2)
0
0
0
0
0
0
0
0
α1 + b(2)
p
p
p
p
Next we find the eigenvalues ∆i . The energy of the state | ii shifts by ∆i .
3 3.1
0
Variational method Formalism
Let’s think about that helium calculation again. We approximated the wave function for the Helium atom with two electrons as the product of the wave functions of two singly ionized atoms. Then we included the interaction of the two electrons as a perturbation.
18
The total hamiltonian is H=
p21 p2 e2 Ze2 Ze2 + 2 − + − 2m r1 2m r2 |r1 − r2 |
with Z = 2. The approximate wave function is Z3 πa30
| ψ1 i| ψ2 i =
!1/2
Z3 πa30
!1/2
e−Zr1 /a0 e−Zr2 /a0
Then we calculated Eapprox = E10 + E20 + ∆ = hψ1 ψ2 | H | ψ1 ψ2 i = hψ1 ψ2 | H0 | ψ1 ψ2 i + hψ1 ψ2 | HH | ψ1 ψ2 i Now it is possible to write our trial wave function as a linear combination of the complete set of exact eigenkets. | ψ1 ψ2 i =
X
an | ni,
H| ni = En | ni
n
Then Eapprox =
X
|an |2 En ≥ E0
X
n
|an |2
n
So we know for sure that Eapprox ≥ E0 and in fact that will be true for any trial wave function that we choose. It’s expectation value will always overestimate the ground state energy. So why not try to come up with a better approximation. One adjustment that we might make is to account for the shielding effect of one electron on the other. That is, each electron does not really see the full coulomb attraction because it is shielded by the other. The effective Z is a little less than 2. We can incorporate this approximation in our wave function by writing | ψ1 i| ψ2 i =
Z 03 πa30
!1/2
Z 03 πa30
!1/2
0
0
e−Z r1 /a0 e−Z r2 /a0 =
Z0 πa0
3
0
e−Z (r1 +r2 )/a0
Or define a00 = a0 /Z 0 and we get that
ψ1 ψ2 =
1 πa0 3
1/2
e−r/a
0
1 πa0 3
1/2
e−r/a
0
Note that our new wave function is normalized. This is very important. Now we compute *
hψ1 ψ2 | H | ψ1 ψ2 i =
p2 ψ1 ψ2 | | ψ1 ψ2 2m
+
*
−Ze2 −Ze2 + ψ1 ψ2 | + | ψ1 ψ2 r1 r2
19
+
*
e2 + ψ1 ψ2 | | ψ1 ψ2 |r1 − r2 |
+
Note that ψ1 and ψ2 just look like the hydrogen ground state. Therefore *
p2 ψ1 ψ2 | | ψ1 ψ2 2m
+
Z 0 e2 =2 2a0
Z 0 2 e2 a0
!
=
The second term *
−Ze2 −Ze2 ψ1 ψ2 | + | ψ1 ψ2 r1 r2
+
= −2Z
e2 a0
!
= −2
Z 0 Ze2 a0
The third term we calculated before 1
2
E =e
Z 03 πa3
!2 Z
Therefore
0
e−2Z (r1 +r2 )/a q
r12 + r22 − 2r1 r2 cos θ2
r12 dΩ1 r22 dΩ2 = e2
5 e2 5 = Z0 0 8a 8 a0
5 e2 2 hψ1 ψ2 | H | ψ1 ψ2 i = (Z 0 − 2Z 0 Z + Z 0 ) 8 a0
It will be a function of the parameter Z 0 . Next, we minimize the expectation value with respect to Z 0 . This is OK because no matter what we choose for Z 0 , we know that we are overestimating the energy. Set dhHi 5 5 27 = (2Z 0 − 2Z + ) = 0 → Z 0 = Z − = 0 dZ 8 16 16 And hHi = −2.85
3.2
4 4.1
e2 = −77.5ev a0
Examples
Time dependent perturbation theory Formalism
Now e want to consider time dependent perturbations. The Hamiltonian is written in the form H = H0 + V (t) We suppose that we know the complete set of eigenkets of H0 , namely | ni and as usual H0 | ni = En | ni. We note that for a time dependent hamiltonian, the time evolution operator is not simply e−iHt/¯h . Typically the time dependent part of the hamiltonian is something that turns on and then off so it is reasonable to ask about the probability of
20
finding the system in one of the uperturbed states. An arbitrary state at t = 0 can be written X | αi = cn (0)| ni n
We want | α, t0 = 0; ti =
X
cn (t)e−iEn t/¯h | ni
n
We work in the interaction picture, which is somewhere between the Schrodinger picture, in which operators have no time dependence, but the states evolve in time, and the Heisenberg picture in which the states are fixed and the operators evolve. In the interaction picture, we take the time dependence associated with the unperturbed Hamiltonian from states and attach it to operators. And we attach the time dependence from the perturbation to the states. So in the interaction picture | α, t0 ; tiI = eiH0 t/¯h | α, t0 ; tiS Observables are AI = eiH0 t/¯h As e−iH0 t/¯h According to the Schrodinger equation i¯ h i¯ h
∂ | α, t0 ; tiS = (H0 + V )| α, t0 ; tiS ∂t
∂ −iH0 t/¯h e | α, t0 ; tiI = (H0 + V )e−iH0 t/¯h | α, t0 ; tiI ∂t ∂ i¯ he−iH0 t/¯h | α, t0 ; tiI = eiH0 t V e−iH0 t | α, t0 ; tiI ∂t ∂ i¯ h | α, t0 ; tiI = VI | α, t0 ; tiI ∂t
(14)
Then we can also write that | α, t0 ; ti =
X
cn (t)| ni
(15)
n
Substitute that last into Equation ?? and multiply from the left by hn | and we have i¯ h
X ∂ hn | α, t0 ; ti = hn | VI | mi hm | α, t0 ; tiI ∂t m
Also from Equation ?? we see that cn = hn | α, t0 ; ti. Then i¯h
X d cn (t) = Vnm eiωnm t cm (t) dt m
where ωmn =
En − Em ¯h
21
(16)
and we used
D
E
n | eiH0 t/¯h V (t)e−iH0 t/¯h | m = Vnm (t)ei(En −Em )t/¯h
We expand the time evolution operator | α, t0 ; tiI = UI (t, t0 )| α, t0 ; t0 iI i¯ h
d UI (t, t0 ) = VI (t)UI (t, t0 ) dt
The initial condition is U (t, t0 )|t=t0 = 1 and we integrate to get i UI (t, t0 ) = 1 − ¯h
Z t
VI (t0 )UI (t0 , t0 )dt0
t0
We get an approximate solution by iteration: i UI (t, t0 ) = 1 − h ¯
Z t
i h ¯ = +...
Z t
= 1−
4.1.1
"
i VI (t ) 1 − ¯h t0 0
VI (t0 ) +
t0
−i ¯h
Z t0
# 00
00
VI (t )UI (t , t0 )dt
00
dt0
t0
2 Z t
dt0
Z t0
t0
dt00 VI (t0 )VI (t00 )
t0
Transition
Suppose we have a system in an eigenket | ii. Then | i, t0 ; ti = UI (t, t0 )| ii =
X
| ni hn | UI (t, t0 ) | ii =
n
X
cn (t)
n
We see that hn | UI | ii is the amplitude that a system initial in state | ii will be in state | ni as a function of time. We note that | iiI
= eiH0 t/¯h | α, t0 ; tiS
| iiI
= eiH0 t/¯h U (t, t0 )| α, t0 ; t0 iS
| iiI
= eiH0 t/¯h U (t, t0 )e−iH0 t0 | α, t0 ; t0 iI → UI (t, t0 ) = eiH0 t/¯h US e−iH0 t0 /¯h
Therefore hn | UI | ii = ei(En t−Ei t0 )/¯h hn | U | ii
22
Then as long as keti and | ni are eigenkets of H0 , | hn | UI | ii | = | hn | U | ii | So c0n (t) = δni Z i t
1 cn (t) = − n | VI (t0 ) | i dt0 ¯h t0 Z i t
c1n (t) = − n | V (t0 ) | i ei(En −Ei )t/¯h dt0 ¯h t0 Z 0 Z
i 2 X t 0 t 00
2 cn (t) = − dt dt n | VI (t0 ) | m m | VI (t00 ) | i h ¯ t0 m t0
=
4.1.2
−
i h ¯
2 X Z t m
dt0
t0
Z t0
00
dt00 Vnm (t0 )eiωnm t Vmi (t00 )eiωmi t
t0
Constant perturbation
We turn on a constant perturbation at t0 . Then the amplitude to make a transition from state i to n to first order is i cn (t) = − h ¯
Z t
Vni eiωni t =
t0
−i Vni iωni t e −1 ¯h iωni
Then |cn (t)|2 = 4
|Vni |2 |Vni |2 |Vni |2 2 2 sin (ω t/2) = 4 sin (ω t/2) = 4 sin2 ((En − Ei )t/2¯h)2 ni ni 2 (En − Ei )2 (En − Ei )2 h2 ωni ¯
If En = Ei , then |cn (t)|2 = ¯h12 |Vni |2 t2 If there is a finite number of final states close to | ni then the total transition probability is X
|cn |2 .
n,En =Ei
and if there are many final states, more or less nearby we replace X
|cn |2 →
Z
dEρ(En )|cn |2 = 4
Z
sin2
2 (En − Ei )t |Vni ρ(E)dE 2¯h |En − Ei |2
Next we see that 2
lim sin
t→∞
(En − Ei )t |Vni |2 πt ρ(E) = δ(En − Ei ) 2 2¯h |En − Ei | 2¯h
23
Finally 2
|cn (t)| =
Z
2
ρ(E)dE|cn (t)| =
The rate is
wi→n = or
wi→n =
2π |Vni |2 ρ(En )t ¯h
2π |Vni |2 ρ(En ) ¯h
2π |Vni |2 δ(En − Ei ) ¯h
where integration over final states is implied.
4.1.3
Second order perturbation
We found earlier that c2n (t)
Z Z 0 1 X t 0 t 00 0 00 = − 2 dt dt Vnm (t0 )eiωnm t Vmi (t00 )eiωmi t h m 0 ¯ 0
If V (t) is constant in time then c2n (t)
0
!
1 X t 0 eiωmi t − 1 0 = − 2 dt Vnm eiωnm t Vmi iωmi h m 0 ¯ Z t 1 X 1 0 0 dt0 Vnm eiωni t − eiωmn t Vmi = − 2 iωmi h m 0 ¯ Z t i X Vnm Vmi 0 0 eiωni t − eiωmn t dt0 = h m Em − Ei 0 ¯ Z
∼
i X Vnm Vmi h m Em − Ei ¯
Z t
0
eiωni t
dt0
0
Now it looks just like the first order term with −i i X Vnm Vni Vni → ¯h ¯ m Em − Ei h So to second order
wi→n
2
X Vnm Vmi 2π 2π = Vni + ρ(En ) → h ¯ ¯h m Ei − Em
Z
X Vnm Vmi 2 dEn Vni + ρ(En )δ(En − Ei ) m Ei − Em
The average is over all final states n with energy En ∼ Ei .
24
4.1.4
Scattering
Scattering from a fixed potential is a fine example of a perturbation that turns on, stays on for awhile and then turns off. In our study of time dependent perturbation theory we determined the the transition probability from initial state ψa to final state ψb is given by the absolute square of the amplitude i cb (t) = − ¯h
Z t/2
−t/2
0
ψb | H(t0 ) | ψa eiωab t dt0
b where ωab = Ea −E . To apply the theory to a scattering process we imagine that the h ¯ 0 perturbation H(t ) turns on at −t/2 and off at t/2 and while it is turned on it has constant value H. Then we can integrate and we get
cb (t) = −
i Hab 2i sin ωab t/2 i 1 h iωab t/2 − e−iωab t/2 Hab = − e h iωab ¯ ¯h ωab
and the transition probability is |cb |2 =
2 1 1 sin ωab t/2 2 4 sin ωab t/2 = 2 |Hab |2 2 2 |Hab | ωab t/2 ω h ¯ ¯ h ab
2
t2
Let’s examine the ωab dependent piece
f (ω) ≡
sin ωt/2 ωt/2
2
t2
The first zero of f (ω) occurs when ω = 2π/t. Its maximum value (at ω = 0) is t2 . In the limit of large t, f (ω) → 2πtδ(ω). To check that assertion we integrate over all ω Z ∞
Z
f (ω)dω = −∞
sin2 x 2 2 dx t = 2πt = x2 t
Z
2πtδ(ω)dω
In terms of the energies of initial and final states, 2πtδ(ω) → 2πt¯hδ(Eb − Ea ) and we can write |cb |2 =
1 |Hab |2 2πtδ(Eb − Ez ) ¯h2
The transition rate is
2π |Hab |2 δ(Eb − Ea ) ¯h In scattering experiments, the detector always has some finite acceptance. And what we measure is a sum over all final states consistent with that acceptance. R=
R=
2π h ¯
Z
|Hab |2 δ(Eb − Ea )ρ(Eb )dEb
25
ρ(Eb ) is the density of final states, the number of final states per unit energy. Well, we have figured this out before. The number of states between k and k + dk is dN =
V k 2 dkdΩ V p2 dpdΩ = 8π 3 (2π¯h)3
Using E = p2 /2m and 2pdp/2m = dE we have dN =
V pmdEdΩ (2π¯h)3
dN V pm = ρ(E) = dΩ dE (2π¯h)3 so R=
2π 1 V pm|Hab |2 dΩ ¯h (2π¯h)3
Now how do we connect to the cross section? The incoming particle is represented by a plane wave 1 ψa = √ eika ·r V and the outgoing wave by 1 ψb = √ eikb ·r V √ The V in the denominator is so that the wave function is normalized. The particle density in the incoming wave is |ψa |2 = 1/V and the flux of incoming particles is Vv = p/m V . And (dσ)Flux = dN = R dσ =
R V 2 m2 = | hψb | H | ψa i |2 dΩ Flux (2π¯h2 )2 dσ = dΩ
Vm 2π¯h2
2
|Hab |2
That means that
mV hψb | H | ψa i 2π¯h2 The negative sign is a convention. Suppose that H = V (~r). Then f (θ) = −
mV 1 1 ~ ~ √ e−ikb ·~r V (~r) √ eika ·~r d3 r 2 2π¯h V V Z m ~ ~ = − ei(ka −kb )·~r V (~r)d3 r 2π¯h2
f (θ) = −
Z
The result is equivalent to that of the Green’s function analysis in the first Born approximation. (See Griffiths p. 368)
26
4.2
Harmonic perturbation
Now suppose that the perturbation is V (t) = V eiωt + V † e−iωt We will get that i t † i(ωni −ω)t Vni ei(ωni +ω)t + Vni e h 0 ¯ ! i(ωni −ω)t − 1 i ei(ωni +ω)t − 1 † e = − Vni + Vni h ¯ i(ωni + ω) i(ωni − ω)
cn (t) = −
Z
If ωni ∼ ω then the second term is the significant one. En > Ei and there is absorption. If Ef < Ei then ωni < 0 and the first term counts. Then |cn (t)|2 =
2 4 4π 2 sin (ωni ± ω)t/2 = lim → 2 |Vni |2 δ(ωni ± ω)t/2 |V | 2 ni 2 t→∞ (ωni ± ω) h ¯ ¯h
and just like for the constant perturbation wi→n =
4.2.1
π2 2π 2 |Vni |2 δ(En − Ei ± ¯hω) 2 |Vni | πδ(ωni ± ω) = ¯ h h ¯
En = Ei ± ¯hω
Interactions with radiation field
Let
e A·p mc legit as long as ∇ · A = 0 which is true for a plane wave in empty space. For that very same plane wave V (t) = −
A = 2A0 cos(ωt − k · x) = A0 ei(ωt−k·x) + e−i(ωt−k·x)
Consider absorption. Then we take the second term. We can expand the exponential for small k. Then A ∼ 2A0 (1 + ik · x + . . .) In the long wavelength limit, xk → 2πx/λ is small. Small that is compared to the size of the wave function which is perhaps the size of an atom. The typical transition energy for an atom is Z 2 e2 E< 2a0
27
The corresponding wavelength is λ∼
¯c h 2a0 2a0 ¯hc = 2 2 = 2 a0 E Z e Z α
The transition rate in the dipole approximation is wi→n =
2π e | A0 hn | · p | ii |2 δ(En − Ei ± ¯hω) h mc ¯
The energy density of the radiation field is U = gives us wi→n = → = = = = =
1 2
2 Emax 8π
+
2 Bmax 8π
=
1 ω2 2 2π c2 |A0 | .
Substitution
(2π)2 e U| hn | · p | ii |2 δ(En − Ei ± ¯hω) h ¯ mω e (2π)2 ρ(ω)dω| hn | · p | ii |2 δ(En − Ei ± ¯hω) h ¯ mω (2π)2 e 2 2 ρ(ω)| mω hn | · p | ii | h ¯ e (2π)2 2 2 ρ(ω)| mω · hn | p | ii | h ¯ e m (2π)2 · hn | [H, x] | ii |2 2 ρ(ω)| mω ¯ h h ¯ (2πe)2 ρ(ω)|· hn | x | ii |2 h2 ¯
In the last step we use hn | [H, x] | ii =
4.2.2
−¯h hn | p | ii = h ¯ ωni hn | x | ii m
Absorption vs emission
The rate that we just computed for absorption from a radiation field with energy density ρ(ω) is identically the rate we would find for the emission in that same radiation field. In the event of spontaneous emission, we can replace ρ with an expression for the density of final states available to the photon.
4.2.3
Spontaneous emission
An alternative strategy for determining the spontaneous emission rate is to begin with the expression for the stimulated rate, and then substitute the ground state radiation density of the vacuum. Let’s try to come up with a more systematic development.
28
The rate wi→n =
2π h ¯
e A0 mc
2 D
E
| n | · peik·r | i |2 δ(En − Ei ± ¯hω)
We want to replace vector potential with energy density. We use U= Then wi→n
2π = h ¯
2πc2 U ω2
!
e mc
1 ω2 |A0 |2 2π c2
2 D
E
| n | · peik·r | i |2 δ(En − Ei ± ¯hω)
U has some frequency dependence. The total energy density is Z
ρ(ω)dω = U We replace U with ρ(ω)dω and integrate with the δ function. We get wi→n
2π = 2 h ¯
2πc2 ρ(ωni ) ω2
!
e mc
2 D
E
| n | · peik·r | i |2
What do we mean by ρ(ω)? It is the energy density per unity volume per unit frequency. The density at the frequency corresponding to the transition energy is the only part that counts. The transition rate from n → i or in the other direction is exactly the same, in one case it is absorption and in the other it is stimulated emission. How do we get from here to spontaneous emission? It must be that there is some vacuum fluctuation or vacuum radiation density. We need second quantization for a mathematically complete theory. But we can develop an expression for spontaneous emission based on a heuristic argument. We simply replace the expression for the radiation density with the energy density of available photon states. The total number of states with energy E < ¯hkc, where kx = πnx /L, etc. is N
= = =
1 dN V dω
=
14 π(n2x + n2y + n2z )3/2 83 3 14 L π (k 2 )3/2 83 π π L 3 ω 3 ( ) 6 π c 2 1 ω number of states per unit volume per unit frequency 2π 2 c3
29
Finally to get the energy density we multiply by two to include the two polarization states and multiply by ¯hω, the energy of the photon in the state and we end up with ρ(ω) =
¯ ω3 h π 2 c3
The spontaneous transition rate becomes wi→n =
4π ¯h2
2πc2 ¯h ω 3 ω 2 2π 2 c3
!
e mc
2 D
E
| n | · peik·r | i |2
E e 2 D | n | · peik·r | i |2 mc E α 1 D ik·r = 4 (ω) | n | · pe | i |2 mc2 m
= 4
ω ¯hc
which sure enough is a rate. In the dipole approximation we get α 1 |mω hn | · x | ii |2 2m mc α = 4 ω 3 2 | hn | · x | ii |2 c
wi→n = 4 (ω)
4.2.4
Average over polarization and angle
To get the total spontaneous emission rate we average that last over all directions k for the outgoing photon and all polarizations . Suppose that x is along the z-direction and the outgoing photon is at an angle θ, φ = 0. First note that if θ = 0, then · x = 0. So project onto the x-axis. Then the polarization is in the y-z plane and the average of R 2 1 R 1 4 2 h · ˆ zi = 4π sin sin θdθ cos2 αdα = 4π ( 3 )π = 13 . The total spontaneous decay rate into all angles and polarizations is wi→n =
4.2.5
4 3 α ω | hn | x | ii |2 3 c2
Angular distribution
Suppose we know that the transition is E1 and ∆m = 1. Since it is E1 that means that ∆l = ±1 by Wigner Eckart. Then hf | x | ii = hl ± 1, m ± 1 | x | lmi
30
will be a vector in the x-y plane. No z-component. That means that if the photon is headed in the z-direction · x = 1 for all polarizations. If it is headed in the x-direction then we need to average. The angle θ is that of the polarization with respect to the y-axis. 1 2π
Z
cos2 θdθ =
1 2
For a photon headed in the θ direction with respect to the z-axis, it has a component cos θ in the z-direction and and component sin θ in the x-direction. We get all of the z-direction piece and half of the x-direction piece so the probability to go into the angle θ is P (θ) = cos2 θ +
1 1 sin2 θ = (1 + cos2 θ) 2 2
which happens to be the same |d11,1 |2 + |d11,−1 |2 =
4.2.6
1 (1 + cos θ) 2
2
+
2
1 (1 − cos θ) 2
1 = (1 + cos2 θ) 2
Absorption cross section
Or instead of writing the rate in terms of the energy density of the incident radiation, we could define the cross section. Then the rate is the incoming flux times the cross section. The incoming energy flux is the energy density times the velocity. The flux of photons is the energy flux/photon energy. So Rate = σFlux
and
Flux = cU/¯hω
Then
wi→n ¯ hω (2π)2 e2 = | hn | · p | ii |2 δ(En − Ei ± ¯hω) cU c m2 ω Again we replace hpi with m/¯ hh[H, x]i and get that σabs =
σabs =
(2π)2 ωe2 (2π)2 ωe2 2 | hn | · x | ii | δ(E − E ± h ¯ ω) = | hn | · x | ii |2 δ(ωni − ω) n i c ¯ c ¯h h2
If we suppose the polarization is in the x direction then The total absorption cross section is Z X σ(ω)dω = 4π 2 ωni α| hn | x | ii |2 n
4.3
Photo electric effect
Imagine a hydrogen atom at rest at the origin. Electromagnetic radiation in the form of a plane wave is propagating in the +z direction (k = |k|ˆ z ) with E-field polarized along the x-direction. The radiation field interacts with the atom and frees the electron. The electron flys off in the (θ, φ) direction with momentum p = h ¯ kf
31
4.3.1
Golden rule
The transition rate is given by the golden rule R=
| hf | H 0 | ii |2 2πδ(Ef − Ei − ¯hω) ¯h
The initial state is the ground state of hydrogen 2 e−r/a h~x | ii = √ 4πa3 The final state is a free electron h~x | f i =
1 L3/2
eikf ·r
which we represent as a plane wave. We imagine the plane wave in a box with sides of length L so that it is normalized over the volume. We assume that the energy of the electron is so high that we can neglect the effect of the coulomb attraction of the hydrogen nucleus in the final state.
4.3.2
Interaction Hamiltonian
The interaction hamiltonian is
q A·p mc We write the vector potential as a plane wave, H0 =
A = 2A0 cos(k · r − ωt)
= A0 e
i(k·r−ωt)
+e
−i(k·r−ωt)
(17)
(18)
where is the unit vector that gives the polarization of A and E = − ∂A ∂t After application of the energy conserving δ-function we are left with A = A0 ei(k·r−ωt) Then H0 =
−i¯hq A0 eik·r · ∇ mc
32
4.3.3
Density of states
The free electron in the final state is traveling at angle θ, φ into solid angle dΩ. In order to determine the total rate for the transition we need to integrate over all possible final state electrons that are directed into the solid angle. The electrons are free particles confined to a box with side L. The electron wave function at the boundaries vanishes so kx =
2nx π 2ny π 2nz π , ky = , kz = L L L
and kf2 = Imagine a sphere of radius the volume of the sphere,
q
2π L
2
(n2x + n2y + n2z )
n2x + n2y + n2z . The total number of states with |k| < |kf | is
4 4 N = π(n2x + n2y + n2z )3/2 = π 3 3
L k 2π
2 !3/2
The number of states per unit k into solid angle dΩ are dN 4 L = π dk 3 2π
3
3k 2
dΩ 4π
The number of states per unit energy into solid angle dΩ are dN dE
where we have use E =
4.3.4
L 3 2 dk dΩ k 2π dE 4π 3 km L = dΩ 2π ¯h2
=
(19) (20)
¯ 2 k2 h 2m .
Transition rate
The total transition rate is dN | hf | H 0 | ii |2 R = 2π δ(Ef − Ei − ¯hω) h ¯ dE 3 | hf | H 0 | ii |2 L km 2π = dΩ h ¯ 2π ¯h2 3 E qA0 2 D L km ik·r 2 = ¯h | f | e · ∇ | i | 2π dΩ mc 2π ¯h2 Z
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(21) (22) (23)
4.3.5
Absorption cross section
We define a differential cross section dσ Rate for photoelectrons = dΩ Flux of incident photons The energy flux for the incoming plane wave is cu =
c 2 1 ω2 2 Emax = A 8π 2π c 0
and since each photon has energy ¯hω, the photon flux is Flux =
1 1 ω 2 A20 = ωA20 2π¯hωc 2π¯hc
Then dσ dΩ
q 2 A2
=
D
E
h (mc)02 | f | eik·r · ∇ | i |2 2π ¯
3
km h2 ¯
(ωA20 /2π¯hc) D
E
2π(q¯h)2 | f | eik·r · ∇ | i |2 2π =
L 2π
L 2π
3
mE2 cω q 2 | f | eik·r · ∇ | i |2 L3 k
km h2 ¯
(24)
(25)
D
=
4.3.6
(26)
2πmcω
Expectation value
All that remains is to compute the expectation value. Note that it is not appropriate to work in the electric dipole limit. We are assuming that the energy of the final state electron is big compared to the binding energy. Therefore it is not safe to assume that k · r 1 and we will attempt to do the integral exactly. We have that D E 1 1 f | eik·r · ∇ | i = 3/2 1/2 3/2 I L π a where Z I = e−ikf ·r eik·r · ∇e−r/a r2 drdΩ Integrating by parts we get that I = · − · − ·
Z Z Z
∇ e−ikf ·r eik·r e−r/a r2 drdΩ
∇ e−ikf ·r eik·r e−r/a r2 drdΩ
e−ikf ·r ∇ eik·r e−r/a r2 drdΩ
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(27) (28) (29)
Equation 11 can be written as a surface integral over a volume that we can take to infinity. The wave function falls off exponentially so Equation ?? is zero. On taking the gradient Equation ?? and pulling k outside of the integral, we get i · k
Z
e−ikf ·r eik·r e−r/a r2 drdΩ = 0
since polarization and propagation vector k are orthogonal. Finally I = ·
Z
∇ eikf ·r eik·r e−r/a r2 drdΩ
= −i · kf = −i · kf = −i · kf = −i · kf
Z Z Z Z
(30)
e−ikf ·r eik·r e−r/a r2 drdΩ
(31)
e−i(kf −k)·r e−r/a r2 drdΩ
(32)
e−iq·r e−r/a r2 drdΩ
(33)
e−iq cos θr e−r/a r2 dr sin θdφ
(34)
where q = kf − k. First we do the angular integral I = i2π · kf
Z
(e−iqr − eiqr ) −r/a 2 e r dr iqr
(35)
Then integrate over r. (e−iqr−r/a − eiqr−r/a ) rdr iq 1 1 2π · kf − q(−iq − 1/a)2 q(iq − 1/a)2 1 1 2π · kf − q(−q 2 + 1/a2 − 2iq/a) q(−q 2 + 1/a2 + 2iq/a) −4iq/a 2π · kf q((−q 2 + 1/a2 )2 + 4(q/a)2 )2 −4i 2π · kf 2 a(q + (1/a)2 )
I = 2π · kf = = = =
Z
(36) (37) (38) (39) (40)
Finally D
f | eik·r · ∇ | i
E
= =
1
1
L3/2
π 1/2 a3/2
2π · kf
a(q 2
−8iπ 1/2 · kf L3/2 a3/2 a(q 2 + (1/a)2 )2
−4i + (1/a)2 )2
(41) (42) (43)
35
and dσ dΩ
4.3.7
=
α¯ hL3 kf 2πmω
=
32α¯hkf ( · kf )2 mω a5 (q 2 + (1/a)2 )4
8π 1/2 · kf L3/2 a5/2 (q 2 + (1/a)2 )2
!2
(44) (45)
Angular distribution
The E-field vector k is in the z-direction and the electron in the final state has kf in the θ, φ direction. Then q 2 = kf2 + k 2 − 2kf k cos θ. And the polarization is in the x-direction so that · kf = kf sin θ cos φ and 32α¯ hkf (kf sin θ cos φ)2 dσ == dΩ mω a5 (kf2 + k 2 − 2kf k cos θ + (1/a)2 )4 The differential cross section is peaked for the outgoing electron in the θ = π/2, φ = 0 direction, namely parallel to the polarization of the field.
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