Perturbation Theory. D. Rubin. December 2, Stationary state perturbation theory

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1 Perturbation Theory D. Rubin December, 010 Lecture 3-41 November 10- December 3, Stationary state perturbation theory 1.1 Nondegenerate Formalism We have a Hamiltonian H = H 0 + V and we suppose that we have determined the complete set of solutions to H 0 with ket n 0 so that H 0 n 0 = E 0 n n 0. And we suppose that there is no degeneracy. The eigenkets of H satisfy H n = E n n H 0 + V ) n = E 0 n + n ) n 1) E 0 n H 0 ) n = V n ) n ) where E n = E 0 n + 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 E 0 n H 0 ) n = λv n ) n 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 n = 1 E 0 n H 0 λv n ) n 1

2 To make sure we don t do something crazy like divide by zero we project out the n 0 ket by introducing a projection operator φ = k 0 k 0 k n Then we make the equation safe from nan. n = φ E 0 n H 0 λv n ) n We are not losing anything here since Equation?? says that n 0 λv n n = 0 4) Finally, so we have something sensible as λ 0 we add the solution to the unperturbed part and write n = n 0 + φ E 0 n H 0 λv n ) n One last thing to note. Normalization. It seems that n n 0 = 1. We will use this fact. n is not normalized by itself. Now Equation?? says that Now we suppose that and n = λ n 0 V n n = n 0 + λ n 1 + λ n +... n = λ 1 n + λ n +... Then equation powers of λ. Our energy equation becomes So λ 1 n + λ n +... = λ n 0 V n 0 + λ n 1 + λ n +... ) 1 n = n 0 V n 0 n = n 0 V n 1 3 n = n 0 V n

3 And for the wave function n 0 +λ n 1 +λ n φ +... = n 0 + λv λ 1 En 0 H n λ n+...)) n 0 +λ n 1 +λ n Equating stuff linear in λ we get n 1 = Next everything quadratic in λ gives n = φ E 0 n H 0 n = V φ E 0 n H 0 V n 0 5) φ V n 1 1 En 0 H n n 1 ) 0 And we can use our solution for n 1 to see that ) φ φ V n En 0 0 n 0 V n 0 V n H 0 En 0 0 H 0 φ n = n 0 V V n En 0 0 = n 0 V k 0 H 0 k n En 0 Ek 0 It is evident that degeneracy could get us in trouble. 1. Examples 1..1 Helium To first approximation, the energy of the ground state of helium is ) e Z E 0 = Z = 4) 13.6ev) = 108.8ev a The next approximation would be to include the interaction of the electrons as a perturbation. e V = r 1 r The unperturbed wave function is taken to be ψr 1, r ) = ψ 0 r 1 )ψ 0 r ) = 8 πa 3 e r 1+r )/a 3

4 where Set the polar axis along r 1 and ) 8 E 1 = e πa 3 Then the total energy is E = 4) ) e a a = hc αmc e 4r 1+r )/a r1dω 1 rdω = e 5 r1 + r r 1 r cos θ + 5e 4a = 11e 4a 1.. Stark effect for rigid rotator The hamilonian for the rigid rotator is H = L I 4a = ev) = 74.8ev 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 and eigenvalues are ψ = l, m and E l = h ll+1). In the presence of a uniform I electric field in the z direction we introduce V = eze = ere cos θ We suppose that e r = d is the electric dipole moment of the rotator. All states have the same radial wave function). We can write V = T 1 0 = de 4π 3 Y 1,0θ) We need to evaluate l, m T 1 0 l 1, m 1 The matrix element is non zero if m 1 = m and l l 1 1, Wigner Eckart), l 1 l, parity). Then the non zero matrix element can be written l ± 1, m T 1 0 l, m = de 4π 3 Y l±1,my 1,0 Y l,m dω 4

5 4π = de 1) + 1)l + 1) 1, 0, l, m l ± 1, m 1, 0, l, 0 l ± 1, 0 3 4πl ± 1) + 1) = de l + 1) 1, 0, l, m l ± 1, m 1, 0, l, 0 l ± 1, 0 l ± 1) + 1) l m = de 4l 1 The unperturbed levels are all degenerate. But since H 0 and V both commute with L z, 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, l, m T 1 0 l, m = 0. The second order correction reduces to the two terms corresponding to l = ±1. Finally ɛ lm = ) de ll + 1) 3m l 1)l + 3) E 0 l The degeneracy is only partly removed, since it depends on m. Some symmetry remains. We could compute the electric dipole moment. We have that Then µ k n n n 0 + k n µ e n z n n 0 z k 0 k 0 T 1 0 n 0 E 0 n E 0 k ) = 1 E 1..3 Stark effect in hydrogen k 0 T0 1 n 0 k En 0 Ek 0 0 k n n 0 T 1 0 k 0 k 0 T 1 0 n 0 E 0 n E 0 k ) As in the case of the rigid rotator, the perturbation commutes with L z 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 L, it does commute with L z. So there is no change to the z-component and therefore no mixing of states with 5

6 m m. Parity eigenstates have zero electric dipole moment. The electric dipole moment is d = α, l, m x α, l, m = x ψx) d 3 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 The energy shift is second order = e E k 0 = 1 α E k 0 z 1, 0, 0 E 0 0 E 0 k ) 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 k 0 z 1, 0, 0 = k 0 z 1, 0, 0 k 0 all k since 1, 0, 0 z 1, 0, 0 = 0. The sum over all k becomes Therefore, k 0 z 1, 0, 0 = 1, 0, 0 z 1, 0, 0 = 1 all k 3 r = a 0 e E a 0 E 0 0 E 0 k ) = e E a 0 e /a e /a 0 ) Degenerate perturbation theory = E 8a3 0 3 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

7 .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 i, i = 1,, 3 for example. Now if i V j = V δ ij 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 1 V 1 1 V 1 V 3 I V I 0 0 V = V 1 V V 3 0 II V II 0 3 V 1 3 V 3 V III V III The new basis is related to the old by a unitary transformation. I = a 1 + b + c 3 The states I, II, III 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 n 1 = k D k 0 V n 0 E 0 n E 0 k ) k0 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..1.1 Projection operators We could more formally use the projection operator approach. We go back to the Schrodinger equation 0 = E H 0 λv ) l = E E D λv )P 0 l + E H 0 λv )P 1 l Projecting from the left by P 0 and P 1 we get P 0 E E D λv )P 0 l + E H 0 λv )P 1 l ) = E E D λp 0 V P 0 l λp 0 V P 1 l 6) P 1 E E D λv )P 0 l + E H 0 λv )P 1 l ) = λp 1 V P 0 l + E H 0 λp 1 V P 1 ) l 7) 7

8 We solve Equation?? for l and then project out the states not part of the degenerate subspace and then the denominator is safe P 1 l = P 1 1 E H 0 λp 1 V P 1 λp 1 V P 0 l P 1 l 1 k 0 k0 V l 0 k D ED 0 Ek 0 Then substitute Equation?? into Equation?? for P 1 l multiply by P 0 and we have E E D λp 0 V P 0 λp 0 V P 1 1 E H 0 λp 1 V P 1 λp 1 V )P 0 l E E D λp 0 V P 0 ) l = 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 l 0 kets. l = a i l i Then the Schrodinger equation including the perturbation is Substitute the expansion for l and 8) E 0 l H 0 ) l = λv n ) l 9) l = l 0 + λ l 1 + λ l Here l 0 = i=d a i i l 0 is a normalized linear combination of the degenerate states in the basis in which the perturbation matrix is diagonalized. Multiply by m 0 from the left m0 E 0 n H 0 l = λ m 0 V l l m 0 l and keep the first order in λ, when m 0 = l D and we get that 0 = l D V l 0 1 l l D l l D V l D = i D When m 0 l D En 0 Em) m 0 0 l = λ m 0 V l l m 0 l 8

9 To first order in λ we have E 0 n E 0 m) m 0 l 1 = m 0 V l 0 and and m0 l 1 m 0 V l 0 l 1 = l D E 0 l E 0 m m 0 V l 0 m En 0 Em 0 0 Note that we can always add states with eigenvalue E 0 l. The above gives no information about them. We define l 1 and all higher order contributions to include no l 0 or for that matter any of the l D degenerate states. Then the exact solution is l = i=d a i i + l 1 + l +... and l 0 = i=d a i i l 0 l j = 0, for j 0. We suppose that the a i are chosen so that l 0 is normalized. We see that in this way l is a solution to Equation?? independent of the coefficients a i. And the second order energy shift l = l 0 V n 1 =.1. Second order degenerate perturbation theory Another strategy 1 for getting second order correction of degenerate states is to write the true state a = c α α + d µ µ α µ where the sum over α includes the degenerate states and the sum over µ is all the others. We know that 1 Gottfried and Yan H E a ) a = H 0 + λh 1 E a ) a = 0 9

10 Then 0 = α c α H E a ) α + µ d µ H E a ) µ 0 = α c α E α E a + λh 1 ) α + µ d µ E µ E a + λh 1 ) µ Then multiply from the left first by β which is an element of the degenerate space and then by ν which is outside of the degenerate space. 0 = c β E β E a ) + λ α c α β H 1 α + λ µ d µ β H 1 µ 10) 0 = α c α λ ν H 1 α + d ν E ν E α ) + λ µ d µ ν H 1 µ 11) Drop the last term in Equation?? as it is higher order in λ. Justification for this step is that the state a is mostly a linear combination of the states in the degenerate space. The coefficients d µ 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 β H 1 µ is more important that ν H 1 µ since it connects to the relevant states. Finally, we could keep the last term in Equation??, solve for d ν, substitute into Equation?? and then drop the highest order term. That would be equivalent to solving for α c α ν H 1 α d ν = λ 1) E ν E D ) Substitution into?? gives 0 = c β E β E α ) + α c α λ β H 1 α + λ µ ) µ H 1 α β H 1 µ E µ E D which is the eigenvalue problem in the subspace D for the effective Hamiltonian H eff 13) β H eff α = λ β H 1 α + λ µ β H 1 µ µ H 1 α E D E µ. So we diagonalize β H 1 α. Define P to be the projection operator onto the degenerate subspace D, P = α α α. Then we can write H eff = λp H 1 P + λ 1 P P H 1 H 1 P E H 0 10

11 .1.3 Example Suppose we have a system with 3 states. The first two are degenerate. The third is at energy above the two. The H 1 = The energies and eigenstates are 0 0 λm 0 0 λm λm λm H eff = λm) ) E S = λm), E S = ) and E A = 0, E S = 1 1 ) 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 each of the three states is given by the eigenvalues of H 1 which are, E 0 = 0, E ± = 1 ± + 8λM) ). The eigenvectors are 0 = 1 1 ), ± = E ± 3 ) λm + E±/λ M What if we went back to the weak perturbation case where M, but we had started with states S = ) and A = 1 1 ) and III = 3. Then 0 0 M V = M 0 Now H eff = M ) /

12 . Fine Structure..1 Spin orbit coupling The magnetic field at the electron due to its motion through the E field of the nucleus is B = v E c The electric field is radial so E = E r and p = mv and then r Next, note that E = 1 e V = 1 e B = r p E mc r V r = L E mc r and we get that B = The magnetic moment of the electron is Finally and if V = e/r then L 1 V mec r r µ e = e mc S H = µ B = 1 V m c r r L S H = e L S m c r e L S 3 m c r 3 Then a factor of 1 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 B = JdV ) r c r 3 For a current loop, we find the magnetic field at the center is B = I cr 3 eλdlv r = er p cmr 3 The unperturbed states have orbital angular momentum and spin. We can use l, m l s, m s as base kets with eigenvalues l, m, s, and m s, or we can take the linear 1

13 combinations that would be eigenkets of j, j z, l, s. Since L S = 1 J L S ), it is clear that it will be more convenient to use the latter. Then En 1 = e h 4m c 1 jj r + 1) ll + 1) 3 ) 3 4 E SO = h e { } 1 [jj + 1) ll + 1) 3/4] 4m c n 3 a 3 0 ll + 1 )l + 1) where For j = l ± 1 1 r 3 = 1 n 3 a 3 0ll + 1/)l + 1) E SO = h e 4m c 1 a 3 0n 3 = mc α 4 4 [ ] l l + 1) ll + 1 )l + 1) [ ] l l + 1) n 3 ll + 1 )l + 1) We remember that Substitution into the above gives E 0 = 1 α mc = e a 0 a 0 = e α mc = h me En 1 e h α mc ) 3 = h c α mc ) 3 = 1 m c e 6 e 4 mc ) α mc α ) jj + 1) ll + 1) 3 ) 4 No dependence on m l. No mixing of levels. No need for degenerate formalism. Each level is split into j = l ± 1... Relativistic correction Really K = mc ) + pc) mc p m p4 8m 3 c 13

14 And ) p 4 ψ ψ m p = m ψ p m ψ = E 0 V )ψ E 0 V )ψ = ψ E0 E 0 V + V ψ We use the fact that p is Hermitian in the first step. ) E0 1 m) 8m 3 c E n V r)) 4m E 8m 3 c n E n e r + e4 r 4m E e e 4 ) 8m 3 c n E n + n a 0 l + 1 )n3 a 3 0 4m 8m 3 c E n n ) l + 1) 1 mc E n 3 + 4n ) l + 1) 1 mc α n ) 4n 4 l + 1) 5 mc E 0 1 α mc ) = 5 4 E 0α Note that 1, 0, 0 1 1, 0, r 0 = 1 a 0 and 1, 0, 0 1 1, 0, 0 =.) Depends on n, l, r a 0 not on j. Again there is no mixing of degenerate levels. In general..3 Darwin term 1 r = 1 n a 0, 1 r = 1 l + 1 )n3 a 0 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) + i V r i δr i + 1! i j V r i r j δr i δr j + Oδr 3 ) = V r) δr) V + Oδr 3 ) 14

15 If δr h/mc then H D 3 π e h m c δ3 r) Note that according to Merzbacher p. 06 and 08) Then ψ n,l,m 0) = n00 H D n00 = 3 π e h m c 1 na 0 ) 3 = π 3 1 n! na) 3 nn 1)! = 1 na 0 ) 3 e h m c 1 n 3 m 3 e 6 h 6 = π 3 1 n 3 mc α 4 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...4 Fine Structure Combining all order α corrections we have E SO = E n mc { } n[jj + 1) ll + 1) 3/4] ll + 1 )l + 1) The relativistic correction is E 1 r = E n mc [ 4n l ] The total α correction is E 1 n [ 1 + α n n j )].3 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 = in hydrogen. There are l+1 = 3 levels with l = 1 and 1 level with l = 0. Spin is not relevant. The candidate 15

16 set of base states are the orbital angular momentum kets. The perturbation matrix elements i V j, where V = ez E are 0, 0, 0 V, 1, 0 0 0, 1, 0 V, 0, V = Only the upper left is relevant. All of the other states are not coupled. The eigenvectors of the upper left X matrix are 1 ± = ± e E, 0 r, 1, 0, 0 cos θ, 1, 0 = ±3ea 0 E The eigenvectors are ± = 1, 0, 0 ±, 1, 0 ) These two states are shifted positively and negatively with an electric field. The other states are not effected by the E-field..4 Zeeman effect We have and then H = p m + V cr) e mc B L z + H Z = e mc B L z + S z ) e 8mc B x + y ) H spin = µ B = e mc S B = e mc B S z and H spin orbit = 1 1 dv r) L S m c r dr Assuming B is very weak so that the spin-orbit piece dominates, we need to evaluate j, m, l, s L z + S z j, m, l, s = j, m, l, s J z + S z j, m, l, s Since l ± 1 l ± m + 1, m = ± l + 1 l, m 1 1, 1 l m l + 1 l, m + 1 1, 1 we can compute l ± 1, m S z l ± 1, m 16 = ± m h l + 1

17 Meanwhile l ± 1, m J z l ± 1, m = hm. Then.4.1 Paschen-Back limit 1 B = e hb mc m 1 ± 1 ) l + 1 If the magnetic field is very strong, then total angular momentum is not conserved. There is an external torque). But the L z and S z are fixed. So now it is better to use the basis, m l s, m s. It is easy to evaluate L z + S z but not quite as simple to compute L S for the spin orbit coupling. But if we write L S = L z S z + 1 L +S + L S + ) and note that L ± = S ± = 0 then we get that.4. Intermediate field L S = h m l m Let s examine the effect of an external magnetic field on the n = states of hydrogen. And we suppose that H spin orbit H z. 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, j z, l, s, l = 1, j = ±3/, ±1/,or l = 0, j = ± 1 or eigenstates of l, m l, s, m s, l = 1, m l = ±1, 0, m s = ±1/, l = 0, m s = ±1/). We have to construct the perturbation matrix elements i V j in whatever basis we have chosen and V = e mc BL z + S z ) + e 1 m c r L S 3 Let s pick the basis of eigenstates of j, j z, 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, 0, 1 = 0, 0 1, 1 = 1, 1, 0, 1 = 0, 0 1, 1 3 = 1, 1, 1, 1 = 1 3 1, 0 1, 1 3 1, 1 1, 1 17

18 4 = 1, 1, 1, 1 = 1 3 1, 0 1, , 1 1, 1 5 = 3, 1, 1, 1 = 3 1, 0 1, , 1 1, 1 6 = 3, 1, 1, 1 = 3 1, 0 1, , 1 1, 1 7 = 3, 3, 1, 1 = 1, 1 1, 1 8 = 3, 3, 1, 1 = 1, 1 1, 1 Now we can compute the matrix elements α 0 b α 0 + b α 1 b 1/3 0 b / α 1 + b 1/3) 0 b / b /3) 0 α 1 b /3) b /31)) 0 α 1 + b /3) α 1 + b ) α 1 + b) Next we find the eigenvalues i. The energy of the state i shifts by i. 3 Variational method 3.1 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

19 The total hamiltonian is H = p 1 m Ze + p r 1 m Ze e + r r 1 r with Z =. The approximate wave function is Then we calculated ψ 1 ψ = ) Z 3 1/ ) Z 3 1/ e Zr 1/a 0 e Zr /a 0 πa 3 0 πa 3 0 E approx = E E 0 + = ψ 1 ψ H ψ 1 ψ = ψ 1 ψ H 0 ψ 1 ψ + ψ 1 ψ H H ψ 1 ψ Now it is possible to write our trial wave function as a linear combination of the complete set of exact eigenkets. ψ 1 ψ = n a n n, H n = E n n Then E approx = n a n E n E 0 a n So we know for sure that E approx E 0 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. We can incorporate this approximation in our wave function by writing ) Z 3 1/ ) Z 3 1/ ψ 1 ψ = πa 3 0 πa 3 e Z r 1 /a 0 e Z r /a 0 Z ) 3 = e Z r 1 +r )/a 0 0 πa 0 Or define a 0 = a 0/Z and we get that ψ 1 ψ = ) 1 1/ ) 1 1/ πa 3 e r/a πa 3 e r/a Note that our new wave function is normalized. This is very important. Now we compute p ψ 1 ψ H ψ 1 ψ = ψ 1 ψ m ψ 1ψ + ψ 1 ψ Ze + Ze e ψ 1 ψ + ψ 1 ψ r 1 r r 1 r ψ 1ψ n 19

20 Note that ψ 1 and ψ just look like the hydrogen ground state. Therefore p Z ψ 1 ψ m ψ e ) 1ψ = a = Z e a 0 The second term ψ 1 ψ Ze r 1 + Ze ψ 1 ψ = Z r ) e a = Z Ze a 0 The third term we calculated before ) Z 3 E 1 = e e Z r 1 +r )/a πa 3 r1dω 1 rdω = e 5 r1 + r r 1r cos θ 8a = Z 5 8 Therefore ψ 1 ψ H ψ 1 ψ = Z Z Z + Z 5 8 ) e a 0 It will be a function of the parameter Z. Next, we minimize the expectation value with respect to Z. This is OK because no matter what we choose for Z, we know that we are overestimating the energy. Set e a 0 And d H dz = Z Z ) = 0 Z = Z 5 16 = 7 16 H =.85 e a 0 = 77.5ev 3. Examples 4 Time dependent perturbation theory 4.1 Formalism Now e want to consider time dependent perturbations. The Hamiltonian is written in the form H = H 0 + V t) We suppose that we know the complete set of eigenkets of H 0, namely n and as usual H 0 n = E n n. 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 0

21 finding the system in one of the uperturbed states. An arbitrary state at t = 0 can be written α = c n 0) n n We want α, t 0 = 0; t = n c n t)e ient/ h 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 Observables are According to the Schrodinger equation α, t 0 ; t I = e ih 0t/ h α, t 0 ; t S A I = e ih 0t/ h A s e ih 0t/ h i h t α, t 0; t S = H 0 + V ) α, t 0 ; t S i h ) e ih0t/ h α, t 0 ; t t I = H 0 + V )e ih0t/ h α, t 0 ; t I i he ih 0t/ h t α, t 0; t I = e ih 0t V e ih 0t α, t 0 ; t I i h t α, t 0; t I = V I α, t 0 ; t I 14) Then we can also write that α, t 0 ; t = n c n t) n 15) Substitute that last into Equation?? and multiply from the left by n and we have i h t n α, t 0; t = m n V I m m α, t 0 ; t I 16) Also from Equation?? we see that c n = n α, t 0 ; t. Then i h d dt c nt) = m V nm e iωnmt c m t) where ω mn = E n E m h 1

22 and we used n e ih0t/ h V t)e ih0t/ h m = V nm t)e ien Em)t/ h We expand the time evolution operator α, t 0 ; t I = U I t, t 0 ) α, t 0 ; t 0 I i h d dt U It, t 0 ) = V I t)u I t, t 0 ) The initial condition is and we integrate to get Ut, t 0 ) t=t0 = 1 U I t, t 0 ) = 1 ī h We get an approximate solution by iteration: [ Transition U I t, t 0 ) = 1 ī h = 1 ī h = +... t t t 0 V I t )U I t, t 0 )dt t V I t ) 1 ī V I t )U I t, t 0 )dt dt t 0 h t 0 ) i t t V I t ) + dt dt V I t )V I t ) t 0 h t 0 t 0 t Suppose we have a system in an eigenket i. Then ] i, t 0 ; t = U I t, t 0 ) i = n n n U I t, t 0 ) i = n c n t) We see that n U I i is the amplitude that a system initial in state i will be in state n as a function of time. We note that i I = e ih 0t/ h α, t 0 ; t S i I = e ih 0t/ h Ut, t 0 ) α, t 0 ; t 0 S i I = e ih 0t/ h Ut, t 0 )e ih 0t 0 α, t 0 ; t 0 I U I t, t 0 ) = e ih 0t/ h U S e ih 0t 0 / h Therefore n U I i = e ient E it 0 )/ h n U i

23 Then as long as keti and n are eigenkets of H 0, So c 0 nt) = δ ni c 1 nt) = ī h t n U I i = n U i t 0 n VI t ) i dt t c 1 nt) = ī n V t ) i e ien Ei)t/ h dt h t 0 c nt) = ī ) t t dt dt n V I t ) m m V I t ) i h m t 0 t 0 = ī ) t t dt dt V nm t )e iωnmt V mi t iωmit )e h t 0 m t Constant perturbation We turn on a constant perturbation at t 0. Then the amplitude to make a transition from state i to n to first order is c n t) = ī t V ni e iωnit = i V ) ni e iωnit 1 h t 0 h iω ni Then c n t) = 4 V ni h ωni sin V ni V ni ω ni t/) = 4 E n E i ) sin ω ni t/) = 4 E n E i ) sin E n E i )t/ h) If E n = E i, then c n t) = 1 h V ni t If there is a finite number of final states close to n then the total transition probability is n,e n=e i c n. and if there are many final states, more or less nearby we replace cn deρe n ) c n = 4 [ ] sin En E i )t Vni h E n E i ρe)de Next we see that lim t sin [ ] En E i )t h V ni πt ρe) = E n E i h δe n E i ) 3

24 Finally c n t) = ρe)de c n t) = π h ) V ni ρe n )t The rate is w i n = π h ) V ni ρe n ) or w i n = π h ) V ni δe n E i ) where integration over final states is implied Second order perturbation We found earlier that c nt) = 1 h m t 0 t dt dt V nm t )e iωnmt V mi t )e iω mit 0 If V t) is constant in time then c nt) = 1 h m = 1 h m = ī h m ī h m t 0 t 0 V nm V mi E m E i V nm V mi E m E i ) e dt iω mi t 1 V nm e iωnmt V mi iω mi dt V nm e iω nit e iωmnt ) V mi 1 t 0 t 0 e iω nit e iωmnt ) dt e iω nit ) dt iω mi ) Now it looks just like the first order term with i h V ni ī V nm V ni h m E m E i So to second order w i n = π h V ni + m V nm V mi ρe E i E m n ) π h de n V ni + m V nm V mi ρe E i E m n )δe n E i ) The average is over all final states n with energy E n E i. 4

25 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 ψ a to final state ψ b is given by the absolute square of the amplitude c b t) = ī h t/ t/ ψb Ht ) ψ a e iω ab t dt where ω ab = Ea E b h. To apply the theory to a scattering process we imagine that the perturbation Ht ) turns on at t/ and off at t/ and while it is turned on it has constant value H. Then we can integrate and we get c b t) = ī 1 [ ] e iωabt/ e iω abt/ H ab = H ab h iω ab h and the transition probability is c b = 1 h H ab 4 sin ω ab t/ ωab = 1 h H ab Let s examine the ω ab dependent piece fω) ) sin ωt/ t ωt/ i sin ω ab t/ ω ab ) sin ωab t/ t ω ab t/ The first zero of fω) occurs when ω = π/t. Its maximum value at ω = 0) is t. In the limit of large t, fω) πtδω). To check that assertion we integrate over all ω fω)dω = sin x x dx t t = πt = In terms of the energies of initial and final states, πtδω) πt hδe b E a ) πtδω)dω and we can write The transition rate is c b = 1 h H ab πtδe b E z ) R = π h H ab δe b E a ) 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 = π h H ab δe b E a )ρe b )de b 5

26 ρe b ) 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 dkdω 8π 3 = V p dpdω π h) 3 Using E = p /m and pdp/m = de we have so dn de R = π h dn = V pmdedω π h) 3 = ρe) = V pm π h) 3 dω 1 π h) 3 V pm H ab dω Now how do we connect to the cross section? The incoming particle is represented by a plane wave ψ a = 1 e ika r V and the outgoing wave by ψ b = 1 V e ik b r The V in the denominator is so that the wave function is normalized. The particle density in the incoming wave is ψ a = 1/V and the flux of incoming particles is v V = p/m V. And dσ)flux = dn = R dσ = R Flux = V m π h ) ψ b H ψ a dω ) dσ V m dω = π h H ab That means that fθ) = mv π h ψ b H ψ a The negative sign is a convention. Suppose that H = V r). Then fθ) = mv π h = m π h 1 e i k b r V r) 1 e i k a r d 3 r V V e i k a k b ) r V r)d 3 r The result is equivalent to that of the Green s function analysis in the first Born approximation. See Griffiths p. 368) 6

27 4. Harmonic perturbation Now suppose that the perturbation is V t) = V e iωt + V e iωt We will get that t c n t) = ī h 0 = ī h V ni e iω ni+ω)t + V ni eiω ni ω)t ) V ni e iω ni+ω)t 1 iω ni + ω) + V ni e iωni ω)t ) 1 iω ni ω) If ω ni ω then the second term is the significant one. E n > E i and there is absorption. If E f < E i then ω ni < 0 and the first term counts. Then c n t) = 4 h V ni sin ω ni ± ω)t/ ω ni ± ω) = lim t 4π h V ni δω ni ± ω)t/ and just like for the constant perturbation w i n = π h V ni πδω ni ± ω) = π h V ni δe n E i ± hω) E n = E i ± hω 4..1 Interactions with radiation field Let V t) = e mc A p legit as long as A = 0 which is true for a plane wave in empty space. For that very same plane wave A = A 0 ɛ cosωt k x) = A 0 ɛ e iωt k x) + e iωt k x)) Consider absorption. Then we take the second term. We can expand the exponential for small k. Then A A 0 ɛ 1 + ik x +...) In the long wavelength limit, xk πx/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 E < Z e a 0 7

28 The corresponding wavelength is λ hc E = a 0 hc Z e = a 0 Z α a 0 The transition rate in the dipole approximation is w i n = π h e mc A 0 n ɛ p i δe n E i ± hω) The energy density of the radiation field is U = 1 gives us E max 8π ) + B max 8π = 1 ω π A c 0. Substitution In the last step we use w i n = π) h U e mω n ɛ p i δe n E i ± hω) = π) h ρω)dω e mω n ɛ p i δe n E i ± hω) = π) h = π) h = π) h e ρω) n ɛ p i mω e ρω) ɛ n p i mω e ρω) mω m ɛ n [H, x] i h = πe) h ρω) ɛ n x i 4.. Absorption vs emission n [H, x] i = h m n p i = hω ni n x i The rate that we just computed for absorption from a radiation field with energy density ρω) 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 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. 8

29 The rate w i n = π h ) e mc A 0 n ɛ pe ik r i δe n E i ± hω) We want to replace vector potential with energy density. We use Then w i n = π h U = 1 ω π c A 0 ) πc ) e ω U n ɛ pe ik r i δe n E i ± hω) mc U has some frequency dependence. The total energy density is ρω)dω = U We replace U with ρω)dω and integrate with the δ function. We get w i n = π ) πc ) e h ω ρω ni) n ɛ pe ik r i mc What do we mean by ρω)? 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 k x = πn x /L, etc. is N = πn x + n y + n z) 3/ = 1 ) 4 L π k ) 3/ π = π ) L 3 ω 6 π c )3 1 V dn dω = 1 ω π c 3 number of states per unit volume per unit frequency 9

30 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 The spontaneous transition rate becomes w i n = 4π πc h ω = 4 ω hc = 4 ω) ρω) = h π ω 3 h ω 3 π c 3 c 3 ) e mc ) n ɛ pe ik r i ) e mc α 1 mc m n ɛ pe ik r i ) n ɛ pe ik r i which sure enough is a rate. In the dipole approximation we get α 1 w i n = 4 ω) mc mω n ɛ x i m = 4 ω 3) α n ɛ x i c 4..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 ɛ. 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 ɛ x = 0. So project onto the x-axis. Then the polarization is in the y-z plane and the average of ɛ ẑ = 1 4π sin sin θdθ cos αdα = 1 4π 4 3 )π = 1 3. The total spontaneous decay rate into all angles and polarizations is w i n = 4 3 ω 3) α n x i c 4..5 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 f x i = l ± 1, m ± 1 x lm 30

31 will be a vector in the x-y plane. No z-component. That means that if the photon is headed in the z-direction ɛ 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 π cos θdθ = 1 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 which happens to be the same P θ) = cos θ + 1 sin θ = cos θ) d 1 1,1 + d 1 1, 1 = cos θ) ) cos θ) ) = cos θ) 4..6 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 σ abs = w i n hω = π) e cu c m ω n ɛ p i δe n E i ± hω) Again we replace p with m/ h [H, x] and get that σ abs = π) c ωe h n ɛ x i δe n E i ± hω) = π) ωe c h n ɛ x i δω ni ω) If we suppose the polarization is in the x direction then The total absorption cross section is σω)dω = 4π ω ni α n x i 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 ẑ) 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 = hk f 31

32 4.3.1 Golden rule The transition rate is given by the golden rule R = f H i πδe f E i hω) h The initial state is the ground state of hydrogen The final state is a free electron x i = 4πa 3 e r/a x f = 1 L 3/ eik f 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 Interaction Hamiltonian The interaction hamiltonian is H = q mc A p We write the vector potential as a plane wave, A = A 0 ɛ cosk r ωt) 17) = A 0 ɛ e ik r ωt) + e ik r ωt)) 18) where ɛ is the unit vector that gives the polarization of A and E = A t of the energy conserving δ-function we are left with After application A = A 0 ɛe ik r ωt) Then H = i hq mc A 0e ik r ɛ 3

33 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 and Imagine a sphere of radius the volume of the sphere, k x = n xπ L, k y = n yπ L, k z = n zπ L ) π kf = n x + n y + n L z) n x + n y + n z. The total number of states with k < k f is N = 4 3 πn x + n y + n z) 3/ = 4 3 π L π k ) ) 3/ The number of states per unit k into solid angle dω are dn dk = 4 ) L 3 3 π 3k dω π 4π The number of states per unit energy into solid angle dω are where we have use E = h k m Transition rate The total transition rate is R = f H i h = f H i = h h qa0 mc ) dn L 3 de = k dk π de ) L 3 km = π π dω 4π 19) dω 0) h δe f E i hω) dn de ) L 3 km π π h ) f e ik r ɛ i 33 1) dω ) ) L 3 km π π dω 3) h

34 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 8π E max = 1 ω π c A 0 and since each photon has energy hω, the photon flux is Then dσ dω = h Flux = 1 π hωc ω A 0 = 1 π hc ωa 0 q A 0 mc) ) 3 f e ik r ɛ i π L km π h ωa 0 /π hc) 4) = m cω πq h) f e ik r ɛ i = q Expectation value f e ik r ɛ i πmcω L 3 k π L π ) 3 km h 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. where We have that f e ik r ɛ i = 1 L 3/ 1 π 1/ a 3/ I I = e ik f r e ik r ɛ e r/a r drdω 5) 6) Integrating by parts we get that I = ɛ ɛ ɛ e ik f r e ik r e r/a) r drdω 7) ) e ik f r e ik r e r/a r drdω 8) e ik f r e ik r) e r/a r drdω 9) 34

35 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 iɛ k e ik f r e ik r e r/a r drdω = 0 since polarization ɛ and propagation vector k are orthogonal. Finally ) I = ɛ e ik f r e ik r e r/a r drdω 30) = iɛ k f e ik f r e ik r e r/a r drdω 31) = iɛ k f e ik f k) r e r/a r drdω 3) = iɛ k f e iq r e r/a r drdω 33) = iɛ k f e iq cos θr e r/a r dr sin θdφ 34) where q = k f k. First we do the angular integral Then integrate over r. e iqr e iqr ) I = iπɛ k f e r/a r dr 35) iqr Finally e iqr r/a e iqr r/a ) I = πɛ k f rdr 36) iq ) 1 = πɛ k f q iq 1/a) 1 qiq 1/a) 37) ) 1 = πɛ k f q q + 1/a iq/a) 1 q q + 1/a 38) + iq/a) ) 4iq/a = πɛ k f q q + 1/a ) + 4q/a) ) 39) 4i = πɛ k f aq + 1/a) 40) ) f e ik r ɛ i = 1 1 L 3/ π 1/ a 3/ πɛ k 4i f aq + 1/a) ) 41) = 8iπ 1/ ɛ k f L 3/ a 3/ aq + 1/a) ) 4) 43) 35

36 and dσ dω = α hl3 k f 8π 1/ ) ɛ k f πmω L 3/ a 5/ q + 1/a) ) 44) = 3α hk f mω ɛ k f ) a 5 q + 1/a) ) 4 45) Angular distribution The E-field vector k is in the z-direction and the electron in the final state has k f in the θ, φ direction. Then q = k f + k k f k cos θ. And the polarization is in the x-direction so that ɛ k f = k f sin θ cos φ and dσ dω == 3α hk f k f sin θ cos φ) mω a 5 kf + k k f k cos θ + 1/a) ) 4 The differential cross section is peaked for the outgoing electron in the θ = π/, φ = 0 direction, namely parallel to the polarization of the field. 36