Physics 505 Homework No. 9 Solutions S9-1 1. As Promised, Here

Physics 505 Homework No. 9 Solutions S9-1

1. As promised, here is the trick for summing the matrix elements for the Stark eﬀect for the ground state of the hydrogen atom.

Recall, we need to calculate the correction to the ground state energy to second order in the perturbation due to an external ﬁeld. This correction is ∞ m, 1, 0 z 1, 0, 0 2 ∆E = e2E2 |h | | i| . 1 E E mX=2 1 − m

(a) Suppose we had an operator A such that z 0 =(AH H A) 0 , (1) | i 0 − 0 | i where H0 is the unperturbed Hamiltonian for the hydrogen atom, p2 e2 H = , 0 2m − r where m is the reduced mass of the electron and proton. Show that n z 0 =(E E ) n A 0 . h | | i 0 − n h | | i Also show that ∞ n z 0 2 |h | | i| = 0 zA 0 0 z 0 0 A 0 = 0 zA 0 . E E h | | i−h | | ih | | i h | | i nX=1 0 − n Solution

n z 0 = n AH 0 n H A 0 =(E E ) n A 0 , h | | i h | 0 | i−h | 0 | i 0 − n h | | i Since H0 is Hermitian. Also, ∞ n z 0 2 0 z n (E E ) n A 0 |h | | i| = h | | i 0 − n h | | i E0 En E0 En nX=1 − nX=16 − = 0 z n n A 0 h | | ih | | i nX=16 = 0 z n n A 0 0 z 0 0 A 0 h | | ih | | i−h | | ih | | i Xn = 0 zA 0 0 z 0 0 A 0 h | | i−h | | ih | | i = 0 zA 0 , h | | i

where the second term has been dropped, since 0 z 0 = 0. h | | i End Solution

(b) So, if we knew A, we could get the answer just by calculating one matrix element. If we assume A is a function only of coordinates, then equation (1) is an inhomogeneous diﬀerential equation for A. If you’re really good at diﬀerential equations, you could solve it. The result is ma r A = 2 + a z. − ¯h 2 Show that this expression does, in fact, solve equation (1). (Note that the normalization of 0 cancels out, so you can just take 0 = exp( r/a).) | i | i − Solution We need to calculate

¯h2 ∂2 2 ∂ ¯h2 1 ∂ ∂ e2 + sin θ − − 2m ∂r2 r ∂r − 2mr2 sin θ ∂θ ∂θ − r ma r2 + ar cos θ e r/a . − ¯h2 2 −

This is the second term on the right hand side of equation (1). There may be some ways to simplify this, but in the end, it appears brute force is required, so, we’ll just evaluate each term. First of all, the second derivative term is

a ∂2 r2 a ∂ 1 r2 cos θ + ar e r/a = cos θ (r + a) + ar e r/a − 2 ∂r2 2 − − 2 ∂r − a 2 − a 2 1 r2 = cos θ 1 (r + a) + + ar e r/a − 2 − a a2 2 − a r r2 = cos θ + e r/a . 2 2 − 4a −

The ﬁrst derivative term is a ∂ r2 a 1 r2 cos θ + ar e r/a = cos θ (r + a) + ar e r/a − r ∂r 2 − − r − a 2 − r a2 = cos θ e r/a . 2 − r −

The angular derivative term is

a 1 ∂ ∂ r2 a r2 sin θ cos θ + ar e r/a = cos θ + ar e r/a −2r2 sin θ ∂θ ∂θ 2 − r2 2 − a a2 = cos θ + e r/a . 2 r −

The potential term is

2 2 e ma r r/a r r/a 2 + ar cos θ e− = cos θ a e− . − r ¯h 2 −2 −

The ﬁrst term on the right of equation (1) is considerably easier to evaluate,

e2 AH 0 = A e r/a 0 | i − 2a − e2 ma r2 = + ar cos θ e r/a −2a − ¯h2 2 − r2 r = cos θ + e r/a . 4a 2 −

We now add up the results of the last 5 calculations to ﬁnd

a r r2 r a2 a a2 r r2 r (AH H A) 0 = cos θ + + + + a + + e r/a 0 − 0 | i 2 2 − 4a 2 − r 2 r − 2 − 4a 2 − r/a = cos θ re− = z 0 , | i which is what was to be shown! End Solution

(c) Calculate the Stark eﬀect energy shift for the ground state of hydrogen to second order in the applied ﬁeld. Solution We need to evaluate 0 zA 0 . Here, the normalization of 0 matters, so we use 0 = 2(a)−3/2(4π)−1/2 exp(h |r/a).| i Also, zA is proportional to z2|.i As far as evaluating the| i matrix element, we can use− symmetry to replace z2 by r2/3. So,

∞ 2 2 r r 4 2r/a 2 e 0 zA 0 = + a 3 e− r dr h | | i − Z0 2 3 a 9 = a3 . −4

Finally, we need to include E2 in the energy shift, so

9 ∆E = a3E2 . 1 −4

End Solution

2. A particle in a 2D box. (Based on a problem from Merzbacher.) A particle is conﬁned to a square box, 0 x L and 0 y L. We are not interested in the z-motion, so this is a 2D problem. ≤ ≤ ≤ ≤

(a) Obtain the energies and eigenfunctions. What is the degeneracy of the few lowest levels? Solution The wave function must vanish at the boundaries of the box. This means 2 nm = sin(nπx/L) sin(mπy/L) , | i L where n and m are integers greater than 0. The energy is

¯h2π2 E = n2 + m2 . nm 2mL2 The degeneracy has to do with how many ways one can choose n and m to give the same n2 + m2. The ground state is non-degenerate with n = m = 1. The ﬁrst excited state has a degeneracy of 2 with n = 2, m = 1 or n = 1, m = 2. The next excited state is non-degenerate with n = m = 2. The third excited state is doubly degenerate with n =3, m =1 or n =1, m = 3. That’s enough! End Solution

(b) A small perturbation V = Cxy, where C is a constant, is applied. Find the energy change for the ground state and the ﬁrst excited state in the lowest non-vanishing order. Construct the appropriate eigenfunctions in the case of the ﬁrst excited state. Solution We calculate

L π 2 n1πx n2πx 2L x sin sin dx = 2 sin n1x sin n2xxdx Z0 L L L π Z0 L π = 2 (cos((n1 n2)x) cos((n1 + n2)x)) xdx π Z0 − −

The integrals can be evaluated with an integration by parts. If n = n , the result is 1 6 2 L 2 n πx n πx L ( 1)n1−n2 1 ( 1)n1+n2 1 x sin 1 sin 2 dx = − − − − . Z L L L π2 (n n )2 − (n + n )2 0 1 − 2 1 2

If n1 and n2 are both odd or both even, the result is 0. If one is odd and the other is even, the result is L 2 n πx n πx 8L n n x sin 1 sin 2 dx = 1 2 . Z L L L − π2 (n2 n2)2 0 1 − 2

If n1 = n2, the result is L/2.

So, the ground state changes energy by

∆E = 11 Cxy 11 = CL2/4 , 11 h | | i since the expectation value is the product of two of the integrals just discussed with n1 = n2 = m1 = m2 = 1.

The first excited state is degenerate, so we need to choose a basis which diagonalizes the perturbation. We calculate the matrix elements for all the states: 21 V 21 21 V 12 1/4 256/81π4 h | | i h | | i = CL2 . 12 V 21 12 V 12 256/81π4 1/4 h | | i h | | i The eigenvalues of this matrix are 1 256 ∆E = CL2 , first excited 4 ± 81π4 with eigenvectors 1 first excited± = ( 21 12 ) . | i √2 | i±| i End Solution

3. Hyperfine splitting of the hydrogen ground state. As you know, the spatial part of (−3/2) (−1/2) the hydrogen ground state is very simple: ψ100(r, θ, φ) = exp( r/a) a π . Since there is no orbital angular momentum, there is no spin orbit e−ffect. However, the ground state has a degeneracy of 4 since both the proton and the electron have spin 1/2. The spins can align, giving a triplet state, or anti-align, giving a singlet state. Since there are magnets associated with the spins, we expect that there should be a difference in energy between the triplet and singlet states. The nuclear spin is often denoted by I and produces a magnetic moment egp µp = I , 2mpc

where µp is the magnetic moment of the proton, gp is its g-factor and mp is the mass of the proton. (Note: to consider other nuclei, we would use the appropriate Z, g, and m.) We take the proton as fixed at the origin and it produces a magnetic field, 3e (e µ ) µ 8π B(r) = r r · p − p + µ δ(r) , r3 3 p (Jackson, Classical Electrodynamics, 2nd ed., p. 184). The interaction energy of this field and the magnetic moment of the electron is 3(e µ )(e µ ) µ µ 8π H = µ B = r · e r · p − e · p (µ µ )δ(r) . HF − e · − r3 − 3 e · p

Aside: If we were to consider other than s-states, the hyperﬁne Hamiltonian would also include a spin orbit term due to the interaction of the magnetic moment of the nucleus with magnetic ﬁeld produced by the moving electron(s).

Evaluate the hyperfine Hamiltonian above for the ground state of hydrogen. How does it depend on the proton and electron spins? Or, what is the energy difference between the singlet and triplet states? Which is the actual ground state: triplet or singlet? What are the wavelength and frequency of the radiation emitted or absorbed in the transition between these states? Hint: can you show that the first term in HHF vanishes for s-states? Solution The first term in the hyperfine Hamiltonian is zero by the following argument. The numerator times r2 is 3x µ x µ r2µ µ . When we take the expectation value with i ei j pj − e · p an s-state, the angular integral will eliminate cross terms xixj with i = j. The squared terms averaged over angles become r2/3. This leaves 3µ µ r2/3 r26µ µ = 0. e · p − e · p The δ-function in the second term is a three dimensional delta function and it just picks out the value of the integrand at the origin. So,

100 8πµ µ /3 100 = 8µ µ /3a3 . h | − e · p | i − e · p The product of the magnetic moments is

ege egp µe µp = − S I · 2mec 2mpc ·

gegp 1 2 2 2 = µBµN F S I , − ¯h2 2 − − 2 2 where µB and µN are the Bohr and nuclear magnetons, and F is the total spin. S = I = 3¯h2/4. For the singlet state F = 0, F 2 = 0 and the quantity in brackets is 3¯h2/4. For a triplet state, F = 1, F 2 = 2¯h2. and the quantity in brackets is +¯h2/4. −

Putting everything together, we wind up with

2g g µ µ H = + e p B N h HF ti 3a3 2g g µ µ H = e p B N h HFsi − a3 8g g µ µ H H = + e p B N . h HF ti−h HFsi 3a3

We see that the singlet is the ground state.

−20 −1 For the numerical evaluation, we use ge = 2, gp =5.59, µB =0.927 10 erg G , µ =0.505 10−23 erg G−1, a =0.529 10−8 cm. We ﬁnd, × N × × ∆H =9.43 10−18 erg = 5.89 10−6 eV HF × × ν =1.42 109 Hz HF × λHF = 21 cm . This transition is the famous 21 cm line of neutral Hydrogen which is seen all over the sky. It’s one of the principal ways to study our galaxy and other galaxies with radio telescopes! End Solution

4. Zeeman splitting. We consider an atom with a single valence electron, subject to a magnetic ﬁeld B = Bez in the z-direction. The Hamiltonian for the electron is

H = H0 + Hso + HB , where P 2 H = + V (r) , 0 2m accounts for the dominant electric interaction of the electron (for Hydrogen, V (r) = e2/r, for alkali metals, V (r) takes account of the ﬁlled shells in an approximate way). The spin orbit interaction is 1 1 dV H = L S = f(r) L S . so 2m2c2 r dr · · The interaction with the applied magnetic ﬁeld is eB eB eB H = (L + gS ) = (L +2S ) = (J + S ) , B 2mc z z 2mc z z 2mc z z where the term proportional to B2 has been dropped. Also, some other small terms, for example, the relativistic correction to the momentum, have been dropped since they don’t give a splitting dependent on j, l and s. In the calculations below, we are interested in Hso and HB; H0 determines the zeroth order energies and states which are used in computing expectation values of, for example, f(r), but can otherwise be ignored.

(a) Suppose the magnetic field is very weak. What are the appropriate basis states and what are the spin-orbit and Zeeman splittings? Solution If the magnetic field is completely turned off, there is only the spin-orbit interaction. The appropriate states are those of total angular momentum, z-component of angular momentum, and orbital and spin angular momentum: njmjls . These states diagonalize the spin-orbit interaction which is assumed to be larger| than thei Zeeman interaction.

¯h2 l j = l +1/2 H = f(r) h soi h inl 2 × (l + 1) j = l 1/2 − −

Using these basis states we need to calculate the matrix elements of the Zeeman term,

eB H = njm ls (J + S ) njm ls . h Bi j 2mc z z j

The matrix element of Jz is just mj¯h. The matrix element of Sz requires more work. In particular, the states njmjls must be written in terms of nlml sms . Recall the expansions given in lecture,| i | i| i

l + m +1/2 l m +1/2 j = l +1/2, m ,l = + j lm 1/2 + − j lm +1/2 | j i r 2l +1 | j − i|↑i r 2l +1 | j i|↓i l m +1/2 l + m +1/2 j = l 1/2, m ,l = − j lm 1/2 + j lm +1/2 . | − j i −r 2l +1 | j − i|↑i r 2l +1 | j i|↓i

With these expansions given in lecture, we ﬁnd

¯hmj 2l +2 j = l +1/2 J + S = . h z zi 2l +1 × 2l j = l 1/2 −

Both cases are covered by

2j +1 J + S =¯hm =¯hm g, h z zi j 2l +1 j

where the g-factor varies from 2 for l =0 to 1 for l . Then the Zeeman term is →∞ H = gm µ B. h Bi j B Each spin-orbit level is split into 2j + 1 equi-spaced levels by the Zeeman eﬀect. End Solution

(b) Now suppose the magnetic ﬁeld is very strong so the Zeeman term is larger than the spin-orbit term. What are the appropriate states and what are the Zeeman and spin-orbit splittings? Solution The Zeeman term is diagonal in the basis lm m and | li| si eB eB H = L +2S = ¯h(m +2m )=(m +2m )µ B. h Bi 2mc h z zi 2mc l s l s B In this basis, the spin-orbit term is easy to evaluate,

L S = L S + L S + L S = L S =¯h2m m . h · i h x x y y z zi h z zi l s

Then H = m m ¯h2 f(r) . h soi l s h inl Note that in this limit there are some degeneracies. For example, if l = 1, then states m = 1, m = +1/2 and m = +1, m = 1/2 are degenerate. l − s l s − End Solution

(c) Suppose that neither the spin-orbit nor the Zeeman eﬀect is appreciably larger than the other. How would you determine the level splittings in this case? (This is a short essay question, no calculations are required!) Solution Pick a convenient basis for the levels that are degenerate in the absence of the spin- orbit and Zeeman interactions. Calculate the matrix elements in this basis. The eigenvalues of this matrix are the energy shifts in the states that correspond to the eigenvalues. End Solution

5. Virial theorem for a particle in a ﬁxed potential. (See Schwabl, chapter 12.) Consider x p and a Hamiltonian H = p2/2m + V (x). · (a) Show that p2 [H, x p] = i¯h x V (x) . · − m − · ∇ Solution We consider p2 ﬁrst:

[p2, x p] = p p x p x p p2 · j j i i − i i = p (x p i¯hδ )p x p p2 j i j − ij i − i i = pjx p p i¯hp2 x p p2 i j i − − i i =(x p i¯hδ )p p ¯hp2 x p p2 i j − ij j i − − i i = x p p2 2i¯hp2 x p p2 i i − − i i = 2i¯hp2 . − Now V (x): [V (x),x p ] = V (x)x p x p V (x) i i i i − i i ¯h ∂ = V (x)xipi xi V (x) + V (x)pi − i ∂xi = +i¯hx V (x) . · ∇ So, p2 [H, x p] = i¯h x V (x) . · − m − · ∇

End Solution

(b) If ψ is a stationary state of H, H ψ = E ψ , show that | i | i | i p2 ψ ψ ψ x V (x) ψ =0 , m −h | · ∇ | i

and therefore, for the Coulomb potential, Ze2 2 ψ H ψ + ψ ψ =0 . h | | i r

Solution

ψ [H, x p] ψ = ψ Hx p x pH ψ h | · | i h | · − · | i = ψ Ex p x pE ψ h | · − · | i = E ψ x p x p ψ h | · − · | i =0 . With the Coulomb potential, we have Ze2 Ze2x Ze2 x = x + = + . · ∇ − r · r3 r So, p2 0 = ψ ψ ψ x V (x) ψ m −h | · ∇ | i

p2 Ze2 = ψ ψ ψ ψ m − r

p 2 Ze 2 Ze2 =2 ψ ψ 2 ψ ψ + ψ ψ 2m − r r

Ze2 =2 ψ H ψ + ψ ψ . h | | i r

End Solution

(c) Determine 1/r for the hydrogen atom. h inl Solution From the above, we have 1 Ze2 E = , n −2 r and we know, E = Z2e2/2an2, so n − 1 Z = 2 . r nl an End Solution