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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 ∆E1 = e2 E 2 ∞ X | hm, 1, 0 | z | 1, 0, 0i |2 . E − E 1 m m=2 To simplify the notation, lets call |1, 0, 0i = |0i, the ground state with energy E0 and call |m, 1, 0i = |ni with energy En and n ≥ 1. So, we want to compute ∞ X | hn | z | 0i |2 ∆E0 = e E . E − E 0 n n=1 2 2 (a) Suppose we had an operator A such that z |0i = (AH0 − H0 A) |0i , where H0 is the unperturbed Hamiltonian for the hydrogen atom, e2 p2 − , 2m r where m is the reduced mass of the electron and proton. Show that H0 = hn | z | 0i = (E0 − En ) hn | A | 0i . Also show that ∞ X | hn | z | 0i |2 = h0 | zA | 0i − h0 | z | 0i h0 | A | 0i = h0 | zA | 0i . E0 − En n=1 Solution hn | z | 0i = hn | AH0 | 0i − hn | H0 A | 0i = (E0 − En ) hn | A | 0i , Since H0 is Hermitian. Also, ∞ X h0 | z | ni (E0 − En ) hn | A | 0i X | hn | z | 0i |2 = E0 − En E0 − En n=1 n6=1 X = h0 | z | ni hn | A | 0i n6=1 = X n h0 | z | ni hn | A | 0i − h0 | z | 0i h0 | A | 0i = h0 | zA | 0i − h0 | z | 0i h0 | A | 0i = h0 | zA | 0i , c 2012, Edward J. Groth Copyright (1) Physics 505 Homework No. 9 Solutions S9-2 where the second term has been dropped, since h0 | z | 0i = 0. 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. 2 h̄ Show that this expression does, in fact, solve equation (1). (Note that the normalization of |0i cancels out, so you can just take |0i = exp(−r/a).) Solution We need to calculate 2 h̄2 h̄2 e2 ∂ 2 ∂ ∂ 1 ∂ − − − − + sin θ 2m ∂r 2 r ∂r 2mr 2 sin θ ∂θ ∂θ r 2 ma r −r/a . + ar cos θ e − 2 2 h̄ 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 2 a ∂2 a ∂ 1 r2 r −r/a −r/a − cos θ = − cos θ (r + a) − + ar e + ar e 2 ∂r 2 2 2 ∂r a 2 2 1 r2 a 1 − (r + a) + 2 + ar e−r/a = − cos θ 2 a a 2 r2 a r e−r/a . + − = cos θ 2 2 4a The first derivative term is 2 a ∂ r a 1 r2 −r/a − cos θ = − cos θ (r + a) − + ar e + ar e−r/a r ∂r 2 r a 2 r a2 e−r/a . − = cos θ 2 r The angular derivative term is 2 2 a r 1 ∂ ∂ a r −r/a − 2 cos θ sin θ + ar e = 2 cos θ + ar e−r/a 2r sin θ ∂θ ∂θ 2 r 2 a a2 e−r/a . + = cos θ 2 r c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Solutions S9-3 The potential term is e2 − r ma h̄2 r r2 −r/a = cos θ − − a e−r/a . + ar cos θ e 2 2 The first term on the right of equation (1) is considerably easier to evaluate, 2 e AH0 |0i = A − e−r/a 2a 2 ma r 2 e −r/a − 2 + ar cos θ e = − 2a 2 h̄ 2 r r e−r/a . + = cos θ 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 (AH0 − H0 A) |0i = cos θ + − + − + + − −a+ + 2 2 4a 2 r 2 r 2 4a 2 = cos θ re−r/a e−r/a = z |0i , 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 h0 | zA | 0i. Here, the normalization of |0i matters, so we use |0i = 2(a)−3/2 (4π)−1/2 exp(−r/a). Also, zA is proportional to z 2 . As far as evaluating the matrix element, we can use symmetry to replace z 2 by r 2 /3. So, 2 e h0 | zA | 0i = − 9 = − a3 . 4 Z 0 ∞ r2 4 +a e−2r/a r 2 dr 3 2 3 a r Finally, we need to include E 2 in the energy shift, so 9 ∆E1 = − a3 E 2 . 4 End Solution c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Solutions S9-4 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 |nmi = 2 sin(nπx/L) sin(mπy/L) , L where n and m are integers greater than 0. The energy is Enm h̄2 π 2 2 2 = n + m . 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 Z L Z 2 n1 πx n2 πx 2L π x sin sin dx = 2 sin n1 x sin n2 x x dx L L L π 0 0 Z L π (cos((n1 − n2 )x) − cos((n1 + n2 )x)) x dx = 2 π 0 The integrals can be evaluated with an integration by parts. If n1 6= n2 , the result is Z 0 L n1 πx n2 πx L 2 sin dx = 2 x sin L L L π (−1)n1 −n2 − 1 (−1)n1 +n2 − 1 − (n1 − n2 )2 (n1 + n2 )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 Z L 2 n1 πx n2 πx 8L n1 n2 x sin sin dx = − 2 2 . L L L π (n1 − n22 )2 0 c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Solutions S9-5 If n1 = n2 , the result is L/2. So, the ground state changes energy by ∆E11 = h11 | Cxy | 11i = CL2 /4 , 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: 1/4 256/81π 4 h21 | V | 21i h21 | V | 12i 2 . = CL 256/81π 4 1/4 h12 | V | 21i h12 | V | 12i The eigenvalues of this matrix are ∆Efirst excited = CL with eigenvectors 2 256 1 ± 4 81π 4 , 1 |first excited± i = √ (|21i ± |12i) . 2 End Solution 3. Hyperfine splitting of the hydrogen ground state. As you know, the spatial part of the hydrogen ground state is very simple: ψ100 (r, θ, φ) = exp(−r/a) a(−3/2) π (−1/2) . Since there is no orbital angular momentum, there is no spin orbit effect. 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 I, µp = 2mp c 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, B(r) = 3er (er · µp ) − µp 8π + µp δ(r) , r3 3 (Jackson, Classical Electrodynamics, 2nd ed., p. 184). The interaction energy of this field and the magnetic moment of the electron is HHF = −µe · B = − 3(er · µe )(er · µp ) − µe · µp 8π − (µe · µp )δ(r) . 3 r 3 c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Solutions S9-6 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 r 2 is 3xi µei xj µpj − r 2 µe · µp . When we take the expectation value with an s-state, the angular integral will eliminate cross terms xi xj with i 6= j. The squared terms averaged over angles become r 2 /3. This leaves 3µe · µp r 2 /3 − r 2 µe · µp = 0. 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, h100 | −8πµe · µp /3 | 100i = −8µe · µp /3a3 . The product of the magnetic moments is −ege egp S ·I 2me c 2mp c ge gp 1 2 2 2 = − 2 µB µN , F −S −I 2 h̄ µe · µp = where µB and µN are the Bohr and nuclear magnetons, and F is the total spin. S 2 = I 2 = 3h̄2 /4. For the singlet state F = 0, F 2 = 0 and the quantity in brackets is −3h̄2 /4. For a triplet state, F = 1, F 2 = 2h̄2 . and the quantity in brackets is +h̄2 /4. Putting everything together, we wind up with 2ge gp µB µN 3a3 2ge gp µB µN hHHF s i = − a3 8ge gp µB µN hHHF t i − hHHF s i = + . 3a3 hHHF t i = + We see that the singlet is the ground state. c 2012, Edward J. Groth Copyright Physics 505 µN Homework No. 9 Solutions S9-7 For the numerical evaluation, we use ge = 2, gp = 5.59, µB = 0.927 × 10−20 erg G−1 , = 0.505 × 10−23 erg G−1 , a = 0.529 × 10−8 cm. We find, ∆HHF = 9.43 × 10−18 erg = 5.89 × 10−6 eV νHF = 1.42 × 109 Hz 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 P2 + V (r) , 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 Hso = L · S = f (r) L · S . 2 2 2m c r dr The interaction with the applied magnetic field is H0 = HB = eB eB eB (Lz + gSz ) = (Lz + 2Sz ) = (Jz + Sz ) , 2mc 2mc 2mc where the term proportional to B 2 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: |njmj lsi. These states diagonalize the spin-orbit interaction which is assumed to be larger than the Zeeman interaction. h̄2 l j = l + 1/2 hHso i = hf (r)inl × −(l + 1) j = l − 1/2 2 c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Solutions S9-8 Using these basis states we need to calculate the matrix elements of the Zeeman term, eB (Jz + Sz ) njmj ls . hHB i = njmj ls 2mc The matrix element of Jz is just mj h̄. The matrix element of Sz requires more work. In particular, the states |njmj lsi must be written in terms of |nlml i |sms i. Recall the expansions given in lecture, r l + mj + 1/2 l − mj + 1/2 |lmj − 1/2i |↑i + |lmj + 1/2i |↓i |j = l + 1/2, mj , li = + 2l + 1 2l + 1 r r l − mj + 1/2 l + mj + 1/2 |j = l − 1/2, mj , li = − |lmj − 1/2i |↑i + |lmj + 1/2i |↓i . 2l + 1 2l + 1 r With these expansions given in lecture, we find h̄mj hJz + Sz i = × 2l + 1 2l + 2 2l j = l + 1/2 . j = l − 1/2 Both cases are covered by hJz + Sz i = h̄mj 2j + 1 = h̄mj g , 2l + 1 where the g-factor varies from 2 for l = 0 to 1 for l → ∞. Then the Zeeman term is hHB i = gmj µB 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 |lml i |ms i and hHB i = eB eB hLz + 2Sz i = h̄(ml + 2ms ) = (ml + 2ms )µB B . 2mc 2mc In this basis, the spin-orbit term is easy to evaluate, hL · Si = hLx Sx + Ly Sy + Lz Sz i = hLz Sz i = h̄2 ml ms . c 2012, Edward J. Groth Copyright Physics 505 Homework No. 9 Then Solutions S9-9 hHso i = ml ms h̄2 hf (r)inl . Note that in this limit there are some degeneracies. For example, if l = 1, then states ml = −1, ms = +1/2 and ml = +1, ms = −1/2 are degenerate. 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 spinorbit 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 [H, x · p] = −ih̄ p2 − x · ∇V (x) . m Solution 2 We consider p first: [p2 , x · p] = pj pj xi pi − xi pi p2 = pj (xi pj − ih̄δij )pi − xi pi p2 = pjxi pj pi − ih̄p2 − xi pi p2 = (xi pj − ih̄δij )pj pi − h̄p2 − xi pi p2 = xi pi p2 − 2ih̄p2 − xi pi p2 = −2ih̄p2 . Now V (x): [V (x), xipi ] = V (x)xi pi − xi pi V (x) h̄ ∂ V (x) + V (x)pi = V (x)xi pi − xi i ∂xi = +ih̄x · ∇V (x) . So, [H, x · p] = −ih̄ p2 − x · ∇V (x) m c 2012, Edward J. Groth Copyright . Physics 505 Homework No. 9 Solutions End Solution (b) If |ψi is a stationary state of H, H |ψi = E |ψi, show that 2 p ψ − hψ | x · ∇V (x) | ψi = 0 , ψ m and therefore, for the Coulomb potential, 2 hψ | H | ψi + ψ Ze2 r ψ =0. Solution hψ | [H, x · p] | ψi = hψ | Hx · p − x · pH | ψi = hψ | Ex · p − x · pE | ψi = E hψ | x · p − x · p | ψi =0. With the Coulomb potential, we have Ze2 Ze2 x Ze2 x·∇ − =x· + 3 =+ . r r r So, p2 ψ − hψ | x · ∇V (x) | ψi m Ze2 p2 ψ ψ − ψ m r 2 2 p ψ − 2 ψ Ze ψ + ψ 2m r Ze2 ψ . = 2 hψ | H | ψi + ψ r 0 = ψ = ψ =2 ψ End Solution (c) Determine h1/rinl for the hydrogen atom. Solution From the above, we have 1 En = − 2 and we know, En = −Z 2 e2 /2an2 , so Ze2 r , 1 Z = . r nl an2 End Solution c 2012, Edward J. Groth Copyright Ze2 r ψ S9-10