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 To simplify the notation, lets call 1, 0, 0 = 0 , the ground state with energy E0 and call m, 1, 0 = n with energy E and|n 1.i So,| i we want to compute | i | i n ≥ ∞ n z 0 2 ∆E = e2E2 |h | | i| . 0 E E nX=1 0 − n (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 Copyright c 2012, Edward J. Groth Physics 505 Homework No. 9 Solutions S9-2 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 normaliza- tion 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 − Copyright c 2012, Edward J. Groth Physics 505 Homework No. 9 Solutions S9-3 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 Copyright c 2012, Edward J. Groth 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 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 Copyright c 2012, Edward J. Groth Physics 505 Homework No. 9 Solutions S9-5 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.