216 - PH 216 Final 2011 Instructions The are three...

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Unformatted text preview: PH 216, Final, 2011 Instructions: The are three questions worth 100 points each. Feel free to consult 2 sides each of two sheets of notes. 1. Calculate the first-order relativistic correction to all of the energy levels (enumerated by n) of the simple harmonic oscillator of classical frequency (i.e. first order in /mc2 ). Start out by finding the lowest-order relativistic correction to the kinetic energy in terms of the momentum operator (the zeroth order term being p2 /2m) and then use perturbation theory based on the solution to the 1D Schrodinger equation with potential energy V (x) = 1 m 2 x2 . 2 Solution: The kinetic energy is E - E0 , where E0 = mc2 is the energy at rest. Thus T = p2 c2 + m2 c4 - mc2 = mc2 (1 + p2 /m2 c2 )1/2 - mc2 . Then, since 1 1+1+ - 2 4 1 2 p p2 T - mc = 2m 8 m 4 c4 H = - 1 2 , 8 p2 p4 - . 2m 8m3 c2 we have Thus the first correction to the classical Hamiltonian is p4 . 8m3 c2 The first-order correction to the Hamiltonian will be given in first-order timeindependent perturbation theory by 1 En = n |H n . For the SHO, we can write m a -a 2 in terms of creation and annihilation operators, and then p=i (a - a)2 = (a )2 - a a - aa + a2 . Now, thinking ahead a bit, we want to square this but we will be evaluating it between n | and |n , so only terms with the same number of creation and annihilation operators will give a nonzero result. Keeping just these terms, we have: (a - a)4 = (a )2 a2 + a aaa + aa a a + aa aa + a aa a + a2 (a )2 . 1 Now, we have [a , a] = 1, a |n = n + 1|n+1 , a|n = n|n-1 , and their conjugate equations, and we can use these to obtain: n |(a )2 a2 |n = nn-1 |a a|n-1 = n(n - 1) n |a2 (a )2 |n = (n + 1)n+1 |aa |n+1 = (n + 1)(n + 2) n |a(a )2 a|n = n(n + 1)n+1 |(a )2 |n-1 = n(n + 1) n |a a2 a |n = n(n + 1)n-1 |(a2 |n+1 = n(n + 1). n |a aa a|n = n2 , n |a aa a|n = (n + 1)2 2 2 and similarly Thus n |p |n = = Thus 4 m 2 m 2 n(n - 1) + (n + 1)(n + 2) + n2 + (n + 1)2 + 2n(n + 1) (6n2 + 6n + 3) 1 En = - 1 2 m 2 2 3 ()2 (6n2 + 6n + 3) = - (2n2 + 2n + 1). 8m3 c2 4 32 mc2 2. Calculate the Born approximation to the differential and total cross sections for scattering a partical of mass m off the -function potential V (r) = g 3 (). r Solution: In the Born approximation, m f () = - 22 r r e-ik V (r )eik d r where and are, respectively, the wave vectors of the incident and scattered waves. k k Let = - Then q k k. m m mg qr f () = - e-i (r )d = - r exp(-i 0 q) = . 2 2 2 2 22 (Note that I used here the regular 3D -function. You have to be careful if you use the scattering formulation appropriate to spherical symmetry, since going from 3 () in cartesian to spherical coordinates has a Jacobian factor in it: 3 ( - ) = r r r 1 (x - x )(y - y )(z - z ) = r2 (r - r )(cos - cos )( - ), so if you integrate a spherically symmetry function with value V0 at the origin, you do not get zero, but rather V0 . This contracted my statement that there were no hidden subtleties in the problem, always a foolhardy one to make in physics...) 2 Given this, the differential cross section is d m2 g 2 = |f ()|2 = 2 4 , d 4 and since this is angle-independent, we have = 4 = m2 g 2 . 4 3. A hydrogen atom (with spinless electron and proton, for our purposes) in its ground state is subjected to a uniform weak electric field E = E0 e-t (t), where is real and (t) is the step function. Find the probability for the atom to be in any of the n = 2 states after a long time. Some hydrogenic wave functions in spherical coordinates are: 1 1 r 100 = 3 e-r/a0 , 210 = e-r/2a0 cos , a0 a0 32a3 0 1 r 1 r = 1- e-r/2a0 , 211 = e-r/2a0 sin ei . 2a0 a0 8a3 64a3 0 0 200 A useful integral is: xn e-ax dx = n! 0 an+1 , 0 and a useful thing to remember is that as a tensor operator, z = T1 , where the bottom number is the total, and the top azimuthal, angular momentum. Solution: Take the direction of the electric field as the z direction. Then the perturbation Hamiltonian is H = e E(t) = ezE(t). r SInce z is parity-odd, the non-vanishing matrix elements of H are those between the states of opposite parities. Thus P (1s 2s) = 0. Consider P (1s 2p). The 2p state is three-fold degenerate, i.e., |2p, m, with m = 0, 1. For m |z|m not to vanish, the rule is m = 0, since the total angular momentum part of the selection rule is satisfied by l = 1. Thus 2p, 1|H |1s, 0 = 2p, -1|H |1s, 0 = 0, and finding the probability of the transition 1s 2p is reduced finding that of |1s, 0 |2p, 0. 3 Now, because the perturbation is effectively of a finite duration, we can't really use Fermi's Golden Rule. But we can use the full expression as a time intergral. Then, as 2p, 0|H (t)|1s, 0 = eE(t) 210 r cos 100 dr (1) 2 eE(t) = exp(-3r/2a0 )r4 cos2 sin dr d d 4 4 2a0 0 0 0 4! 27 2a0 e eE(t) 2 2 = E(t), = 35 4 2a4 3 (3/2a0 )5 0 the probability amplitude to first order in H is 1 + 1 2p, 0|H |1s, 0ei12 t dt c2p0,1s0 = i - 1 27 2a0 e = E0 e-t ei12 t dt i 35 0 1 27 2a0 e 1 = E0 , 5 i 3 - i21 (2) where 21 = E2 -E1 as usual. Hence the probability of the transition |1s, 0 |2p, 0 is 215 a2 e2 E 2 P = |c1 |2 = 10 2 02 0 2 . 2p0,1s0 3 ( + 21 ) Note that 21 e2 = 2a0 1 1 - 2 2 1 = 3e2 . 8a0 4 ...
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