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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 firstorder 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 lowestorder 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 firstorder correction to the Hamiltonian will be given in firstorder 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 + 1n+1 , an = nn1 , and their conjugate equations, and we can use these to obtain: n (a )2 a2 n = nn1 a an1 = n(n  1) n a2 (a )2 n = (n + 1)n+1 aa n+1 = (n + 1)(n + 2) n a(a )2 an = n(n + 1)n+1 (a )2 n1 = n(n + 1) n a a2 a n = n(n + 1)n1 (a2 n+1 = n(n + 1). n a aa an = n2 , n a aa an = (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 eik 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 () =  ei (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 angleindependent, 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 et (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 er/a0 , 210 = er/2a0 cos , a0 a0 32a3 0 1 r 1 r = 1 er/2a0 , 211 = er/2a0 sin ei . 2a0 a0 8a3 64a3 0 0 200 A useful integral is: xn eax 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 parityodd, the nonvanishing 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 threefold degenerate, i.e., 2p, m, with m = 0, 1. For m zm 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, 1H 1s, 0 = 2p, 1H 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, 0H (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, 0H 1s, 0ei12 t dt c2p0,1s0 = i  1 27 2a0 e = E0 et 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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 Spring '09
 Energy

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