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# solset4 - PH 216 Problem set 4 Due 4/27 1 A particle of...

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Unformatted text preview: PH 216, Problem set 4, Due 4/27 1. A particle of mass M is in a one-dimensional harmonic oscillator potential V1 = 1 kx2 . 2 (a) It is initially in its ground state. The spring constant is suddenly doubled (k 2k) so that the new potential is V2 = kx2 . The particle's energy is then measured. What is the probability for finding that particle in the ground state of the new potential V2 ? (b) The spring constant is suddenly doubled as in part (a), so that V1 suddenly becomes V2 , but the energy of the particle in the new potential is not measured. Instead, after a time T has elapsed since the doubling of the spring constant, the spring constant is suddenly restored back to the original value. For what values of T would the initial ground state in V1 be restored with 100% certainty? Solution: (a) The wave function of the system before k change is 1/4 1 1 M 0 2 (x) = e- 2 M 0 x /. Suppose that the particle is also in the ground state of the new potential well after k change. Thus the new wave function is 1/4 1 1 M 1 2 (x) = e- 2 M 1 x /. The transition matrix element is 1/2 1 1 M 2 | = (0 1 )1/4 e- 2 M (0 +1 )x /dx 1/2 1 M = (0 1 )1/4 1 M (0 +1 ) 2 (0 1 )1/4 = . 1 (0 + 1 ) 2 When k changes into 2k, 0 changes into 1 = 20 , thus 2 2 (0 1 )1/2 ( 20 )1/2 | || = 1 = 1 = 2 21/4 ( 2 - 1). (0 + 1 ) ( 2 + 1)0 2 2 Hence the probability that the particle is in the state (x) is 25/4 ( 2 - 1). 1 (b) The quantum state is not destroyed as the energy is not measured. At t = 0, (x, 0) = 0 , n (x) being the eigenstates of V1 . We expand (x, 0) in the set of eigenstates of V2 : (x, 0) = m |0 |m (x). Here and below we shall use the convention that a repeated index implies summation over that index. Then (x, t) = e-iH2 t/(x, 0) = m |0 |m (x)e-iEm t/, where H2 is the Hamiltonian corresponding to V2 . Since 0 (x) has even parity, parity conservation gives 0, m = 2n + 1, m (x)|0 (x) = = 0, m = 2n, and so |(x, ) = |2m (x)2m |0 e-iE2m /. Hence |(x, ) = |0 (x) can be expected only if E2m / = 2N + c, where N is a natural number and c is a constant, for any m. As 1 1 , E2m = 2m + 2 we require Setting 1 c = 1 , 2 we require 2m1 = 2N , 1 2m + = 2N + c. 2 1 or 21 = 2N i.e., = where N = 0, 1, 2, ... Thus only if = N N , 1 M , 2k will the state chance into 0 (x) with 100% certainty. 2 2. An electron is in the n = 1 eigenstate of a one-dimensional infinite square-well potential which extends from x = -a/2 to x = a/2. At t = 0 a uniform electric field E is applied in the x-direction. It is left on for a time and then removed. Use time-dependent perturbation theory to calculate the probabilities P2 and P3 that the electron will be, respecively, in the n = 2 and n = 3 eigenstates at t > . Assume that is short in the sense that /(E1 - E2 ), where En is the energy of the eigenstate n. Specify any requirements on the parameters of the problem necessary for the validity of the approximations made in the application of time-dependent perturbation theory. Solution: The electron in the n = 1 eigenstate of the potential well V = 0 |x| a/2, otherwise 2 n a sin +x , a a 2 has wavefunctions and corresponding energies n (x) = n = 1, 2, 3, ... The uniform electric field E^x has potential = - Edx = -Ex . The potential energy e of the electron (charge -e) due to E, H = eEx, is considered as a perturbation. We have Hn2 n1 = n2 |H |n1 En = 2 2 n2 /2ma2 , 2 = a eE = a a 2 a 2 -a 2 n1 a n2 a sin x+ sin x+ eExdx a 2 a 2 -a 2 eE = a a2 a2 n1 -n2 n1 +n2 [(-1) - 1] - [(-1) - 1] (n1 - n2 )2 2 (n1 + n1 )2 2 = 4eEa n1 n2 [(-1)n1 +n2 - 1], 2 (n2 - n2 )2 1 2 1 2 2 = (En2 - En1 ) = (n2 - n2 ), 1 2 2ma Hk k eik k t dt = (n1 - n2 ) a (n1 + n2 ) a cos x+ - cos x+ xdx a 2 a 2 n2 n1 1 Ck k (t) = i 0 1 1 Hk k (1 - eik k ) . k k 3 For the transition 1 2, H21 = 2|H |1 = - 16eEa , 9 2 21 = 3 2 /2ma2 , and so the probability of finding the electron in the n = 2 state at t > is P2 = |C21 (t)|2 = = for . E1 -E2 1 2 2 21 2 H21 (1 - ei21 )(1 - e-i21 ) 16a2 9 2 3 eEm sin 2 3 2 4ma2 2 16 eEa 9 2 2 For the transition 1 3, H31 = 3|H |1 = 0, and so P3 = |C31 (t)|2 = 0. The validity of the time-dependent perturbation theory requires the time during which the perturbation acts should be small. The perturbation potential itself should also be small. 3. In general, the Heisenberg position operators X(t), evaluated at different times, do not commute. Therefore, a product of Heisenberg position operators, where each operator is evaluated at a different time, will depend on the order in which the operators appears. One particular order is one in which the largest time appears to the left and all the other operators appear in time order (i.e., time decreases as we move from left to right). This particular order corresponds to the time-ordered product, T [X(t1 )X(t2 ) X(tn )] = X(ti1 )X(ti2 ) X(tin ) , Prove that x , t | T [X(t1 )X(t2 ) X(tn )] |x , t = where ti1 > ti2 > > tin . D[x(t)] x(t1 )x(t2 ) x(tn ) eiS[x(t)]/ , where T is the time-ordered product symbol, S[x(t)] is the action [which depends on the path x(t)], X(t) is the position operator in the Heisenberg picture, and x(t) is the eigenvalue of X(t) when acting on the position eigenstate |x , t. Assume that t ti and t ti for all i = 1, 2, . . . , n. Solution: Consider the derivation of the path integral for x, t|T [X(t1 )X(t2 ) X(tn )]|x , t = x, t|X(ti1 )X(ti2 ) X(tin )|x , t . 4 The problem states that t ti1 and t tin . We can begin the derivation of the path integral in the same way as presented in the class lecture, by dividing up the time interval between t and t into slices: t tin tin-1 ti2 ti1 t . In contrast to the derivation presented in class, we do not assume that tik = t + k (i.e., uniform time slices), since the tik (k = 1, 2, . . . , n) are fixed by the product of the Heisenberg position operators. Next, we insert a complete set of states, |xik , tik xik , tik | dxik , - immediately following each Heisenberg position operator (for k = 1, 2, . . . , n). Using X(tik ) |xik , tik = x(tik ) |xik , tik , where x(tik ) is the (c-number) eigenvalue of X(tik ), it follows that: T [X(t1 )X(t2 ) X(tn )] = dxi1 dxi2 dxin x(ti1 )x(ti2 ) x(tin ) - - - x, t | xi1 , ti1 xi1 , ti1 | xi2 , ti2 xin , tin | x , t . (1) Since the x(tik ) are commuting c-numbers, we can rewrite eq. (1) as: T [X(t1 )X(t2 ) X(tn )] = dxi1 dxi2 dxin x(t1 )x(t2 ) x(tn ) - - - x, t | xi1 , ti1 xi1 , ti1 | xi2 , ti2 xin , tin | x , t . To make further progress, we now divide up the entire interval from t to t into N equally spaced time intervals (each interval of size ), TJ = t + J , where J = 1, 2, . . . , N - 1 , where N n. If N is large enough, then each one of the times (t1 , t2 , . . . , tn ) that appears in the time ordered product of Heisenberg position operators will approximately coincide with one of the TJ (up to an error that can be made arbitrarily small in the N limit). Then insert the appropriate complete set of states, |XJ , TJ XJ , TJ | dXJ , - at each time TJ (excluding those TJ that best approximate the times appearing in the Heisenberg position operators, since these have already been dealt with). Repeating the rest of the derivation of the path integral treated in class, we take the N , 0 limit to obtain the final result: x , t | T [X(t1 )X(t2 ) X(tn )] |x , t = D[x(t)] x(t1 )x(t2 ) x(tn ) eiS[x(t)]/ , (2) 5 where S[x(t)] = t dt L(x(t ), x(t ), t ) . t Note that eq. (2) is independent of the order of the tk (k = 1, 2, . . . , n). In particular, on the left hand side of eq. (2), the time ordered product is independent of the order of the operators X(tk ), since by definition the time ordered product yields operators that appear in time order (independently of the order of the operators appearing in the argument of the T -product). On the right hand side of eq. (2), the order of x(t1 )x(t2 ) x(tn ) is irrelevant as these are commuting (c-number) quantities. 6 ...
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