Midterm 7 Solutions

# N part b the mean excitation number is given by xn

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Unformatted text preview: +1 an cn = give the recurrence relation c0 n! where we determine c0 through normalization as Ha* Lm Xa a\ = B⁄¶ = 0 m m! = ⁄¶, n = 0 m = c0 2 = c0 2 Xa a\ = c0 c* Xm FB⁄¶= 0 n 0 Ha* Lm an ⁄¶, n = 0 m 2 Ha* Lm an a2 n! ⁄¶= 0 n 2 c0 m! n! m! n! an n! c0 n\F Xm n\ dmn n 2 ‰a Setting equal to 1 and choosing the overall phase to be real gives c0 2 2 ‰a =1 2 c0 = ‰- a ê2 Therefore a\ = ‰- a 2 ê2 ⁄¶= 0 n an n! n\ ü Part b The mean excitation number is given by ` XN \ = Za N a^ `† ` = Za a a a^ XN \ = a 2 For XN \ = N0 , a = Xn a\ Xn a\ 2 2 ¬༼ ` a a] = a a] `† Za a = Xa a* = = N0 . The probability of measuring n excitations in such a state is thus given by ‰- a n! ‰ -N0 n! 2 a 2n ¬༼ a = n N0 This is a Poisson distribution with the expectation value of n being N0 . Printed by Mathematica for Students N0 MidtermExam7Solutions.nb ü Part c We have aHtL\ = ‰ - - aHtL\ = ‰ - aH0L\ — = ‰- a =‰ aH0L\ = ‰- a ` ÂH t 2 Âwt 2 Âwt 2 ⁄¶= 0 n ê2 ¬༼ ê2 ⁄¶= 0 n ` H n^ = — w In + 1 M 2 an n\ n! n\ 1 an ‰ n! a ‰-Â w t 2 ê 2 ‰- 2 -Â w t Jn + N n\ 2 n Ia ‰-Â w t M ⁄¶= 0 n n\ n! therefore a ‰-Â w t ] and ` ` ` ÂH t ` ÂH t ` XE\ = ZaHtL H aHtL^ = XaH0L ‰ — H ‰ — aH0L\ ÂH t ` ¬༼ H commutes with ‰ — ` = XaH0L H aH0L\ ` = XaH0L — w JN + XE \ = — w I a 2 ` ` ¬༼ H = — w JN + 1 N 2 1 N 2 ` ¬༼ ZaH0L N aH0L^ = aH0L\ 1 + 2M Similarly `2 ` YE2 ] = Xa H...
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## This note was uploaded on 02/01/2014 for the course PHYC 491/496 taught by Professor Akimasamiyake during the Fall '13 term at New Mexico.

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