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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.
 Fall '13
 AkimasaMiyake

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