Density Matrix for Harmonic Oscillator
Density Matrix for Harmonic Oscillator
Density Matrix for Harmonic Oscillator
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and it follows that〈a † 〉 = 1 Z Tr[a† ρ] = 0〈a〉 = 1 Z Tr[aρ] = 0 (50)Due to this we have a quite simple expression <strong>for</strong> the average position〈X 〉 = 1 )(〈a † 〉 + 〈a〉 = 02〈K 〉 = i)(〈a † 〉 − 〈a〉 = 02(51)and similarly <strong>for</strong> the mean square deviations〈X 2 〉 = 1 2 〈(a† + a) 2 〉 = 1 ()〈a † a † 〉 + 〈a † a〉 + 〈aa † 〉 + 〈aa〉2〈K 2 〉 = − 1 2 〈(a† − a) 2 〉 = 1 () (52)−〈a † a † 〉 + 〈a † a〉 + 〈aa † 〉 − 〈aa〉2because of 〈n|a † a † n〉 = 0 and 〈n|aan〉 = 0 we have only〈X 2 〉 = 〈K 2 〉 = 〈a † a〉 + 〈aa † 〉 (53)and it remains to evaluate 〈a † a〉 and 〈aa † 〉. To do this I shall use two things.1. Baker-Campbell-Haussdorf theorem e βH ae −βH = ae −βħω2. Cyclic property of the trace Tr[ABC ] = Tr[BC A] = Tr[C AB]Using these we can find the following relationship:〈a † a〉 = Z −1 Tr[a † aρ] = Z −1 Tr[a † ae −βH ] = Z −1 Tr[a † e} −βH {{e βH}ae −βH ]=I= Z −1 Tr[a † e −βH e} βH ae{{ −βH}] = Z −1 Tr[a † e −βH a]e −βħω = Z −1 Tr[aa † e −βH ]e −βħωae −βħω= 〈aa † 〉e −βħω (54)Now using the commutation relation [a, a † ] = 1:1 = 〈[a, a † ]〉 = 〈aa † 〉 − 〈a † a〉 = 〈a † a〉(e βħω − 1) (55)or〈a † a〉 = 1/(e βħω − 1) and 〈aa † 〉 = −1/(e −βħω − 1) (56)8