Midterm 7 Solutions

Nb ye2 2 w2 xa jn n ye2 2 w2 i a 4 2 a xe2

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Unformatted text preview: a\ = Xa —2 w2 JN + 12 N 2 a\ where `2 `† ` `† ` Za N a^ = Za a a a a a^ `† ` `† ` = Za a a a a a^ ¬༼ = a 2 ` `† Za a a a^ = a 2 `† ` Za Ja a + 1N a^ `2 Za N a^ = a 2 Ia ` a a] = a a] `† Za a = Xa a* 2 ` `† ¬༼ Ba, a F = 1 + 1M Thus `2 ` YE2 ] = —2 w2 Xa JN + N + YE2 ] = —2 w2 I a 4 +2 a XE\2 = —2 w2 I a 4 + 1 N 4 2 a\ = —2 w2 A a 2 Ia 2 + 1M + a 1 + 4M and Printed by Mathematica for Students Therefore a 2 1 + 4M 2 1 + 4E a 2 from (b) 7 Thus 8 `2 MidtermExam7Solutions.nb YE2 ] = —2 w2 Xa JN ` +N+ YE2 ] = —2 w2 I a 4 +2 a XE\2 = —2 w2 I a 4 + 1 N 4 2 a\ = —2 w2 A a 2 Ia 2 + 1M + a 2 1 + 4E 1 + 4M and a 2 1 + 4M Therefore DE = YE2 ] - XE\2 = — w a To summarize 2 XE \ = — w I a 1 + 2M DE = —w a ü Part d From the defining property of the coherent state ` a a] = a a] `† Za a = Xa a* and ` x= `† ` Ja + aN — 2mw ` px = Â m—w 2 `† ` Ja - aN we have, for an arbitrary coherent state a\ — 2mw Xx\ = X px \ = Â Ha* + aL = m—w 2 `* Ia - aM = 2— mw ReHaL 2 m — w Im...
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