Re: What does < a^{\dagger} a > mean exactly?

meyr2_at_web.de
Date: 12/23/04 

Date: Thu, 23 Dec 2004 11:46:58 +0000 (UTC)

Hi,

when looking at the second formula, that looks like the left side of a
commutator between a^dagger and a, it looks like that

<a^+ a> is just the expectation value and that someone just calculated

< a^+ a> = Tr (rho a^+ a)
At least a^+ a is the number operator for some mode.

Rene.