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

From: Arnold Neumaier (Arnold.Neumaier_at_univie.ac.at)
Date: 12/27/04 

Date: Mon, 27 Dec 2004 13:32:36 +0000 (UTC)

Yi-Zen Chu; Yiren Qu wrote:
> Arnold Neumaier wrote:
>
>>There are no bras and kets in this formula.
>>a^{\dagger} a is the number operator written in second-quantized form,
>>and <.> is the vacuum expectation. Similar for your other example,
>>except that the operator is more complicated.
>
> When you say vacuum expectation, don't you mean <0| a^{\dagger}_j[p]
> a_i[p'] |0>, where |0> means 0 particles at a given momentum? If so
> wouldn't this just be identically zero?

Sorry for my nonsense. I wasn't paying close enough attention.
<.> is the expectation of an _arbitrary_ state. Of course, in the
vacuuum state, the number of particles is zero, as you just observed.

> Is it possible to take expectation values with respect to non-ground
> states? Say with respect to the 1 muon neutrino state at a given momentum?

Yes, of course. And also to mixed states.

Arnold Neumaier