The expectation of an expection (relating to Wick's Theorem)

[vertices]

Hi:

If we want to work out the expectation of:

<0|T(φ₁φ₂)|0>

ie. <0|<0|T(φ₁φ₂)|0>|0>

apparently it is acceptable to pull out the <0|T(φ₁φ₂)|0>:

So <0|<0|T(φ₁φ₂)|0>|0>=<0|T(φ₁φ₂)|0><0|I|0>

I do realize this is a really stupid question, but I want to be 100% sure. Is this simply because an expectation is always a constant, not an operator which acts on a state? Can you always pull out an expectation in this way?

Thanks.

 

[Matterwave]

As far as I know, you can always pull out the expectation that way. I know of no cases where the expectation value is not simply just a number (or value). I don't see any point in finding the "expectation of the expectation value" tho...It's always just the same as the expectation value itself.

 

[vertices]

As far as I know, you can always pull out the expectation that way. I know of no cases where the expectation value is not simply just a number (or value). I don't see any point in finding the "expectation of the expectation value" tho...It's always just the same as the expectation value itself.

thanks matterwave.

Yes indeed, the expectation of an expectation is the expectation itself. A better example is the following:

Apparently this expression equals the one below it:

<0|(<0|T(φ₁φ₂)|0>)φ₃|0>

=<0|T(φ₁φ₂)|0><0|φ₃|0>

So you're saying one can always pull out the <0|T(φ₁φ₂)|0> in this way?

 

[SpectraCat]

thanks matterwave.

Yes indeed, the expectation of an expectation is the expectation itself. A better example is the following:

Apparently this expression equals the one below it:

<0|(<0|T(φ₁φ₂)|0>)φ₃|0>

=<0|T(φ₁φ₂)|0><0|φ₃|0>

So you're saying one can always pull out the <0|T(φ₁φ₂)|0> in this way?

Yes ... it is essentially a constant, and can be treated like any other constant. You are right to be cautious though, since with expressions Dirac notation you need to be conscientious about re-ordering bra's and ket's, since it can change the meaning of the expression in general. However in this case you are fine.

 

[vertices]

Thanks SpectraCat.