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

1. Feb 25, 2010

### 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 realise 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.

2. Feb 25, 2010

### 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.

3. Feb 25, 2010

### vertices

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?

4. Feb 25, 2010

### SpectraCat

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.

5. Feb 26, 2010

### vertices

Thanks SpectraCat.