Expectation value of Normal ordered Stress-energy tensor

[LAHLH]

Hi,

In Birrel and Davies ch4 they write:

$$\langle \psi|:T_{ab}:|\psi \rangle =\langle \psi|T_{ab}|\psi \rangle -\langle 0|T_{ab}|0 \rangle$$

this is for the usual Mink field modes and vac state. Why does normal ordering reduce to this expression, could anybody point me the way to deriving this?

 

[fzero]

In terms of free fields, the stress tensor is quadratic, so the difference with the normal-ordered expression is just a (divergent) c-number. This c-number is the only contribution to the vacuum expectation value, so you find the expression in question.

An analogous expression holds for any operator quadratic in the fields, as some playing around with annihilation and creation operators will reveal. You can already see how it works for a single harmonic oscillator if you consider the operator $\hat{\mathcal{O}}= \alpha a^\dagger a + \beta a a^\dagger$ and expectation values in the $|n\rangle$ state.

 

[LAHLH]

In terms of free fields, the stress tensor is quadratic, so the difference with the normal-ordered expression is just a (divergent) c-number. This c-number is the only contribution to the vacuum expectation value, so you find the expression in question.

An analogous expression holds for any operator quadratic in the fields, as some playing around with annihilation and creation operators will reveal. You can already see how it works for a single harmonic oscillator if you consider the operator $\hat{\mathcal{O}}= \alpha a^\dagger a + \beta a a^\dagger$ and expectation values in the $|n\rangle$ state.

Thanks for the reply.

Taking your harmonic oscillator example I think I see what you're getting at:

$\langle n|:\hat{\mathcal{O}}:|n\rangle =\langle n|\alpha a^{\dagger}a+\beta a^{\dagger}a|n\rangle$

then use $aa^{\dagger}-a^{\dagger}a=1$

to give:

$\langle n|:\hat{\mathcal{O}}:|n\rangle =\langle n|\alpha a^{\dagger}a+\beta (aa^{\dagger}-1)|n\rangle=\langle n|\hat{\mathcal{O}}|n\rangle-\beta\langle n|n\rangle=\langle n|\hat{\mathcal{O}}|n\rangle-\beta$

But actuallythe VEV $\langle 0|\hat{\mathcal{O}}|0\rangle =\langle 0|\alpha a^{\dagger}a+\beta aa^{\dagger}|0\rangle=\beta\langle 0| aa^{\dagger}|0\rangle=\beta\langle 1|1\rangle =\beta$

So we could have wrote

$\langle n|:\hat{\mathcal{O}}:|n\rangle=\langle n|\hat{\mathcal{O}}|n\rangle-\langle 0|\hat{\mathcal{O}}|0\rangle$

I just need to convince myself this works for an object like $\langle \psi|:\phi^2:|\psi\rangle$ now...