Computing Thermal Average for a Free Real Scalar Field

[parton]

Homework Statement

Homework Equations

Hi!

I read a book where a free real scalar field with Hamiltonian

$$H = \int \dfrac{\mathrm{d}^{3}p}{(2 \pi)^{3}} \, E_{\vec{p}} a_{\vec{p}}^{\dagger} a_{\vec{p}}$$

is beeing discussed.

Note that:

$$E_{\vec{p}} = \sqrt{\vert \vec{p} \vert^{2} + m^{2}}$$

and

$$\left[ a_{\vec{p}}, a_{\vec{q}}^{\dagger} \right] = (2 \pi)^{3} \delta^{3}(\vec{p} - \vec{q})$$.

Now there is the statement that the thermal average of the operator
$$a_{\vec{p}}^{\dagger} a_{\vec{q}}$$ is given by:

$$\left \langle a_{\vec{p}}^{\dagger} a_{\vec{q}} \right \rangle = \dfrac{1}{Tr \left[ \mathrm{e}^{-\beta H} \right]} Tr\left[\mathrm{e}^{-\beta H} a_{\vec{p}}^{\dagger} a_{\vec{q}} \right] = \dfrac{1}{\mathrm{e}^{\beta E_{\vec{p}}} -1} (2 \pi)^{3} \delta^{3}(\vec{p} - \vec{q})$$

I don't know how to obtain this average value.

The Attempt at a Solution

When I first start to compute $$Tr \left[ \mathrm{e}^{-\beta H} \right]$$ I end up with an ill-defined expression:

$$Tr \left[ \mathrm{e}^{-\beta H} \right] = \int \dfrac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \left \langle \vec{k} \right| \mathrm{e}^{-\beta H} \left| \vec{k} \right \rangle = \int \dfrac{\mathrm{d}^{3} k}{(2 \pi)^{3}} (2 \pi)^{3} \delta^{3}(0)$$

where I used

$$H \left| \vec{k} \right \rangle = \int \dfrac{\mathrm{d}^{3} k}{(2 \pi)^{3}} E_{\vec{p}} a_{p}^{\dagger} a_{\vec{p}} a_{\vec{k}}^{\dagger} \vert 0 \rangle = \int \dfrac{\mathrm{d}^{3} k}{(2 \pi)^{3}} E_{\vec{p}} \delta^{3}(\vec{p} - \vec{k}) (2 \pi)^{3} \vert \vec{p} \rangle = E_{\vec{k}} \vert \vec{k} \rangle$$

I think the problem is that I used momentum states as eigenstates of the Hamiltonian, but they are not normalizable.

Does anyone have an idea how to compute this thermal average?

(If I have the simple Hamiltonian $$H = \omega a^{\dagger} a$$ and consider a complete set of eigenstates |n> it is very easy to compute this average. But in the case described above I don't know how to do that).

I hope someone can help me.

Note:

The same problem is also described here (page 5, eq. (3.10))
http://www.actaphys.uj.edu.pl/vol38/pdf/v38p3661.pdf

 

[mark726]

Thank you for sharing your question and for seeking help in understanding this concept. It seems like you are struggling with computing the thermal average of an operator in the context of a free real scalar field with a given Hamiltonian. I will try my best to explain the steps to obtain this average value.

First, let's start with the definition of the thermal average of an operator A in a system with a given Hamiltonian H:

\left \langle A \right \rangle = \dfrac{1}{Z} Tr\left[\mathrm{e}^{-\beta H} A \right]

where Z is the partition function given by:

Z = Tr\left[ \mathrm{e}^{-\beta H} \right]

Now, in your case, the operator A is given by:

a_{\vec{p}}^{\dagger} a_{\vec{q}}

Substituting this in the above equation, we get:

\left \langle a_{\vec{p}}^{\dagger} a_{\vec{q}} \right \rangle = \dfrac{1}{Z} Tr\left[\mathrm{e}^{-\beta H} a_{\vec{p}}^{\dagger} a_{\vec{q}} \right]

Next, let's focus on computing the partition function Z. We can expand it as follows:

Z = Tr\left[ \mathrm{e}^{-\beta H} \right] = \int \dfrac{\mathrm{d}^{3}p}{(2 \pi)^{3}} \left \langle \vec{p} \right| \mathrm{e}^{-\beta H} \left| \vec{p} \right \rangle

= \int \dfrac{\mathrm{d}^{3}p}{(2 \pi)^{3}} \left \langle \vec{p} \right| \mathrm{e}^{-\beta \int \dfrac{\mathrm{d}^{3}p}{(2 \pi)^{3}} E_{\vec{p}} a_{\vec{p}}^{\dagger} a_{\vec{p}} } \left| \vec{p} \right \rangle

= \int \dfrac{\mathrm{d}^{3}p}{(2 \pi)^{3}} \mathrm{e