Energy Momentum Tensor in Phi^3 Theory

[Diracobama2181]

Homework Statement

If I have an energy-momentum tensor
$$\hat{T}^{\mu \nu} (x)=\partial^{\mu}\Phi\partial^{\nu}\Phi-\\g^{\mu \nu}\mathcal{L}$$
,with
$$
\mathcal{L}=(\frac{1}{2}(\partial_{\mu}\Phi)^2-\frac{1}{2}m^2\Phi^2-\frac{\lambda \Phi^3}{3!})
$$
, how would I go about calculating
$$
\bra{ \vec{ p'}} T^{\mu,\nu} \ket{ \vec {p}}
$$?

Relevant Equations

$\ket{\vec{p}}=\hat{a}^{\dagger}(\vec{p})\ket{0}$ for a free field with $[\hat{a}({\vec{k})},\hat{a}^{\dagger}({\vec{k'})}]=2(2\pi)^3\omega_k\delta^3({\vec{k}-\vec{k'}})$

$$ \bra{ \vec{ p'}} T_{\mu,\nu} \ket{ \vec {p}}=\bra{\Omega}\hat{a}(\vec{p'})(\partial^{\mu}\Phi\partial^{\nu} \Phi-g^{\mu \nu}\mathcal{L})\hat{a}^{\dagger}(\vec{p})\ket{\Omega}=\bra{\Omega}(\hat{a}(\vec{p'})\partial^{\mu}\Phi\partial^{\nu} \Phi\hat{a}^{\dagger}(\vec{p})\\ -\hat{a}(\vec{p'})g^{\mu \nu}(\frac{1}{2}(\partial_{\mu}\Phi)^2-\frac{1}{2}m^2\Phi^2-\frac{\lambda \Phi^3}{3!}))\hat{a}^{\dagger}(\vec{p})\ket{\Omega}$$
In this instance, would this imply that, for example, $$\bra{\Omega}(\hat{a}(\vec{p'})(\frac{1}{2}(\partial_{\mu}\Phi)^2-\frac{1}{2}m^2\Phi^2-\frac{\lambda \Phi^3}{3!})\hat{a}^{\dagger}(\vec{p}))\ket{\Omega}=lim_{t \to \infty(1-i\epsilon)}\frac{\bra{0}[T(\hat{a}(\vec{p'})(\frac{1}{2}(\partial_{\mu}\Phi)^2-\frac{1}{2}m^2\Phi^2-\frac{\lambda \Phi^3}{3!})\hat{a}^{\dagger}(\vec{p}))e^{-i\int \lambda \frac{\Phi^3}{3!}dt}]\ket{0}}{\bra{0}T e^{-i\int \lambda \frac{\Phi^3}{3!}dt}\ket{0}}$$?
Also, how would I go about calculating terms like $$\bra{\Omega}\hat{a}(\vec{p'})(\partial^{\mu}\Phi\partial^v \Phi )\hat{a}^{\dagger}(\vec{p})\ket{\Omega}$$ using Feynman diagrams? Thanks in advance.
Note: $$\ket{\Omega}$$ is the ground state in the interacting theory.

 

[BigJDubs]

A:I'm going to give a somewhat abbreviated answer here, because it's probably beyond the scope of this site to provide a full textbook-level explanation.The correct way to calculate matrix elements of the energy-momentum tensor is to perform a Wick contraction; in other words, you need to express the operators inside the expectation value using time-ordered products, and then take the appropriate number of derivatives of the resulting expression with respect to the metric tensor. This approach is explained in some detail in chapter 3 of Peskin and Schroeder's "An Introduction to Quantum Field Theory".In the case at hand, we have an expectation value of the form$$\langle \Omega | T_{\mu \nu} | \Omega \rangle = \langle \Omega | \partial_\mu \Phi \partial_\nu \Phi - g_{\mu \nu} \mathcal{L} | \Omega \rangle$$where $\mathcal{L}$ is the Lagrangian density of the theory. To evaluate this, we use the fact that the vacuum state $|\Omega\rangle$ (or more accurately, the vacuum-to-vacuum expectation value) is annihilated by all annihilation operators, and so$$\langle \Omega | T_{\mu \nu} | \Omega \rangle = \langle \Omega | \partial_\mu \Phi(x) \partial_\nu \Phi(x) - g_{\mu \nu} \mathcal{L}(x) | \Omega \rangle \\= \langle 0 | T \left[ \partial_\mu \Phi(x) \partial_\nu \Phi(x) - g_{\mu \nu} \mathcal{L}(x) \right] | 0 \rangle$$where $T$ denotes the time ordering operator.The second equality follows from the fact that the vacuum-to-vacuum expectation value in an interacting theory is equal to the time ordered expectation value of the same operators in the free theory, which is much easier to calculate.Now, the left-hand side can be expressed as a sum of diagrams, where each diagram is a product of propag