Phi 4 Theory Propagator Question

[Diracobama2181]

Homework Statement

Find $$\bra{\Omega}\partial^{\mu}\Phi\partial^{\nu}\Phi\ket{\Omega}$$ in $$\Phi^4$$ theory.

Relevant Equations

$$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\Phi)^2-\frac{1}{2}m^2\Phi^2-\lambda\frac{\Phi^4}{4!}$$

I know in the Heisenburg picture,
$$\Phi(\vec{x},t)=U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)U(t,t_0)$$
where $$\Phi_{0}$$ is the free field solution, and
$$U(t,t_0)=T(e^{i\int d^4x \mathcal{L_{int}}})$$. Is there a way I could solve this using contractions or Feynman diagrams?
Because otherwise, it would appear I would have to solve this by taking the derivative of $$U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)U(t,t_0)$$
and this would give $$\partial^{\mu}\Phi(\vec{x},t)=\partial{\mu}U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)U(t,t_0)+U^{\dagger}(t,t_0)\partial^{\mu}\Phi_{0}(\vec{x},t)U(t,t_0)+U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)\partial^{\mu}U(t,t_0)$$
which would mean I would have to calculate
$$\partial^{\mu}U(t,t_0)=T(\partial^{v}(i\int d^4x \mathcal{L}_{int}))U(t,t_0)$$
which would make this calculation messy. Thanks in advance.

 

[mark726]

Thank you for your question. Yes, there is a way to solve this using contractions or Feynman diagrams. In fact, this is the standard way to approach quantum field theory calculations.

First, let's rewrite the expression in terms of creation and annihilation operators:
$$\Phi(\vec{x},t)=U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)U(t,t_0)=U^{\dagger}(t,t_0)\sum_{\vec{k},\lambda}a_{\vec{k},\lambda}e^{i(\vec{k}\cdot\vec{x}-\omega_{\vec{k}}t)}+a^{\dagger}_{\vec{k},\lambda}e^{-i(\vec{k}\cdot\vec{x}-\omega_{\vec{k}}t)}U(t,t_0)$$

Next, let's expand the time-evolution operator using the Dyson series:
$$U(t,t_0)=\mathcal{T}\left[e^{i\int_{t_0}^{t}d\tau \mathcal{L}_{int}(\tau)}\right]=1+i\int_{t_0}^{t}d\tau\mathcal{L}_{int}(\tau)+\frac{i^2}{2!}\int_{t_0}^{t}d\tau_1\int_{t_0}^{t}d\tau_2\mathcal{T}\left[\mathcal{L}_{int}(\tau_1)\mathcal{L}_{int}(\tau_2)\right]+\cdots$$

Using the definition of the time-ordered product, we can write the second term in the series as:
$$\int_{t_0}^{t}d\tau\mathcal{L}_{int}(\tau)=\int_{t_0}^{t}d\tau\int d^3x \mathcal{H}_{int}(\vec{x},\tau)$$
where $\mathcal{H}_{int}(\vec{x},\tau)$ is the interaction Hamiltonian density.

Using the same procedure, we can expand the third term in the series as:
$$\int_{t_0}^{t}d\tau_1\int_{t_0}^{t}d\tau_