Phi 4 Theory Propagator Question

  • #1
Diracobama2181
74
2
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.
 

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