Calculating Terms in $\bra{P'}\phi^4\ket{P}$

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The calculation of the term $\bra{P'} \phi^4 \ket{P}$ involves integrating over multiple momentum variables and includes a significant number of terms. The focus is on evaluating expressions like $\bra{0}a_{P'}a_{k_1}a^{\dagger}_{k_2}a_{k_3}a^{\dagger}_{k_4}a^{\dagger}_{P}\ket{0}$, which requires commuting annihilation and creation operators to simplify the expectation value. It's noted that the integrand scales as $\sim k^{-2}$, indicating that the integral diverges even with a single integration measure. This divergence is evident without fully executing the commutation process, as the scaling behavior of the integrand suffices to demonstrate the issue. Overall, the discussion highlights the complexity and challenges in calculating these terms in quantum field theory.
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Homework Statement
I am trying to show that ## \bra{P'}\phi^4 \ket{P}##is divergent, where ##\ket{P}## is a free spin 0 boson.
Relevant Equations
##\ket{P}=a^{\dagger}(k)\ket{0}##
##\phi=\int \frac{d^3k(\hat{a}e^{-ikx}+\hat{a}^{\dagger}e^{ikx})}{2\omega_{k}(2\pi)^3}##
##\omega=\sqrt{k^2+m^2}##
From this, I find
$$\bra{P'} \phi^4 \ket{P} = \int \frac {d^3 k_1 d^3 k_2 d^3 k_3 d^3 k_4} {16 \omega_{k_1}\omega_{k_2}\omega_{k_3}\omega_{k_4} (2\pi)^{12} }\bra{0} a_{P'}(a_{k_1} a_{k_2} a_{k_3}a_{k_4}e^{i(k_1+k_2+k_3+k_4)x}+...)a^{\dagger}_{P}\ket{0}$$ (a total of 16 different terms)
Right now, I am trying to figure out how to calculate terms like
## \bra{0}a_{P'}a_{k_1}a^{\dagger}_{k_2}a_{k_3}a^{\dagger}_{k_4}a^{\dagger}_{P}\ket{0} ##. Any examples would be greatly appreciated.
 
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Commute an annihilation or creation operator to the right or left of the product to render the expectation value zero, repeat this with the new terms generated by the commutator. Although to show that this quantity is divergent you don't have to do this, just look at how the integrand scales with ##k##. After doing all the commutators you will find that the integrand scales as ##\sim k^{-2}##, which means that even with one integration measure ##d^3 k## the integral will still be divergent since ##\int \frac{d^3 k}{k^2} \sim \Lambda##.
 
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