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

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SUMMARY

The calculation of the expectation value $\bra{P'} \phi^4 \ket{P}$ involves integrating over four momentum variables and applying creation and annihilation operators. Specifically, the expression is given by $$\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}$. The calculation of terms such as $\bra{0}a_{P'}a_{k_1}a^{\dagger}_{k_2}a_{k_3}a^{\dagger}_{k_4}a^{\dagger}_{P}\ket{0}$ requires commuting operators to evaluate the expectation value. The integrand scales as $\sim k^{-2}$, indicating that the integral diverges as $\int \frac{d^3 k}{k^2} \sim \Lambda$.

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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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