A Does x affect the value of [a^2,(a^†)^2×e^2ikx]

[Sciencemaster]

TL;DR

In terms of QFT commutators involving raising and lowering operators and the mode expansion, does [a^2,(a^†)^2×e^2ikx] get changed by the last positionally-dependent term?

What is the commutator between a^2 (lowering operator squared) and the squared mode expansion from QFT (the integral of a^2e^2ikx, the conjugate, and the cross term I don't feel like writing out)? My instinct is to try and divide the mode expansion into its two parts since integration is linear, such as with just [a^2,(a^†)^2×e^2ikx]. However, even if I was sure I could do that, I’m not sure how to go about taking care of the positional dependence. For example, if we just had [a^2,(a^†)^2], it would just be 2(1+(a^†)a), but I'm having trouble with the extra e^ikx term in the new commutator. My thought is that it's really just a number so it can be pulled out by linearity leaving just [a^2,(a^†)^2], but I'm unsure if the ladder operators have any impact on the e^whatever operator (such as switching modes), which would then change the commutator. Also, is it any different if the first lowering operator is instead a raising operator? Any insight into the math here would be appreciated!

 

[jambaugh]

In the treatments I've seen the x and k are c-numbers i.e. parameters and not operators. All operators are expressed in terms of the creators and annihilators. Typically however the creators and annihilators are themselves indexed e.g. by momentum k. Note from the texts how e.g. the momentum operator is expressed.
\hat{p}= \int k a^\dagger(k)a(k) dk= \int k \hat{n}(k) dk
(suitably normalized.)

 

[Sciencemaster]

In the treatments I've seen the x and k are c-numbers i.e. parameters and not operators. All operators are expressed in terms of the creators and annihilators. Typically however the creators and annihilators are themselves indexed e.g. by momentum k. Note from the texts how e.g. the momentum operator is expressed.
\hat{p}= \int k a^\dagger(k)a(k) dk= \int k \hat{n}(k) dk
(suitably normalized.)

I know that in the quantum harmonic oscillator, you can have the ladder operators explicitly written in terms of x and k (vice versa). However, I'm not sure if I can use that explicit form since I'm working with a QFT system closer to a particle bath.
At any rate, I'm inclined to treat the e^ikx term as a scalar that's simply parameterized by the value of k and x if we can treat x as a parameter in the ladder operators rather than itself an operator. For example, if [(a^†)^2,a]=2a^†, we could just pull out our exponential to get [(a^†)^2,ae^ikx]=2(e^ikx)a^† rather than worrying about what implicit commutation the ladder operator might have with e^ikx. I'm still not sure, does this make sense?