Sciencemaster
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- TL;DR Summary
- 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!