I Fourier Transform of Photon Emission Hamiltonian

thatboi
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Hey all,
I just wanted to double check my logic behind getting the Fourier Transform of the following Hamiltonian:
$$H(x) = \frac{ie\hbar}{mc}A(x)\cdot\nabla_{x}$$
where $$A(x) = \sqrt{\frac{2\pi\hbar c^2}{\omega L^3}}\left(a_{p}\epsilon_{p} e^{i(p\cdot x)} + a_{p}^{\dagger}\epsilon_{p} e^{-i(p\cdot x)}\right)$$
and ##\epsilon_{p}## is the polarization vector and ##a_{p},a_{p}^{\dagger}## are the photon creation/annihilation operators for a photon with momentum ##p##. Also, we can treat the ##p## and ##x## as 4-vectors. To transform ##H(x)## to the momentum basis, I insert an integral of ##d^{4}x## and multiply by ##e^{ik\cdot x}##. Doing this leaves me with $$H(k) \propto \delta(p-k)k$$ since the ##\nabla_{x}## drags down a factor of ##k## and the Fourier Transform of an exponential function is just a Dirac delta. This result seems too simple to me so I was wondering if I made a mistake somewhere.
Thanks.
 
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What happens to ##A(x)## when you do the integral? It seems like you're just leaving it out.
 
PeterDonis said:
What happens to ##A(x)## when you do the integral? It seems like you're just leaving it out.
I thought the ##\delta(p-k)## factor came from the ##e^{(ip\cdot x)}## term in ##A(x)##, in which case I also forgot a ##\delta(p+k)*k## factor in my above response.
 
Let me try to say what @PeterDonis said but slightly more direct. Where are ##a_p## and ##\epsilon_p##? Also, the expression you give for ##H(k)## is proportional to ##k##, and ##k## is generally a vector. Your sanity check alarms should be going off.
 
Perhaps you should fill in the steps (using, say, mathematics

thatboi said:
A(x)=2πℏc2ωL3(apϵpei(p⋅x)+ap†ϵpe−i(p⋅x))
and ϵp is the polarization vector and ap,ap† are the photon creation/annihilation operators for a photon with momentum p. Also, we can treat the p and x as 4-vectors. To transform H(x) to the momentum basis, I insert an integral of d4x and multiply by eik⋅x. Doing this leaves me with

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