Undergrad Fourier Transform of Photon Emission Hamiltonian

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The discussion revolves around the Fourier Transform of the Hamiltonian for photon emission, specifically examining the expression H(x) and its transformation to momentum space. The user initially derives H(k) and concludes it is proportional to the Dirac delta function, but questions the simplicity of the result. Participants highlight the omission of the A(x) term in the integral and emphasize the need to include the photon creation and annihilation operators, as well as the polarization vector. They stress that a more thorough mathematical approach is necessary to validate the transformation and ensure all components are accounted for. The conversation underscores the importance of detailed calculations in quantum mechanics.
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

Hand waving is not enough. It is good that your radar went off. It is bad that you didn't do the physics.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

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