Fourier Transform of Photon Emission Hamiltonian

In summary, the conversation discusses the process of obtaining the Fourier Transform of a Hamiltonian, specifically one involving the momentum and position operators. The individual's initial approach is questioned and they realize they may have made a mistake in their calculations. They are reminded to include all necessary factors and fully explain the steps in their mathematical approach. The importance of thoroughness and accuracy in physics is emphasized.
  • #1
thatboi
121
18
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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  • #2
What happens to ##A(x)## when you do the integral? It seems like you're just leaving it out.
 
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  • #3
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.
 
  • #4
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.
 
  • #5
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.
 

1. What is the Fourier Transform of Photon Emission Hamiltonian?

The Fourier Transform of Photon Emission Hamiltonian is a mathematical operation that decomposes a function describing the dynamics of photon emission into its constituent frequencies. It is used in quantum mechanics to analyze the energy levels and transitions of a system of particles.

2. How is the Fourier Transform of Photon Emission Hamiltonian calculated?

The Fourier Transform of Photon Emission Hamiltonian is calculated by taking the integral of the function describing the dynamics of photon emission over all possible frequencies. This integral can be solved using various mathematical techniques, such as integration by parts or the use of complex numbers.

3. What does the Fourier Transform of Photon Emission Hamiltonian tell us about a system?

The Fourier Transform of Photon Emission Hamiltonian provides information about the energy levels and transitions of a system of particles. It can reveal the frequencies at which the system emits or absorbs photons, as well as the relative strengths of these emissions or absorptions.

4. How is the Fourier Transform of Photon Emission Hamiltonian used in practical applications?

The Fourier Transform of Photon Emission Hamiltonian is used in a variety of practical applications, such as in spectroscopy and quantum optics. It is also an essential tool for understanding the behavior of particles in quantum systems, and is used in the development of quantum technologies.

5. Are there any limitations to using the Fourier Transform of Photon Emission Hamiltonian?

Like any mathematical model, the Fourier Transform of Photon Emission Hamiltonian has its limitations. It assumes that the system is in a stable state and does not account for any external factors that may affect the system's behavior. Additionally, it may not accurately describe systems with complex or nonlinear dynamics.

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