I'm having trouble understanding the interaction of radiation with matter in (elementary non-relativistic QED) in Coulomb gauge ($\nabla\cdot\boldsymbol{A}=0$). We saw how to quantize the free electromagnetic field in a cavity of finite dimension, by writing the hamiltonian in terms of creation and annihilation operators, i.e.: \begin{equation} \hat{H}_{\mathrm{em}}^{\mathrm{free}}=\sum_{\lambda=1}^{2}\sum_{\boldsymbol{k}}\hbar\omega_{\boldsymbol{k}}\left(\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\hat{a}_{\lambda\boldsymbol{k}}+\frac{1}{2}\right) \end{equation} where $\lambda$ runs on the two polarization of the field. Thus the hamiltonian for field-particle system in the non-relativistic approximation is: \begin{equation} \begin{split} \hat{H}&=\sum_{a}\left(\frac{\hat{\boldsymbol{p}}_{a}}{2m_{a}}+V_{a}\right)+\sum_{\lambda=1}^{2}\sum_{\boldsymbol{k}}\hbar\omega_{\boldsymbol{k}}\left(\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\hat{a}_{\lambda\boldsymbol{k}}+\frac{1}{2}\right)+\sum_{a}\left(-q_{a}\frac{\hat{\boldsymbol{A}}\cdot\hat{\vec{p}}_{a}}{m_{a}c}\right)=\\ &=\hat{H}_{\mathrm{particle}}+\hat{H}_{\mathrm{em}}^{\mathrm{free}}+\hat{H}_{\mathrm{int}} \end{split} \end{equation} The problem I have is that to compute the transition probability we used Fermi's Golden Rule: \begin{equation} P_{i\rightarrow f}=\frac{2\pi}{\hbar}\left\vert W_{fi}\right\vert^2\rho\left(E_f\right)\end{equation} (f and i being the final and initial state) while this formula (in all books I read about non relativistic quantum mechanics) is usally derived upon considering the perturbation hamiltonian time independent or a perturbation of the form: \begin{equation} \hat{H}_{\mathrm{int}}=\hat{T}_{+}e^{-i\omega t}+\hat{T}_{-}e^{i\omega t} \end{equation} (where $\hat{T}_{-}=\hat{T}_{+}^{\dagger}$), while our perturbation hamiltonian is \begin{equation} \hat{H}_{\mathrm{int}}\propto\sum_{\lambda\boldsymbol{k}}\left[\hat{\boldsymbol{p}}_{a}\cdot\boldsymbol{e}_{\lambda\boldsymbol{k}}\left(\hat{a}_{\lambda\vec{k}}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\right)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \right] \end{equation} ($\boldsymbol{e}_{\lambda\boldsymbol{k}}$ being the two polarizations vector), and the $\hat{a},\hat{a}^{\dagger}$'s are time dependent. I hope I made my point clear enough.

P.S.: What we did in order to quantize the EM field was to write a Fourier series expansion of the vector potential \begin{equation}\boldsymbol{A}\left(\boldsymbol{x},t\right)=\sum_{\boldsymbol{k}}\boldsymbol{a}_{\boldsymbol{k}}\left(t\right)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}\end{equation} and by plugging this expression in $\Box \boldsymbol{A}=0$ one find $\ddot{\boldsymbol{a}}_{\boldsymbol{k}}+\omega_{\boldsymbol{k}}\boldsymbol{a}_{\boldsymbol{k}}=0$ so that $\boldsymbol{a}_{\boldsymbol{k}}{}_{\pm}\propto e^{\pm i\omega_{\boldsymbol{k}}t}$


1 Answer 1


The answer is that you are confused.

We need only explicit time-independence of the perturbation Hamiltonian i.e. that it were the time-independent combination of the canonical momenta and coordinates. If there is no explicit time-dependence the only evolution is due to the evolution of the momenta and coordinates (which appears only in the Heisenberg picture) and is given by the commutator of the perturbation Hamiltonian with free one.

If the commutator is zero then the perturbation Hamiltonian is constant even in the Heisenberg picture. However you may say that in such a case the perturbation would be trivialized - you can find an eigenbasis common for the free and perturbation Hamiltonian that therefore will not change under perturbation at all. If you are able to represent all the states of your interest in terms of such an eigenbasis you solve your problem exactly! Usually it is assumed that the perturbation and free Hamiltonians don't commute.

You say that the creation and annihilation operators depend on time but in what sense? They actually are time-independent combinations of the canonical coordinates and momenta, \begin{equation} \hat{a}_{\lambda\boldsymbol{k}}=\frac{1}{\sqrt{2}}\boldsymbol{e}_{\lambda\boldsymbol{k}}\cdot\Big(\hat{\boldsymbol{A}}_{\lambda\boldsymbol{k}}(\boldsymbol{k})+i\hat{\boldsymbol{\pi}}_{A,\lambda\boldsymbol{k}}(\boldsymbol{k})\Big),\quad \hat{a}^\dagger_{\lambda\boldsymbol{k}}=\frac{1}{\sqrt{2}}\boldsymbol{e}_{\lambda\boldsymbol{k}}\cdot\Big(\hat{\boldsymbol{A}}_{\lambda\boldsymbol{k}}(\boldsymbol{k})-i\hat{\boldsymbol{\pi}}_{A,\lambda\boldsymbol{k}}(\boldsymbol{k})\Big), \end{equation} where $\hat{\boldsymbol{A}}_{\lambda\boldsymbol{k}}(\boldsymbol{k})$ and $\hat{\boldsymbol{\pi}}_{A,\lambda\boldsymbol{k}}(\boldsymbol{k})$ are given by the spatial Fourier transforms and hence contain only $A$ and $\pi_A$ from the same time slice (just as we wish from the Hamiltonian p.o.v.)

They do evolve in the sense that they don't commute with the Hamiltonian but it's only due to the evolution of the canonical coordinates and momenta and not due to the explicit time-dependence. So your perturbation Hamiltonian is ok.

UPD: I think this stems from the way the $a$ and $a^\dagger$ are introduced in many QFT courses. They usually Fourier transform the potentials with respect both to time and spatial coordinates already at the classical level stressing the time dependence and then boldly put the hats at the top. Accurate way to do this canonically is to Fourier transform the potentials only with respect only to spatial coordinates treating then the Fourier components as variables (they are just linear combinations of the initial field variables from the same time) This way you get the harmonic oscillator for each Fourier component that can be quantized canonically and then be rewritten in terms of the corresponding creation and annihilation operators.

  • $\begingroup$ I'm quite certain that there should be $|\boldsymbol{k}|$ normalization factors in the equations for creation and annihilation operators but can't write them off the top of my head without a doubt. I'll check it later. $\endgroup$
    – OON
    Mar 29, 2016 at 6:42
  • $\begingroup$ Thank you for your answer. By the way, I updated my post in order to explain better the starting point. $\endgroup$
    – Mitch
    Mar 29, 2016 at 7:22

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