Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to quantize the electromagnetic field by solving the vector potential wave equation, that is:

$$\nabla^{2} \mathbf{A} = \dfrac{1}{c^{2}} \dfrac{\partial ^{2} \mathbf{A}}{\partial t^{2}}, $$

combined with the boundary conditions:

$$ \mathbf{A}(\mathbf{x},t) = \mathbf{A}(\mathbf{x}+\mathbf{L},t), $$

where $\mathbf{L}$ is A vector that represent the width, lenght and height of a cubic box of size L.

Well, I suppose that the solution of the above equation could be separable as:

$$ \mathbf{A}(\mathbf{x},t)=u(\mathbf{x})T(t); $$

Then, solving the $\mathbf{x}$ and the $t$ dependence, the solution is:

$$ A(\mathbf{x},t) = \sum_{\mathbf{k},s}\ \ \mathbf{e}_{\mathbf{k},s} \left( A\exp[i \mathbf{k}\cdot \mathbf{x}] + B\exp[-i \mathbf{k}\cdot \mathbf{x}] \right) \left( C\exp[i \omega t] + D\exp[-i \omega t] \right), $$

were A,B,C and D are constants, $\mathbf{k} = 2n\pi \mathbf{x}/\mathbf{L}$, $\omega = |\mathbf{k}|c$, $\mathbf{e}_{\mathbf{f},s}$ its the polarization vector and $s$ represent the two independent polarization directions.

The solution, according to Sakurai (Advanced Quantum Mechanics, ) is:

$$ A(\mathbf{x},t) = \sum_{\mathbf{k},s}\ \ \mathbf{e}_{\mathbf{k},s} \left( A\exp[-i \omega t]\exp[i \mathbf{k}\cdot \mathbf{x}] + B\exp[i \omega t]\exp[-i \mathbf{k}\cdot \mathbf{x}] \right) $$

What I did wrong or what other condition I have to impose to get the correct solution? I have imposed the reality condition on $\mathbf{A}$, but I got nothing useful.

share|cite|improve this question
up vote 1 down vote accepted

You forgot the sume over $k$, as any linear combination of solutions is again a solution. And you do not need to treat the $-k$ case separately from the $k$ case as otherwise you'd get each term twice. Then -after multiplying out the reamining product - your solution will be identical with Sakurai's (also amended by summing over $k$).

share|cite|improve this answer
I wrote the wrong index; it is "k" instead of "f". +1 to your answer because it help me to see the problem with "other eyes". Thanks. – Rodrigo Thomas Mar 8 '12 at 4:13

There is a straightforward way of finding the radiation field you are looking for. For the wave equation $\Delta_{\mathbf{r}}\mathbf{A}-\dfrac{1}{c^2}\partial^2_{t}\mathbf{A}=\mathbf{0}$

the solution are plane waves, normalized in your box of volume $V=L^3$

$\psi_{\mathbf{k},s}=\dfrac{1}{\sqrt{V}}\mathbf{e}_s e^{i(\mathbf{k}\cdot\mathbf{r}-\omega_{\mathbf{k}}t)}$

where $s=1,2$ is the polarization. Next, your vector potential is a superposition of plane waves (a real one) with complex coefficients $A_{\mathbf{k},s}$

$\mathbf{A}(\mathbf{r},t)=\dfrac{1}{\sqrt{V}} \sum_{\mathbf{k}} \sum_{s=1,2}\left(A_{\mathbf{k},s}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega_{\mathbf{k}}t)}+\text{c.c} \right)\mathbf{e}_s$

where c.c. means complex conjugate.

It seems to me that you are not taking into account that in the separation of variables the coefficients $A,B,C,D$ are not independent. Solve first Helmholtz equation for $u_{\mathbf{k}}(\mathbf{r})$ with expansion coefficients $A_{\mathbf{k},s}(t)$ (in your case ($A$ and $B=A^\ast$), insert into the equation of the harmonic oscillator and it should be fine.


share|cite|improve this answer

An alternative way to think of this is that the last expression in your Question is no more than a presentation of $A_\mu(x)$ as a Fourier transform on Minkowski space, subject to the twin constraints that there are only light-like components and that the observables are the electromagnetic field (not the potential), to which only components in the 2-space determined by the polarization vectors at a given wave-number contribute.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.