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With a view to quantising the EM field, consider a classical free field in the absence of charge and currents, we can take a coulomb gauge, $\phi=0, \partial_kA_k=0$. The physical fields in terms of the potentials are then:

  • $E_k=-\dot{A_k}$
  • $B_k = \epsilon_{klm}\partial_lA_m$

So the dynamics of the field:

$$\partial_j\partial_jA_k=\frac{1}{c^2}\ddot{A_k}$$

are determined by the vector potential $A_k(x_k,t)$. The spatial component of the potential has periodic boundary conditions, which are satisfied by exponential functions:

$$A(x_j) \propto e^{ik_jx_j}$$

with $k_j=\frac{2\pi}{L}n_j$

We can construct a form of $A$ which takes into account both propagation directions $k$, and polarisations (for plane waves) $\sigma$, in each dimension. The (un-normalised) potential $A$ can now be expressed as a Fourier series in terms of the wave modes:

$$\vec{A}(x_j,t)=\sum\limits_{k_j}\sum\limits_\sigma \vec{u}_{k_j,\sigma}\left(a_{k_j,\sigma}(t)e^{ik_jx_j}+a^*_{k_j,\sigma}(t)e^{-ik_j}\right)$$

Here, $\vec{u}$ is a unit vector normal to $\vec{k}$, the time dependence is separated into $a(t)$, and we explicitly choose that $\vec{A}(x_j,t)$ be real by taking both $\pm ik_jx_j$ for the plane waves.

I am told that we need to impose an additional restriction on the summation over $k_j$, such that $k_3>0$ (or equally with any one of the $k_j$) to prevent double counting. I fail to see the justification for this argument. Furthermore, could we have changed anything in the way we chose to write $A(x_j)$ or $k_j$ to prevent this?

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So just one of the $k_i$ has to be bigger than zero or all of them? I would have expected all of them... –  gatsu May 20 '13 at 8:30
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