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Suppose that we have the four-vector potential of the electromagnetic field, $$A^i$$

The wave equation is given by $$\Box A^i=\frac {\partial^2}{\partial x^k \partial x_k } A^i= \left(\frac {1}{c^2} \frac {\partial^2}{\partial t^2}-\nabla^2 \right) A^i=0$$

Now the solution, for a purely spatial potential vector, is given by

$$\mathbf{A}(t, \mathbf{x})=\mathbf{a_k} \exp{i(\pm \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}}); \mathbf{k}.\mathbf{a}=0$$

To find the general solution we write the superposition as

$$\mathbf{A}(t, \mathbf{x})=\int (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}})) \frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

My question is that where this $$\frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

comes from? Shouldn't it be $$d^3\mathbf{x}$$

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  • $\begingroup$ a general solution depends (non-trivialy) on the boundary conditions of the problem, for some usual cases general solutions exist $\endgroup$ – Nikos M. Nov 10 '14 at 11:24
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It depends on your definition of the Fourier transform. Sometimes it is written with $1/(2\pi)$ for each one-dimensional variable; sometimes the coefficient is chosen $1/\sqrt{2\pi}$, and sometimes it is just $1$. The "inverse" Fourier transform has also some coefficient at the integral and it is a matter of taste which direct/inverse transform's coefficients one chooses.

(I think you meant $d^3\mathbf{k}$.)

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  • $\begingroup$ Why we do not say that $$\mathbf{A}(t, \mathbf{x})=\sum_{\mathbf{f(k)}\mathbf{g(k)}} (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}})) $$ Isn't superposition simply given by the above sum? Why and how we transform to the integral? $\endgroup$ – user56963 Nov 10 '14 at 10:25
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    $\begingroup$ Yes, it is often written like a Fourier sum $\sum_i$ over discrete $\mathbf{k}_i$ without any coefficient at it. But when you consider a continuous $\mathbf{k}$, the Fourier transform is used with some coefficient. You can consider it as a numerical redefinition of $\mathbf{f}$ and $\mathbf{g}$. $\endgroup$ – Vladimir Kalitvianski Nov 10 '14 at 10:32
  • $\begingroup$ where does $\mathbf{a_k}$ go when we change from discrete to the continuous $\mathbf{k}$? $\endgroup$ – user56963 Nov 10 '14 at 10:47
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    $\begingroup$ $\mathbf{a}_{\mathbf{k}}$ are those $\mathbf{f}(\mathbf{k})$ and $\mathbf{g}(\mathbf{k})$. In particular, when you apply the condition of reality to the vector-potential $\mathbf{A}=\mathbf{A}^*$, its relates $\mathbf{f}(\mathbf{k})$ and $\mathbf{g}(\mathbf{k})$ like this: $\mathbf{f}(\mathbf{k})=\mathbf{g}(\mathbf{k})^*$. $\endgroup$ – Vladimir Kalitvianski Nov 10 '14 at 11:46

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