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When we are solving the electromagnetic wave equation,

\begin{equation} \frac{\partial^{2} E}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}, \end{equation}

by separation of variables, that is, by assuming that the solution has the form $E(x,t)=\chi(x)T(t)$, we have to introduce a constant $k$:

\begin{equation} \frac{\partial^{2} \chi(x)}{\partial x^2}-k^2\,\chi(x)=0 \end{equation}

\begin{equation} \frac{\partial^{2} T(t)}{\partial t^2}-k^2c^2\,T(t)=0 \end{equation}

and we end up with solutions of the form $\exp\bigg[ik(x\pm ct)\bigg]$. This constant $k$ is given by $k=2\pi/\lambda$ is related to the energy $\epsilon$ of the wave through

\begin{equation} k=\frac{\epsilon}{\hbar c}. \end{equation}

How is this relation established? What additional information do you have to bring in to relate this constant you just brought up as a helping hand in solving the wave equation to the momentum and the energy of the wave?

Thank you very much.

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    $\begingroup$ Not sure what you mean. In classical EM, it is not established at all. You need QM (de Broglie's ansatz at the very least) to make that connection. $\endgroup$
    – NickD
    Commented Jul 3, 2017 at 3:56
  • $\begingroup$ How is the connection made in quantum mechanics? $\endgroup$
    – Gabu
    Commented Jul 3, 2017 at 5:31
  • $\begingroup$ They are connected in QM via Planck's Law: $E=\hbar c k$. $\endgroup$ Commented Jul 3, 2017 at 8:05

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