# How to relate the wavenumber to the momentum and the energy of a electromagnetic wave?

When we are solving the electromagnetic wave equation,

$$\frac{\partial^{2} E}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2},$$

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 $k$:

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

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

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

$$k=\frac{\epsilon}{\hbar c}.$$

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.

• 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. Commented Jul 3, 2017 at 3:56
• How is the connection made in quantum mechanics?
– Gabu
Commented Jul 3, 2017 at 5:31
• They are connected in QM via Planck's Law: $E=\hbar c k$. Commented Jul 3, 2017 at 8:05