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In waveguide $u_{phase}u_{group}=c^2$ was a constant in classical electromagnetism.

However, the phase, although mentioned in a classical sense, seemed to have a strong quantum mechanical indication, where the group velocity $<=c$ may be well seen as a constraint from special relativity. Yet these two concepts were connected through the equation $u_{phase}u_{group}=c^2$.

My question was this: Is there a quantum mechanical or relativistic implication to group velocity times phase velocity being constant in a waveguide?

xray0
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ShoutOutAndCalculate
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1 Answers1

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This is generic feature of systems with relativistic energy-momentum relations, meaning $$E=\sqrt{p^{2}+m^{2}},$$ where time and space are rescaled to have the same units (i.e. $c=1$), and $m^{2}$ is any nonnegative constant. The phase and group velocities are $$u_{ph}=\frac{E}{p},$$ from $E=\hbar\omega$ and $\vec{p}=\hbar\vec{k}$ for a wave $\psi\propto\exp\left(i\vec{k}\cdot\vec{r}-\omega t\right)$; and $$u_{g}=\frac{\partial E}{\partial p}.$$

Setting these two speeds to be reciprocals (since the constant $c$ has been set to unity) gives a differential equation for the energy as a function of momentum, $E(p)$, $$\frac{dE}{dp}=\frac{p}{E}.$$ This is a first-order differential equation, so it has a one-parameter family of solutions. (That is, the general solution will have one undetermined constant.) In fact, this is a very easy separable equation, and a direct integration shows that the general solution is $E=\sqrt{p^{2}+m^{2}}$, with $m^{2}$ as the arbitrary constant.

So any system with relativistic dispersion relations will have $u_{ph}u_{g}=1$ (in these units).

Buzz
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