# Wavelength as an observable in quantum mechanics?

Your student is correct, and there's no problem with the "wavelength" observable. The wavelength of a state $|p\rangle$ of definite momentum $p$ is just $h/p$. So we can define the wavelength operator by $$\hat{\lambda} |p \rangle = \frac{h}{p} |p \rangle.$$ Mathematically, this is equally as (il)legitimate as the momentum operator. In other words, it can't be the case that mathematical formalities prevent us from introducing it in quantum mechanics courses, because we already do plenty of things that are just as mathematically "wrong".

The physical reason we don't care much about it is just what you said: our classical theories are built around momentum, not wavelength, so upon quantization it's the momentum that shows up everywhere. It's the momentum that is squared in the kinetic energy, and which is affected by force, and so on.

$\lambda=h/p$ (which is the absolute of the momentum, so there is no sign) and I think that this is a perfectly fine operator. In **momentum space**, it holds that

$$\hat{\mathbf p}=\mathbf p$$

and thus

$$\hat{\mathbf \lambda}=\frac h {|\mathbf p|}.$$

I think the historical aspect is relevant here that Heisenberg started matrix mechanics with the explicit intention of only describing measureable values (as input as well as output of the calculation) in opposition to other quantum mechanical theories that introduced unobservable "structure".

Another explanation why this is not as present as other observables might be that $\lambda=2\pi/|\mathbf k|$ and $\mathbf k$ is universally present in quantum mechanics and de facto equivalent, even better (it adds direction, which is important).