Recently I was discussing a problem with one of my students in which she found that two states of the particle in a box were orthogonal and was then asked to give an example of an observable that would make these two states perfectly distinguishable. She thought of the wavelength. This took me by surprise, since I don't think a trained physicist would have ever come up with this answer, and yet it was hard for me to specify anything wrong with it.

The answer I came up with at the time was sort of a "meta" answer. I told her that usually when we talk about observables in quantum mechanics, we have in mind classical quantities like position, energy, momentum, or angular momentum, which can then be taken over into the microscopic context. Classically, an electron doesn't have a wavelength, so wavelength isn't this type of quantity.

I'm also wondering whether there is some purely mathematical answer. We want an observable to be representable by a linear operator that is hermitian (or maybe just normal). The wavelength operator would sort of be the inverse of the momentum operator, but it would be signed, whereas a sign isn't something we normally associate with a wavelength. In a finite-dimensional space, the inverse of a hermitian matrix is also hermitian. It's not clear to me whether there are new issues that arise in the infinite-dimensional case, or whether it matters that there is some kind of singular behavior as we approach zero momentum.

Is there any clear physical or mathematical justification for excluding wavelength from the full rights and privileges of being a quantum-mechanical observable?

  • 1
    $\begingroup$ A duplicate of physics.stackexchange.com/questions/278353/…. However, the linked question doesn't have an answer. :) $\endgroup$
    – user87745
    May 16, 2020 at 19:53
  • $\begingroup$ So you made it sound like there are many choices. Smart. But the only expected answer was 'energy'!? In the 1d box case $E(\lambda)$ is obv. monotone, so $\lambda$ makes sense as well. In most other cases, except obviously the free one, there just is no $\lambda$. So it does not make general sense as an observable. $\endgroup$
    – user257090
    May 17, 2020 at 9:05

2 Answers 2


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.

  • $\begingroup$ Wouldn't the two operators be defined over different domains? I'm not sure but naively I'd expect the conditions $\int p^2|\psi(p)|^2 dp < \infty$ and $\int p^{-2}|\psi(p)|^2 dp < \infty$ to be valid for distinct (although not disjoint) sets of functions $\psi(p)$. $\endgroup$
    – user87745
    May 17, 2020 at 9:13
  • $\begingroup$ @DvijD.C. No clue. I've gone through life always assuming everything was defined on every domain, and never gotten anything concrete wrong as a result. Certainly there may be technicalities here, but one could just as easily say that $\hat{p}$ is "illegitimate", whatever that means. $\endgroup$
    – knzhou
    May 17, 2020 at 18:40
  • $\begingroup$ If we now plug in $\hat{p}=-i\hbar\frac{\partial}{\partial x}$ and solve for $\hat{\lambda}$, we see that $\hat{\lambda}=2\pi i\int dx$. $\endgroup$
    – Kexanone
    Apr 1, 2021 at 3:22

$\lambda=h/p$ (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).


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