"However, eigenvalues are naturally complex numbers, without any units" why do you say that?

I don't really see how this is a problem at all. Why can't observables have units, like J, m, etc.?

Why does an eigenvalue have to be unitless? The operator "1 kilogram" is proportional to the identity matrix and has $1$ eigenvalue, which is exactly "1 kilogram". Of course you could add another postulate saying this isn't allowed, but then you can't describe real observables, so it would be pointless.

@DanielSank because that's literally the definition of being an eigenvalue. In particular you need the multiplication by the eigenvalue to be a vector in $V$, what does it even mean to multiply a vector by one meter?

Shay, I think you should remember that the operators themselves have units too, which is where the units of the observables come from. At least that's how I think it works.

@FelisSuper I don't see how you can do it naturally. You could say something like: an observable is an Hermitian operator plus units (which you interpret the numbers as having these units). However if you change the units the numbers must change, thus the operator itself must depend on the units. This means that you can have something like "Hamiltonian" in units of joules, but not just a bare hamiltonian.

Shay, let's take a look at a particular operator, namely the spin operator S_z. Its eigenfunctions are either states with spin up or states with spin down. Now, if you look at the form of the operator as a matrix, then you'll see that h bar is a factor in front of the pauli matrix. In other words, the operator has units of J*s.

@knzhou I don't agree that the operator "1 kilogram" exists, how do you define it mathematically as an operator on a Hilbert space? I don't need more postulates to disallow this, I just don't see how to make sense of it.

@FelisSuper Your example is very similar to what I said about momentum. I understand how you can see the units in these kinds of specific examples. I don't see how that arises mathematically from the definition. What does it even mean to multiply a vector by hbar (which has units)?

Energy, the eigenvalues of the TISE, don't have units? You learn something everyday! Siiighgh...

Shay, I think you need to remember that all an operator does to an eigenfunction, is that it recoveres the eigenfunction itself, times some number, which can have any unit. I don't think there is anything that tells us that this product of a vector and a number has to be a vector in the space itself.

If you're against having operators with units, let's back up a bit. The only thing you need to define operators with units, is to define scalars with units. Are you against scalars with units too, because they're not "rigorous"? If so, how do you distinguish 1 kilogram from 1 meter?

If you do believe that scalars with units can exist, then they obviously form a field. Just define your vector space as over that field and you can have operators with units.

@FelisSuper "I don't think there is anything that tells us that this product of a vector and a number has to be a vector in the space itself", it does if $\hat A$ is an operator from $V$ to $V$, which is by definition.

@knzhou I have no problem with numbers with units, but they definitely do not form a field. You can't add 1 kilogram and 1 meter.

@ShayBenMoshe In this picture, dimensional analysis is an additional constraint you can impose to forbid those kinds of operators as physically meaningless. Of course you could do other things. You could say that every dimensionful operator is a combination of a dimensionless operator and an element of $\mathbb{R}^3$, where e.g. 1 kg corresponds to the identity matrix and $(1, 0, 0)$. I mean, this is really trivial stuff, you could formalize it a million different ways and none of them make any difference whatsoever, which is why textbooks don't bother talking about it.

Shay, alright, that's actually a good point, and I am not sure about any of this, but I think that definition requires that the A operator is unitless, because if it is not, then as you say, this doesn't make much sense.

Unless, of course, the abstract vectors (in other words the ket vectors) can have any unit, which I think it actually might.

@knzhou Yes I agree that one can cook up some definition that take into account the dimensions of operators, but I think that it is not so trivial and that is why I ask the question. In particular I am afraid that if you want an operator with dimensions, you have to choose units to these dimensions.

@FelisSuper I am glad that you agree that there is a point here. And also abstract vector can't have any units, because how can you sum one with meters units and one with kilogram units?

Related Question: Do bras and kets have dimensions?

@BioPhysicist that's a very useful link, thank you.