Non-hermiticity of fourth power of momentum operator in L=0 state of hydrigen atom

[jungljim77]

Non-relativistic QM.

I have read in many places that the fourth power of the momentum operator (P^4) is not hermitian on the L=0 states of the Hydrigen atom. Indeed, I was able to pove this using Fnlm for the nlm state of hydrogen, computing <Fn00 | P^4 Fm00> and doing the integration by parts. Some of the boundary terms do not disappear.

However, I was unable to reconcile this with the theorem (which I have also seen in many a book) that if T is a hermitian operator then so are all finite powers of T. If this theorem holds, then since P^2 is hermitian on L=0 states, then so is (P^2)^2. I considered the sequence <Fn00 | P^4 Fm00> = <P^2 Fn00 | P^2 Fm00> = <P^4 Fn00 | Fm00> and looked back through my integration by parts calculation to see if I could find a step that just didn't jive. The only suspicious step I see is moving the first P^2 from the ket over to the bra. Perhaps P^2 is not hermitian when acting on the function (P^2 Fm00) although I couldn't say specifically why that might be the case -- P^2 Fm00 is normalizable (although just 'barely').

If this is the case, it seems the theorem needs to be adjusted to include some sort of 'closure' for the operator in the sense that all powers of the operator acting on a function must produce functions that are in the domain of hermiticity of the operator. (I don't know the technical terms for this sort of closure because that is not my area, but hopefully that makes sense).

Can anyone confirm or deny my statement and possibly enlighten me a bit more?

Is there a better more complete version of the theorem that might apply to the L2(f) space used for states in QM? I am starting to use powers of operators frequently in my work and it could be tedious if I have to constantly prove which powers of an operator acting on some function are hermitian and which are not. (Does this have anything to do with eigenfunctions of the free particle not being normalizable?) Are questions of hermiticity and powers of operators handled in a cleaner way in relativistic QM or a QFT?

Thanks

 

[DrDu]

I would guess that the theorem about the hermiticy of powers of an hermitian operator only holds for bounded operators. P^2 does not belong to that class. Hence, without having checked explicitly, I suppose that P^2|Fm00> is not in the domain on which the operator P^2 is defined.

 

[Bill_K]

The L=0 wavefunctions are extremely well behaved. They are all like (polynomial in r) times e^-r, which means they are finite at the origin and converge at infinity. I can't see anything wrong with applying pⁿ to them as many times as you like, or anything that would cause pⁿ to be non-Hermtian.

 

[Bill_K]

Wait, never mind, I take that back. There's a kink at the origin, that's the problem. e^-r has a discontinuous slope.