Is momentum operator p is Hermitian only for a normalized wave function?What is the case for the box normalization as done for a free particle?

Actually when we prove the Hermiticity of the momentum operator, we do simple by parts integartion and use the scalar products.I never bothered about whether the wave function is normalized or not.

Can anyone suggest anything?

Hi kolahal_b,

Here is the text of the more detailed question you sent to me, so that everyone can see both the question and the answer.



The proof that the momentum operator is hermitian, which uses integration by parts, DOES FAIL when the wave functions do not vanish at infinity. However, that by itself does not mean that the conclusion is false, only that this proof did not work.

It is not correct to say that the momentum operator is hermitian only for normalized functions, because that assumption is too strong. The proof using integration by parts works as long as the term evaluating the product of the functions vanishes. This is certainly true for normalized functions, since if this limit did not vanish, they could not have a finite square integral. Of course, if you take a normalized function and multiply it by an arbitrary constant, it will no longer be normalized, but it will still work in the proof. So your professor is certainly wrong on that point, although that is not the major question here.

I think it IS true that if all the eigenvalues are real then the operator is hermitian. (Write both sides of <Φ* | Ψ> as sums of eigenstates of the operator, so it becomes a doubly indexed sum. Then each term is either zero or < Ψ_i | Ψ_j>. Inserting the operator on either the right or the left of each term produces the same value, since the eigenvalue is real. Therefore the operator can be in either the bra or the ket, with the same result, so it is hermitian.)

The real question here is about the application of operators to functions that do not vanish at infinity. The wave functions in quantum mechanics are supposed to be in a Hilbert space, which requires that the inner product be defined. Since this inner product is computed by integration, the requirement is that wave functions must be L2 functions. The Hilbert space of quantum mechanics is basically a space of L2 functions. The trouble is that the eigenfunctions of the momentum operator are NOT L2 functions, because their norm is infinite. They are limits of L2 functions, in the sense that you can take exp(ikx) and multiply it by a gaussian function to make it have a finite norm, which puts it inside the Hilbert space, then take the limit as the width of the gaussian goes to infinity, to get back the original exp(ikx).

Should these functions be excluded from quantum mechanics then? I don't think so; they simply require more careful handling than functions with a finite norm. One way to do this is to treat them not as functions but as "distributions," which basically means to think of them as operators on a DIFFERENT set of functions called the "test functions." That way, if you can prove a certain property is true when you try it out on every test function, then you can say it is true for your wave function. This is very similar to the way you prove things about operators: you show that the property is true for any function that the operator acts on, and then from that you conclude that the property is true of the operator itself. This is just how hermiticity is usually checked.

For the question you have right now, let's try this: suppose you insert the function exp(-x^2 / 2s^2) next to each of Φ and Ψ in the integral. Here, s is the standard deviation of the gaussian function. Now do the integration by parts calculation again, taking care to use the product rule. You will now find extra terms that involve the derivative of the gaussian, and these terms will have a factor of 1/s^2 in them. Consider the case where s is taken very large (not infinity). Then the terms with 1/s^2 become very small compared to the other two terms; all the terms are growing if the Φ and Ψ do not vanish at infinity, but these terms become negligible compared to the terms with ∂Φ*/∂x and ∂Ψ/∂x. Therefore, for very large s, the hermiticity condition is approximately true, and as s grows, the condition is satisfied more and more closely. Therefore, it is not unreasonable to say that in the limit, even though the integrals themselves diverge, there is a sense in which hermiticity is true.

So you see it is really a mathematical question more than a physical one. Dealing with function spaces and distributions is rather technical, and your professor perhaps does not want to spend the time to go into it deeply.

Hope that helps!

--Stuart Anderson