In Griffiths 2nd Edition Introduction to Quantum Mechanics, it is stated that "Although none of the eigenfunctions of p lives in Hilbert space, a certain family of them (those with real eigenvalues) resides in the nearby suburbs, with a kind of quasi-normalizability." However, my understanding is that observables are represented by hermitian operators.

So my question is:

For an operator to yield an observable, the wave function should be square integrable and thus belong to a Hilbert space (a complete inner-product space). So if the eigenfunctions of the momentum operator do not belong to a Hilbert space, how is the operator Hermitian?


1 Answer 1


An operator being self adjoint or not depends greatly on the Hilbert space upon which it acts. The momentum operator is self adjoint on functions defined over $\mathbf R^3$ when acting upon functions that are square integrable (I.e $L^2$ functions).

The problem begins when one looks for eigen states or the operator because it turns out that plane waves are not normalisable (they can be though of as the limit of Gaussian wave packets as the width of th frequency envelope tends to zero, hence why they "live in the suburbs") - so momentum eigen states are not really in the Hilbert space, but that doesn't mean it is not Hermitian when acting on normalisable states.


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