# Domain of the infinite square well hamiltonian

I am reading the book by Gitman et al. 'self-adjoint extensions in quantum mechanics'.

In the book, they give a precise definition of the domain of the hamiltonian of an infinite square well.

For me, the point is that, if a wave function belongs to the domain, then itself and its first derivative should be absolutely continuous. This means the tent function below is not in the domain---its first derivative is discontinuous.

This is somehow surprising to me. As I remember, we had excercises in which we were asked to expand the function in terms of the eigenstates of the well. We can then evolve the state in time. Everything seems okay.

So, what is the problem with this state? Why should we rule it out from quantum mechanics?

• What's the energy of this state? Apr 19 '19 at 10:03
• it is finite. But the second moment of the hamiltonian diverges Apr 19 '19 at 10:48
• Can you quote precisely what they say in the book? Do they require this property to hold almost everywhere or sth like that? Otherwise, we are debating about your interpretation of what the book says, rather about what the book says. Apr 19 '19 at 10:55
• As the function does not belong to the domain of the Hamiltonian operator, the energy variance is not defined in the said state. The expectation value could be defined in any cases (it is enough to decompose the state along the eigenvectors of the Hamiltonian and sum the series of the energies...to check it). Apr 19 '19 at 12:49
• However, all that has nothing to do with the time evolution of the state, which is always defined, even if the state does not beleng to the Hamiltonian domain, as it is implemented by a unitary operator. You should not rule out that state from QM! T Apr 19 '19 at 12:51

The simple answer to this problem is that the second partial derivative of one of the canonical variables appears in the Schrödinger equation when working with a continuous basis. In most problems this is the spatial coordinate as in $$i\hbar\frac{\partial}{\partial t} \psi(x, t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) + \Phi(x)\psi(x,t)$$ for the case of one dimension, which is applicable to your example, but it can also be the electric field among other things.