Why is E not equal to p^2/2m for a particle in a box?

[stevendaryl]

No, the Hilbert space is L^2 ([0,a], dx), that is the set of all (equivalence classes of) functions with finite norm, that is the Hilbert space is:

$H = \{\psi (x) | \int_{0}^{a} |\psi^2 (x)| dx <\infty \}$

The necessity of boundary conditions on the wavefunction stems from the fact that the Hamiltonian and the momentum are (in coordinates representation) derivative operators, hence their spectral equations are (Sturm-Liouville) differential equations whose unique solutions need boundary conditions (Dirichlet or Neumann).

But what in the definition of a "Hilbert space" prevents you from making a Hilbert space that consists of only the functions satisfying the boundary conditions?

 

[dextercioby]

Completion.

 

[vanhees71]

Well, but on your space, i.e., simply $\mathrm{L}^2([0,a])$ without any boundary conditions, $-\mathrm{i} \mathrm{d}_x$ is not even Hermitean, because
$$\int_0^a \mathrm{d} x \psi_1^*(x) (-\mathrm{i}) \psi_2'(x)=\psi_1^*(a) \psi_2(a)-\psi_1^*(0) \psi_2(0) +\int_0^a \mathrm{d} x [(-\mathrm{i}) \psi_1'(x)]^* \psi_2(x).$$
So even for getting the operator Hermitean you need boundary conditions to ensure that the integral-free part vanishes.

 

[stevendaryl]

Completion.

Completion in this sense is something like:

If $\psi_1, \psi_2, ...$ is a converging infinite sequence of elements, then the limit is also an element?

So a counterexample might be if $\psi_n(x)$ is the function that is equal to 1 in the interval $1/n < x < 1 - 1/n$, but is equal to 0 outside that interval. Each $\psi_n$ is in the space (of functions that are 0 on x=0 and x=1) but the limit is the constant function 1, which is not in that space.

Okay. Everything is illuminated now.