11
$\begingroup$

I'm studying an undergraduate Quantum Mechanics course and I have some doubts about the solution of the Schroedinger equation by the separation of variables method. If we suppose that the solutions have the form $$\Psi(x,t)=T(t)\psi(x)$$ we obtain two equations, the first one give the time evolution phase factor $T_n(t)=e^{-iE_nt/\hbar}$ and the other one the "spatial wave function" $\psi_n(x)$.

So all the separable solutions have the form $$\Psi_n(x,t)=e^{-iE_nt/\hbar}\psi_n(x)$$ and these represents the stationary states.

If we sum these solutions we can obtain even non-separable solutions $$\Psi=\sum C_n\psi_nT_n.$$

But I can't find any postulate or theorem which states that every solution of the Schroedinger equation can be expressed in this form.

Are all the possible solutions of the equation expressible by (infinite) sum of separable solutions?

If I recall correctly math courses this can be expressed asking if the Hamiltonian operator eigenvectors are a complete set (basis) of the Hilbert space.

And in the case of continuous spectrum?

$\endgroup$
11
$\begingroup$

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent.

Sadly, very few interesting Hamiltonians satisfy that property (an example being the harmonic oscillator). In general, the solution of the Schrödinger equation exists for any initial condition $\psi_0\in\mathscr{H}$ (the Hilbert space), using the unitary one-parameter strongly continuous group $e^{-it H}$ associated to the self-adjoint Hamiltonian $H$ by Stone's theorem. The solution at any time $t\in \mathbb{R}$ is then simply written $$\psi(t)=e^{-itH}\psi_0\; .$$ Such solution is continuous in time, and with respect to initial data, but it is differentiable in time only if $\psi_0\in D(H)$, where $D(H)$ is the domain of the self-adjoint operator $H$. In the language of analysis of PDEs, that means that the Schrödinger equations, for self-adjoint Hamiltonians, are globally well posed on $\mathscr{H}$.

$\endgroup$
  • $\begingroup$ A minor comment: If the Hamiltonian is compact actually $0$ may be a point (the only one) of the continuos spectrum, even if it does not produce any problem with the expansion with respect to the Hilbert basis of eigenvectors... $\endgroup$ – Valter Moretti May 11 '16 at 9:02
  • $\begingroup$ @ValterMoretti what do you mean by "$0$ may be a point"? Zero energy? Or $\psi=0$? Or something else? $\endgroup$ – Ruslan May 11 '16 at 11:54
  • $\begingroup$ I mean zero energy, but no corresponding bound state. $\endgroup$ – Valter Moretti May 11 '16 at 12:10
  • $\begingroup$ Thanks, with a little more digging I understand a bit more the topic. However, since I attend an introductory course and I hadn't the occasion to thoroughly study functional analysis, could you describe in a simple way how to tell if the Hamiltonian operator of my physics problem is either compact or not? $\endgroup$ – skdys May 11 '16 at 19:59
  • $\begingroup$ @skdys Sadly, I don't know a simple way to tell whether an operator is compact. However, in all likelihood the Hamiltonian operator of your problem will be unbounded (because, roughly speaking, there is a term $\hat{p}^2$ in the kinetic energy, and it is unbounded). No unbounded operator is compact; but it may have compact resolvent (like the harmonic oscillator). A good candidate in order to have compact resolvent is to have a confining potential (that, roughly speaking, goes to infinity when $x\to\infty$); however this is not a strict rule... $\endgroup$ – yuggib May 12 '16 at 8:20
1
$\begingroup$

Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix.

In case of a partially continuous spectrum one gets the same, except that the sum must be replaced by an integral over a set of labels that resolves the complete spectrum. The continuous spectrum is labelled by the momenta of the possible scattering states. The transformation to the diagonalized operator is given by the so-called Moeller operator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.