In my physics class we always solve the Schrödinger Equation using separation of variables. However, this makes me wonder if we are really getting all solutions to the Schrödinger Equations in this way. In the case of the harmonic oscillator we where able to show that the solution was a basis of the Hilbert space $L^2$, so no other solutions could exist.
However, we solved the wave equation not assuming that the solutions where separable, we solved it by using the Fourier transform. This is of course under the assumption that the solution and its Fourier transform is at least $L^1$, so there would be other solutions that are not in $L^1$.
So, when I am using separation of variables I am automatically assuming that the solution has the form $\Psi(x,t) = A(x)B(t)$. But can't it be that I am missing some solution $\Psi(x,t)$ that is not separable?
Best Answer
The idea is that the (assumed time-independent) Hamiltonian $H$ is a self-adjoint operator on the Hilbert space $\mathscr H$, which means that any vector $\psi\in\mathscr H$ can be expanded in terms of its (possibly generalized) eigenstates $\phi_n$ as $$\psi= \sum_n c_n \phi_n$$ Now let $\psi_t$ be the state vector which represents the state of the system at time $t$. For each value of $t$, we can expand $\psi$ in terms of the $\phi_n$'s, with the expansion coefficients of course being different at different times. As a result, we have that $$\psi_t = \sum_n c_n(t) \phi_n$$ Therefore, we see that any $\psi_t$ can be expanded as a (time-dependent) linear combination of eigenstates of $H$. Indeed we need not even use eigenstates of $H$ specifically; any self-adjoint operator possesses a basis of (possibly generalized) eigenvectors, and any such basis would do. However, the use of the energy eigenbasis is convenient because application of the Schrodinger equation yields
$$i\hbar \frac{d}{dt}\psi_t = H\psi_t \implies \sum_n i \hbar \dot c_n(t) \phi_n = \sum_n c_n(t) H\phi_n = \sum_n E_n c_n(t) \phi_n $$ $$\implies \dot c_n(t) = -\frac{iE_n}{\hbar} c_n (t) \iff c_n(t) = e^{-iEt/\hbar} c_n(0)$$
where we used the fact that $H\phi_n = E_n\phi_n$.
The situation becomes more complex if $H$ is permitted to be time-dependent. In that case, $H(t)$ and $H(t')$ are different operators, which means the sets of eigenstates $\phi_{n,t}$ and $\phi_{n,t'}$ are also generally different. In principle, any set $\phi_{n,t}$ spans the Hilbert space and so the previous expansion could be used, but the coefficients are not so trivial to find because $\phi_{n,t}$ is generally not an eigenstate of $H(t')$. Still, the principle remains the same - given any basis $\phi_n$ which spans the Hilbert space, we can always write $$\psi_t = \sum_n c_n(t) \phi_n$$ though in the general case, the $c_n(t)$'s might be very difficult to find.