I'm just starting to learn quantum mechanics, and the book I'm reading (Griffiths) states that every solution to the Schrödinger equation can be written as a linear combination of the separable solutions:
$$
\Psi(x,t) = \sum_{n=1}^{\infty}c_n\psi_n(x)e^{-iE_nt/\hbar}.
$$
However, it does not provide a proof that the set of $\psi_n(x)$ is a complete basis, even for arbitrary $V(x)$. I understand that completeness can be proven in specific cases such as the infinite square well, simple harmonic oscillator, et cetera, by solving for the separable solutions first. I also know that if $V$ is not time-independent the whole separation of variables scheme falls apart. My question is, do the separable solutions always form a complete basis, even for arbitrary $V(x)$? If so, what does the proof look like?
Are separable solutions to the Schrödinger equation always complete
hilbert-spacequantum mechanicsschroedinger equationsuperpositionwavefunction
Related Question
- [Physics] Why is the general solution of Schrodinger’s equation a linear combination of the eigenfunctions
- [Physics] Energy Eigenfunction Completeness
- [Physics] Using separation of variables to solve Schrödinger equation for a free particle
- Separability of Solutions of Schrödinger Equation
- Check that any linear combination of solutions is itself a solution to the time-dependent Schrödinger equation
Best Answer
This result is called the spectral theorem. For a finite-dimensional Hilbert space $\mathscr H$, the statement is that given any self-adjoint$^\ddagger$ operator $H$, there exists an orthonormal basis $\{\hat e_i\}$ consisting of eigenvectors of $H$, and that all of the corresponding eigenvalues are real.
The proof of this statement goes as follows.
For an infinite-dimensional Hilbert space $\mathcal H$, this situation becomes more complicated because the spectrum $\sigma$ of an arbitrary operator can consist of discrete points (called the point spectrum, $\sigma_p$) as well as a continuum (called the continuous spectrum, $\sigma_c$).
If the spectrum of $H$ is pure point (so $\sigma_c = \emptyset$), then the proof is similar in spirit to the finite-dimensional case, but there are technicalities which come into play if $H$ is not bounded; nevertheless, the conclusion is the same except for the fact that the basis in question does not have a finite number of elements. If the spectrum of $H$ contains a continuous part, then even more technicalities arise, and the full machinery of functional analysis is required; in physics, this operationally corresponds to the appearance of non-normalizable (or generalized) eigenstates, such as the ones which appear for the free particle Hamiltonian $H:= \frac{\hat P^2}{2m}$.
$^\ddagger$It's easy to show that if $H\neq H^\dagger$ but $[H,H^\dagger]=0$, then $H=A + i B$ where $A,B$ are commuting self-adjoint operators. This allows us to generalize this proof to so-called normal operators, and the only thing that changes is that the spectrum of $H$ may be complex.