# Are separable solutions to the Schrödinger equation always complete?

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?

• This follows from the fact that any Hermitian operator is diagonalizable with orthonormal eigenvectors. Griffiths discusses this fact in chapter 3 of his textbook. The Hamiltonian is a Hermitian operator, so it admits a complete orthonormal basis which spans the Hilbert space. Oct 6, 2021 at 19:17

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.

1. By the fundamental theorem of algebra, $$\mathrm{det}(H-\lambda \mathbb I)=0$$ has at least one solution - say, $$\lambda_1$$. This implies that there exists at least one non-zero vector $$\hat e_1$$ (which we normalize for convenience) such that $$(H-\lambda_1 \mathbb I) \hat e_1 = 0 \iff H\hat e_1 = \lambda_1 \hat e_1.$$
2. Because $$H$$ is self-adjoint, we have $$\lambda_1 = \langle \hat e_1,H\hat e_1\rangle = \langle H \hat e_1 ,\hat e_1 \rangle = \overline{\lambda_1} \implies \lambda_1\in \mathbb R$$
3. Let $$\{\hat e_1\}^\perp$$ denote the orthogonal complement of $$\hat e_1$$ - that is, the set of all vectors $$v\in\mathscr H$$ such that $$\langle \hat e_1,v\rangle = 0$$. Because $$H$$ is self-adjoint, we have that $$\langle \hat e_1,Hv\rangle = \langle H\hat e_1,v\rangle = \lambda_1 \langle \hat e_1,v\rangle = 0 \implies Hv \in \{\hat e_1\}^\perp$$ We say that $$\{\hat e_1\}^\perp$$ is invariant under the action of $$H$$. As a result, if we let $$\hat e_1$$ be the first element of our orthonormal basis, then $$H$$ takes the form $$H = \pmatrix{\lambda_1 & \matrix{0 &\cdots&0}\\\matrix{0\\\vdots\\ 0} & H'}$$ where $$H'$$ is an $$(n-1)\times (n-1)$$-dimensional self-adjoint matrix. This process can be repeated for $$H'$$ and so on, eventually yielding a diagonal matrix with real entries and the claimed basis of eigenvectors.

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.

• Why is it obvious that the TIS is always self-adjoint? I mean if you take a circular waveguide and fill it with dielectric only partially you get complex eigenvalues. When I first learned of this I could not believe it but then I saw with my own eyes ... Oct 7, 2021 at 1:25
• @hyportnex Indeed, the self-adjointness of an unbounded operator is rarely obvious (as opposed to its Hermiticity, which is easy to check). Checking for self-adjointness requires a careful analysis of operator domains and boundary conditions, and is typically beyond the scope of an undergraduate course in QM. However, one can show that if $V$ is locally square-integrable and bounded from below then $H := -\frac{1}{2}\nabla^2 + V$ is [...] Oct 7, 2021 at 2:20
• @hyportnex [...] essentially self-adjoint on the space of smooth, compactly supported functions, and therefore has a unique self-adjoint extension. If the Hilbert space in question is not $L^2(\mathbb R^d)$ - e.g. the waveguide you reference - then boundary conditions may play an important role in deciding whether the operator is self-adjoint (or even Hermitian). Oct 7, 2021 at 2:23