quantum mechanics - On linearly dependent solutions of the Schrödinger equation

Can someone explain the following statement?

Let $\psi(x)$ and $\psi(x+a)$ represent two solutions of the Schrödinger equation with a periodic potential, $V(x)=V(x+a)$ so that these two solutions are representing the same physical electron state. Then $\psi(x)$ and $\psi(x+a)$ differ only by a constant, i.e., they are linearly dependent.

I already know that this constant has to have an absolute value equal to one, but I could not see the linear dependence.

Answer

The claim as you've stated it is false:

• The quote correctly points out that if $\psi(x)$ is an eigenfunction of $H=\frac1{2m}p^2+V(x)$ with eigenvalue $E,$ where $V(x+a)=V(x)$, then the translation $\psi(x+a)$ is also an eigenfunction of $H$ with eigenvalue $E$.


• The quote then implicitly assumes that the spectrum of $H$ is non-degenerate to conclude that $\psi(x+a)$ must be linearly dependent with $\psi(x)$.

However, that implicit assumption is false, in general, as is the result. (Moreover, the text is ambiguous about what it actually means by "solution of the Schrödinger equation", but that's probably a minor sin.)

For a simple counter-example, consider the eigenfunction $\psi(x) = \sin(kx)$ of the periodic hamiltonian $H=p^2/2m$, where the potential $V(x)\equiv 0$ is periodic under any real displacement $a$, so the theorem as claimed by the quote should in principle apply to it. However, setting $a=\pi/2k$ we obtain $$ \psi(x+a) = \cos(kx), $$ which is linearly independent with $\psi(x)$; this is inconsistent with the claim as quoted.

The second bullet point above also shows why this isn't more obvious in practice - you require a degenerate hamiltonian to get around the restrictions, and 1D hamiltonians don't generally have a lot of degeneracy. However, once you have that in mind, it is perfectly easy to construct non-translation-invariant eigenfunctions of translation-invariant hamiltonians.

More generally, your text has the Bloch theorem backwards:

• The theorem proves that, because $H$ and $T_a$ commute, then there exists at least one shared eigenbasis between the two, i.e. an eigenbasis of the hamiltonian that is translation invariant, i.e. the Bloch-wave basis.


• The theorem does not prove that all possible eigenbases of the hamiltonian are translation invariant. That's because that result is false unless the hamiltonian is degenerate.

There are very similar considerations in more depth in my answer to Translationally invariant Hamiltonian and property of the energy eigenstates.