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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.

Emilio Pisanty
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R. Ferreira
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    The claim as you've stated it is false (for a simple counterexample, consider any two plane-wave eigenfunctions of the periodic potential $V(x)\equiv 0$); either your text is wrong, or you're misquoting it. There's no way to tell without an explicit reference, though. – Emilio Pisanty Nov 28 '17 at 20:00
  • @EmilioPisanty $e^{ikx}$ and $ie^{ikx}$ do represent the same state even though they are two "different" eigenfunctions. I put different in quotes because, of course, they are only off by a overall constant, and so they represent the same state. To maybe answer the question, you can see they are linearly dependent because $e^{ikx} + i \left(ie^{ikx}\right)=0$ – Brian Moths Nov 28 '17 at 20:05
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    @R.Ferreira To be clear: is that quote your own understanding of the material (which you're asking other people to explain to you?), or are you directly quoting from somewhere else? All the usual rules of scholarship apply on this site, which means that you need to appropriately source all the material that you present. (I.e.: this answer does not meet the standards, in either its original or its current form.) – Emilio Pisanty Nov 29 '17 at 18:23
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    Could you give a source for the claim? – Brian Moths Nov 29 '17 at 18:28
  • @EmilioPisanty, I don't have a source for that claim, it was just a discussion with a friend. But we didn't know things like Bloch's theorem. We just were trying to study the translational symmetry in terms of a potential in the Schrödinger equation. – R. Ferreira Nov 29 '17 at 22:18
  • @EmilioPisanty, but as you remarked, and I said here today, if both functions are eigenfunctions with the same eigenvalue, then using the symmetry of the potential we can see that those functions have the same absolute value so that they differ by a complex phase, which means that they are linearly dependent. – R. Ferreira Nov 29 '17 at 22:25
  • @R.Ferreira as I said, the argument is flawed, it implicitly assumes that the energy level is non-degenerate, and that is not true in any case of interest in this context. I've already explained in my answer - you are trying to prove the converse of Bloch's theorem, which is false -, as well as the link at the end, and I don't see the point in explaining it again. – Emilio Pisanty Nov 29 '17 at 22:32
  • Regarding the claim, if that is your understanding and you want to offer it as a quote, then you still need to label it as such. I'm downvoting due to that. – Emilio Pisanty Nov 29 '17 at 22:33
  • @EmilioPisanty all right, you're very rigorous about such rules here! I still consider yor answer cause it helped me! – R. Ferreira Dec 03 '17 at 03:05

2 Answers2

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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.

Emilio Pisanty
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    The quoted text assumes that $\psi$ and its translation represent the same state, the quoted text does not assert that this is always the case. This is seen from the use of the word "let" at the beginning. The OP will probably find this information useful nonetheless. – Brian Moths Nov 28 '17 at 20:32
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    @NowIGetToLearnWhatAHeadIs The claim remains false, and a misunderstanding of the structure of Bloch's theorem; any attempt at explaining it will only be misleading at best. The only way to interpret the quote as ruling out the counterexample is by reading it as "Let $\psi(x)$ be an eigenfunction of the hamiltonian with periodic potential $V(x)=V(x+a)$ such that $\psi(x)$ and $\psi(x+a)$ represent the same physical electron state"; in that case, the conclusion that $\psi(x)$ and $\psi(x+a)$ are linearly dependent is synonymous with the premise and you can just cut out the eigenfunction bit. – Emilio Pisanty Nov 28 '17 at 20:36
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    Your quote is exactly how I read the claim, I figure the source is more likely to make a trivial true claim than a false claim. I also assume that this quoted text is supposed to be aimed at people who are learning the subject so naturally it would say seemingly simple and obvious things. – Brian Moths Nov 28 '17 at 20:40
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    @NowIGetToLearnWhatAHeadIs Then you have a much lower tolerance for misleading claims than I do; I'm not sure that's a good thing, but it's your prerogative. For me it's quite obvious that the text is attempting to teach Bloch's theorem without any real understanding of how it works, which can only lead to confused readers (like OP) or to ingrained misconceptions that are much harder to root out (which is worse). I stand by my downvote of your answer, as it only serves to reinforce those misconceptions. – Emilio Pisanty Nov 28 '17 at 20:43
  • @EmilioPisanty, yes, thanks, I proved the linear dependence with the hypothesis that they are eigenfunctions with the same energy. On the other hand, when I claim that both solutions are representing the same physical state of the electron, I realized that this is the same as saying they have the same energy E, but now I think I was wrong because the word "state" is not yet "measured" so that it is related to some energy E. A state isn't necessarily an eigenstate! This was the "mistake" of the claim, right? – R. Ferreira Nov 29 '17 at 17:12
  • @R.Ferreira If the claim is your own understanding, then it is ambiguous and it makes it unclear what you mean, i.e. you say "solution of the SE" like it's enough to unambiguously determine what you mean (which it isn't. TDSE or TISE? if TISE, at what energy?). Moreover, it does not follow from "they are eigenfunctions of the same energy" that they are linearly dependent (because in general your hamiltonian won't be non-degenerate, which is the only condition in which the implication holds), i.e. same energy and same state are not synonymous. – Emilio Pisanty Nov 29 '17 at 18:21
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In any vector space, two vectors $\mathbf{v}$ and $c\mathbf{v}$ that are different by a multiplicative constant, in this case $c$, are linearly dependent because $\mathbf{v} - c^{-1}\left(c \mathbf{v}\right) = \mathbf{0}$, (if $c=0$, then simply $0\mathbf{v}+c \mathbf{v} = \mathbf{0}$).

For example, consider two plane wave solutions to the shroedinger equation with constant potential: $e^{ikx}$ and $ie^{ikx}$. In this case $c=i$ so $c^{-1}=-i$, and we have $e^{ikx} + i \left(ie^{ikx}\right)=0$, so the functions are linearly dependent.

Brian Moths
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  • This isn't incorrect, but it's also not relevant. Yes, some states are translation-invariant; the claim, however, is that all eigenfunctions are translation invariant, which is false. – Emilio Pisanty Nov 28 '17 at 20:32
  • We have different interpretations of the claim; see my comment on your answer. – Brian Moths Nov 28 '17 at 20:33