# Non-commuting operators can't share any eigenvector

In an introductory Quantum Mechanics textbook, I found the following statement:

For two Hamiltonians $H$ and $H'$, non commuting with each other, but commuting with the same group of translations ${\cal{T}} (\vec{R})$ an eigenvector of $H$ can't be an eigenvector of $H'$.

But I don't see how $[H,H']\neq 0$ implies that $[H,H']$ cannot vanish for a specific eigenvector $\alpha$ of $H$, making it a shared eigenvector with $H'$.

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What textbook are you referring to? –  Qmechanic Oct 18 '12 at 12:47
French QM textbook by Aslangul: amazon.fr/… The chapter on electrons in a crystal. –  Just_a_wannabe Oct 18 '12 at 13:38

By the following example you see that indeed $[H,H']\ne 0$ doesn't imply that there is no eigenvector-kernel.

The context of the question is still not completely clear but I assume the translation is supposed to be onto and so the situation is as such: If there were a translation $T$ communing with both operators, it would also commute with the commutator. And then by $$[H,H'](T\alpha)=T([H,H']\alpha)=0,$$

you would spread the kernel via $\alpha \rightarrow T\alpha$ to cover all of space, hence making $[H,H']$ annihilate everything. That means the Hamiltonians commute - a contradition.

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I think you got the point at the end (plus thanks for the numerical example). The context is about Bloch theorem and Bloch wave-functions, to describe for example electrons in a periodic static crystal. –  Just_a_wannabe Oct 18 '12 at 13:57
Leaving the part about the translations aside, one could formulate the eigenvalue equation $H\alpha = \epsilon\alpha$, while $H^{\prime}$, applied to $\alpha$, generally equals some function $f(\alpha)$. The commutator would then be:
$$[H,H^{\prime}]\alpha = HH^{\prime}\alpha - H^{\prime}H\alpha =Hf(\alpha) - \epsilon f(\alpha) \neq 0 \\ \Rightarrow Hf(\alpha) \neq \epsilon f(\alpha)$$
This apparently only holds true if $f(\alpha)$ is not just $\alpha$ times a constant; thus, $\alpha$ cannot be an eigenvector to $H^{\prime}$.
Hum.. I'm not sure about your view on the problem, you're postulating from the beginning that $\alpha$ is not annihilated by the commutator, which is what I am questioning. Why should it annihilate every eigenvectors of $H$? –  Just_a_wannabe Oct 18 '12 at 13:54
Ah, I think I understand your question now. You want to know if there are for the "general" non-vanishing commutator $[H,H^{\prime}] \neq 0$ any "special" eigenfunctions $\alpha$ which make the expression $[H,H^{\prime}] \alpha$ vanish. However, the commutator is indepedent of the function you are applying it to; in other words, $\alpha$ may be just any function. (If I am not mistaken, it won't even necessarily have to be an eigenfunction of $H$ or $H^{\prime}$). The only case where the expression would indeed equal zero is $\alpha$ being the zero vector, which is a non-physical solution. –  Antimon Oct 18 '12 at 15:22