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 $H'$.
 A: 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.
A: 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 $H^{\prime}$.
