The following problem is out of my quantum mechanics textbook.
Assume that two operators $H$ and $\Gamma$ commute. Show that if $|\psi\rangle$ is a non-degenerate eigenvector of $H$, that is, $H|\psi\rangle = \epsilon|\psi\rangle$ and $|\psi\rangle$ is the only eigenvector belonging to eigenvalue $\epsilon$, then $|\psi\rangle$ is also an eigenvector of $\Gamma$, that is $\Gamma|\psi\rangle = \gamma|\psi\rangle$. Note, however, that $\gamma \ne \epsilon$ that is the eigenvalues are not the same.
Before moving on, this question is in regards to symmetry in quantum mechanics, and is based on an introduction to group theory. Here, $\Gamma$ represents a rotation matrix and $H$ is the system's Hamiltonian. But, we can consider this as two general operators if we wanted too.
If we perform a rotation on the Hamiltonian system, we have
\begin{equation}
\Gamma H|\psi\rangle = \Gamma \epsilon|\psi\rangle
\Rightarrow
H\Gamma|\psi\rangle = \epsilon\Gamma|\psi\rangle
\Rightarrow
H\Gamma|\psi\rangle = \epsilon\gamma|\psi\rangle
\Rightarrow
\Gamma|\psi\rangle = \gamma|\psi\rangle \quad \text{IFF } \epsilon \ne \gamma.
\end{equation}
The above work is the correct solution to the problem, but I could use some help with understanding what is going on. Firstly, I'm looking for an explanation to
…then $|\psi\rangle$ is also an eigenvector of $\Gamma$, that is $\Gamma|\psi\rangle = \gamma|\psi\rangle$.
I'm also looking for an explanation of the note,
Note, however, that $\gamma \ne \epsilon$ that is the eigenvalues are not the same.
My professor stated that $\Gamma$ is 'just' a scaled version of $H$, and therefore it has its own eigenvalue but the same eigenvector. I know the argument for this is subtle, I would like a different perspective on this question.
Please let me know if I need to elaborate any further.
Thank you!
Best Answer
Your second $\implies$ assumes $|\psi\rangle$ is an eigenvector of $\Gamma$; don't do that. Instead, note that $$H|\psi\rangle=\epsilon|\psi\rangle\implies H\Gamma|\psi\rangle=\epsilon\Gamma|\psi\rangle\implies\exists\gamma(\Gamma|\psi\rangle=\gamma|\psi\rangle),$$since the $\epsilon$-eigenspace of $H$ is $1$-dimensional. The statement $\gamma\ne\epsilon$ cannot be proved; it's certainly not true, for example, if you take $\Gamma=H$. The statement
was a little clumsy, but was intended to mean that $\gamma$ is not in general equal to $\epsilon$, i.e. you shouldn't conflate them as concepts. Similarly,
isn't how I would have said what your professor was getting at, which is that we have the operator equation $\gamma H=\epsilon\Gamma$ on this $1$-dimensional eigenspace of $H$. Obviously a choice such as $\Gamma=H^2$ doesn't in general allow this to extend to the full Hilbert space.