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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!

NoVa
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2 Answers2

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  1. Suppose two operators commute: $\Gamma H = H\Gamma$.
  2. If $H$ has an eigenvector $|\psi\rangle$, we have $H|\psi\rangle = \epsilon |\psi\rangle$ for some multiple $\epsilon$.

  3. Apply $\Gamma$ to both sides of (2).

    $\Gamma H |\psi\rangle = \Gamma \epsilon |\psi\rangle$.

  4. On the left side, use the fact (1) that the operators commute so you can exchange their order $\Gamma H = H \Gamma$. On the right side, pull out the scalar $\epsilon$ (it's a number).

    $H\Gamma |\psi\rangle = \epsilon \Gamma |\psi\rangle$

  5. Note that statement (4) says that the vector $\Gamma |\psi\rangle$ is an eigenvector of $H$ with eigenvalue $\epsilon$. After all, when you apply $H$, you get an $\epsilon$ multiple of it back.

    $H\;\; \Gamma |\psi\rangle = \epsilon \;\; \Gamma |\psi\rangle$

  6. Because we assume that $|\psi\rangle$ is a non-degenerate eigenvector of $H$, the only eigenvectors of $H$ with eigenvalue $\epsilon$ are other multiples of $|\psi\rangle$. Hence (5) proves that $\Gamma |\psi\rangle$ is some multiple of $|\psi\rangle$.

    We can use $\gamma$ to denote the multiple: $\Gamma |\psi\rangle \equiv \gamma |\psi\rangle$.

  7. But this proves that $|\psi\rangle$ is an eigenvector of $\Gamma$ with eigenvalue $\gamma$, which is what we were trying to prove.

  8. Nothing in our argument requires the eigenvalues $\gamma$ and $\epsilon$ to be equal. For some operators, they will, and for some operators, they won't. A simple case where they'll be equal: $\Gamma = H$. A simple case where they'll be different: $\Gamma = 2H$.

  9. Finally, if all of the eigenvectors of $H$ are non-degenerate, our proof establishes that $H$ and $\Gamma$ have all the same eigenvectors. This means, roughly, that their eigenbases have "components in the same direction". But because of (8), each of these individual axes might be scaled differently.

user326210
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  • Thank you, for the very good explanation. By non-generate eigenvector, we mean an eigenvector to which only one eigenvalue exists and vice versa. Say that we have an eigenvector $\psi$ belonging to the eigenvalue $\lambda$. Why is $\lambda$ also the eigenvalue of an integer multiple of $\psi$? – Rasmus Andersen Jul 31 '23 at 16:28
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    $H|\psi\rangle = \lambda |\psi \rangle$. Because $H$ is linear, $H (m |\psi\rangle) = m H|\psi\rangle$, which because $|\psi\rangle$ is an eigenvector means that $H(m|\psi\rangle) = m(H|\psi\rangle) = m(\lambda |\psi\rangle) = (m\lambda) |\psi\rangle$. So the multiple $m|\psi\rangle$ is also an eigenvector of $H$ with eigenvalue $(m\lambda)$. ($m$ can be any scalar; does not have to be an integer.) – user326210 Aug 03 '23 at 00:17
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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

Note, however, that $\gamma\ne\epsilon$ that is the eigenvalues are not the same.

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,

$\Gamma$ is 'just' a scaled version of $H$, and therefore it has its own eigenvalue but the same eigenvector

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.

J.G.
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