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So, one of my homework problems reads the following

Let $A$ and $B$ be commuting operators and $| \psi_i \rangle$ denote the eigenfunctions of $A$. Show that $\langle \psi_i |B| \psi_j \rangle=\delta_{ij}$

I have tried to approach the solution in the following way:

Since $A$ and $B$ commute, we have, for any vectors $| \psi_i \rangle$ and $| \psi_j \rangle$, $$\langle \psi_i |AB| \psi_j \rangle=\langle \psi_i |BA| \psi_j \rangle$$ Let $\lambda_i$ be the eigenvalues of $A$ for each vector $|\psi_i \rangle$. Therefore, solving the above equation, we get- $$\lambda_i^*\langle \psi_i |B| \psi_j \rangle=\lambda_j\langle \psi_i |B| \psi_j \rangle$$ This leaves us with- $$(\lambda_i^*-\lambda_j)\langle \psi_i |B| \psi_j \rangle=0$$ From here, I have certain questions on proceeding further-

  1. Is it to be assumed that $A$ is hermitian? If so, then since $\lambda_i^*=\lambda_i$, so for $i \neq j$, we have $\langle \psi_i|B|\psi_j \rangle=0$. But which theorem states that $A$ can,be considered Hermitian here.
  2. If $i=j$, how do I go further and prove $\langle \psi_i|B|\psi_j \rangle=1$
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    $\begingroup$ The problem looks to be incorrectly stated. For a general operator $\hat{B}$, you don't expect its matrix elements to be $\delta_{ij}$. Perhaps they mean $B_{ij}$? Also, in your point 2 you have identified a very important point: if you have degenerate eigenvalues, then the prove becomes significantly more involved. $\endgroup$
    – ProfM
    Jul 9, 2020 at 16:56
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    $\begingroup$ Yes, the problem statement seems incorrect. Even if hermitian operators $A$ and $B$ commute, it does not mean that any eigenbasis of $A$ is an eigenbasis of $B$. Commutation only guarantees the existence of a common eigenbasis: $(1,1,0)$ is an eigenvector of $\mathrm{diagonal}(1,1,-1)$ but not an eigenvector of $\mathrm{diagonal}(1,-1,1)$, despite the fact that diagonal matrices commute. $\endgroup$
    – secavara
    Jul 9, 2020 at 17:22
  • $\begingroup$ The only way I can make sense of the question is perhaps if it meant to say that the operator $B$ commutes with every operator $A$, in which case $B$ would have to be the identity. $\endgroup$
    – Philip
    Jul 9, 2020 at 19:29
  • $\begingroup$ A lot of details seems amiss. If this problem is from a book, kindly post the reference. I do not see how else this can be true. Also, at least there should be a factor with the delta function (if at all it can be true) since nothing fixes the normalisation of the B expectation values. $\endgroup$
    – Lelouch
    Jul 9, 2020 at 19:31

3 Answers 3

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The equation $\langle\psi_i|B|\psi_j\rangle=\delta_{ij}$ does not seem right (unless $B$ is the identity, which would make this a strange question). To answer your questions in order:

  1. Not necessarily, although if this is a homework on quantum mechanics it would make sense that the operators in question correspond to observables and so must be Hermitian.

  2. This will only be true if $B$ is the identity matrix, since every state vector is an eigenvector of the identity operator with eigenvalue $\lambda=1$. In the event that this is what the question is asking, you will just find $$\langle\psi_i|B|\psi_i\rangle=1\cdot\langle\psi_i|\psi_i\rangle=1\cdot 1=1,$$ since $\langle\psi_i|\psi_i\rangle$ is the norm of $\psi_i$, which, if normalised, is just $1$.

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A short answer to your question and to the confusion in the comments is that since $A$ and $B$ commute they are simultanously diagonalizable. Therefore your question is true up to some constants $\lambda_j$ in front of the $\delta_{ij}$. Your proof is essential correct and does not require hermiticity.

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There are at least two problems with the problem as stated:

  • as @secavara mentioned, having a common basis of eigenvectors is not enough. We would certainly require the $|\psi_i\rangle$ to be common eigenvectors.
  • Demanding $\langle\psi_i|B|\psi_j\rangle = \delta_{ij}$ is certainly too much, probably it was meant to be that $B$ is diagonal in this basis.

As an example of where things go horribly wrong, take $A$ the identity matrix. This commutes with any $B$, and every vector is an eigenvector. Let's just take the standard basis and denote it $|i\rangle$. In this basis, $\langle\psi_i|B|\psi_j\rangle = \langle i|B|j\rangle$ is just the $i,j$ component $B_{ij}$ of $B$, which can be whatever you want.

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