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Say, an operator is being expressed in a matrix form, using its orthonormal eigenstates as a basis. We have $ A|n\rangle = k|n\rangle$.

To get the matrix representation, what we do is :

$ A_{mn} = \langle m|A|n\rangle = k\langle m|n\rangle$.

However, here comes my confusion. We know that $m$ represents the row and $n$ represents the column, and moreover, the eigenstates are orthonormal, and so their product must be the Kronecker delta. The indexing of the Kronecker delta is not clear to me.

$ A_{mn} = \langle m|A|n\rangle = k\langle m|n\rangle = k\delta_{m,n}$ or $\space k\delta_{n,m}$ ? In either case, only the diagonal terms remain.

This seemingly trivial fact is unclear to especially when we have slightly complex matrices. For example, consider the creation operator in the harmonic oscillator, expanded using the oscillator eigenstates.

$ a^\dagger_{mn} = \langle m|a^\dagger|n\rangle = \sqrt{n+1}\langle m|n+1\rangle = \sqrt{n+1}\delta_{m,n+1}$ or $\space \sqrt{n+1}\delta_{n+1,m}\space ?$

What do the indices of the Kronecker delta refer to ?

Is the element non-zero when $m^{th}$ row is equal to $(n+1)^{th}$ column, for example, $ A_{01} , A_{12}, A_{23}$ ?

Or, is the element non-zero when $m^{th}$ column is equal to $(n+1)^{th}$ row, for example, $ A_{10} , A_{21}, A_{32}$ ?

A third doubt came to my head about this. What if both are exactly equivalent, and its just saying that instead of $m$, we just need $n+1$. So the column remains the same $n$, the row becomes $n+1$ instead. Unlike the previous interpretations, here the column number is fixed from the beginning, and we are just choosing the row. Thus in the representation only $a^\dagger_{mn} = \sqrt{n+1}\langle n+1|a^\dagger|n\rangle = a^\dagger_{(n+1) n} $ elements survive.

Can someone point out which of my interpretations of the Kronecker delta function indices are correct ? I think it is the third one, I just want to be sure.

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    $\begingroup$ Convince yourself that, on the diagonal, you may transpose the Kronecker delta indices. Illustrate with a small, e.g. 3x3 matrix. $\endgroup$ Jul 2, 2021 at 20:40
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    $\begingroup$ $m$ and $n$ refer to indices of the original matrix in your chosen basis, so both $\delta_{m,n}$ and $\delta_{n,m}$ encode the same constraint on the original matrix. In your third interpretation, when $n$ varies, both the row and the column seem to be varying... $\endgroup$ Jul 2, 2021 at 20:50
  • $\begingroup$ What is the difference with this other question of yours? physics.stackexchange.com/questions/649182/… $\endgroup$ Jul 3, 2021 at 3:26

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I hope this helps. As you already said you have: \begin{equation} A_{mn}= <m|a^{\dagger}|n> = \sqrt{n+1}<m|n+1> = \sqrt{(n+1)}\delta_{m,n+1} \end{equation} As a first thing notice that the overlap matrix for the real solutions of the harmonic oscillator is symmetric: \begin{equation} S_{m,n+1} = \delta_{m,n+1} = \delta_{m+1,n} = S_{n+1,m} \end{equation} but your matrix is not because of the $sqrt{(n+1)}$ in front. Now when you define $A_{mn}$ as before you're not fixing either one index nor the other, what the delta is telling you, in this case, is that for a matrix element to be different from zero the $m$ (row) index must be equal to the column index (n) plus one. If that is not the case you simply get zero. What you were writing before is that \begin{equation} A_{mn} = 0\;\; if \;\;m\neq n+1 \;\;\;\;\;\; while \;\;\;\;A_{mn} = \sqrt{(n+1)} \;\; if\;\;m = n+1 \end{equation} Basically, the order of the indexes is fixed at the beginning when you say that you want to compute $A_{mn}$ so that m will always be the row and n the column giving you the matrix. Then the $\delta$ is only telling you in short the condition for the element not to be zero. Moreover, since the who is the row and who is the column is fixed, you will have that the non zero elements according to what I wrote before are: \begin{equation} A_{10}\;\;A_{21}\;\;A_{32}... \end{equation} I hope this helped, have a nice day.

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  • $\begingroup$ This is exactly what I was looking for, thank you so much ! $\endgroup$ Jul 2, 2021 at 21:31
  • $\begingroup$ In my third interpretation, I tried to explain something similar to what you wrote here. That, $n$ remains the column, $m$ just becomes $n+1$ for it to be non-zero. Then we can vary $n$ to get the row and the column number. I might have wrongly used the word 'n' is fixed. However, thank you for this explanation, so much ! $\endgroup$ Jul 2, 2021 at 21:36

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