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I've recently seen the expression of the creation and annihilation operator in the harmonic oscillator in terms of matrices. What I've understood is that:

$$ A_{ij} = \langle i | A | j\rangle. $$

The $i$ and $j$ here are the basis states. Using this, I've managed to find the Matrix representation of position, momentum, creation, and destruction operators using the basis spanned by the harmonic oscillator eigenstates. However, I've seen a slightly different notation which doesn't make sense to me intuitively.

Take the creation operator for example:

$$ A_{nm} = \langle n | A | m\rangle = \sqrt{m+1} \langle n | m+1\rangle = \sqrt{m+1}\delta_{n,m+1}.$$

I'm not being able to understand intuitively, the Kronecker delta at the end of the above expression. My intuition leads me to think that $\delta_{n,m+1}$ refers to the $n^{th}$ row and the $(m+1)^{th}$ column of the matrix. For example $A_{12} , A_{23} ... etc,$ should be the elements of this matrix.

However, doing the calculation tells me the exact opposite. It appears that $\delta_{n,m+1}$ refers to the $n^{th}$ column and the $(m+1)^{th}$ row of the matrix instead. I'm unable to make sense of this intuitively. Note that, I've found the correct answer directly, by calculating each element using the first method. My problem is not understanding how the elements of the Kronecker delta relate to the elements of the matrix.

Any help in understanding and visualizing this intuitively, would be deeply appreciated.

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2 Answers 2

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Remember that when an operator acts on a state, the matrix goes behind the state, not in front. Insert a complete set $$ {\rm Identity}=|m\rangle\langle m| $$ into $A|n\rangle$ to get $$ A|n\rangle= |m\rangle\langle m|A|n\rangle, $$ so the $\langle m|A|n\rangle$ goes to the right.

This is just as it is in linear algebra. If you expand a vector as $$ {\bf x}= x^n{\bf e}_n $$ and want ${\bf y}=A{\bf x}$ when written in components to be the usual matrix multipication $$ y^n= {A^n}_m x^m $$ then $A$ acts on the basis vectors as $$ A({\bf e}_n)= {\bf e}_m {A^m}_n. $$ So again if a matrix is to act from the left on a column vector of components, it must act from the right on the basis vectors.

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  • $\begingroup$ Okay, so you are trying to say that in my example, since $A$ acts on $|m \rangle $ from the left and $|\rangle n$ from the right, in the Kronecker delta, $m+1$ is representing the row and similarly, $n$ represents the column. Okay, so in my first representation, n and m are the components, but in the Kronecker delta representation, they are the basis , and so they act differently. Right ? $\endgroup$ Jul 2, 2021 at 17:14
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    $\begingroup$ In any matrix the entry ${A^n}_m$ the index $n$ labels the row, and $m$ the column. For components $x^m$ the matrix acts to the right as ${A^n}_m x^m$, for basis vectors ${\bf e}_n$ you have ${\bf e}_n {A^n}_m$. It's easy as long as you rember upstairs (contravariant) indices must be paired with downstairs (covariant) indices in any sum. $\endgroup$
    – mike stone
    Jul 2, 2021 at 19:28
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In the context of the equation $A_{nm} = \sqrt{m+1} \delta_{n,m+1}$, the statements

  1. "$\delta_{n,m+1}$ refers to the $n$th row and the $(m+1)$th column of the matrix"
  2. "$\delta_{n,m+1}$ refers to the $n$th column and the $(m+1)$th row of the matrix"

are both incorrect; I would try to forget about them.

Instead let's take a more literal look at the equation $A_{nm} = \sqrt{m+1} \delta_{n,m+1}$. It is telling you that $A_{nm}$ is zero unless $n = m +1$, in which case it takes on the value $\sqrt{m+1}$. Thus we learn that $A_{nm}$ is zero unless $$(n,m) = (1,0), (2, 1), (3, 2), ...$$ in which case it takes on the value $$\sqrt{m+1} = \sqrt{0+1}, \sqrt{1+1}, \sqrt{2+ 1}, ...$$ That is, the nonzero matrix elements are $ A_{10} = \sqrt{1}, A_{21} = \sqrt{2}, A_{32} = \sqrt{3}, ...$

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