Could someone explain/expand equation \eqref{7.16} from Susskind's "Quantum Mechanics-The Theoretical Minimum" in particular, what are the indexes $a',a$ in the operator $\,\mathbf L\,$ running over? Why is there a comma between them?:

So far, this is entirely general. However, if the observable $\,\mathbf L\,$ is associated only with $\,A$, then it acts trivially on the $\,b-$index and we can write the expectation value as \begin{equation} \langle\mathbf L\rangle =\sum\limits_{a,b,a'}\Psi^*\left(a'b\right)L_{a',a}\Psi\left(ab\right). \tag{7.16}\label{7.16} \end{equation}

  • $\begingroup$ To clarify - any information expanding this equation is helpful! $\endgroup$
    – David
    Jul 23, 2022 at 0:05
  • 1
    $\begingroup$ The comma in $L_{a, a'}$ is to emphasize that $a$ and $a'$ are two different indices, and are not being multiplied together. It is a common way to represent indices of a matrix. The two commas on the summation symbol are a shorthand way of saying you really have a triple sum, with one sum over $a$, one over $a'$, and one over $b$. Beyond that, there is not enough information in your question to give further details on the notation, for someone without access to that book. It may help if you provide further details from the source about how the notation is defined. $\endgroup$
    – Andrew
    Jul 23, 2022 at 1:35
  • $\begingroup$ By the way, it's frowned upon to post images of books and equations, because images cannot be searched by users with a similar question. You should type out the relevant passage(s) you would like to quote, using MathJax for the mathematical symbols. $\endgroup$
    – Andrew
    Jul 23, 2022 at 1:38

1 Answer 1


Let $\{|a\rangle\}_{a=1}^n$ and $\{|b\rangle\}_{b=1}^n$ be the basis states of the individual system (in this case they have the same dimension). You can build a basis for the composite system using the tensor product $$\{|ab\rangle\}_{a,b=1}^n\qquad |ab\rangle:=|a\rangle\otimes|b\rangle \quad\forall a,b=1,2...n\tag{1}$$ Your states con be now labeled by $ab$; summing over $ab$ means summing over all possible values of $a$ and $b$ i.e. $$\sum_{ab}=:\sum_a\sum_b=\sum_{a,b}\tag{2}$$ I'll clarify the notation further:

  • The first sum $\sum\limits_{ab}$ is a sum over all basis states $|ab\rangle$;
  • By construction of the basis states (see $(1)$), this means summing over all possible values of $a$ and $b$, that is $\sum\limits_a\sum\limits_b$;
  • Finally, the last sum is just a shorthand notation for the previous one. The comma means we're summing separately over $a$ and over $b$.

Any state of the composite system can be expressed as a linear combination of the basis elements $$|\Psi\rangle=\sum_{ab}\psi(a,b)|ab\rangle$$ So, given a generic operator $\hat{L}$, the average value is given by $$\langle\hat{L}\rangle:=\langle\Psi|\hat{L}|\Psi\rangle=^1\sum_{ab}\sum_{a'b'}\psi(a,b)\underbrace{\langle ab|\hat{L}|a'b'\rangle}_{L_{ab,a'b'}}\psi(a',b')=^2\sum_{ab,a'b'}\psi(a,b)L_{ab,a'b'}\psi(a',b')\tag{3}$$.

Ok, now your question can be answered. If the observable is only associated with $A$, its action on the $b$ part is trivial. More formally, such an operator will be of the form $$\hat{L}=\hat{L}_a\otimes\hat{I}_b$$ Where $\hat{L}_a$ is an operator acting on the $a$-space and $\hat{I}_b$ is the identity operator on the $b$-space. In such case: $$L_{ab,a'b'}=\langle ab|\hat{L}|a'b'\rangle=\langle a|\hat{L}_a|a'\rangle\langle b|b'\rangle=L_{aa'}\delta_{bb'}$$ Using this in $(3)$, the sum over $b'$ collapses due to Kronecker delta $$\sum_{ab,a'b'}\psi(a,b)L_{ab,a'b'}\psi(a',b')=\sum_{a}\sum_{a'}\sum_{b}\sum_{b'}\psi(a,b)L_{aa'}\delta_{bb'}\psi(a',b')=\\ \sum_{a}\sum_{a'}\sum_{b}\psi(a,b)L_{aa'}\psi(a',b)=\sum_{a,a',b}\psi(a,b)L_{aa'}\psi(a',b)$$ And we're done. Hope this makes things clearer.

$^1$ The author separates the row index and the column index using a comma. Do not confuse this with the shorthand notation for sum discussed above.

$^2$ See what I did here? I used the shorthand notation for the double sum, separating indexes with a comma. Each sum is a sum over $ab$, so each of them is really a double sum.


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