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Given two compatible observables $A$ and $B$ with a common eigenbasis, the completeness relation is: $\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle#1|}$ $$ \sum_{i,j}\ket{a^i,b^j}\bra{a^i,b^j} = 1 $$

Since $\ket{a^i,b^j}$ is not guaranteed to exist for all combinations of $i$ and $j$, does the sum imply we simply ignore the terms which don't exist?

cyanic-selkie
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  • "Since $|a^i,b^j\rangle$ is not guaranteed to exist for all combinations of $i$ and $j$" - can you elaborate on this? Right now this seems to come from nowhere. – probably_someone Sep 15 '20 at 00:51
  • Take a concrete example. Say the operators can be represented by 2x2 matrices and they have an identical eigenbasis set of size 2, each with their own eigenvalues (an example of which isn't too hard to find). That would mean that the total number of eigenkets given by combinations of indices is 4. Which is the confusing part since there are only 2 eigenkets. – cyanic-selkie Sep 15 '20 at 01:02
  • Ok, let's back up a bit. What exactly is the ket $|a^i,b^j\rangle$? What does this notation mean in this context? – probably_someone Sep 15 '20 at 01:40
  • It is a notation that represents an eigenket of both $A$ and $B$ where the eigenvalue for $A$ is $a^i$ and for $B$ is $b^j$. – cyanic-selkie Sep 15 '20 at 01:45
  • Ok. What's the source for this relation? Where did you see it? – probably_someone Sep 15 '20 at 01:48
  • It is from one of the problems in J.J. Sakurai's Modern quantum mechanics. Also, it is talked about in this answer: https://physics.stackexchange.com/a/331329/263659 – cyanic-selkie Sep 15 '20 at 01:52
  • In that answer, the notation is ambiguous but the intention is clear: sum over all allowed pairs $(l,m)$. Obviously you don't add a term with $l=0$ and $m=100$ to that sum; the symbol $\sum_{l,m}$ is inappropriate if you remove that context. – probably_someone Sep 15 '20 at 01:56

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There is no stipulation that the set of $b$’s have same cardinality as the set of $a$’s, and the sum is limited to those sets where so there is no state for which $\vert a^i,b^j\rangle$ exist if $b^j$ is not in the set of $b$ and same for $a^i$. The indices $i$ and $j$ are independent so need not range over the same index set.

Thus, for instance, assuming that $b^i$ takes one of two values in $\{+,-\}$, and $a^i$ are energies of a harmonic oscillator, we would have $$ \sum_{i=0}^\infty\sum_{j=-,+}\vert E_i,j\rangle\langle E_i,j\vert = \sum_{i=0}^\infty\vert E_i,+\rangle\langle E_i,+\vert + \sum_{i=0}^\infty\vert E_i,-\rangle\langle E_i,-\vert=\mathbb{1} $$

ZeroTheHero
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