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This question has caught my attention because I was unaware of the fact that the position-momentum canonical commutation relations could be derived out of the only assumption for $\langle x | p\rangle$ to be a plane wave $\textrm{e}^{ipx}$. It seems that the trick follows from the properties of the exponential function and the Dirac delta to be proportional to their derivatives times the exact accidental factors appearing in the game.

I was wondering whether the same could apply to any other pair of operators: namely let $A, B$ be two non-commuting operators with $\{|a\rangle\}, \{|b\rangle\}$ being the corresponding set of eigenstates. Along the same lines as in the answer one has: $$ \langle a |[A,B]| a' \rangle = (a-a')\int\textrm{d}b'\,b'\,\langle a | b'\rangle\langle b' | a'\rangle $$ or any other sum replacing the integral according to what the eigenvalues look like. In the above general case nothing can be said (can it?); in the other very special case of the angular momentum, with the eigenfunctions being the spherical harmonics, do we still have any accidental simplification of the wave functions in order to gain back the $\textrm{SU}(2)$ commutation relations (maybe due to some special recursive properties of those functions)?

Generalising, the question can be put as: given $A, B$ two self-adjoint operators, does the scalar product between elements of the corresponding complete sets of eigenstates fully determine the commutator (i. e. the action) of the two operators on the entire Hilbert space?

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  • $\begingroup$ 1. I'm pretty sure to say that $\langle x \vert p \rangle$ fixes the canonical commutation relations is just a non-rigorous way to state that the Weyl relations define $x$ and $p$ uniquely up to unitary equivalence, cf Stone-von Neumann. 2. Unless you assume $A,B$ to have purely continuous spectrum, the r.h.s. of your equation does not make sense since you can't integrate over $b'$ then, so this case is not actually very general (how many operators with purely continuous spectrum besides $x,p$ and their powers do you know?) $\endgroup$
    – ACuriousMind ♦
    Commented Dec 14, 2015 at 21:36
  • $\begingroup$ I wouldn't worry too much about the continuous spectrum because one can always replace the rhs with the corresponding sums on the domains where it make sense (or else, one can always expand anything onto $|x\rangle$, see the spherical harmonics). I do see the Stone-von Neumann, in fact I answered the question exactly addressing that issue; yet, this seems very special of $x,p$, which makes it very unfair to all the other operators; what do you think? $\endgroup$
    – gented
    Commented Dec 14, 2015 at 21:48
  • $\begingroup$ @GennaroTedesco Glad you found that answer useful. The reality is that knowing the $\langle x| p \rangle$-s means knowing a whole lot, including $[\hat x, \hat p]$, the differential form of $\hat p$ and of $\hat L$, etc. The latter also implies eigenfunctions and eigenvalues, more CCR-s etc. So I'd settle for equivalence between fixing the $\langle x| p \rangle$-s and fixing the CCR-s. As for the general case of knowing a set of $\langle a| b \rangle$-s, we'd also have to know the eigenvalues, which basically means knowing $B$ in the eigenbasis of $A$. So CR-s would follow necessarily. $\endgroup$
    – udrv
    Commented Dec 15, 2015 at 7:19
  • $\begingroup$ As for the case of $\hat \bf L$, would be nice to redo the argument using eigenfunctions of $\hat L_x$, $\hat L_y$ alongside the trivial ones for $\hat L_z$, but I'm a bit hung on finding them without resorting to CR-s o_o. $\endgroup$
    – udrv
    Commented Dec 15, 2015 at 7:30

2 Answers 2

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That was an interesting question, but I am not sure about the validity of the answer given there. In particular, in the fourth line of his/her response @udrv replaces $p_1$ with $-i \partial/\partial x$, but once you are allowed to do that, the commutation relation follows by itselve as an identity, as has been shown countless times in this forum.

EDIT: The remark about @udrv's original answer is inaccurate. As @ShKol has pointed out in a comment, $p_1$ was replaced with $-i \partial/\partial x$ not in "symbolic" context, but as a standard differentiaion-under-integration trick.

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  • $\begingroup$ This was not done in the answer you mention - that integral is over the momentum and the position derivative brings down the momentum from the exponential. This is not the same as using the position space representation of the momentum (which would, of course, lead to the commutator mentioned by the OP). $\endgroup$
    – ShKol
    Commented Sep 18, 2023 at 15:36
  • $\begingroup$ @ShKol I agree. Thanks for pointing that out. $\endgroup$
    – John
    Commented Sep 18, 2023 at 17:38
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Perhaps a useful property to consider when given two non-commuting Hermitian operators is the relationship between their eigenbases. In the case of $\hat{x}$ and $\hat{p}$, we know that $\langle x| p\rangle=\exp(ixp)$. What it means is that these bases are mutually unbiased. In other words, the magnitude $|\langle x| p\rangle|$ does not depend on either $x$ or $p$. In that sense, one can say that the two operators are maximally non-commuting.

In general, two operator can be non-commuting without being maximally non-commuting if their eigenbases are not mutually un-biased. (In other words, for $\hat{A}$ and $\hat{B}$, we have $\langle a| b\rangle\neq\exp(iab)$.) Remember, for two operators to commute, their eigenbases must be identical. There is a lot of variation between two bases being identical and being mutually unbiased.

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