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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?

Qmechanic
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gented
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2 Answers2

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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.

John
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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.

flippiefanus
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