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Well, I'm a little confused. Suppose we have three Hermitian operators

$\widehat A = \widehat A^{\dagger}$

$\widehat B = \widehat B^{\dagger}$

$\widehat C = \widehat C^{\dagger}$.

We know that $[\widehat A, \widehat B] = i \widehat C $ and $[\widehat A,\widehat C] = 0$.

So, we know that $\widehat A f(a) = a f(a) $ because it's its own representation.

But what about $\widehat B f(a) = ?$

I am sure that I can make some statement from above commutation properties. But I can't write smth expect well-known bacics.

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  • $\begingroup$ Is $f(a)$ an eigenvector of $\hat A$ with eigenvalue $a$? $\endgroup$ – Kenny Wong May 4 '17 at 14:57
  • $\begingroup$ @KennyWong yes, for sure $\endgroup$ – Lust_For_Love May 4 '17 at 14:58
  • $\begingroup$ Could you please say where the commutation relations come from? I'm struggling to see any nice properties of $\hat B f(a)$ (but this might be because I'm being dumb). $\endgroup$ – Kenny Wong May 4 '17 at 14:58
  • $\begingroup$ In the case that $A$ is not degenerate (only simple eigenvalue), the result is somewhat nice since $A$ and $C$ share eigenvectors, say $Cf(a) = cf(a)$, as the second answer addresses [here][1]. Simply starts with $[A,B]f(a) = iCf(a)$. note that $A$ commutes with $C$, further assume $A$ is invertible, then: $$(I - aA^{-1})B f(a) = i A^{-1}Cf(a) = icA^{-1}f(a) = i\frac{c}{a}f(a)$$ Assume $(I - aA^{-1})$ is invertible, $$Bf(a) = \frac{ic}{a}(I - aA^{-1})^{-1}f(a)$$ Still :(. [1]: math.stackexchange.com/questions/1227031/… $\endgroup$ – Paichu May 4 '17 at 15:48
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    $\begingroup$ @Paichu however the right answer is $i C(a) \frac{d}{da}$, where $C(a) = \widehat C f(a)$. I don't get it. $\endgroup$ – Lust_For_Love May 5 '17 at 17:16
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First off, let us skip the obnoxious carets, since capital letters suffice to denote operators.

Secondly, and most importantly, you must also assume $[B,C]=0$ as well, so its eigenvalues in any representation must be constants ($\hbar$), since the magic that leads to Weyl's braiding relation in QM will not work, and the relevant Campbell-Baker-Haussdorf expression will be hellishly complicated, in general, depending on the functional dependence of C(A). So I assume it forthwith.

Thirdly, your major snagging point seems to be in the setup of your representation. Dirac (Principles of QM, Ch III.20) spends a long time in his book doing it right, but most people never get to use it much, since it is mostly "monkey-see-monkey-do" in 99.9% of the applications.

In any case, using Dirac's "standard ket", $|\varpi\rangle$, the translationally invariant vacuum, you define your rep as follows, $$ f(a)=\langle a| F(A)|\varpi \rangle , $$ leading to the statement you express informally, $$ \langle a| A F(A)|\varpi \rangle =a~ f(a), $$ with normalization $\langle a|\varpi\rangle =1$, and $|a\rangle \propto \delta(A-a)|\varpi\rangle $. F is the power series of f, essentially f(A). All functions of a in this picture are eigenvectors of A with the same eigenvalue a. Improper appreciation of this statement leads to ambiguities and no end of grief.

Then, the standard argument of QM unfolds easily, upon recognition of the fact that, C being a constant commuting with everything, B acts like a derivative on all functions of A, $$ [B,F(A)]=-iC ~ F'(A), $$ readily derivable from your commutation relation by considering any power of A, and hence a power-series definition of F. But recall C must be a constant, not an elaborate function of A.

Now, it is standard to assume the standard ket to be in the kernel of B, ($B|\varpi\rangle=0$, "translational invariance of the vacuum"), whence $$ \langle a| B F(A)|\varpi \rangle =\langle a| F(A) B -iCF'|\varpi \rangle = \langle a|-iCF'(A)|\varpi \rangle = -iC\partial_a f(a), $$ the familiar representation of the momentum operator.

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