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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.

Ben Sheller
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  • Is $f(a)$ an eigenvector of $\hat A$ with eigenvalue $a$? – Kenny Wong May 04 '17 at 14:57
  • @KennyWong yes, for sure –  May 04 '17 at 14:58
  • 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). – Kenny Wong May 04 '17 at 14:58
  • 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.

    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 :(.

     – Paichu May 04 '17 at 15:48
  • I can do it if $C$ is a constant – Kiryl Pesotski May 04 '17 at 16:35
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    @Paichu however the right answer is $i C(a) \frac{d}{da}$, where $C(a) = \widehat C f(a)$. I don't get it. –  May 05 '17 at 17:16
  • Let me finish some grading first, and I'll try working that out this afternoon. – Paichu May 05 '17 at 17:32
  • Similar to before, $$Bf(a) = \frac{iC}{a}(A - aI)^{-1}f(a)$$

    Since $Af(a) = af(a)$, then $A^{-1}f(a) = \frac{1}{a} f(a)$. And, $(A-aI)^{-1}$ will have eigenvalue of something like $\frac{1}{a - a}$, not really, so here is the very informal part, but it will help with the intuition. Note the strange eigenvalue of $(A-aI)^{-1}$, this can be interpreted like this $(A-aI)^{-1}f(a) = f(a) \lim_{\Delta a \rightarrow 0}\frac{...}{\Delta a}$. In a way the right part after $f(a)$ is a derivative, thus you can obtain your answer. This is not at all rigorous, but I hope it provides some insight.

     – Paichu May 06 '17 at 06:23
  • @Paichu I can tell you that I tried to get it with methods like how it is shown for operators $x$ and $ p_x$. Note that $[x,p_x]=i\hbar$, $[x,\hbar] = 0$ and $\widehat xf(x) = xf(x)$. But it is common knowledge that $\widehat p f(x) = -i \hbar, \frac{d}{dx} f(x)$. But something wrong with me. I can't get it with analogical methods for common $A,B,C$ –  May 06 '17 at 10:54

1 Answers1

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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.

Cosmas Zachos
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