What I often encounter in books, is the following:

Say I have an operator $\hat{o}$, which I want to transform to $\hat{o}'= \hat{M}^{-1} \hat{o} \hat{M}$. When I now consider a "function" $f(\hat{o})$, its transformation is usualy expressed as $f(\hat{o}') = \hat{M}^{-1} f(\hat{o}) \hat{M}$.

My Question: Why can I express it that way? I'd have to make the critical assumption that I can express $f(\hat{o})$ as a sum of products of $\hat{o}$, which means that I can expand $f$ as a taylor series. Is that true?

Edit : I know how to solve this, if f is expandable. The question is, does the equation only hold for f being expandable? I'm thinkin for for example about the Potential operator in the hydrogen atom, which contains a singularity, and cearly isn'texpandable. Is it sufficient that it is expandable nearly everywhere?

  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Sep 3 '17 at 18:18
  • $\begingroup$ It probably would. I might consider reposting it there. $\endgroup$ – Quantumwhisp Sep 3 '17 at 20:18
  • $\begingroup$ Or perhaps migrate? If you crosspost, be sure to mention it on both sites. $\endgroup$ – Qmechanic Sep 3 '17 at 20:22
  • $\begingroup$ I will consider this the better option. $\endgroup$ – Quantumwhisp Sep 3 '17 at 20:28

Let us assume as we always do in physics that your function $f(o)$ admits a taylor expansion in powers of $o$. Now $f(o') = f(M^{-1} o M)$ by definition of $o'$.

Expanding this using the taylor series for $f(o)$ it's easy to see that at the nth term is proportional to $M^{-1}o_1MM^{-1}o_2..........MM^{-1}o_{n}M$.

But $MM^{-1} = M^{-1}M = I$

Thus this nth term collapses to $M^{-1}o_{1}o_{2}.....o_nM$.

Using linearity of the operators $M$ and $M^{-1}$, you can pull $M^{-1}$ out from every term in the series on the left and $M$ out from every term in the series on the right, such that

$f(o') = f(M^{-1} o M) = M^{-1}f(o)M$.


  • $\begingroup$ I think the question was precisely about whether or not the assumption of Taylor expansion is necessary, since defining functions of operators works for any measurable function by Borel functional calculus. $\endgroup$ – ACuriousMind Jul 7 '17 at 11:29
  • $\begingroup$ @ACuriousMind yes, that is the Problem. $\endgroup$ – Quantumwhisp Jul 20 '17 at 12:47

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