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I've seen in some books that given a differential operator $$\frac{d}{dx}$$

under a similarity transformation we get

$$\frac{d}{dx}\rightarrow T\frac{d}{dx}T^{-1}=\left(\frac{d}{dx}-\frac{\dot{T}}{T}\right)$$

for some function $T(x)$. I'd like to know why this is true. Why are the two operators equivalent?

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By "similarity transformation" you mean something that in the one-dimensional case would be $x\mapsto ax+b$? ${}\qquad{}$ – Michael Hardy Jun 2 '14 at 2:39
@MichaelHardy no actually, I'm referring to the type of similarity transformations one would see in a beginner's linear algebra course. But, I'm referring to the case with linear operators and not matrices – Millardo Peacecraft Jun 2 '14 at 2:42

It makes sense if 1) $T^{-1}=\frac{1}{T}$ and 2) $Tf=T(x)\cdot f(x)$. Then for any function $f$ we have

$$ T\frac{d}{dx}T^{-1}f= T\left(\left(\frac{d}{dx}T^{-1}\right)f + \left(\frac{d}{dx}f\right)T^{-1}\right)=T\left(-\frac{\dot{T}}{T^2}f + \frac{1}{T}\left(\frac{d}{dx}f\right)\right) = \left(\frac{d}{dx}-\frac{\dot{T}}{T}\right)f. $$

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