Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I was looking at a QM textbook exercise dealing with dilatations, the transformations are $x \rightarrow x' = \lambda x$ transforming $|\psi\rangle$ into $|\psi'\rangle = \cal{D}_{\lambda}|\psi\rangle$. The conservation of the norm gives $\psi'(x)=\frac{1}{\sqrt{\lambda}}\psi(\lambda^{-1}x)$. And I am supposed to derive the following transformation $$\cal{D}^{\dagger}_{\lambda} x \cal{D}_{\lambda} = \lambda x$$ from the invariance of the mean $\langle \psi'|x'|\psi'\rangle = \langle\psi|x|\psi\rangle$ but I am only able to derive it from the semi-intuitive $x'= \cal{D}^{\dagger}_{\lambda} x \cal{D}_{\lambda}$ which is how an operator is supposed to transform and then equaling with $x'=\lambda x$ which is not an operator equation, at least not at first.

The invariance of the mean gives me $$\langle\psi'|x'|\psi'\rangle = \langle\psi|\cal{D}^{\dagger}_{\lambda}x'\cal{D}_{\lambda}|\psi\rangle = \langle\psi|x|\psi\rangle$$ for any $|\psi\rangle$, giving $$\cal{D}^{\dagger}_{\lambda}x'\cal{D}_{\lambda} = x$$ which is the opposite of the relation I am looking for

share|improve this question
    
The oppositeness is due to a combination of different conventions concerning $D$ and $D^\dagger$, which one is put first, due to the passive vs active transformation, and so on. In all such cases, it's better to avoid words (and arrows) and use equations only. –  LuboŇ° Motl Oct 19 '12 at 10:38
    
Ok, I didn't think about this passage from active to passive transformations with dilations, seems a bit weird to "shrink" a system but clearly not enough to stop a theoretical description ;) –  Just_a_wannabe Oct 19 '12 at 12:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.