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Suppose I have $\hat{O}_{1}=-i\hbar\partial_{x}$ then

\begin{eqnarray} e^{-i\gamma\hat{O}_{1}/\hbar}x\,e^{i\gamma\hat{O}_{1}/\hbar}=x+\gamma \end{eqnarray}

and

\begin{eqnarray} e^{-i\gamma\hat{O}_{1}/\hbar}y\,e^{i\gamma\hat{O}_{1}/\hbar}=y \end{eqnarray}

Where $x$ and $y$ are orthogonal coordinates. This is clear, $\hat{O}_{1}$ must be the linear momentum operator for $x$, it generates translations in $x$, is associated with invariance under $x$-translation. It has nothing to do with the $y$-coordinate so it generates nothing there.

But if I take $\hat{O}_{2}=-i\hbar\left(\partial_{x}+\partial_{y}\right)$ then I would still have

\begin{eqnarray} e^{-i\gamma\hat{O}_{2}/\hbar}x\,e^{i\gamma\hat{O}_{2}/\hbar}=x+\gamma \end{eqnarray}

Which might lead me to naively believe that $\hat{O}_{1}$ and $\hat{O}_{2}$ represented the same symmetry if I never think to check the other coordinates.

How do I know what to check? What do I stick in there? A linear combination of all non-commuting observables?

I have in mind a more complicated operator, where it's not clear what I should be doing.

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  • $\begingroup$ $O_2$ represent translation invariance along $\vec{h} = \vec{x}+\vec{y}$ while $O_1$ is along $\vec{x}$, not sure to understand the question $\endgroup$ – agemO May 4 '15 at 16:58
  • $\begingroup$ @quantum_loser Isn't the second relation at the beginning different with the second operator? Doesn't gives you a translation on y? $\endgroup$ – Constantine Black May 4 '15 at 17:47
  • $\begingroup$ The question is, how do I know in advance what coordinates (in position or momentum space) to check. For the example I gave you can see what it is just by looking at it. But for more complicated situations it's not obvious. What if the operator has several things it does not commute with? It would be sensible to say $J_{z}$ generates rotations about $z$ but it doesn't commute with other operators - like $J_{x}$ for example... $\endgroup$ – quantum_loser May 4 '15 at 17:47
  • $\begingroup$ @EmilioPisanty Ha! Someone edited my title. You just edited it back to they way it was originally :-p $\endgroup$ – quantum_loser May 4 '15 at 17:49
  • $\begingroup$ @agemO If you have an explicit form for $O_2$ you can say that, but if the explicit form is too complicated to analyse and all you know is $e^{-i\gamma{O}_{2}/\hbar}x\,e^{i\gamma{O}_{2}/\hbar}=x+\gamma$, how can you rule out that it's doing other things in other dimensions? The post is asking for a criterion on when one can stop calculating $e^{-i\gamma{O}_{2}/\hbar}A\,e^{i\gamma{O}_{2}/\hbar}$ for operators $A$ to convince oneself that one has caught all the relevant action of $O_2$. (If I understand the post correctly, that is.) $\endgroup$ – Emilio Pisanty May 4 '15 at 17:54

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