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I'm having some trouble with a derivation in Sakurai's Modern Quantum Mechanics (specifically Derivation 1 on §3.2, p. 159), where he computes $$ \exp\left(\frac{iS_Z\phi}{\hbar}\right) \;S_x \; \exp\left(\frac{-iS_Z\phi}{\hbar}\right). $$ I don't understand how to go from $$ (\hbar/2)\exp\left(\frac{iS_Z\phi}{\hbar}\right) \; \{|+\rangle \langle-| + |-\rangle\langle+|\} \; \exp\left(\frac{-iS_Z\phi}{\hbar}\right) $$ to $$ (\hbar/2)\left( e^{i \phi/2}|+\rangle\langle-|e^{i \phi/2} + e^{-i \phi/2}|-\rangle\langle+| \; e^{-i \phi/2}\right) . $$

Is it just a matter of expanding out the Taylor series of $\exp\left(\frac{iS_Z\phi}{\hbar}\right)$?

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  • $\begingroup$ Well, what is $S_z|+\rangle$ equal to, for example? $\endgroup$ – Aaron Stevens Jul 26 at 23:50
  • $\begingroup$ It would be $\hbar/2 |+ \rangle$ ? $\endgroup$ – Snop D. Jul 26 at 23:54
  • $\begingroup$ Ok. Do you also understand that if $A|a\rangle=a|a\rangle$ then $e^A|a\rangle=e^a|a\rangle$? $\endgroup$ – Aaron Stevens Jul 26 at 23:56
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    $\begingroup$ I see that now, thanks! $\endgroup$ – Snop D. Jul 26 at 23:58
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    $\begingroup$ @SnopD. If you've now understood the problem, I would encourage you to write up your solution as an answer for future visitors. $\endgroup$ – Emilio Pisanty Jul 27 at 13:12
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It is quite generally true that for any two $n\times n$ matrices $A,B\in\mathbb R^{n\times n}$ $$ \exp(A) B \exp(-A)=\sum_{n=0}^{\infty}\frac{1}{n!}{\rm ad}_A^n B $$ where I define $$ {\rm ad}_A B= [A,B]=AB-BA. $$

This is proven by replacing $A\to \varepsilon A$ for $\varepsilon \in \mathbb R$ and Taylor-expanding in $\varepsilon$. The formula is also true quite generally when $A,B$ are any elements of the universal enveloping algebra of any Lie algebra.

Since you know the commutation relations of $S_X,S_Y,S_Z$ you can then directly calculate the result

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In general, if $\vert \ell m\rangle$ is an eigenstate of $S_z$, then $$ e^{i\phi S_z/\hbar}\vert \ell m\rangle =\left(I+i \phi \hbar m/\hbar + \frac{1}{2}(i \phi \hbar m/\hbar)^2+ \ldots\right)\vert \ell m\rangle=e^{im\phi}\vert \ell m\rangle $$ by definition of the exponential of an operator, and likewise $$ \langle \ell m\vert e^{i\phi S_z/\hbar}= \left(e^{i\phi S_z/\hbar}\vert \ell m\rangle\right)^\dagger =\langle \ell m\vert e^{-im\phi} $$ so that $$ e^{i\phi S_z/\hbar}\vert \ell m\rangle\langle \ell m'\vert e^{-i\phi S_z/\hbar} = e^{im\phi} \vert \ell m\rangle\langle \ell m'\vert e^{im'\phi} $$ and you can work the rest of the calculation that way.

There is a geometric interpretation to a conjugation like $U \hat A U^\dagger$, where $\hat A$ is an operator: the transformation $U$ is just a change of basis. In your case, $e^{i\phi S_z/\hbar}$ is a change of basis obtained by rotation about $\hat z$ so you would expect under this $S_x$ to go to a linear combination of $S_x\cos\phi\pm S_y\sin\phi$ and $S_y$ since the $\hat x$ axis rotates to a combination $\hat x\cos\phi\pm \hat y \sin\phi$. The difficulty is with the sign, or alternatively, to understand if $e^{i\phi S_z/\hbar}$ produces a clockwise or anticlockwise rotation.

This is fixed easily enough since, by expanding \begin{align} e^{i\phi S_z/\hbar}S_xe^{-i\phi S_z/\hbar} &=S_x+i\phi [S_z,S_x]+\frac{1}{2!}(i\phi)^2 [S_z,[S_z,S_x]]+\ldots\\ &=S_x+i\phi(iS_y)-\frac{1}{2!}(\phi)^2[S_z,iS_y]+\ldots\\ & =S_x-\phi S_y-\frac{1}{2!}\phi^2 S_x+\ldots \end{align} which matches the expansion of $S_x\cos\phi-S_y\sin\phi$.

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