Trouble understanding Sakurai's calculation of $\exp\left(\frac{iS_Z\phi}{\hbar}\right) \;S_x \; \exp\left(\frac{-iS_Z\phi}{\hbar}\right)$

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 $$\exp\left(\frac{iS_Z\phi}{\hbar}\right)$$?

• Well, what is $S_z|+\rangle$ equal to, for example? – Aaron Stevens Jul 26 at 23:50
• It would be $\hbar/2 |+ \rangle$ ? – Snop D. Jul 26 at 23:54
• Ok. Do you also understand that if $A|a\rangle=a|a\rangle$ then $e^A|a\rangle=e^a|a\rangle$? – Aaron Stevens Jul 26 at 23:56
• I see that now, thanks! – Snop D. Jul 26 at 23:58
• @SnopD. If you've now understood the problem, I would encourage you to write up your solution as an answer for future visitors. – Emilio Pisanty Jul 27 at 13:12

2 Answers

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

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 $$S_x\cos\phi-S_y\sin\phi$$.