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I'm reviewing my quantum mechanics by going through Sakurai and Napolitano again and working out all of the derivations. I'm stumped (though I probably shouldn't be) on some algebra in the finite rotation of spin-1/2 systems. The derivation (Sakurai and Napolitano, pg 164) begins by applying the finite rotation operator about the z axis to the $S_x$ operator in the basis of $S_z$ eigenkets:

$\left(\frac{\hbar}{2}\right) \exp \left(\frac{iS_z\phi}{\hbar}\right) \left( \left|+\right>\left<-\right|+\left|-\right>\left<+\right| \right) \exp \left(\frac{-iS_z\phi}{\hbar}\right)$

So far so good - the spectral decomposition of $S_x$ is obvious from the Pauli matrices, and the finite rotation operator is easy to get by taking the limit of applying the infinitesimal rotation operator infinitely many times. Substituting $S_z=\frac{\hbar}{2}$ and distributing, I obtain

$\left(\frac{\hbar}{2}\right) \left( e^{i\phi/2} \left|+\right>\left<-\right| e^{-i\phi/2} + e^{i\phi/2} \left|-\right>\left<+\right| e^{-i\phi/2} \right)$

But Sakurai and Napolitano (correctly) obtain

$\left(\frac{\hbar}{2}\right) \left( e^{i\phi/2} \left|+\right>\left<-\right| e^{i\phi/2} + e^{-i\phi/2} \left|-\right>\left<+\right| e^{-i\phi/2} \right)$

From this form it is straightforward to apply Euler's formula and show that the action of the finite rotation on $S_x$ is just

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

which rotates the spin in the xy plane, as expected. What I can't figure out (and I'm sure it's something simple that I'm just overlooking) is how the exponentials are distributed onto the two terms in the spectral decomposition. Is this an algebra trick that I'm missing, or is there something more deeply wrong in my thinking?

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It is straightforward matrix multiplication and the operators enter through from either side.

$exp(iS_zϕ/ℏ)$ would act from the left on the kets while $exp(−iS_zϕ/ℏ)$ would act from the right on the bras. You can see it more clearly if you do it explicitly.

Use the basis $|+⟩$ and $|−⟩$ to represent the operators as matrices

$|+⟩$ = $\begin{bmatrix}1\\0\end{bmatrix}$ $|−⟩$ = $\begin{bmatrix}0\\1\end{bmatrix}$

$⟨+|$ = $\begin{bmatrix}1&0\end{bmatrix}$ $⟨−|$ = $\begin{bmatrix}0&1\end{bmatrix}$

$exp(iS_zϕ/ℏ)$ would have the form $\begin{bmatrix}{e^{i\phi/2}}&0\\0&{e^{-i\phi/2}}\end{bmatrix}$

while $exp(−iS_zϕ/ℏ)$ would have the form $\begin{bmatrix}{e^{−i\phi/2}}&0\\0&{e^{i\phi/2}}\end{bmatrix}$

$exp(iS_zϕ/ℏ)|−⟩$ = $\begin{bmatrix}{e^{i\phi/2}}&0\\0&{e^{-i\phi/2}}\end{bmatrix}$$\begin{bmatrix}0\\1\end{bmatrix}$ = $e^{-i\phi/2}$$|−⟩$

not $e^{i\phi/2}$$|−⟩$ and

$⟨−|exp(−iS_zϕ/ℏ)$ = $\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}{e^{−i\phi/2}}&0\\0&{e^{i\phi/2}}\end{bmatrix}$ = $⟨−|e^{i\phi/2}$

not $⟨−|e^{−i\phi/2}$

You used the same eigenvalue for the operator acting on both $|+⟩$ and $|−⟩$.

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  • $\begingroup$ Yes! That's my problem, I was stuck thinking about the exponentials as scalars rather than entering as 2x2 matrices... which makes sense since they're operators. Thank you! $\endgroup$ – L. Londau Feb 4 '16 at 13:34
  • $\begingroup$ Glad I could help. :-) $\endgroup$ – Abdullah Al-Shafey Feb 4 '16 at 15:18

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