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In Sakurai's Modern Quantum Mechanics Chapter 4, in the discussion about the Time Reversal Operator, the following formula is presented $$\Theta \mathbf{J} \Theta^{-1} = -\mathbf{J} $$ This is a requirement necessary to conserve the canonical commutation relations between the generators of rotations.

Now, when speaking about time reversal in Spin-1/2 systems, we have the following eigenstate of the $\mathbf{S \cdot \hat{n}}$ operator $$|\mathbf{\hat{n}},+ \rangle = \exp\left( -\frac{i}{\hbar}S_z \alpha \right) \exp\left( -\frac{i}{\hbar}S_y \beta \right) |+ \rangle$$ where $\alpha$ and $\beta$ are the azimuthal and polar angle respectively and $|+\rangle $ is the eigenstate of $S_z$ with eigenvalue $\frac{\hbar}{2}$.

If we now consider the action of the time reversal operator $\Theta$ on the above state, the following equality is presented $$\Theta |\mathbf{\hat{n}},+ \rangle = \exp\left( -\frac{i}{\hbar}S_z \alpha \right) \exp\left( -\frac{i}{\hbar}S_y \beta \right) \Theta|+ \rangle$$ My question is : How does the $\Theta$ act directly on $|+\rangle$ without also affecting $S_z$ and $S_y$ like this? $$S_z \rightarrow -S_z $$ $$S_y \rightarrow -S_y $$ which should follow from the first equation I wrote down.

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I am not sure to understand the question. Are you asking why exponent do not change as a consequence of $\Theta S_k \Theta^{-1} = -S_k$?

You are forgetting that $\Theta$ is anti linear, so $$\Theta e^{-i a S_k} \Theta^{-1} =e^{\Theta(-i a S_k)\Theta^{-1}} =e^{-(-i) a \Theta S_k\Theta^{-1}}= e^{-i a S_k}\:.$$

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  • $\begingroup$ You're right, it was an oversight on my part. Thank you! $\endgroup$ – Sreekar Voleti Jan 28 '18 at 9:14

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