# Time Reversal Operator in Sakurai's Book

In Sakurai's Modern Quantum Mechanics, Eq(4.16) says that

\begin{align} |\alpha\rangle &= \sum_{a^\prime}|a^{\prime}\rangle\langle a^{\prime}|\alpha\rangle \overset{K}{\rightarrow} |\tilde{\alpha}\rangle = \sum_{a^\prime} \langle a^{\prime}|\alpha\rangle ^* K|a^{\prime}\rangle \nonumber \\ &= \sum_{a^\prime} \langle a^{\prime}|\alpha\rangle ^* |a^{\prime}\rangle \nonumber \nonumber \end{align} \tag{4.16}

Why is there a complex conjugate in the second line of the equation? I thought $$|\alpha\rangle = \sum_{a^\prime}|a^{\prime}\rangle\langle a^{\prime}|\alpha\rangle = \sum_{a^\prime}\langle a^{\prime}|\alpha\rangle|a^{\prime}\rangle$$

Or is there something I don't understand about anti-unitary operator?

• You mean eq. 4.4.16. I think he means the second line to be the result of the action of $K$ on the right. In other words, the second line is $|\tilde{\alpha}\rangle$ and not $|\alpha\rangle$. – secavara Mar 21 '18 at 14:57
• Thanks. That makes sense. But I have another question. He conclude that $K$ has no effect on the basis vector, but what if the basis vectors are complex? Since $K$ makes anything on its RHS become the complex conjugate, shouldn't the more general form be $K|a^{\prime}\rangle=|a^{\prime}\rangle^{*}$? – LY3000 Mar 21 '18 at 16:49
• Right after that equation, Sakurai discusses precisely your concern with a couple of examples. The final conclusion is that the effect of $K$ depends on the basis. – secavara Mar 21 '18 at 17:02