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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?

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  • $\begingroup$ 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 $|\alpha\rangle$. $\endgroup$ – secavara Mar 21 '18 at 14:57
  • $\begingroup$ 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 $K|a^{\prime}\rangle=|a^{\prime}\rangle^{*}$? $\endgroup$ – LY3000 Mar 21 '18 at 16:49
  • $\begingroup$ 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. $\endgroup$ – secavara Mar 21 '18 at 17:02

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