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I have another question on the notation in Shankar. I think it's sloppy, but I also may just be misunderstanding it. Again, this is at the very beginning of the math intro.

He has:

$$a\left| V \right\rangle \to \left[ {\begin{array}{*{20}{c}} {a{v_1}}\\ {a{v_2}}\\ \vdots \\ {a{v_n}} \end{array}} \right] \to \left[ {\begin{array}{*{20}{c}} {{a^*}v_1^*,}&{{a^*}v_2^*,}& \cdots, &{{a^*}v_n^*} \end{array}} \right] \to \left\langle V \right|{a^*}$$ It is customary to write $aV\rangle $ as $a\left| V > \right\rangle$ and the corresponding bra as $\left\langle aV \right|$. What we have found is that $\left\langle aV \right|=\left\langle V \right|{a^*}$.

The * means conjugate. This doesn't look correct to me unless I'm missing something. First it would seem that the LHS of the final equation should be a ket not a bra. Then it also seems that it's not really "equals". From what I've seen if it is a bra on the LHS, commuting the scalar shouldn't cause it to result in taking its conjugate. Is the text correct or am I not understanding something?

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up vote 1 down vote accepted

This is OK. We define the scalar multiplication in the Ket vector space to be complex linear, $a\left|V\right>=\left|aV\right>$. The inner product $\left<W|V\right>$ is complex linear on the Ket vector space, but complex anti-linear on the Bra space, $$a\left<W|V\right>=\left<W|aV\right>=\left<a^*W|V\right>.$$ That's almost a matter of definition for the inner product of a complex Hilbert space, where in an alternative notation we would write $(a^*W,V)=(W,aV)=a(W,V)$. In particular, this means that $\left<aW|aV\right>$ is a positive multiple of $\left<W|V\right>$, $\left<aW|aV\right>=|a|^2\left<W|V\right>$.

Perhaps the notation would be slightly less worrying if we wrote $\left<aV\right|=a^*\left<V\right|$?

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Peter - thank you. I guess I was being lazy - no excuse, I'll go back and work it through using the inner product. –  Mitchell Kaplan Mar 8 '12 at 3:09
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It just means that $\langle aV |$ is the conjugate of $| aV \rangle$. The latter equals $a|V\rangle$ and so has conjugate $\langle V| a^\ast$. Hence, $$ \langle aV | = \langle V| a^\ast\,. $$ I think that's all there is to it.

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Josh - thanks. As I told Peter (above) I realize I was being lazy, I should have worked this through. I knew I was uncomfortable with the notation. I appreciate the help. –  Mitchell Kaplan Mar 8 '12 at 3:11
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