TITLE: Another question about Shankar's notation
QUESTION [2 upvotes]: 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?

REPLY [1 votes]: 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.

REPLY [1 votes]: 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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