bra-ket with adjoints identity

[nomadreid]

TL;DR

I thought that, if B is the adjoint of A, <v|A|w>=<v|(A|w>)=<(v|B)|w>. But a simple example with real matrices trips me up.

Continuing the summary: the example in question is

Obviously I am understanding some extremely elementary point incorrectly. What? Many thanks!

 

[PeroK]

The correct equation is:
$$\langle v|A|w\rangle = \langle w|A^{\dagger}|v \rangle^*$$

 

[fresh_42]

TL;DR Summary: I thought that, if B is the adjoint of A, <v|A|w>=<v|(A|w>)=<(v|B)|w>. But a simple example with real matrices trips me up.

Continuing the summary: the example in question is
View attachment 357730
Obviously I am understanding some extremely elementary point incorrectly. What? Many thanks!

The formula is
$$
\bigl\langle A(x),y \bigr\rangle = \bigl\langle x, A^\dagger (y) \bigr\rangle .
$$

This means for your example with $x=(0,1)$ and $y=(3,4)$ that
\begin{align*}
\bigl\langle A(x),y \bigr\rangle &=\bigl\langle \begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\, , \,\begin{pmatrix}3\\4\end{pmatrix} \bigr\rangle =\bigl\langle \begin{pmatrix}2\\4\end{pmatrix}\, , \,\begin{pmatrix}3\\4\end{pmatrix} \bigr\rangle =22\\[12pt]
\bigl\langle x , A^\dagger (y)\bigr\rangle &= \bigl\langle \begin{pmatrix}0\\1\end{pmatrix}\, , \,\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}3\\4\end{pmatrix} \bigr\rangle =\bigl\langle \begin{pmatrix}0\\1\end{pmatrix}\, , \,\begin{pmatrix}15\\22\end{pmatrix} \bigr\rangle =22
\end{align*}
or the other way around
$$
\bigl\langle A^\dagger (x),y \bigr\rangle = \bigl\langle x, A(y) \bigr\rangle =25 .
$$
What you have calculated was the associativity of matrix multiplication with two different matrices:
$XAY \neq_{i.g.} XA^\dagger Y.$

 

[nomadreid]

Many,many thanks, PeroK and fresh42!