I bra-ket with adjoints identity

nomadreid
Gold Member
Messages
1,748
Reaction score
243
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
adjoint.png

Obviously I am understanding some extremely elementary point incorrectly. What? Many thanks!
 
Physics news on Phys.org
The correct equation is:
$$\langle v|A|w\rangle = \langle w|A^{\dagger}|v \rangle^*$$
 
nomadreid said:
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.##
 
  • Like
Likes nomadreid and PeroK
Many,many thanks, PeroK and fresh42!
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top