# Working with Hermitian-Adjoint - Sakurai Problem 1.4b

1. Jul 19, 2010

### bjnartowt

1. The problem statement, all variables and given/known data
Not relevant, but I have some work that reaches incorrect conclusions, and I can't see the mistake "in the middle".

2. Relevant equations

3. The attempt at a solution
$$\begin{array}{l} {(XY)^\dag } = {\left( {\left\langle {a'} \right|X\left| {a''} \right\rangle \left\langle {a''} \right|Y\left| {a'''} \right\rangle } \right)^\dag } \\ = \left( {\left\langle {a'} \right|X{{\left| {a''} \right\rangle }^\dag }} \right)\left( {\left\langle {a''} \right|Y{{\left| {a'''} \right\rangle }^\dag }} \right) \\ = \left( {\left\langle {a''} \right|Y{{\left| {a'''} \right\rangle }^\dag }\left\langle {a'} \right|X{{\left| {a''} \right\rangle }^\dag }} \right) = {Y^\dag }{X^\dag } \\ \end{array}$$

...Somewhere, here, is a mistake. I distributed the adjoint operation "+" (dagger) across the two scalars, "<||>", but if I did that, this would imply:

$${\left( {XY} \right)^\dag } = \left\langle {a'} \right|X{\left| {a''} \right\rangle ^\dag }\left\langle {a''} \right|Y{\left| {a'''} \right\rangle ^\dag } = {X^\dag }{Y^\dag } = {\rm{uh - oh}}{\rm{.}}$$

2. Jul 19, 2010

### qbert

Your first equality doesn't make sense to me.
when you expand out a matrix product you
should have a sum over a'',
and when you get rid of the sum (using completeness)
it tells you the correct order of the factors.