Bra Ket Question: Outer Product Complex Conjugate

[malawi_glenn]

Is the complex conjugate to the outer product this? :

( |a> <b| ) * = ( |b> <a| )

?

 

[Dick]

If you think of it as a matrix that's like saying A_ij*=Aji. Looks like an Hermitian conjugate to me.

 

[Meir Achuz]

The cc does not interchange a and b.

 

[malawi_glenn]

so

(<a|b>) * = (<b|a>)

but not

( |a> <b| ) * = ( |b> <a| )

?

 

[George Jones]

so

(<a|b>) * = (<b|a>)

but not

( |a> <b| ) * = ( |b> <a| )

?

First, note that <a|b> is a complex number, while |a><b| is a linear operator on a vector space.

The answer to your question depends on what you mean by *. What do you mean by *?

If you mean complex conjugation, then I don't know how to take (in a basis-independent way) the complex conjugate of a linear operator.

If you mean adjoint, then (after changing * to ${}^\dagger$)

$$\left< a | b \right>^\dagger = \left< b | a \right>$$

$$\left( \left| a \right> \left< b \right| \right)^\dagger = \left| b \right> \left< a \right|.$$

You should verify this from the definition of adjoint.

 

[malawi_glenn]

hmm yeah i must have confused those two things.. thanks for all help =)