- #1

Shiz

- 9

- 0

## Homework Statement

Linear combination is [itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]. It's given that it is not Hermitian already.

## Homework Equations

∫ψ

_{i}* [itex]\hat{Ω}[/itex] ψ

_{j}= (∫ψ

_{j}* [itex]\hat{Ω}[/itex] ψ

_{i})*

## The Attempt at a Solution

∫ψ

_{i}* ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψ

_{j}= (∫ψ

_{j}* ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψ

_{i})*

I chose to work with the right hand side of the equation first.

∫ψ

_{i}* ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψ

_{j}= {∫ψ

_{j}* [itex]\hat{A}[/itex] ψ

_{i}+ i(∫ψ

_{j}* [itex]\hat{B}[/itex] ψ

_{i})}*

So I have to take the complex conjugate of the right hand side (not sure if that's the proper way to say it). What I don't understand is why the operator would become ∫ψ

_{j}* [itex]\hat{A}[/itex] ψ

_{i}- i(∫ψ

_{j}* [itex]\hat{B}[/itex] ψ

_{i}) and then ∫ψ

_{i}* ([itex]\hat{A}[/itex] - i[itex]\hat{B}[/itex]) ψ

_{j}.

What are the mathematical reasons? The complex conjugate of + i[itex]\hat{B}[/itex] is -i[itex]\hat{B}[/itex]. I would just be replacing the operator with its complex conjugate? That would give me the answer, but it doesn't seem that simple. Clarification at this would help! Thank you!