Show linear combination is not Hermitian

[Shiz]

Homework Statement

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

Homework Equations

∫ψ_i * $\hat{Ω}$ ψ_j = (∫ψ_j * $\hat{Ω}$ ψ_i)*

The Attempt at a Solution

∫ψ_i * ($\hat{A}$ + i$\hat{B}$) ψ_j = (∫ψ_j * ($\hat{A}$ + i$\hat{B}$) ψ_i)*

I chose to work with the right hand side of the equation first.
∫ψ_i * ($\hat{A}$ + i$\hat{B}$) ψ_j = {∫ψ_j * $\hat{A}$ ψ_i + i(∫ψ_j * $\hat{B}$ 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 * $\hat{A}$ ψ_i - i(∫ψ_j * $\hat{B}$ ψ_i) and then ∫ψ_i * ($\hat{A}$ - i$\hat{B}$) ψ_j.

What are the mathematical reasons? The complex conjugate of + i$\hat{B}$ is -i$\hat{B}$. 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!

 

[\Tau_{\mu\ni}]

The reason is the wave functions $\Psi_{i,j}$ also get affected by conjugating. And when you want the hermitian conjugate of a producat of operators, you have to reverse the order of factors, since the hermitian conjugate (for example) of a matrix is the complex conjugate of the transpose of the original matrix: $(A^*)_{ij}=\bar{A_{ji}}$

If that's what you mean

 

[hilbert2]

Are $\hat{A}$ and $\hat{B}$ hermitian operators? If they are, it's not generally true that $\hat{A} + i\hat{B}$ is not hermitian (consider the case where $\hat{B}$ is the zero operator).

 

[Shiz]

Ahat and Bhat are hermitian operators, yes. But (Ahat -/+ iBhat) is not hermitian. I think I understand but not sure. We have to take the complex conjugate to show that it is not hermitian. So (iBhat)* is (-iBhat). I was wondering if there was more to that step than what I am thinking.

Apologies for no code. I am on the iPhone app and am not sure how to do it.