Simplifying Operator and Dirac Algebra for Kets

[MikeBuonasera]

Hi Guys, I am facing a problem playing around with some operators and Kets, would like some help!

I have $$\langle \Psi | A+A^\dagger | \Psi \rangle .A$$

Could someone simplify it? Especially is there a way to change the last operator A into A^\dagger?

The way I thought about this is:
$$=(\langle \Psi |A | \Psi \rangle + \langle \Psi | A^\dagger | \Psi \rangle).A<br /> =(\langle \Psi |A A | \Psi \rangle + \langle \Psi | A^\dagger A | \Psi \rangle)<br /> =\langle \Psi | I | \Psi \rangle$$

 

[bhobba]

$$=(\langle \Psi |A | \Psi \rangle + \langle \Psi | A^\dagger | \Psi \rangle).A<br /> =(\langle \Psi |A A | \Psi \rangle + \langle \Psi | A^\dagger A | \Psi \rangle)<br /> =\langle \Psi | I | \Psi \rangle$$

That's obviously wrong.

You have a scalar times an operator and you get a scalar.

You can't take the operator inside the bra-ket.

Off the top of my head you might like to expand A in terms of the spectral theorem ie ∑ ai |bi><bi| - assuming it applies of course.

Thanks
Bill

 

[MikeBuonasera]

That's obviously wrong.

You have a scalar times an operator and you get a scalar.

You can't take the operator inside the bra-ket.

Off the top of my head you might like to expand A in terms of the spectral theorem ie ∑ ai |bi><bi| - assuming it applies of course.

Thanks
Bill

Thanks Bill. Actually there is a detail that I omitted which may help:
(⟨Ψ|A|Ψ⟩+⟨Ψ|A†|Ψ⟩).A|Ψ⟩
Does this make any difference?

thanks

 

[bhobba]

Thanks Bill. Actually there is a detail that I omitted which may help:
(⟨Ψ|A|Ψ⟩+⟨Ψ|A†|Ψ⟩).A|Ψ⟩
Does this make any difference?

Same problem - only you have a scalar times a vector.

Thanks
Bill