Operator acts on a ket and a bra using Dirac Notation

[Viona]

Summary:: Operator acts on a ket and a bra using Dirac Notation

Please see the attached equations and help, I Think I am confused about this​

 

[PeroK]

We should treat this as a homework problem. Can you make an attempt at answering it?

 

[Viona]

 

[Viona]

 

[PeroK]

View attachment 286721

Looks good. Can you say anything more if $\hat A$ is Hermitian?

 

[PeroK]

View attachment 286722

Can you prove it?

 

[Nugatory]

Summary:: Operator acts on a ket and a bra using Dirac Notation

Please see the attached equations and help, I Think I am confused about this​

View attachment 286719​

Please take a few minutes to learn how to use this site's Latex feature... There's a guide in the help section at https://www.physicsforums.com/help/latexhelp/

 

[Viona]

Looks good. Can you say anything more if $\hat A$ is Hermitian?

If A is Hermitian then the eigenvalue (a) is a real number.

 

[Viona]

Can you prove it?

I am not sure if this is true or not.

 

[PeroK]

I am not sure if this is true or not.

I don't think it is true in general for any operator. Certainly for Hermitian operators and also for normal operators (these are operators that commute with their Hermitian conjugate), but not in general.

 

[PeroK]

Switching to normal linear algebra notation. If $A$ commutes with $A^{\dagger}$ and $v$ is an eigenvector of $A$ with eigenvalue $a$, then:
$$A(A^{\dagger}v) = A^{\dagger}(Av) = A^{\dagger}(av) = a(A^{\dagger}v)$$ and we see that $A^{\dagger}v$ is an eigenvector of $A$ with eigenvalue $a$. Which means that $A$ and $A^{\dagger}$ share eigenspaces.

And it's easy to show that the eigenvalues are complex conjugates.

You can use that to prove the identity in your question, but I think you need that condition.

 

[Viona]

I don't think it is true in general for any operator. Certainly for Hermitian operators and also for normal operators (these are operators that commute with their Hermitian conjugate), but not in general.

Can we say that if the operator is Hermitian then: <ψ| A |Φ> =<Φ| A |ψ>^*= a <Φ | ψ>^* = a <ψ | Φ> ?

 

[PeroK]

Yes, and if we assume that $A$ and $A^{\dagger}$ share eigenvectors with cc eigenvalues, then: $$\langle \psi |A| \phi \rangle = \langle \phi |A^{\dagger}| \psi \rangle^* = \langle \phi |a^*| \psi \rangle^* = a\langle \psi |\phi \rangle$$ So, that's slightly more general than Hermitian, with $A$ normal and non-degenerate.

 

[Viona]

Yes, and if we assume that $A$ and $A^{\dagger}$ share eigenvectors with cc eigenvalues, then: $$\langle \psi |A| \phi \rangle = \langle \phi |A^{\dagger}| \psi \rangle^* = \langle \phi |a^*| \psi \rangle^* = a\langle \psi |\phi \rangle$$ So, that's slightly more general than Hermitian, with $A$ normal and non-degenerate.

It is clear now. It seems to me that I need to educate myself and study more in linear algebra. Thank you for your help!

 

[PeroK]

It is clear now. It seems to me that I need to educate myself and study more in linear algebra. Thank you for your help!

PS ultimately it's simply this equality you need: $$A^{\dagger}| \psi \rangle = a^*| \psi \rangle$$ And that holds for Hermitian operators, some other operators, but not all operators.