# Operator acts on a ket and a bra using Dirac Notation

Viona
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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​

## Answers and Replies

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

Viona

Viona
Homework Helper
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Looks good. Can you say anything more if ##\hat A## is Hermitian?

Homework Helper
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Can you prove it?

Mentor
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​
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Viona
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.

PeroK
Viona
Can you prove it?
I am not sure if this is true or not.

Homework Helper
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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.

Viona
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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.

docnet and Viona
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
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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
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!

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.