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Thread moved from the technical forums to the schoolwork forums
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
Operator acts on a ket and a bra using Dirac Notation
- Thread starter Viona
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In summary, operator acts on a ket and a bra using Dirac Notation. If the operator is Hermitian then: <ψ| A |Φ> =<Φ| A |ψ>*= a <Φ | ψ>* = a <ψ | Φ>
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We should treat this as a homework problem. Can you make an attempt at answering it?
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Looks good. Can you say anything more if ##\hat A## is Hermitian?Viona said:
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Can you prove it?Viona said:
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Nugatory
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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 said:
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If A is Hermitian then the eigenvalue (a) is a real number.PeroK said:
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I am not sure if this is true or not.PeroK said:
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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 said:
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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.
$$A(A^{\dagger}v) = A^{\dagger}(Av) = A^{\dagger}(av) = a(A^{\dagger}v)$$ 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.
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Can we say that if the operator is Hermitian then: <ψ| A |Φ> =<Φ| A |ψ>*= a <Φ | ψ>* = a <ψ | Φ> ?PeroK said:
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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.
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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 said:
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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.Viona said: