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I Inner products and adjoint operators

  1. Apr 25, 2016 #1
    I'm trying to prove the following relation $$\langle\psi\lvert \hat{A}^{\dagger}\rvert\phi\rangle =\langle\phi\lvert \hat{A}\rvert\psi\rangle^{\ast}$$ where ##\lvert\phi\rangle## and ##\lvert\phi\rangle## are state vectors and ##\hat{A}^{\dagger}## is the adjoint of some operator ##\hat{A}## acting on those states.

    As far as I understand, an Hermitian adjoint ##\hat{A}^{\dagger}## of an operator ##\hat{A}## is defined by $$\left(\hat{A}\lvert\psi\rangle ,\lvert\phi\rangle\right)=\left(\lvert\psi\rangle ,\hat{A}^{\dagger}\lvert\phi\rangle\right)$$ and the inner product of two kets defined such that $$\left(\lvert\psi\rangle ,\lvert\phi\rangle\right)\equiv\langle\psi\lvert\phi\rangle ,\qquad\langle\psi\lvert\phi\rangle =\langle\phi\lvert\psi\rangle^{\ast}$$ but using these two definitions I fail to arrive at the expression that I'm trying to prove, as it would appear from this that $$\langle\psi\lvert \hat{A}\rvert\phi\rangle =\langle\psi\lvert \hat{A}^{\dagger}\rvert\phi\rangle$$ If someone could explain my misunderstanding and the correct way to do it, I'd be grateful.
     
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  3. Apr 25, 2016 #2

    stevendaryl

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    I think it's simple:
    [itex]\langle \psi| A^\dagger \phi\rangle = \langle A \psi | \phi\rangle = \langle \phi| A \psi \rangle^*[/itex]

    That doesn't imply that [itex]A = A^\dagger[/itex], as can be seen from the trivial example: [itex]A = i[/itex].
     
  4. Apr 25, 2016 #3
    Thanks, I thought I was just being stupid!

    Just to check though, is ##\lvert A\psi\rangle =A\lvert\psi\rangle## and its dual given by ##\langle A\psi\rvert =\langle\psi\rvert A^{\dagger}## ? I'm still slightly confused by what went wrong with my definition for the inner product?!
     
  5. Apr 25, 2016 #4

    stevendaryl

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    I think that [itex]A |\psi\rangle[/itex] means the same thing as [itex]|A \psi \rangle[/itex]. The only reason to write the [itex]A[/itex] sandwiched in between, as in [itex]\langle \phi|A|\psi \rangle[/itex] is because you can equally well think of [itex]A[/itex] as acting to the left, on [itex]\phi[/itex], or to the right, on [itex]\psi[/itex]. But that notation should (in my opinion) only be used when [itex]A[/itex] is self-adjoint.
     
  6. Apr 25, 2016 #5
    Would it not be in this case then that $$\langle\phi\rvert A\lvert\psi\rangle =\langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle$$ such that $$ \langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle =\langle A\psi\lvert\phi \rangle^{\ast}$$

    I'm still a bit confused as to why my definition of the adjoint, i.e. $$\left(A^{\dagger}\lvert\psi\rangle ,\lvert\phi\rangle\right) =\left(\lvert\psi\rangle ,A\lvert\phi\rangle\right)$$ didn't work?!

    Is it simply that I misinterpreted it slightly. Instead, would this be correct?!

    $$(A\lvert\psi\rangle ,\lvert\phi\rangle)\equiv\left(\langle\psi\rvert A^{\dagger}\right)\lvert\phi\rangle =\langle\psi\rvert A^{\dagger}\lvert\phi\rangle$$ and $$(\lvert\psi\rangle ,A\lvert\phi\rangle)\equiv\langle\psi\rvert\left( A\lvert\phi\rangle\right) =\langle\psi\rvert A\lvert\phi\rangle$$ and furthermore, $$(\lvert\psi\rangle ,\lvert\phi\rangle)=(\lvert\phi\rangle ,\lvert\psi\rangle)^{\ast}.$$ We then have that $$(A\lvert\psi\rangle ,\lvert\phi\rangle)=\langle\psi\rvert A^{\dagger}\lvert\phi\rangle =(\lvert\psi\rangle ,A^{\dagger}\lvert\phi\rangle)=(A^{\dagger}\lvert\phi\rangle ,\lvert\psi\rangle)^{\ast}=(\lvert\phi\rangle ,A\lvert\psi\rangle)^{\ast}=\langle\phi\rvert A\lvert\psi\rangle^{\ast}$$ i.e. $$\langle\psi\rvert A^{\dagger}\lvert\phi\rangle =\langle\phi\rvert A\lvert\psi\rangle^{\ast}$$

    I find it confusing that in so many quantum mechanics lecture notes/text books they pretty much exclusively use this notation.
     
    Last edited: Apr 25, 2016
  7. Apr 25, 2016 #6

    stevendaryl

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    I don't see the difference. Relating your inner product with Dirac expressions,

    [itex](\phi, \psi) = \langle \phi|\psi \rangle[/itex]

    So

    [itex](A^\dagger \phi, \psi) = \langle A^\dagger \phi | \psi \rangle[/itex]

    The definition of [itex]\dagger[/itex] is that

    [itex](A^\dagger \phi, \psi) = (\phi, A \psi) = \langle \phi | A \psi \rangle = \langle \phi | A | \psi \rangle[/itex]

    or conversely

    [itex](A \phi, \psi) = (\phi, A^\dagger \psi) = \langle \phi | A^\dagger \psi \rangle = \langle \phi | A^\dagger | \psi \rangle[/itex]

    I don't see how you derived [itex]A = A^\dagger[/itex]

    I guess it's unambiguous if you always assume that the operator acts to the right.
     
  8. Apr 25, 2016 #7
    Is the updated bit I added to the end of my previous post a correct derivation then?
     
  9. Apr 25, 2016 #8

    stevendaryl

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    It seems correct. Are you retracting your claim that it seems to show that [itex]A = A^\dagger[/itex]?
     
  10. Apr 25, 2016 #9
    Yes. I just applied the definitions incorrectly in the first post and hence arrived at the incorrect result (I knew that it shouldn't be ##A=A^{\dagger}##, what I didn't understand straight away was the mistake that was making me obtain this erroneous result, but I think I've sorted it now).
     
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