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Hermitian operator- Problem

  1. Oct 20, 2013 #1
    Prove the equation

    [itex]A\left|\psi\right\rangle = \left\langle A\right\rangle\left|\psi\right\rangle + \Delta A\left|\psi\bot\right\rangle[/itex]

    where [itex] A[/itex] is a Hermitian operator and [itex] \left\langle\psi |\psi\bot\right\rangle = 0[/itex]


    [itex]\left\langle A\right\rangle[/itex] = The expectation value of A.
    [itex]\Delta A[/itex] = The standard deviation of A.


    My attempt :

    I tried to write [itex]\left|\psi\right\rangle[/itex] as a superposition of the eigenfuncion of the operator [itex] A [/itex] and used the fact that it's a Hermitian operator

    [itex]A\left|\phi_{n}\right\rangle = \lambda_{n}\left|\phi_{n}\right\rangle[/itex] , [itex]\left|\psi\right\rangle = \sum a_{n}\left|\phi_{n}\right\rangle[/itex]

    so that [itex]A\left|\psi\right\rangle = \sum a_{n}\lambda_{n}\left|\phi_{n}\right\rangle[/itex]

    and [itex]\left\langle A\right\rangle = \sum |a_{n}|^{2}\lambda_{n} [/itex]


    [itex]\Delta A = \sqrt{\left\langle A^{2}\right\rangle - \left\langle A\right\rangle ^{2}}[/itex]

    and I wrote [itex]\left|\psi\bot\right\rangle[/itex] as [itex]\left|\psi\bot\right\rangle = \sum b_{n}\left|\phi_{n}\right\rangle[/itex]

    [itex]\sum a^{*}_{n}b_{n} = 0[/itex]

    I don't know how to go on from here...
    any ideas?

    thank you! :)
     
  2. jcsd
  3. Oct 20, 2013 #2

    dextercioby

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    Hmm, using that the s-adj A has a purely discrete spectrum is not ok, it's not assumed in the problem. And I think you left out a square. I get

    [tex] A\psi = \langle A \rangle \psi + (\Delta A)^2 \psi_{\perp} [/tex]

    under the simplifying assumptions [itex] \psi_{\perp} \in D(A)[/itex] and [itex] \mbox{Ran}(A)\subset D(A) [/itex].
     
  4. Oct 20, 2013 #3

    dextercioby

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    The first step I found to resolution is to calculate:

    [tex]\langle \psi,(A-\langle A\rangle ) \psi\rangle [/tex]

    What do you get and what do you conclude ?
     
  5. Oct 20, 2013 #4
    I didn't miss the squar.. this is the question.
    you get wrong units if you put the squar there ...
     
  6. Oct 20, 2013 #5
    this yields zero .. i didn't understand how it can help me

    thank you
     
  7. Oct 20, 2013 #6

    dextercioby

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    If it yields zero, then the vector in the rhs of the scalar product is perpendicular on psi, so it's psi perp times a non-zero constant.
     
  8. Oct 20, 2013 #7
    I don't understand something -
    there are lots of functions that can be perpendicular to psi

    e.g (euclidean space)- the vector z (cartesian coordinates) , x is prep to z and y is prep to z but also the superposition (x+y) is prep to z.

    so by writing psi prep do u mean to the superposition of all the functions that prep to psi or it's enough to choose only one?
     
  9. Oct 21, 2013 #8

    dextercioby

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    Yes, Psi_perp is an arbitrary vector perpendicular to the psi already chosen. You get from ⟨ψ,(A−⟨A⟩)ψ⟩ = 0 that C psi_perp = A psi - <A> psi. All you need to show is that C is related to the standard deviation in square.
     
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