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Observables, Measurements and all that

  1. Mar 2, 2013 #1
    Hi Folks,

    I somehow cannot get the difference and have to admit that I am left confused.

    For a probability of measuring m with the operator M on state [itex]\Psi_i[/itex]
    [itex]p(m|i) = <\Psi_i| M^{+}_m M_m |\Psi_i> = <\Psi_i| M_m |\Psi_i>[/itex].

    The average of an observable is defined as [itex] <O> = <\Psi_i| O |\Psi_i>[/itex].

    So the measurement by M gives me a probability, the measurement with the observable an expectation value? Okay..the observables will be hermitean, the only thing I know about the measurement matrix is, that is not unitary - otherwise [itex] M^{+}_m M_m[/itex] would be equal to the unity matrix.

    Is the difference that an observable doesn't change the system, but a measurement when projective projects the system into one of the states?

    One last question: What is the reasoning for defining a measurement like [itex] M^{+}_m M_m [/itex] and not by M alone directly?

    Thank you so much in advance..I hope the above somehow makes sense ;).
    Steffen
     
  2. jcsd
  3. Mar 2, 2013 #2

    stevendaryl

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    Let [itex]\vert \Phi_m \rangle [/itex] be a complete set of eigenstates of operator [itex]M[/itex]: (for simplicity, let's assume that the eigenstates are nondegenerate; that is, there can't be more than one state with eigenvalue [itex]m[/itex])

    1. [itex]M \vert \Phi_m \rangle = m \vert \Phi_m \rangle[/itex]
    2. [itex]\langle \Phi_m \vert \Phi_m \rangle = 1[/itex]
    3. [itex]\langle \Phi_m' \vert \Phi_m \rangle = 0[/itex], if [itex]m' \neq m[/itex]

    The idea behind a "projection operator" [itex]M_m[/itex] is this:

    [itex]M_m \vert \Phi_m \rangle = \vert \Phi_m \rangle[/itex]
    [itex]M_m \vert \Phi_m' \rangle = 0[/itex], if [itex]m' \neq m[/itex].

    So [itex]M_m[/itex] filters out any eigenstates that do not have eigenvalue [itex]m[/itex].

    If you have such a projection operator, then the action of [itex]M_m[/itex] on an arbitrary wave function [itex]\vert \Psi_i \rangle[/itex] can be computed this way:

    • Write [itex]\vert \Psi_i \rangle[/itex] as a superposition of eigenstates of [itex]M[/itex]:

      [itex]\vert \Psi_i \rangle = \sum_{m'} C_{m'} \vert \Phi_{m'} \rangle[/itex]

      where [itex]C_{m'}[/itex] is just a coefficient. Note that [itex]C_{m'}[/itex] is the amplitude for state [itex]\Psi_i[/itex] to have eigenvalue [itex]m[/itex]. The probability is the square: [itex]P(m'\vert i) = \vert C_{m'} \vert^2[/itex].
    • Now apply the projection operator [itex]M_m[/itex]:
      [itex]M_m \vert \Psi_i \rangle = \sum_m' C_m' M_m \vert \Phi_m' \rangle[/itex]
    • By the assumed properties of [itex]M_m[/itex], all the terms in the sum vanish except one, leaving:
      [itex]M_m \vert \Psi_i \rangle = C_m \vert \Phi_m \rangle[/itex]
    • Now form the quantity [itex]\vert M_m \vert \Psi_i \rangle \vert^2[/itex]:
      [itex]\vert M_m \vert \Psi_i \rangle \vert^2 = \langle \Psi_i M_m^\dagger M_m \vert \Psi_i \rangle = \langle \Phi_m C_m^* C_m \vert \Phi_m\rangle = \vert C_m \vert^2 = P(m \vert i)[/itex]
    • So we conclude: [itex]\langle \Psi_i M_m^\dagger M_m \vert \Psi_i \rangle = P(m \vert i) [/itex]
     
  4. Mar 3, 2013 #3

    tom.stoer

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    This is correct.

    I guess what you mean is that M is a projector on an eigenstate with eigenvalue m (w/o degeneration).

    I would use the letter M for an observable associated with m with appropriate (non-degenerate) eigenstates

    [tex]M |m\rangle| = m |m\rangle[/tex]

    and I would use P to indicate that we have an projector. So we have a clear distinction between the observable M w.r.t. which we define the eigenvalues and eigenstates and the projectors which project an arbitrary state to an eigenstate.

    That means

    [tex]p(m|i) = | \langle m|\Psi_i\rangle|^2 = \langle\Psi_i|m\rangle\langle m|\Psi_i\rangle = \langle\Psi_i|P_m|\Psi_i\rangle = \langle\Psi_i|P_m^\dagger\,P_m|\Psi_i\rangle[/tex]

    where I use

    [tex]|m\rangle\langle m| = P_m[/tex]
    [tex]P_m = P_m^2[/tex]
    [tex]P_m = P_m^\dagger[/tex]
     
    Last edited: Mar 3, 2013
  5. Mar 3, 2013 #4
    Stevendaryl and tom.stoer! Thank you very much for your replies. I think I got it finally, one question that directly follows: is it then right to say that:

    [itex] <M> = \Sigma p(m|i)m [/itex]

    and hence [itex] M = \Sigma P_m m [/itex]
     
  6. Mar 3, 2013 #5

    tom.stoer

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    Yes. For the observable M you can write

    [tex]M = \sum_m m\,|m\rangle\langle m|= \sum_m m\,P_m[/tex]

    Everything else follows from this spectral representation (where I have used the assumptions of descrete and non-degenerate eigenvalues)
     
  7. Mar 3, 2013 #6
    Ah..perfect. So the loop is closed :). Thanks once again!
     
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