Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 ;).
  2. jcsd
  3. Mar 2, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook