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What's the difference between these two equations?

  1. Feb 19, 2016 #1
    say we have some wavefunction |psi> and we want to find the probability of this wavefunction being in the state |q>. I get that the probability is given by P = |<q|psi>|^2 since we're projecting the wavefunction onto the basis state |q> then squaring it to give the probability density.

    However what if we want to measure some observable quantity corresponding to some operator Q with one (of multiple) associated eigenstate |q>, and then find the probability that the system is in a state |q> after being observed. Would probability then be P = |<q|Q|psi>|^2 since I'm collapsing the wavefunction then projecting it onto the basis state |q>?

    But say I didn't make a measurement of the system (not apply Q to |psi>) I get the original result. In these two cases I get a different answer do I not? In both cases I'm finding the probability of the wavefunction being in state |q>. Why are the answers different and what do |<q|psi>|^2 and |<q|Q|psi>|^2 fundamentally mean?
     
  2. jcsd
  3. Feb 20, 2016 #2

    blue_leaf77

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    ##|\langle q_n |\psi\rangle|^2## is the one with probabilistic interpretation.
    Note that ##|\langle q_n |Q|\psi\rangle|^2 = |q_n|^2## where ##q_n## is the eigenvalue corresponding to the eigenstate ##|q_n\rangle## and it's obviously not a value which can be associated to be a probability.
    To emphasize again, ##|\langle q_n |\psi\rangle|^2## is the one with probabilistic interpretation. This follows from the average value of ##Q## (the average value is used here because it connects the theory and experiment) when measured repeatedly over a state ##|\psi\rangle##
    $$
    \langle \psi |Q|\psi\rangle = \sum_n q_n |\langle q_n |\psi\rangle|^2
    $$
    Since the LHS is interpreted as an average, the ## |\langle q_n |\psi\rangle|^2## in the RHS must be the frequency distribution which gives a measure of how often a measurement result with ##q_n## occurs.
     
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