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Finding probability of one state in another

  1. Jun 1, 2013 #1
    if we want to find the probability of a wave function let ψ we use <ψ|ψ> when there is no change in wave function. but when we previously have wave function ψ and we quadruple it then wavefunction becomes ψ' the probability is found by |< ψ'|ψ>|^2. why we take square in this case? i cant understand it.
     
  2. jcsd
  3. Jun 1, 2013 #2
    [itex]\left\langle B \:|\: A\right\rangle[/itex] is the probability amplitude to observe a system initially in state A in state B. To get the actually probability, you need to take the mod squared of the probability amplitude.

    P(A is in B) [itex]=\left|\left\langle B \:|\: A\right\rangle\right|^{2}[/itex]

    Of course [itex]\left\langle A \:|\: A\right\rangle[/itex] is always 1, since the projection of a normalized vector onto itself is unity. So it just follows that the mod square of 1 is 1.

    P(A is in A) [itex]=\left|\left\langle A \:|\: A\right\rangle\right|^{2} = \left|1\right|^{2} = 1[/itex]
     
    Last edited: Jun 1, 2013
  4. Jun 4, 2013 #3
    A kind of projection

    Think of [itex]< \psi^{'}|\psi >[/itex] as a projection onto a "co-ordinate axis which here is [itex] \psi^{'}[/itex]. Just like with vectors, you can have a vector in 3 dimensions and then project onto some basis like [itex] \hat{x} [/itex]. Here we are in the Hilbert space spanned by the eigenstates of the wave function. We project it onto the state we want, but then to find the probability of being in that state we square it.
     
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