1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probabilities for orthonormal wavefunctions

  1. Oct 31, 2007 #1
    Ok I have two orthonormal wavefunctions of a system, [tex]\psi[/tex] 1 and [tex]\psi[/tex] 2 and [tex]\widehat{A}[/tex] is an observable such that

    [tex]\widehat{A}[/tex] |[tex]\phi[/tex] [tex]_{n}[/tex] > = a[tex]_{n}[/tex] |[tex]\phi[/tex] [tex]_{n}[/tex] >

    for eigenvalues a sub n

    what are the probabilities p1(a1) and p2(a2) of obtaining the value a sub n in the state |psi1> and |psi2> respectively in terms of only phi sub i and psi sub 1(or 2)
     
  2. jcsd
  3. Oct 31, 2007 #2
    If I understand your question-

    You can't work out the probabilities from knowledge of the eigenvalues of an operator.
     
  4. Nov 1, 2007 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    But the probability that a measurement yields a particular eigenvalue can be expressed in terms of the associated eigenstate and the state of the system, which, if I have interpreted the original post correctly, kac9 has given.

    kac9: I'm having trouble guiding you to the answer without just writing down the answer. This is a basic postulate of (shut up and calculate) quantum mechanics. It must be in your notes and text. If you're using Griffiths, it's equation [3.43].
     
  5. Nov 1, 2007 #4
    George: Ah yes, the state is either in psi1 or psi2- my mistake.
     
  6. Nov 1, 2007 #5
    Well- a hint would be to write down the identity operator in terms of |phi>

    I=sum_i |phi_i><phi_i|

    |psi_1>=I|psi_1>
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?