1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

QM: Time development of the probability of an Eigenvalue

  1. Apr 26, 2016 #1
    The problem is actually of an introductory leven in Quantum Mechanics. I am doing a course on atomic and molecular physics and they wanted us to practice again some of the basics.
    I want to know where I went conceptually wrong because my answer doesn't give a total probability of one, which of course is wrong.

    1. The problem statement, all variables and given/known data

    [tex]\hat{H}\Psi_1=E_1\Psi_1\quad and\quad\hat{H}\Psi_2=E_2\Psi_2[/tex]

    [tex]\phi_1=(\Psi_1-\Psi_2)/\sqrt2\quad and\quad\phi_2=(\Psi_1+\Psi_2)/\sqrt2[/tex]

    [tex]\hat{A}\phi_1=a_1\quad and\quad\hat{A}\phi_2=a_2[/tex]

    The question goes like this: Assuming you measure A^ and find a1, then how does the probability of finding the same result (a1) change over time?

    2. Relevant equations

    Well, the only important equation is the time development for a wave function in an eigenstate of the Hamilton Operator.


    3. The attempt at a solution

    After measurement the wave function collapses so the state wave function becomes Φ1. Since we know that [tex]\phi_1=(\Psi_1-\Psi_2)/\sqrt2\ [/tex] we know that the time-dependent wave function will be: [tex]\Psi=(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ [/tex]

    Thus if we take [tex]<\Psi|\hat{A}|\Psi> [/tex] it is
    [tex]<(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ |\hat{A}|(\Psi_1\exp(-iE_jt/\hbar)-\Psi_2\exp(-iE_jt/\hbar))/\sqrt2\ > [/tex]
    [tex] = 1/2 (<\Psi_1|\hat{A}|\Psi_1> + <\Psi_2|\hat{A}|\Psi_2> - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar))[/tex]
    [tex] = 1/2 ( 1/2(a_1+a_2) + 1/2(a_2-a_1) - 2<\Psi_1|\hat{A}|\Psi_2>\cos((E1-E2)t/\hbar)))[/tex]
    [tex] = 1/2( a_2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1))[/tex]
    [tex] = a_2/2 - \cos((E_1-E_2)t/\hbar))(a_2-a_1)/2[/tex]

    I don't think this can be true, since when the cosinus is 0 the result is a2/2, thus violating the fact that the sum of the factors should be one... And when one is to insert t=0 one becomes a1/2 and not a1 as expected.

    Could someone tell me where I went conceptually wrong? Thanks in advance :D
  2. jcsd
  3. Apr 26, 2016 #2
    I calculated again and my mistake was with [tex]<\Psi_2|\hat{A}|\Psi_2>[/tex]
    It is 1/2(a2 + a1) not 1/2(a2 - a1)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted