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Quantum system time evolution

  1. Sep 19, 2014 #1
    1. The problem statement, all variables and given/known data

    A quantum system has Hamiltonian H with normalised eigenstates ψn and corresponding energies En (n = 1,2,3...). A linear operator Q is defined by its action on these states:

    1 = ψ2
    2 = ψ1
    n = 0, n>2

    Show that Q has eigenvalues 1 and -1 and find the corresponding normalised eigenstates ζ1 and ζ2, in terms of energy eigenstates. Calculate <H> in each of the states ζ1 and ζ2.

    A measurement of Q is made at time=0, and the result 1 is obtained. The system is then left undisturbed for a time t, at which instant another measurement of Q is made. What is the probability that the result will again be 1? Show that the probability is 0 if the measurement is made after a time T = [itex]\pi[/itex]ħ/(E2 - E1), assuming E2 - E1> 0.

    2. Relevant equations



    3. The attempt at a solution

    I found
    ζ1 = (ψ1 + ψ2)/√2
    ζ2 = (ψ1 - ψ2)/√2
    and <H> = (E1 + E2)/2 for both.

    I have trouble doing the second part.

    Doesnt the system collapse into ζ1 given we know this is the state at time =0?
    so probability will be 1?
     
  2. jcsd
  3. Sep 19, 2014 #2

    ShayanJ

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    Gold Member

    The wave function collapses to [itex] \zeta_1 [/itex] after measurement but after that, it will evolve by Schrodinger equation and so it will change gradually. You should apply the time evolution operator to the state [itex] \zeta_1[/itex] and then calculate the inner product of the evolved state with [itex] \zeta_1 [/itex].
     
  4. Sep 19, 2014 #3
    so ψ(x,0) = (ψ1 + ψ2)/√2,
    and U(t,0) = exp(-iHt/ħ),
    then ψ(x,t) = U(t,0)ψ(x,0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2

    inner product :
    ∫ [exp(iHt/ħ)((ψ1)* + (ψ2)*)/√2][(ψ1 + ψ2)/√2] dx
    = exp(iHt/ħ)

    is this correct? but i dont know what H is here, is it (E1+E2)/2?
     
  5. Sep 19, 2014 #4

    ShayanJ

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    The point you're missing, is how to act on [itex] \psi [/itex]s by the evolution operator.
    [itex]
    e^{\frac{iHt}{\hbar}}\psi_n=\sum_{m=0}^{\infty} (\frac{iH_nt}{\hbar})^m \frac{ \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{it}{\hbar})^m \frac{H^m \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{it}{\hbar})^m \frac{E_n^m \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{iE_nt}{\hbar})^m \frac{ \psi_n}{m!}=e^{\frac{iE_n t}{\hbar}}\psi_n
    [/itex]

    So you don't need to know H!
     
  6. Sep 20, 2014 #5
    so ψ(0) = (ψ1 + ψ2)/√2, ψ(t) = U(t,0)ψ(0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 = [exp(-iE1t/ħ)ψ1 + exp(-iE2t/ħ)ψ]/√2. then find the probability amplitude of (ψ1 + ψ2)/√2 in this time-varying state by taking the inner product ((ψ1 + ψ2)/√2 , ψ(t)), squared it to find the probability = [1 + cos((E2-E1)t/h)]/2, then sub in T = pi*h/(E2-E1) indeed got probability = 0. but why after T it'll stay 0? doesnt it cycle again by the consine term?
     
  7. Sep 20, 2014 #6

    Ray Vickson

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    Why do you say it remains zero? Just because it becomes 0 momentarily does not mean it stays at 0.
     
  8. Sep 20, 2014 #7

    nrqed

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    I think you simply misread the question but I can see how the phrasing may lead to confusion. What they meant is that at the time T = pi*h/(E2-E1) after the initial measurement was made the probability is zero, but they do not mean to say that it will remain zero for all times after that instant. They really just meant at that specific instant, not at all later times. So you are right, it will cycle again.
     
  9. Sep 20, 2014 #8
    ok thanks guys!
     
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