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Basic Two State System Question

  1. May 15, 2009 #1
    A two state system is described by linear superpositions of two stationary states: |E1> and |E2>, with corresponding energies E1 and E2. An observable Q has the eigenstates |Q1> and |Q2>, corresponding to eigenvalues Q1 and Q2 (both real). All states are normalized.

    Given: |Q1> = (cos x)|E1> + (sin x)|E2>

    1) Express |Q2> as a superposition of |E1> and |E2> and determine the coefficients:

    |Q2> = a|E1> + b|E2>

    solution: Since Q is hermitian, the eigenstates are orthogonal ->
    <Q1|Q2> = a (cos x) + b (sin x) = 0

    => a = -sin x; b = cos x (up to an overall phase factor - the minus sign could be switched)

    2) The system is prepared at t = 0 to be in the state |E1>. It is then measured, first by Alice and then by Bob.

    At t = t1, Alice measures Q. What is the probability that the measurement will yield Q1?
    Here I will denote the state of the system as |X>

    solution: |X(t)> = |E1> exp(-iE1 t / h)
    P(Q1) = <Q1|X(t1)><X(t1)|Q1> = (cos x) exp (iE1 t1 /h) <E1|E1> cos (x) exp(-iE1 t1 / h) = (cos x) ^2

    3) At t2 > t1 Bob measures Q. Again, determine the probability a measurement will yield Q1 - but consider the following two cases:

    case 1) Alice reported that her result was Q1
    case 2) Alice did not report the result of her measurement

    Here is where I am confused... any help would be greatly appreciated :)

    If my post has not met the guidelines, please notify me and i will modify where appropriate.
    Thank you
     
  2. jcsd
  3. May 15, 2009 #2
    I smell a cat!! Someone call Mr. Schrodinger and ask him to pick it up.

    Seriously, though - I think you need to consider what your initial state should be in each case, 3.1) and 3.2). You clearly have it for case 3.1), but you don't for 3.2), or rather, you don't know the outcome of Alice's observation, so you can't say which eigenstate it was left in; you only know probabilities for each possibility.
     
  4. May 15, 2009 #3
    belliot, thanks for the response.

    for case 1, is the probability unity?
     
  5. May 15, 2009 #4
    i think the answer to my last question was no.
     
  6. May 15, 2009 #5
    Keep in mind that the Q states are not the stationary states, so they evolve with time.
     
  7. May 15, 2009 #6
    so, tell me if this is absurd:
    for t1 < t < t2:

    |X(t)> = (cos x)|E1>exp(-i E1 (t1 - t) / h) + (sin x)|E2>exp(-i E2 (t1 - t) / h)

    then:

    <Q1|X(t2)> = exp(-iE1 (t2 - t1)/h) [ (cos x)^2 + (sin x)^2 exp(-i(E2 - E1)(t2 - t1)/h)

    so, P(Q1) = (cos x)^4 + (sin x)^4 + 2(cos x)^2 (sin x)^2 cos[(E2 - E1)(t2-t1)/h]

    This is strictly positive, and is <= 1 - but I'm not sure it's correct.
     
  8. May 15, 2009 #7
    Looks good to me ....

    I have to admit that I'm pretty rusty at good ol' QM, but FWIW, I don't see any mistakes. I'm sure someone else will point them out if there are any, however.
     
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