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QM: Extremely fast alternating H and K observations

  1. Jun 19, 2010 #1


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    The question: Consider extremely fast alternating energy and omega-ness measurements—so fast that you can neglect the much slower time evolution produced by the Hamiltonian. Draw and explain the branches of the probability tree for the case where the first measurement is an energy measurement. You can stop at the fourth generation: H, K, H, K. Your probability tree should show the possibilities, the probabilities, and the resulting state vectors for each measurement.

    So we have an observable K = [tex] \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}[/tex]

    and its eigenvectors are k1 = (-i, 1)T/[tex]\sqrt{2}[/tex] and k2 = (i, 1)T/[tex]\sqrt{2}[/tex] corresponding to eigenvalues 1 and -1, respectively.

    We also have an observable H = [tex] \begin{bmatrix} 3 & 0 \\ 0 & -4 \end{bmatrix}[/tex]

    and its eigenvectors are v1 = (1, 0)T and v2 = (0, 1)T corresponding to eigenvalues 3 and -4, respectively.

    The original state vector is N [tex] \begin{bmatrix} -5 \\ 6 \end{bmatrix}[/tex]

    Now, if we take the commutator [H,K], we see that the two observables do not commute. Thus, they cannot share the same eigenvectors.

    So I take an energy measurement first. And then I get a 36/61 chance of getting v1 corresponding to state vector (0, 1)T. Now, I take a K measurement. So that means I multiply k1T with v2 (to get a probability of 1/2 and a state vector of k1) and k2T with v2 to get a probability of 1/2 and a state vector of k2. Now I take an energy measurement again. Now, intuitively, I would expect to multiply the new state vector by v2. If I did, I would only get a probability of 1/2 for v2. But since I've observed v2 to be in the state of v2, and there hasn't been any time for the state vector to evolve, shouldn't I expect v2 to have a probability of 1? How could I explain this discrepancy?
    Last edited: Jun 19, 2010
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