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Help on QM Problem - Dirac Notation

  1. Nov 20, 2004 #1

    AKG

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    Consider the following state vector and Hamiltonian:

    [tex]|\psi (0) \rangle = \frac{1}{5}\left (\begin{array}{cc}3\\0\\4\end{array}\right )[/tex]

    [tex]\hat{H} = \left (\begin{array}{ccc}3&0&0\\0&0&5\\0&5&0\end{array}\right )[/tex]

    • If we measure energy, what values can we obtain and with what probabilities?
    • Find the state of the system at a later time t; express [itex]|\psi (0) \rangle[/itex] in terms of the eigenvectors of [itex]\hat{H}[/itex].
    • Find the total energy at t = 0 and at a later time t.
    Here's what I have:
    • We can get values of either 3 or 5. I don't know how to get the probabilities.
    • The state of the system would be:

      [tex]|\psi _t\rangle = \hat{H}|\psi (0)\rangle = (1/5)(9\ \ 20\ 0)^t[/tex]

      Also:

      [tex]|\psi (0)\rangle = (3/5)(1\ 0\ 0)^t + (4/5)(0\ 0\ 1)^t[/tex]

    • I can only guess that the energy at t=0 is:

      [tex]\langle \psi (0)|\hat{H}|\psi (0)\rangle = \langle \psi (0) | \psi _t\rangle[/tex]

      and the energy at a later time t is:

      [tex]\langle \psi _t |\hat{H}|\psi _t \rangle[/tex]
     
    Last edited: Nov 20, 2004
  2. jcsd
  3. Nov 21, 2004 #2
    (a) You need to find the eigenvectors of the H matrix, solve for the zeros of the characteristic polynomial first.
    (b) Do you know the general solution at time t, given the solution at time 0?
     
  4. Nov 21, 2004 #3

    Galileo

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    The energies you can measure are the eigenvalues of H. 3,5 and -5.
    Write [itex]|\psi (0) \rangle [/itex] as a linear comb. of normalized eigenvectors of H.
    Then the prob. of measuring a certain eigenvalue is equal to the square of the absolute value of the component of the eigenvector corresponding to that eigenvalue (..phew)
     
  5. Nov 21, 2004 #4

    AKG

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    Okay, so we have eigenvalues of 3, 5, and -5 with normalized eigenvectors of (1 0 0), sqrt(1/2)(0 1 1), and sqrt(1/2)(0 1 -1) respectively. I don't know how to find the corresponding probabilities.

    I'm not sure how to find the state at a later time, given the state at t = 0. My notes say that if a measurment gives an eigenvalue A1, then the state after the measurment will be v1, the eigenvector corresponding to eigenvalue A1. So, My guess is that the state at time t > 0 would be P1v1 + P2v2 + P3v3, where v1, v2, v3 are the eigenvectors, and P1, P2, and P3 are their corresponding probabilites (which I can't figure out how to find yet).

    Expressing psi(0) in terms of eigenvectors of H is simple, assuming I have given the right eigenvectors.

    Am I right in the way I started part c, i.e. the total energy at time 0 is <psi(0) | H | psi(0)>?

    Thanks.
     
  6. Nov 21, 2004 #5

    Galileo

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    Now you need to express [itex]|\psi \rangle[/itex] as a linear combination of the eigenvectors you found.
    [tex]\left(\begin{array}{cc}3\\0\\4\end{array}\right )=3\left(\begin{array}{cc}1\\0\\0\end{array}\right )+2\left(\begin{array}{cc}0\\1\\1\end{array}\right )-2\left(\begin{array}{cc}0\\1\\-1\end{array}\right )[/tex]
    This is not exactly the combination you are looking for since i've used non-normalized eigenvectors and omitted a factor of 1/5. If you do this with the normalized eigenvectors then the square of the abs. value of the component of an eigenvector will give the probability of measuring the corresponding eigenvalue.
    That is, the probability of measuring an energy of 3 units (the eigenvalue of (1 0 0 )) is the square of the absolute value of it's coefficient. etc.

    If you know the state of a system at time t=0:
    [tex]|\psi(0) \rangle = \sum_n c_n |\psi_n \rangle[/tex]
    where the basis used is the one consisting of eigenvectors of the hamiltonian. Then:
    [tex]|\psi(t) \rangle = \sum_n c_n \exp\left(-{\frac{iE_n}{\hbar}t\right)|\psi_n \rangle[/tex]

    I`m not sure what the question means by the total energy. Maybe they mean the average value? If so, <H> is independent of time.
     
    Last edited: Nov 21, 2004
  7. Nov 21, 2004 #6

    AKG

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    So I should have, I believe:

    [tex]|\psi (0)\rangle = \frac{3}{5}\left(\begin{array}{cc}1\\0\\0\end{array}\right )+\frac{2\sqrt{2}}{5}\left(\begin{array}{cc}0\\1/sqrt{2}\\1/sqrt{2}\end{array}\right )-\frac{2\sqrt{2}}{5}\left(\begin{array}{cc}0\\1/sqrt{2}\\-1/sqrt{2}\end{array}\right )[/tex]

    From this, the probability that the energy is 3 is 9/25, and the probability of the other two eigenenergies are 8/25.
    So the [itex]c_n[/itex] should be the square roots of the probabilites, and the [itex]E_n[/itex] should be the corresponding energy eigenvalues, right?, i.e:

    [tex]|\psi(t) \rangle = \frac{3}{5} \exp\left(-{\frac{3i}{\hbar}t\right)\left(\begin{array}{cc}1\\0\\0\end{array}\right ) + \dots[/tex]

    I can figure out the other two terms in the sum if the first term I've given is right.
    And can I compute this by [itex]\langle \psi (0) | H | \psi (0) \rangle[/itex]?

    Thanks.
     
  8. Nov 21, 2004 #7

    Galileo

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    Yes, that all looks right.
     
  9. Nov 21, 2004 #8

    AKG

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    Okay, now if <H> is independent of time, then where it asks to find the "total" (I think they mean average) energy at a later time t, then is the the same as [itex]\langle \psi (0) | H |\psi (0) \rangle[/itex]? Or, if it changes, how do I find the changed energy?
     
  10. Nov 21, 2004 #9
    Your hamiltonian matrix isn't time-dependent, so you only get time-dependent phase factors in front of your eigenvectors. That means the [itex]c_n[/itex]'s don't change and your average energy is constant.
     
  11. Nov 21, 2004 #10

    AKG

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    Also, here are my answers quickly, could anyone check if they look right?

    Possible energy values: 3, 5, -5
    probabilities: 9/25, 8/25, 8/25

    [tex]|\psi (0)\rangle = \frac{3}{5}(1\ 0\ 0)^T + \frac{2\sqrt{2}}{5}(0\ 1/\sqrt{2} \ 1/\sqrt{2})^T + \frac{2\sqrt{2}}{5}(0\ -1/\sqrt{2} \ 1/\sqrt{2})^T[/tex]

    [itex]|\psi (t)\rangle[/itex] is the same as above, but each term is multiplied by a factor of:

    [tex]\exp \left (\frac{-E_ni}{\hbar}t \right )[/tex]

    The average energy at t = 0 is 27/25, and if average energy is time-independent, then it is the same at a later time t. However, what if the energy is measured at t=0, would that effect the later energy?
     
  12. Nov 21, 2004 #11

    Galileo

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    The answers look allright to me. I wonder if there is another way you were supposed to do this problem though, since the thread title is 'Dirac Notation'.

    If you measure the energy and you get (say) 5. Then the particle will be in the state 1/sqrt{2}(0 1 1).
    An energy measurement on this state is sure to return 5 and since <H> is independent of time, this will hold for all t (after the measurement). (this does not generally hold for any observable, but it does for the energy)
     
  13. Nov 21, 2004 #12

    AKG

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    The thread is titled Dirac Notation because that's what I assume is going on in class. I don't actually know, but the online note that is most closely related to these questions is called "Dirac Notation" so I guessed that's what it's about. Part c just doesn't sound right, asking for "total" energy. I would think the expected energy is 27/25, and at a later time, it is just whatever was measured the first time. I think I might just ask other people in my class to see if they know what the prof is asking. However, based on the question, is there any reason to believe that the answer for "total energy" at time 0 should be different than the energy at time t?
     
  14. Nov 21, 2004 #13

    AKG

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    Another question on the same topic:

    [tex]|\psi (2) \rangle = \left (\begin{array}{cc}-1\\2\\1\end{array}\right ),\ \ \hat{A} = \frac{1}{\sqrt{2}}\left (\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right ),\ \ \hat{B} = \left (\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right )[/tex]

    • What is the probability that a measurement of A at time t yields -1?

      I found the eigenvalues of A (surely, -1 was one of them) and then found the corresponding normalized eigenvectors. I then expressed [itex]|\psi (t)\rangle[/itex] in terms of those eigenvectors. Then squaring the co-efficient of the eigenvector corresponding to [itex]\lambda _3 = -1[/itex], i.e. [itex]c_3 = \sqrt{2}[/itex] and dividing by the norm of the state vector gives the desired probability.

    • If we measure B first then A right after, what is the probability of obtaining 1 for A, and 0 for B?

      I would do something similar to the above to find the probability of getting 0 for B. Then, the new state vector becomes the eigenvector corresponding the eigenvalue 0 for B. Using that as the new state vector, I do the problem again (i.e. express the state vector in terms of eigenvectors of A, then take the coefficient of the eigenvector corresponding to 1 in the express on of the state vector, and divide by the norm of the state vector).

    • As above, but measure A first.

      If the above is right, I can do this of course.

    • Explain why the two results are different.

      Because measuring the state of a system alters the state of the system. Is that right?
     
    Last edited: Nov 21, 2004
  15. Nov 22, 2004 #14

    Galileo

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    Yes, when a measurement is made, the state vector gets projected onto the eigenspace spanned by the measured eigenvalue.
     
  16. Nov 22, 2004 #15

    Imo

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    Are you by chance at the University of Toronto in Professor Savard's QM class? because I just got those same exact questions on my problem set.

    Also, to check your answers (and mine) for the second question I got that Prob(A=-1)=1/3 and for b) I got 1/3, and c) 1/18.
     
  17. Nov 22, 2004 #16

    AKG

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    I saw your earlier post on the average energy being 3 or 3/5 or something? I'm guessing you got 27/25 now (because that post is gone). Anyways, I got 1/3 for a), for b), I got that the probability that B=0 is 2/3, and the probability that A=1 given B=0 is 1/2, so yes, the probability that B=0 and A=1 is 1/3 (since it's 2/3 * 1/2). For c, I got:
    Prob(A = 1) = 1/3
    P(B = 0 given A = 1) = 1/2
    P(B = 0 AND A = 1) = 1/2 * 1/3 = 1/6 (not 1/18).
     
  18. Nov 22, 2004 #17

    Imo

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    ahh, sorry, I hadn't checked that answer yet. Yes, it's 1/6.
     
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