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Simple quantum exam question

  1. Jan 14, 2005 #1

    I just want to ask a simple question about quantum mechanics...
    This question appears in it's original form here:
    http://www.chriscentral.com/physics/PC3101.PDF (question 2)

    I have had a go and put some notes into the text below... I am stuck on parts c) (ii) and (iii).
    Any help much appreciated! Thanks.

    A quantum mechanical system is described by a two-dimensional state space spanned by the orthonormal vectors |1> and |2>. The Hamiltonian of the system can be defined by:
    H|1> = a|1> + 2a|2>
    H|2> = 2a|1> + a|2>

    with "a" a positive constant.

    (a) Write down the matrix representation of H in the basis {|1> |2>}.

    (a 2a)
    (2a a )

    (b) Find the energy eigenvalues of the systems and associated orthonormal eigenvectors.

    I get {3, 1/root2*(1,-1)} and {-1,1/root2*(1,1)}, where this has been expressed in the form {eigenvalue,eigenvector}.

    (c) If at time t = 0 the system is described by the state vector |2>:

    (i) Find the mean value of the energy <H> at time t = 0.

    I get <H> = a.

    (ii) What is the probability that the system will be found having its highest possible energy at time t = 0?

    (iii) If the highest energy is measured at time t = 0, calculate the state vector of the system at time t.


    Any ideas? Thanks! Chris.
  2. jcsd
  3. Jan 14, 2005 #2


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    Some hints:
    For (ii), what is the state that corresponds to that highest energy ? If you have a given state |psi> and you have an observable A with an eigenstate |a> and eigenvalue a, remember that the probability to measure the eigenvalue a is given by |<a|psi>|^2 (Born's rule).

    For (iii), we know that after we measured an eigenvalue a (no matter what was the probability to find that ; now we HAVE measured it), then the state immediately after that measurement is |a>, the corresponding eigenvector.
    In this case, we have the hamiltonian ; what's special about the time evolution of the eigenstates of a hamiltonian ?

    hope this helps...

  4. Jan 14, 2005 #3
    Hi Patrick, thanks for your speedy reply!

    For ii) I have been attempting use Born's rule however I still find myself at a loss. The key things I ask myself here are:
    The system is in state |2>, which is NOT an eigenstate of the hamiltonian, therefore there is uncertain energy (which we expect), so it is fair to say maybe:
    [tex]H|1> = a_{1}|E_{1}> + a_{2}|E_{2}>[/tex]

    [tex]H|2> = a_{1}|E_{1}> - a_{2}|E_{2}>[/tex]

    where I think [tex]a_{n}[/tex] is the nth eigenvalue of the hamiltonian matrix.

    The plus's and minus's come from the eigenvectors of the hamiltonian.

    Still, I'm lost becaue I'm not sure what to do with it (or even if it's correct).

    For iii) right after measurement the system is in a state [tex]|2>[/tex] which is our boundary condition. I imagine that our system's state vector is going to be something like:

    [tex]|\psi>=|1>exp(iE_{1}t/\hbar) +|2>exp(iE_{2}t/\hbar) [/tex]
    but I'm not sure yet.

    Thanks for the help btw!
  5. Jan 14, 2005 #4
    For part (c)
    You will need to write the vector |2> in terms of the eigenvectors you found in part (b), then you know everything! You know what H will give when it acts on that state. And you know the probabilities of measuring a given energy (in this case it will be the number in front of your eigenvectors squared).

    By the way how did you get part i) without using the eigenvectors?
  6. Jan 14, 2005 #5
    Hey Norman,

    Surely the question gives me |2> as eigenvectors, right? I'm still a little miffed, I think someone is just going to have to post the answer and explain it. My lecturer didn't make me understand it fully either :-(

    For part i):
    [tex]<H> = <2|H|2> = <2|(2a|1>+a|2>)> = a [/tex]
  7. Jan 15, 2005 #6


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    No ! Not that way, you do not have to apply H in the left hand side.

    You just write your actual state, |2> as a combination of the eigenvectors of H:

    [tex] |2> = a_1 |E_1> + a_2 |E_2> [/tex]

    and then [tex] |a_2|^2[/tex] is simply the probability to find, after an energy mesurement, the energy state 2.
    Note also that [tex] a_2 = <E_2 | 2> [/tex] so that's very easy to do !

    No, it is not in state [tex] |2> [/tex] but it is in state [tex] |E_2> [/tex], because you measured the energy to be [tex] E_2 [/tex]. So because of that, the system is now in the corresponding eigenvector.

    NO! In order to write the above, you need not to work with the vectors |1> and |2> but with the ENERGY EIGENSTATES.
    And you ARE (after the measurement) in one single energy eigenstate....

  8. Jan 15, 2005 #7
    Ok I think I've got this. (Credit also to Mehring in this thread http://www.advancedphysics.org/viewthread.php?tid=1374)

    I'm going to go through the question now from the beginning and we'll see if it's all correct...

    a) [tex] H = a\left(\begin{array}{cc}1&2\\2&1\end{array}\right)[/tex]
    b) [tex] \lambda_{1} = -1a, e_{1}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\-1\end{array}\right)[/tex]
    [tex] \lambda_{2} = 3a, e_{2}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right)[/tex]
    i) [tex] <H> = a [/tex]
    [tex] |1> = \left(\begin{array}{c}\lambda_{1}\\\lambda_{2}\end{array}\right).e_{1} = \frac{1}{\sqrt{2}}(|-1a>-|3a>)[/tex]
    |2> = \left(\begin{array}{c}\lambda_{1}\\\lambda_{2}\end{array}\right).e_{2} = \frac{1}{\sqrt{2}}(|-1a>+|3a>)
    We want the state
    [tex]|< 3a | 2 >|^2 = \left|\frac{1}{\sqrt{2}} <3a | (|-1a>+|3a>)\right|^2 = \frac{1}{2}

    The system is in eigenstate [tex] 3a [/tex] at [tex] t=0 [/tex]. We want the time evolution of this state alone, which is
    [tex] |3a(t)> = |3a>\exp^{\frac{i3at}{\hbar}}[/tex]

    Some things I want to check:

    I were to swap around [tex] e_{1} [/tex] for [tex]e_{2}[/tex] and viceversa, in this case it produces the same answer, BUT, how should I be associating these eigenvectors states; does it matter which way around they are in general?

    Thanks to all again,
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