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Homework Help: QM - matrix elements of operators

  1. Apr 23, 2008 #1
    1. The problem statement, all variables and given/known data

    we shall describe a simple model for a linear molecule, say, CO2.
    the states |L>, |C>,|R> are the eigenstates of D operator (corresponds to dipole moment)
    D|L>=-d|L> , D|C>=0 , D|R>= +d|R>.
    When the electron is localized exactly on the carbon atom, its energy is E1 and
    when it is localized on one of the oxygen atoms, it has energy E0 ( assuming
    E1 > E0). In addition, the electron can "jump" from one atom to another, and
    this jump is characterized by kinetic energy a. We can assume that the jumping
    occurs only between nearest neighbors.

    Write matrix representation of the Hamiltonian in the basis of |L>, |C>,|R>.
    2. The attempt at a solution

    we should build the matrix from the matrix elements - <L|H|L>, <L|H|C> etc.

    the matrix should look like that:
    E0 a 0
    a E1 a
    0 a E0

    it seems very logical, but I still don't see the direct connection between the matrix elements and the given data. for example, why does the element <L|H|L> equals E0?
    what does the element <1|H|2> mean in general?
  2. jcsd
  3. Apr 23, 2008 #2

    George Jones

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    What does H|L> equal?
  4. Apr 23, 2008 #3
    well, the problem is that |L> is not an eigenstate of H.
    So all I know is that |L> can be represented as a superposition of the Hamiltonian's eigenfunctions...
    What I understand from the data is that when I measure the energy in the state |L> I get the result E0... so using H will change the state to one of the eigensates of H and yield the result E0...(?)
  5. Apr 23, 2008 #4

    George Jones

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    If the matrix you gave is the answer, then you can write down

    [tex]H\left|L\right> = E_0 \left|L\right> + a \left|C\right>.[/tex]

    Do you see why?

    I also don't find this to be very intuitive. For example, I get the energy eigenvalues of [itex]H[/itex] to be

    [tex]\frac{1}{2}\left( E_1 + E_0 + \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right)[/tex]


    [tex]\frac{1}{2}\left( E_1 + E_0 - \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right)[/tex]

    with associated eigenvectors

    [tex]\left| L \right> + \frac{1}{2a}\left( E_1 - E_0 + \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right) \left| C \right> + \left| R \right>[/tex]

    [tex]\left|L\right> - \left|R\right>[/tex]

    [tex]\left| L \right> + \frac{1}{2a}\left( E_1 - E_0 - \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right) \left| C \right> + \left| R \right>[/tex]
  6. Apr 24, 2008 #5
    first of all - thanks for your reply(:

    secondly - yes, from the matrix I can see very clearly how H acts on any of the states. All I have to do is to multiply it (from the right) with the appropriate column vector; for instance, (1,0,0)T (T denotes "transpose") will represent |L>.
    I also understand how to find the eigenvalues and the eigenstates of H - directly from the matrix.

    But these problems are to be solved after you find the matrix. And this is the part that is unclear to me- how to create the matrix according to the given data. And to clarify my point: the problem is not technical. I know how to solve the exercise.

    My problem is that I don't understand why the following arguments are correct:

    1) from the fact that the energy of the system in the state |L> is E0 ---> we can conclude directly that <L|H|L> = E0.

    2) in a similar way - from the fact that the electron can "jump" from |L> to |C> (and vice versa) and that this jump is characterized by kinetic energy a ---> we deduce that
    <C|H|L> = <L|H|C> = a.

    and so on...
  7. Apr 24, 2008 #6

    George Jones

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    The only way we can know with certainty that the energy of the system in state |L> is E0 is if |L> is an eigenstate of H with corresponding eigenvalue E0. This follows directly from the probabilistic rules of quantum theory. Then, <L|H|L> = E0 follows easily.

    For the given the matrix representation of H, this argument, however, is incorrect.

    I used Maple to solve for the energy eigenvalues and eigenstates, and the solution shows that |L> is a complicated linear combination of three energy eigenstates. So, if the system is in state |L>, then the given matrix implies that there are three possible energies of the system that each have a non-zero probability.

    This clashes with the statement of the question, which makes me wonder whether the question is even consistent. Did the question come from a book? If so, which one? From the prof?

    I have never seen this interpretation.
  8. Apr 24, 2008 #7
    Hi again,

    this is exactly what I thought - that |L> should be an eigenstate of H, but it is not..!!

    the question comes from our tutorial, and we also had a homework assignment with a similar exercise. I believe the question was invented by our prof. or tutor.
    It does seem to be illogical and inconsistent.:grumpy:

    thanks a lot, anyway:smile:
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