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Eigenstates for Hamiltonian

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume a Hilbert space with the basis vectors [itex]\left| 1 \right\rangle[/itex], [itex]\left| 2 \right\rangle[/itex] and [itex]\left| 3 \right\rangle[/itex], and a Hamiltonian, which is described by the chosen basis as:
    [tex]H=\hbar J\left( \begin{matrix}
    0 & 1 & 1 \\
    1 & 0 & 1 \\
    1 & 1 & 0 \\
    \end{matrix} \right),
    where [itex]J[/itex] is a constant.

    Now, show that the linear combination [itex]\left| {{\psi }^{\left( k \right)}} \right\rangle =\sum\nolimits_{n=1}^{3}{c_{n}^{\left( k \right)}\left| n \right\rangle }[/itex], where [itex]c_{n}^{\left( k \right)}={{e}^{in\cdot 2\pi k/3}}/\sqrt{3}[/itex] with [itex]k = 1,2,3[/itex], is eigenstates for [itex]H[/itex].

    2. Relevant equations

    3. The attempt at a solution
    I've been told that for [itex]k = 1[/itex], as an example, I should get:
    [tex]H\left| {{\psi }^{\left( 1 \right)}} \right\rangle =\hbar J\left( \begin{matrix}
    0 & 1 & 1 \\
    1 & 0 & 1 \\
    1 & 1 & 0 \\
    \end{matrix} \right)\left( \begin{matrix}
    {{e}^{2\pi i/3}} \\
    {{e}^{4\pi i/3}} \\
    1 \\
    \end{matrix} \right)\frac{1}{\sqrt{3}}
    And from this you see that it is an actual eigenstate, since I end up with:
    [tex]H\left| {{\psi }^{\left( 1 \right)}} \right\rangle =-\hbar J\left| {{\psi }^{\left( 1 \right)}} \right\rangle[/tex]

    My question is, how did the guy, which I got this from, come up with the vector for [itex]\left| {{\psi }^{\left( 1 \right)}} \right\rangle[/itex].
    I can't seem to figure that out :/

    Thanks in advance.
  2. jcsd
  3. Jun 3, 2013 #2


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    The expression for [itex]\left| {{\psi }^{\left( 1 \right)}} \right\rangle[/itex] just comes from letting ##k=1## in [itex]\left| {{\psi }^{\left( k \right)}} \right\rangle =\sum\nolimits_{n=1}^{3}{c_{n}^{\left( k \right)}\left| n \right\rangle }[/itex]

    Did you try writing that out explicitly and then interpreting as a column matrix?
  4. Jun 3, 2013 #3
    I just figured it out :)

    Thank you.
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