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Eigenvectors of this Hamiltonian

  1. Jul 24, 2014 #1


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    I've got a problem which is asking for the eigenvalues and eigenstates of the Hamiltonian [itex] H_0=-B_0(a_1 \sigma_z^{(1)}+a_2 \sigma_z^{(2)}) [/itex] for a system consisting of two spin half particles in the magnetic field [itex] \vec{B}=B_0 \hat z [/itex].
    But I think the problem is wrong and no eigenstate and eigenvalue exist. Here's my reason:
    -B_0(a_1 \sigma_z^{(1)}+a_2 \sigma_z^{(2)}) \left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2 \end{array}\right) =\lambda\left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2\end{array}\right) \Rightarrow -B_0(a_1 \left(\begin{array}{c}x_1\\-y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2 \end{array}\right)+a_2 \left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\-y_2 \end{array}\right)) =\lambda\left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2\end{array}\right)
    But as far as I know, that is impossible. Am I right?
    I'm asking this because I'm affraid maybe my knowledge on tensor product states and operators contain some holes because I learnt it by reading little things here and there.
  2. jcsd
  3. Jul 24, 2014 #2


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    The final step doesn't look right. I haven't looked at the details, but I would try representing the tensor product state using a 4X1 column vector as in http://www.matfys.lth.se/education/quantinfo/QIlect1.pdf (p10) or http://www.tau.ac.il/~quantum/Reznik/lecture notes in quantum information.pdf (p32). The operators should be 4X4 matrices, and ##\sigma_z^{(1)}## should be ##\sigma_z^{(1)} \otimes \mathbb{I}^{(2)}##.

    Edit: rubi's and stevendaryl's posts are correct. The 4X1 vector should be have unknowns [a b c d]T.
    Last edited: Jul 24, 2014
  4. Jul 24, 2014 #3


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    The problem with the calculation is that a general vector in the tensor product space isn't ##(\sum_{i=1}^2 a_i e_i)\otimes(\sum_{j=1}^2 b_j e_j)## but rather ##\sum_{i,j=1}^2 c_{ij} e_i\otimes e_j##.
  5. Jul 24, 2014 #4


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    The state [itex]\left( \begin{array}\\ x_1 \\ y_1\end{array} \right) \left( \begin{array}\\ x_2 \\ y_2\end{array} \right) [/itex] is not the most general state for two spin-1/2 particles. The most general state looks something like this:

    [itex]\alpha_1 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_2 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right) + \alpha_3 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_4 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right) [/itex]

    Now, operate on this general state with the Hamiltonian to find out what kind of constraints on the [itex]\alpha[/itex]s would make it an eigenstate.
  6. Jul 25, 2014 #5


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    Thanks people.
    Now another question.
    The next part of the problem asks for the first order correction for the perturbation [itex] k \, \vec\sigma^{(1)} \cdot \vec\sigma^{(2)} [/itex]. Is the following correct?

    \vec\sigma^{(1)} \cdot \vec\sigma^{(2)}=(\sigma_x^{(1)*} \,\,\, \sigma_y^{(1)*} \,\,\, \sigma_z^{(1)*}) \left( \begin{array}{c} \sigma_x^{(2)} \\ \sigma_y^{(2)} \\ \sigma_z^{(2)} \end{array} \right)=\sigma_x^{(1)*}\sigma_x^{(2)}+\sigma_y^{(1)*}\sigma_y^{(2)}+ \sigma _z^{(1)*}\sigma_z^{(2)}

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