Eigenvectors of this Hamiltonian

1. Jul 24, 2014

ShayanJ

I've got a problem which is asking for the eigenvalues and eigenstates of the Hamiltonian $H_0=-B_0(a_1 \sigma_z^{(1)}+a_2 \sigma_z^{(2)})$ for a system consisting of two spin half particles in the magnetic field $\vec{B}=B_0 \hat z$.
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. Jul 24, 2014

atyy

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
3. Jul 24, 2014

rubi

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 $\sum_{i,j=1}^2 c_{ij} e_i\otimes e_j$.

4. Jul 24, 2014

stevendaryl

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

$\alpha_1 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_2 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right) + \alpha_3 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_4 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right)$

Now, operate on this general state with the Hamiltonian to find out what kind of constraints on the $\alpha$s would make it an eigenstate.

5. Jul 25, 2014

ShayanJ

Thanks people.
Now another question.
The next part of the problem asks for the first order correction for the perturbation $k \, \vec\sigma^{(1)} \cdot \vec\sigma^{(2)}$. Is the following correct?
$\vec\sigma^{(1)} \cdot \vec\sigma^{(2)}=(\sigma_x^{(1)*} \,\,\, \sigma_y^{(1)*} \,\,\, \sigma_z^{(1)*}) \left( \begin{array}{c} \sigma_x^{(2)} \\ \sigma_y^{(2)} \\ \sigma_z^{(2)} \end{array} \right)=\sigma_x^{(1)*}\sigma_x^{(2)}+\sigma_y^{(1)*}\sigma_y^{(2)}+ \sigma _z^{(1)*}\sigma_z^{(2)}$