- #1

ShayanJ

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## Main Question or Discussion Point

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:

[itex]

-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)

[/itex]

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.

But I think the problem is wrong and no eigenstate and eigenvalue exist. Here's my reason:

[itex]

-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)

[/itex]

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.