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Consider a system of two distinguishable spin-1/2 particles with Hamiltonian \begin{align} H &= \frac{\alpha}{4} \vec{\sigma}_1 \cdot\vec{\sigma}_2.\\ \end{align} where $\vec{\sigma}_1 = (\sigma_x\otimes 1, \sigma_y\otimes 1, \sigma_z\otimes 1)$ and $\vec{\sigma}_2 = (1\otimes \sigma_x,1\otimes \sigma_y,1\otimes \sigma_z)$ . In the uncoupled z-basis, we can write the Hamiltonian as \begin{align} H&= \frac{\alpha}{4}\left(\sigma_{x}\otimes\sigma_{x}+\sigma_{y}\otimes\sigma_{y}+\sigma_{z}\otimes\sigma_{z}\right)\\ &= \frac{\alpha}{4}\left(\sigma_{x}+\sigma_{y}+\sigma_{z}\right)\otimes\left(\sigma_{x}+\sigma_{y}+\sigma_{z}\right)\\ &= \frac{\alpha}{4}\begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix}\otimes \begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix} \end{align} The matrix $$\begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix}$$ has eigenvalues $\pm\sqrt{3}$, so in the uncoupled diagonal-basis $$H = \frac{3\alpha}{4}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\otimes\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$$ which has eigenvectors $$\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}1\\0\end{pmatrix}\hspace{2mm}, \hspace{2mm}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0\\1\end{pmatrix} \\ \begin{pmatrix}0\\1\end{pmatrix}\otimes\begin{pmatrix}1\\0\end{pmatrix}\hspace{2mm}, \hspace{2mm} \begin{pmatrix}0\\1\end{pmatrix}\otimes\begin{pmatrix}0\\1\end{pmatrix}$$ with respective eigenvalues $3\alpha/4, -3\alpha/4, -3\alpha/4, 3\alpha/4$.

We could've rewritten the Hamiltonian as \begin{align} H &= \frac{\alpha}{2}\left[\left(\frac{1}{2}\vec{\sigma}_1+\frac{1}{2}\vec{\sigma}_2\right)^2 - \left(\frac{1}{2}\vec{\sigma}_1\right)^2 - \left(\frac{1}{2}\vec{\sigma}_2\right)^2\right]\\ &=\frac{\alpha}{2}\left[s(s+1) - \frac{1}{2}\left(\frac{1}{2} +1\right) - \frac{1}{2}\left(\frac{1}{2} +1\right)\right]\\ &=\frac{\alpha}{2}\left[s(s+1) - \frac{3}{2}\right] \end{align} where $s$ is the the spin in the coupled basis ($s=0$ or $1$). Therefore the eigenvalues of the Hamiltonian in the coupled basis are $-3\alpha/4$ (with degeneracy 1) and $\alpha/4$ (with degeneracy 3).

The eigenvalues of the Hamiltonian shouldn't depend on your choice of basis, but in the above I get different eigenvalues in the coupled and uncoupled bases. Where am I going wrong?

Solution (thanks to Vadim): In the $|\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle,|\downarrow\downarrow\rangle$ basis the Hamiltonian takes the form \begin{align} H&= \frac{\alpha}{4}\left(\sigma_{x}\otimes\sigma_{x}+\sigma_{y}\otimes\sigma_{y}+\sigma_{z}\otimes\sigma_{z}\right)\\ &= \frac{\alpha}{4}\begin{pmatrix}1&0&0&0\\0&-1&2&0\\0&2&-1&0\\0&0&0&1\end{pmatrix} \end{align} which has eigenvalues $-3\alpha/4$ and $\alpha/4$. This is not the same as \begin{align} \frac{\alpha}{4}\left(\sigma_{x}+\sigma_{y}+\sigma_{z}\right)\otimes\left(\sigma_{x}+\sigma_{y}+\sigma_{z}\right) = \frac{\alpha}{4}\begin{pmatrix}1&1-i&1-i&-2i\\1+i&-1&2&-1+i\\1+i&2&-1&-1+i\\2i&-1-i&-1-i&1\end{pmatrix} \end{align} which has eigenvalues $\pm3\alpha/4$.

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  • $\begingroup$ The most straightforward way is to work in the basis of two-particle states: $|\uparrow, \uparrow\rangle, |\uparrow, \downarrow\rangle, |\downarrow, \uparrow\rangle, |\downarrow, \downarrow\rangle$. The resulting 4-by-4 matrix is actually easily diagonalizable. $\endgroup$
    – Roger V.
    Apr 7, 2020 at 16:31

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The error is in the first approach: $$\sigma_x\otimes\sigma_x + \sigma_y\otimes\sigma_y + \sigma_z\otimes\sigma_z \neq (\sigma_x + \sigma_y + \sigma_z)\otimes(\sigma_x + \sigma_y + \sigma_z),$$ as it is easy to verify by writing down these matrices in 4-by-4 basis $|\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle, |\downarrow\downarrow\rangle.$ Working with 4-by-4 matrices may seem dounting at first, but it is actually quite easy, once you get a grasp of how they nest one within the other, e.g. $$ \sigma_x^{(1)}\otimes\sigma_x^{(2)} =\begin{pmatrix} 0&\sigma_x^{(2)}\\\sigma_x^{(2)}&0\end{pmatrix} = \begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0 \end{pmatrix}$$ $$ \sigma_x^{(1)}\otimes\sigma_y^{(2)} =\begin{pmatrix} 0&\sigma_y^{(2)}\\\sigma_y^{(2)}&0\end{pmatrix} = \begin{pmatrix} 0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0 \end{pmatrix}$$ Incidentally, it is also helpful when dealing with the $\gamma$-matrices in the Dirac equation.

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  • $\begingroup$ Just using distributivity shows one would get $\sigma_x\otimes\sigma_y$ terms which do not appear in the original Hamiltonian. $\endgroup$ Apr 7, 2020 at 19:07

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