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I came across a problem in the context of degenerate perturbation theory.

$ \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\braket}[2]{\langle{#1}|{#2}\rangle} \newcommand{\acomm}[2]{\left\{#1,#2\right\}}$ Consider a pair of two spin-1/2 systems with orthonormal basis $\{ \ket{\uparrow}, \ket{\downarrow} \}$ for each system where $$S_z\ket{\uparrow} = \frac{1}{2}, \quad S_z\ket{\downarrow} = -\frac{1}{2}\ket{\downarrow}$$ Define unperturbed Hamiltonian $H_0 = \hat{S_z}\otimes\hat{S_z}$. What are the eigenstates $\ket{\phi_n}$ and eigenvalues $E_n$ of $H_0$?

I know I should manage to transform the tensor product $H_0 = \hat{S_z}\otimes\hat{S_z}$ to a matrix and solve for the eigenvalues and eigenstates of that matrix. However, I feel confused how to deal with the tensor product. Can someone help me?

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Given the matrix representation of $\hat S_z$: $$\hat S_z\rightarrow \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \tag{1}$$ we can write the tensor product as follows: $$\begin{align} \hat S_z\otimes \hat S_z &\rightarrow \frac{\hbar}{2} \begin{pmatrix} 1 \cdot\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & 0 \cdot\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ 0 \cdot\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & -1 \cdot\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{pmatrix} \\&=\frac{\hbar^2}{4}\begin{pmatrix} 1 &0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align} \tag{2}$$ You can find the above formula here.

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  • $\begingroup$ @AndrewSteane Fixed, thank you! $\endgroup$
    – Charlie
    Nov 20, 2020 at 22:24

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