Matrix representation of the square of the spin angular momentum | Quantum Science Philippines

## Matrix representation of the square of the spin angular momentum

By: Maria Christine L. Lugo, MS Physics I, MSU-IIT

Show that

a.) $(S^{2} = S_{1}^{2}+S_{2}^{2}+ 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})$

b.) $S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)$

Solution:

a.) We can expand $S^{2}$ by its components $S_{1}^{2}$ and S_{2}^{2},

$S^{2} = S_{1}^{2}+S_{2}^{2} + 2(S_{1}\cdot S_{2})$

But we know that,

$S_{1}\cdot S_{2} = \frac{1}{2} (S_{1+}S_{2-}+S_{1-}S_{2+}) +S_{1z}S_{2z}$

Therefore,

$S^{2} = S_{1}^{2} + S_{2}^{2} + 2[\frac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+})+S_{1z}S_{2z}]$

$S^{2} = S_{1}^{2} + S_{2}^{2} +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}$

b.) Using the eigenvalue equation of $S^{2}$ and substituting the basis $(|++>, |+->, |-+>, |- ->)$

$S^{2}|sm> = (S_{1}^{2}+S_{2}^{2} +2S_{1z}S_{2z}+ S_{1+}S_{2-}+S_{1-}S_{2+})|sm>$

We have,

$\begin{array}{lll} S^{2}|++>&=& S_{1}^{2}|++>+ S_{2}^{2}|++>+2S_{1z}S_{2z}|++> +S_{1+}S_{2-}|++>+S_{1-}S_{2+}|++>\\ \\&=&\frac{3}{4}\hbar^{2}|++>+\frac{3}{4}\hbar^{2}|++>+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|++>\\ \\&=&2\hbar^{2}|++>\end{array}$

$\begin{array}{lll} S^{2}|+->&=& S_{1}^{2}|+->+ S_{2}^{2}|+->+2S_{1z}S_{2z}|+-> +S_{1+}S_{2-}|+->+S_{1-}S_{2+}|+->\\ \\&=&\frac{3}{4}\hbar^{2}|+->+\frac{3}{4}\hbar^{2}|+->-\frac{1}{2}|+->+\hbar^{2}|-+>\\ \\&=&\hbar^{2}(|+->+|-+>)\end{array}$

$\begin{array}{lll} S^{2}|-+>&=& S_{1}^{2}|-+>+ S_{2}^{2}|-+>+2S_{1z}S_{2z}|-+> +S_{1+}S_{2-}|-+>+S_{1-}S_{2+}|-+>\\ \\&=&\frac{3}{4}\hbar^{2}|-+>+\frac{3}{4}\hbar^{2}|-+>-\frac{1}{2}|-+>+\hbar^{2}|+->\\ \\&=&\hbar^{2}(|-+>+|+->)\end{array}$

$\begin{array}{lll} S^{2}|->&=& S_{1}^{2}|- ->+ S_{2}^{2}|- ->+2S_{1z}S_{2z}|- -> +S_{1+}S_{2-}|- ->+S_{1-}S_{2+}|- ->\\ \\&=&\frac{3}{4}\hbar^{2}|- ->+\frac{3}{4}\hbar^{2}|- ->+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|- ->\\ \\&=&2\hbar^{2}|- ->\end{array}$

Therefore, the matrix representation will give us

$S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)$