Permutation operator on two spin 1/2 particles

Question

Suppose we have two spin 1/2 particles. Then we have a common eigenbasis $\{|m_{s1},m_{s2}\rangle\}$ to the operators $\hat S_{1z}$ and $\hat S_{2z}$. The permutation operator is defined as:

$$ \hat P_{21}|m_{s1},m_{s2}\rangle=|m_{s2},m_{s1}\rangle $$

My job is to find the eigenvalues of $\hat P_{21}$ and its eigenkets, as well as to prove that:

$$ \hat P_{21}=\hat 1+\frac{\hat 1}{2}+2\hat{\vec{ S_1}}\cdot\hat{\vec{ S_2}}, \tag{1} $$

Here's my attempt:

We know that for some vector $(x,y)^T$ there is a matrix that can put it like $(y,x)^T$ given by $ M= \left( {\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} } \right) $, so I would expect that, after writing the (1) in $\{|m_{s1},m_{s2}\rangle\}$ basis something of the sort would come up. But after some tweeking I get that:

$$ \hat{\vec{ S_1}}\cdot\hat{\vec{ S_2}}=\frac{\hbar^2}{4}\bigg[(\sigma_x \otimes \hat 1)(\hat 1 \otimes \sigma_x)+(\sigma_y \otimes \hat 1)(\hat 1 \otimes \sigma_y)+(\sigma_z \otimes \hat 1)(\hat 1 \otimes \sigma_z) \bigg] $$

which reduces to:

$$ \hat{\vec{ S_1}}\cdot\hat{\vec{ S_2}}=\frac{\hbar^2}{4}\bigg[(\sigma_x \otimes \sigma_x)+(\sigma_y \otimes \sigma_y)+(\sigma_z \otimes \sigma_z) \bigg] $$

As we know that $ \sigma_x \otimes \sigma_x= \left( {\begin{array}{cc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} } \right) $ , $ \sigma_y \otimes \sigma_y= \left( {\begin{array}{cc} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ \end{array} } \right) $ and $ \sigma_x \otimes \sigma_x= \left( {\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right) $, we get that $ \sum_i \sigma_1 \otimes \sigma_1= \left( {\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right) $. Plugging this all back in (1) we see that what we end up with is something like this:

$$ \hat P_{21}= \left( {\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right) + \frac{1}{2}\left( {\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right) + \frac{\hbar^2}{2} \left( {\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right) $$

$$ \hat P_{21}= \left( {\begin{array}{cc} 3/2+\hbar^2/2 & 0 & 0 & 0 \\ 0 & 3/2-\hbar^2/2 & 3/2+\hbar^2 & 0 \\ 0 & 3/2+\hbar^2 & 3/2-\hbar^2/2 & 0 \\ 0 & 0 & 0 & 3/2+\hbar^2/2 \\ \end{array} } \right) $$

But this has no resemblance to the matrix M and tells me nothing. How can one prove (1)? Why is my approach wrong?

Answer 1

I think there's a misprint in your question so I'll go with what I think is right.

Using the coupled basis (which is already permutation symmetric) \begin{align} \vert 11\rangle = \vert +\rangle_1\vert +\rangle_2\, ,\qquad \vert 10\rangle &= \frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2 +\vert -\rangle_1\vert +\rangle_2\right)\, ,\quad \vert 1-1\rangle = \vert -\rangle_1\vert -\rangle_2\, ,\\ \vert 00\rangle &= \frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2 -\vert -\rangle_1\vert +\rangle_2\right) \end{align} it is clear that the permutation operator will have the form \begin{align} P_{12}\mapsto \left(\begin{array}{cccc} 1&&&\\ &1&&\\ &&1&\\ &&&-1\end{array}\right) \end{align} with eigenvalue $+1$ on the symmetric $S=1$ states and $-1$ on the antisymmetric $S=0$ state. The operator $$ \hat{\textbf{J}}\cdot\hat{\textbf{J}}= \hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_1+\hat{\textbf{S}}_2\cdot\hat{\textbf{S}}_2+2\hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2= \frac{3}{2}\hat{\boldsymbol{1}}+ 2\hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2 \quad\Rightarrow 2\hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2=\hat{\textbf{J}}\cdot\hat{\textbf{J}}-\frac{3}{2}\hat{\boldsymbol{1}} $$ using $\hbar=1$, so, given that $\hat{\textbf{J}}\cdot\hat{\textbf{J}}=2$ for symmetric states and $0$ for the antisymmetric state, we see that \begin{align} 2\hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2= \left\{\begin{array}{cc}-\frac{3}{2}& \hbox{if } J=0\\ \frac{1}{2} &\hbox{if } J=1\, ,\end{array}\right. \end{align} Hence, \begin{align} P_{12}=\frac{1}{2}\hat{\boldsymbol{1}}+2\hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2 \end{align}

Note that in the uncoupled basis, with ordering $\vert+\rangle_1 \vert+\rangle_2, \vert+\rangle_1\vert-\rangle_2,\vert-\rangle_1\vert+\rangle_2, \vert-\rangle_1\vert-\rangle_2$, we have \begin{align} \hat{\textbf{S}}_1\cdot\hat{\textbf{S}}_2 =\frac{1}{4}\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \end{align} and then $P_{12}$ takes the form \begin{align} P_{12}\mapsto\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \end{align} which clearly permutes the $\vert \pm\rangle_1\vert \mp\rangle_2$ states. The eigenvalues and (unnormalized) eigenvectors of the $2\times 2$ submatrix are $\pm 1$ and $(1,\pm 1)^\top$, from which one can obtain the properly symmetrized combinations of $\vert \pm\rangle_1\vert \mp\rangle_2$ states.