# Permutation operator on two spin 1/2 particles

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?

• I think $\hbar=1$ is implicit in the eigenvalues of $\hat S_1\cdot\hat S_2$ else this operator is not dimensionally consistent with $\hat 1$. Commented Jan 5, 2020 at 18:42
• Not really, but as it was on one of our exams I took it for granted. How should it be defined then, in a matrix representation (and operators if it exists)? Commented Jan 5, 2020 at 18:55

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.