Physical predictions involving identical particles

We'll treat the problem only for a two-particle system. Consider two identical particles which can be in the individual states $\left|u_{a}\right\rangle ,\left|u_{b}\right\rangle$$\left|u_a\right\rangle, \left|u_b\right\rangle$,which include the spin-dependence and are orthogonal. The physical state of the system is:

$${\begin{aligned}\left|u_{S}\right\rangle ={\frac {1}{\sqrt {2}}}\left(\left|u_{a}(1),u_{b}(2)\right\rangle +\left|u_{a}(2),u_{b}(1)\right\rangle \right)\quad {\mbox{for bosons}}\\\left|u\right\rangle _{A}={\frac {1}{\sqrt {2}}}\left(\left|u_{a}(1),u_{b}(2)\right\rangle -\left|u_{a}(2),u_{b}(1)\right\rangle \right)\quad {\mbox{for fermions}}\end{aligned}}$$

$${\begin{aligned}\left|u_{S}\right\rangle ={\frac {1}{\sqrt {2}}}\left(\left|u_{a}(1),u_{b}(2)\right\rangle +\left|u_{a}(2),u_{b}(1)\right\rangle \right)\quad {\mbox{for bosons}}\\\left|u\right\rangle _{A}={\frac {1}{\sqrt {2}}}\left(\left|u_{a}(1),u_{b}(2)\right\rangle -\left|u_{a}(2),u_{b}(1)\right\rangle \right)\quad {\mbox{for fermions}}\end{aligned}}$$

Suppose we want to measure on each particle the physical quantity $G$$G$

$$G\left|g_{i}\right\rangle =g_{i}\left|g_{i}\right\rangle \quad \left[G,P_{21}\right]=0$$

$$G\left|g_i\right\rangle=g_i\left|g_i\right\rangle \quad \left[G,P_{21}\right]=0$$

What is the probability of finding $g_{n}$$g_n$ for one particle and $g_{n'}$$g_{n'}$ for the other? The final state is:

$${\begin{aligned}\left|g_{n},g_{n'}\right\rangle _{S}={\frac {1}{\sqrt {2}}}\left(\left|g_{n}(1),g_{n'}(2)\right\rangle +\left|g_{n}(2),g_{n'}(1)\right\rangle \right)\quad {\mbox{for bosons}}\\\left|g_{n},g_{n'}\right\rangle _{A}={\frac {1}{\sqrt {2}}}\left(\left|g_{n}(1),g_{n'}(2)\right\rangle -\left|g_{n}(2),g_{n'}(1)\right\rangle \right)\quad {\mbox{for fermions}}\end{aligned}}$$

$${\begin{aligned}\left|g_{n},g_{n'}\right\rangle _{S}={\frac {1}{\sqrt {2}}}\left(\left|g_{n}(1),g_{n'}(2)\right\rangle +\left|g_{n}(2),g_{n'}(1)\right\rangle \right)\quad {\mbox{for bosons}}\\\left|g_{n},g_{n'}\right\rangle _{A}={\frac {1}{\sqrt {2}}}\left(\left|g_{n}(1),g_{n'}(2)\right\rangle -\left|g_{n}(2),g_{n'}(1)\right\rangle \right)\quad {\mbox{for fermions}}\end{aligned}}$$

The desired probability is given by $_{S}\left\langle g_{n},g_{n'}\mid u\right\rangle _{S}$$_S\left\langle g_n,g_{n'}\mid u\right\rangle _S$ for bosons and $_{A}\left\langle g_{n},g_{n'}\mid u\right\rangle _{A}$$_A\left\langle g_n,g_{n'}\mid u\right\rangle_A$ for fermions, that is:

$${\begin{aligned}\underbrace {\left\langle g_{n}\mid u_{a}\right\rangle \left\langle g_{n'}\mid u_{b}\right\rangle } _{\mathbf {J} }+\underbrace {\left\langle g_{n'}\mid u_{a}\right\rangle \left\langle g_{n}\mid u_{b}\right\rangle } _{\mathbf {K} }\quad {\mbox{for bosons}}\\\underbrace {\left\langle g_{n}\mid u_{a}\right\rangle \left\langle g_{n'}\mid u_{b}\right\rangle } _{\mathbf {J} }-\underbrace {\left\langle g_{n'}\mid u_{a}\right\rangle \left\langle g_{n}\mid u_{b}\right\rangle } _{\mathbf {K} }\quad {\mbox{for fermions}}\end{aligned}}$$

$${\begin{aligned}\underbrace {\left\langle g_{n}\mid u_{a}\right\rangle \left\langle g_{n'}\mid u_{b}\right\rangle } _{\mathbf {J} }+\underbrace {\left\langle g_{n'}\mid u_{a}\right\rangle \left\langle g_{n}\mid u_{b}\right\rangle } _{\mathbf {K} }\quad {\mbox{for bosons}}\\\underbrace {\left\langle g_{n}\mid u_{a}\right\rangle \left\langle g_{n'}\mid u_{b}\right\rangle } _{\mathbf {J} }-\underbrace {\left\langle g_{n'}\mid u_{a}\right\rangle \left\langle g_{n}\mid u_{b}\right\rangle } _{\mathbf {K} }\quad {\mbox{for fermions}}\end{aligned}}$$

The term $\mathbf {J}$$\mathbf{J}$ is called direct term, while the term $\mathbf {K}$$\mathbf{K}$ is called exchange term:

$$P=J\pm K$$

$$P= J \pm K$$

This shows the appearance of a sort of interference phenomenon when we're dealing with identical particles.