# 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 $\left|u_{a}\right\rangle ,\left|u_{b}\right\rangle$ ,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}} Suppose we want to measure on each particle the physical quantity $G$ $G\left|g_{i}\right\rangle =g_{i}\left|g_{i}\right\rangle \quad \left[G,P_{21}\right]=0$ What is the probability of finding $g_{n}$ for one particle and $g_{n'}$ 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}} The desired probability is given by $_{S}\left\langle g_{n},g_{n'}\mid u\right\rangle _{S}$ for bosons and $_{A}\left\langle g_{n},g_{n'}\mid u\right\rangle _{A}$ 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}} The term $\mathbf {J}$ is called direct term, while the term $\mathbf {K}$ is called exchange term:
$P=J\pm K$ This shows the appearance of a sort of interference phenomenon when we're dealing with identical particles.