# 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 ${\displaystyle \left|u_{a}\right\rangle ,\left|u_{b}\right\rangle }$,which include the spin-dependence and are orthogonal. The physical state of the system is:

{\displaystyle {\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 ${\displaystyle G}$
${\displaystyle G\left|g_{i}\right\rangle =g_{i}\left|g_{i}\right\rangle \quad \left[G,P_{21}\right]=0}$
What is the probability of finding ${\displaystyle g_{n}}$ for one particle and ${\displaystyle g_{n'}}$ for the other? The final state is:
{\displaystyle {\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 ${\displaystyle _{S}\left\langle g_{n},g_{n'}\mid u\right\rangle _{S}}$ for bosons and ${\displaystyle _{A}\left\langle g_{n},g_{n'}\mid u\right\rangle _{A}}$ for fermions, that is:
{\displaystyle {\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 ${\displaystyle \mathbf {J} }$ is called direct term, while the term ${\displaystyle \mathbf {K} }$ is called exchange term:
${\displaystyle P=J\pm K}$
This shows the appearance of a sort of interference phenomenon when we're dealing with identical particles.