# Identical particles

Two particles are said to be identical if all their intrinsic properties, such as mass, spin, charge, are exactly the same. We can rephrase the definition saying that two particles are identical if no experiment can distinguish between them. An important consequence can be deduced from this definition: a system composed of identical particles must be invariant under the exchange of the particles roles.

Identical particles are of great importance in physics: all the electrons, protons, neutrons, etc in the universe are identical. Since we are dealing with the problem of an exhaustive quantum description of atoms and molecules, where lots of identical particles are concerned, we have to develop a proper quantum mechanical theory about identical particles.

In classical mechanics identical particles are treated just as they were different.In principle, indeed, it is always possible to distinguish between them following their orbits, and for this reason we can label the particles with ${\displaystyle \left(1\right),\left(2\right)\dots }$, labels which act as intrinsic properties that distinguish the particles and make them "no longer identical". The problem of indistinguishability is purely quantum mechanical, since we have learned that particles no longer have definite trajectories. The labels ${\displaystyle \left(1\right),\left(2\right)\dots }$ can't be seen as intrinsic properties. Consider for example a system of two identical particles:when we detect one particle in a region of space in which both of them have a non-zero position probability, we have no way of knowing if the particle detected is the number ${\displaystyle \left(1\right)}$or the number ${\displaystyle \left(2\right)}$.

Indistinguishability creates fundamental difficulties, since it introduces an exchange degeneracy: when identical particles are concerned a complete measurement on each of the particle does not permit the determination of a unique ket of the state space of the system.

Example (Example 1)

Consider a system of two identical particles with spin degrees of freedom only. Each particle spin observable ${\displaystyle S_{i,z}\quad i=1,2}$ constitutes a C.S.C.O in the state space of the single particle, and its eigenkets form a base of that space. We can construct an orthonormal base of the state space of the system taking the tensor product between the bases of the two state spaces.

{\displaystyle {\begin{aligned}{\begin{cases}{\mbox{particle (1) state space spanned by}}\quad \mid +\rangle _{1},\mid -\rangle _{1}\\{\mbox{particle (2) state space spanned by}}\quad \mid +\rangle _{2},\mid -\rangle _{2}\\{\mbox{system state space spanned by}}\quad \mid +\rangle _{1}\mid +\rangle _{2},\mid +\rangle _{1}\mid -\rangle _{2},\mid -\rangle _{1}\mid +\rangle _{2},\mid -\rangle _{1}\mid -\rangle _{2}\\\end{cases}}\end{aligned}}}
Suppose we make a measurement of ${\displaystyle S_{1,z}}$ and ${\displaystyle S_{2,z}}$ and find ${\displaystyle {\frac {+\hbar }{2}},{\frac {-\hbar }{2}}}$. Since the particles are identical we can't know which particle has Oz spin component ${\displaystyle {\frac {+\hbar }{2}}}$ and which has ${\displaystyle {\frac {-\hbar }{2}}}$. The state space of the system is then two dimensional, spanned by
${\displaystyle \mid +\rangle _{1}\mid -\rangle _{2}=\mid +,-\rangle \quad ,\quad \mid -\rangle _{1}\mid +\rangle _{2}=\mid -,+\rangle }$
Each ket of the form ${\displaystyle \alpha \mid +,-\rangle +\beta \mid -,+\rangle ,\quad \mid \alpha \mid ^{2}+\mid \beta \mid ^{2}=1}$ represents the same physical state and in general physical predictions depends on the coefficients ${\displaystyle \alpha }$ and ${\displaystyle \beta }$. This clearly shows the presence of exchange degeneracy (in this example in the initial state), which must be removed in order to make physical predictions.

Example (Example 2)

Consider a system of three identical particles ${\displaystyle \left(1\right),\left(2\right),\left(3\right)}$ and an observable ${\displaystyle B\left(1\right)}$ which form a C.S.C.O in the state space of particle ${\displaystyle \left(1\right)}$. Since the particles are identical the observables ${\displaystyle B\left(2\right),B\left(3\right)}$ exist and form a C.S.C.O in the state space of particle ${\displaystyle \left(2\right),\left(3\right)}$. We can construct an orthonormal base of the state space of the system by taking the tensor product between each base of single-particle state spaces.Since the particles are identical we are not able to distinguish between ${\displaystyle B\left(1\right),B\left(2\right),B\left(3\right)}$ and we can only measure the observable ${\displaystyle B}$ for each of the particles. Suppose we find the values ${\displaystyle a,b,c}$. The final state of the system can be represented by any ket spanned by the six kets

{\displaystyle {\begin{aligned}\mid a(1),b(2),c(3)\rangle \quad \mid b(1),a(2),c(3)\rangle \quad \mid b(1),c(2),a(3)\rangle \\\mid c(1),a(2),b(3)\rangle \quad \mid c(1),b(2),a(3)\rangle \quad \mid a(1),c(2),b(3)\rangle \end{aligned}}}

Examples clearly show the deep connection between exchange degeneracy and permutations. In the next chapter we're going to develop this connection by the introduction of permutation operators