# Hamiltonian for identical particles

When identical particles are concerned it is intuitive that no physical property can be modified when the roles of the particles are permuted:any physical observable must be invariant under all permutations, that is: if $G$ is a physical observable then $\left[G,P_{\alpha }\right]=0\quad \forall \alpha$ . This also means that the spaces $\xi _{S}$ and $\xi _{A}$ are invariant under the action of a physical observable $G$ : if $\left|\psi \right\rangle \in \xi _{S}\left(\xi _{A}\right)\Rightarrow G\left|\psi \right\rangle \in \xi _{S}\left(\xi _{A}\right)$ . The hamiltonian of a system of identical particles is a physical observable (it corresponds to the energy) and must commute with all $P_{\alpha }$ . This restricts the possible forms of the interacting potential.

Let us investigate deeply this fact. Consider a system of 2 non-interacting identical particles. The hamiltonian of this system is:

$H=h(1)+h(2)$ with $h(1)=h(2)$ because of the indistinguishability of the particles. Now suppose that there is an interaction between the particles. In order to preserve the invariance of $H$ under permutations of the particles ($\left[H,P_{21}\right]=0$ ) , that is in order to keep the hamiltonian a physical observable, the interaction potential has to be of the form:
$V\left(\mathbf {r} _{1},\mathbf {r} _{2}\right)=V(\left|\mathbf {r_{1}} -\mathbf {r_{2}} \right|)$ .

## Energy

For the sake of simplicity we are going to consider a system of N non-interacting identical particles. Then

{\begin{aligned}H=h(1)+h(2)+\dots +h(N)\quad h(i)=h(j)\forall i,j=1,\dots ,N\\h(1)\left|u_{n}\right\rangle =\epsilon _{n}\left|u_{n}\right\rangle \quad \left|u_{n}\right\rangle n^{th}{\mbox{ individual state}}\end{aligned}} Suppose we are dealing with $n_{1}$ particles in the state $\left|u_{1}\right\rangle$ , $n_{2}$ particles in the state $\left|u_{2}\right\rangle$ and so on.The total energy of each state is:
$E_{1}=n_{1}\epsilon _{1}\quad E_{2}=n_{2}\epsilon _{2}\quad E_{n}=n_{n}\epsilon _{n}$ and the total energy of the system is:
$E=E_{1}+E_{2}+\dots +E_{n}$ For bosons there are no restrictions on the numbers $n_{k}$ . It follows that the ground state is the one with all the N bosons on the lowest energy level (we suppose it's $\epsilon _{1}$ ), with energy
$E_{\mbox{ground state}}=N\epsilon _{1}$ The situation for fermions is different, since we have learned that $n_{k}$ can only be 0 or 1. Thus if we have N fermions the ground state is the one with one fermion per quantum state. N quantum states will be "full" ($n_{k}=1$ for $k\leq N$ ), the others will be empty ($n_{k}=0$ for $k>N$ ).The situation is the following:
$E_{\mbox{ground state}}=\epsilon _{1}+\epsilon _{2}+\dots +\epsilon _{N}$ The highest individual energy $\epsilon _{N}$ , corresponding to the last full quantum state in the ground state is called Fermi energy

Example (Two identical non-interacting particles in an infinite potential well)

First consider the particles to be spinless. Since $s=0$ we're dealing with bosons. It's known from quantum mechanics that the wave function of a particle in such a system is given by

$u_{n}(x)={\sqrt {\frac {2}{L}}}\sin({\frac {n\pi x}{L}})$ where $L$ defines the width of the well. Energy levels are given by
$E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}$ These expressions define the individual states and the individual energies. Suppose we have a particle on the state $u_{a}$ and a particle on the state $u_{b}$ . Taking into account the nature of the particles we can immediately write down the wave function of the system:
$\psi _{S}(x_{1},x_{2})={\frac {1}{\sqrt {2}}}\left(u_{a}(x_{1})u_{b}(x_{2})+u_{a}(x_{2})u_{b}(x_{1})\right)$ Now suppose that the particles have spin ${\frac {1}{2}}$ .

• Individual states must take into account the spin degrees of freedom
• we're now dealing with fermions

We know that a particle is on the level $u_{a}$ with its spin up $\left(m_{s}={\frac {\hbar }{2}}\right)$ and the other on the level $u_{b}$ with its spin down $\left(m_{s}={\frac {-\hbar }{2}}\right)$ . Individual states are then given by:

$u_{a}(i)\chi ^{+}(i)\quad u_{b}(i)\chi ^{-}(i)\quad i=1,2$ Remembering that a permutation operator acts not only on the orbital variables but also on spin ones, antisymmetrization gives us the following wave function:
$\psi _{A}(x_{1},x_{2})={\frac {1}{\sqrt {2}}}\left(u_{a}(x_{1})\chi ^{+}(1)u_{b}(x_{2})\chi ^{-}(2)-u_{a}(x_{2})\chi ^{+}(2)u_{b}(x_{1})\chi ^{-}(1)\right)$ 