# Difference between Bose-Einstein and Fermi-Dirac statistics

Differentiate between particles that obey Bose-Einstein and Fermi-Dirac statistics, giving one example each. Give examples of particles that obey each of these statistics.

Â© BrainMass Inc. brainmass.com October 2, 2022, 12:37 pm ad1c9bdddfhttps://brainmass.com/physics/energy/difference-between-bose-einstein-fermi-dirac-statistics-185519

## SOLUTION This solution is **FREE** courtesy of BrainMass!

I'll explain the fundamental physics a bit first. In the last few sentences I explain the difference between the two statistics.

According to quantum field theory, particles are considered to be excited states of fields. The Hamiltonian of a field theory is similar to that of coupled harmonic oscillators. One has an harmonic oscillator for each possible momentum. The creation and annihilation operator which map an energy eigenstate to a higher or lower energy eigenstate can be interpreted as adding or removing a particle with some momentum.

It turns out that for integer spin fields, one has to demand that the creation/annihilation operators commute while for half integer spin the operators have to anti-commute (i.e. satisfy a relation like A B + B A = 0). This rule ensures that local observables will commute when they are space like separated. I.e., if you have an observable for some physical quantity at some point x at time t, O1(x,t) and another at some other point x', t', O2(x', t'), then if |x - x'| > c |t-t'|, there is no way that a measurement of O1 could influence the outcome of a measurement of O2. So, it must always be possible to have simultaneous eigenstates of O1 and O2, and thus they have to commute. The structure of quantum theory then still allows for so-called quantum nonlocality (you may have read about entangled states, Bell/Aspect type experiments), but here no information can be transferred in a nonlocal way. Quantum theory is local, but because it is different from classical physics, you would need a nonlocal classical theory to get some of the strange results of the local quantum theory.

Anyway, the bottom line is that for integer spin fields two different creation operators will commute, while for half integer spin they must anti-commute (note that the creation/annihilation operators themselves are not Hermitian and therefore they are not observables themselves). If we create a two particle state by applying two creation operators to the vacuum, then for integer spins, the order in which we do that does not matter. But for half integer spin, changing the order introduces a minus sign. This means that a two particle wavefunction for integer spin must be symmetric w.r.t. interchanging the two particles, but for half integer spin it must be anti-symmetric. We call particles that have symmetric wavefunctions Bosons and particles that have anti-symmetric wavefunctions, Fermions. As explained above, Quantum field theory implies that integer spin particles are Bosons and half integer spin particles are Fermions.

Two Fermions cannot be in the same state. This is because if we interchange the two particles, then that state, like any other two particle Fermion state, would have to be multiplied by a minus sign. But if the two particles are indeed in the same state, then interchanging them would leave the state invariant. So, many particle fermion states will have either zero or one particle in the single particle states. In statistical mechanics we call the restriction to at most one particle in each single particle state Fermi-Dirac statistics. In case of Bosons, there is no problem to get a state in which many particles are in the same state. We call allowing an unrestricted number of particles in the single particle states, Bose-Einstein statistics.

An example of a Boson is the photon. An example of a Fermion is the electron.

Â© BrainMass Inc. brainmass.com October 2, 2022, 12:37 pm ad1c9bdddf>https://brainmass.com/physics/energy/difference-between-bose-einstein-fermi-dirac-statistics-185519