What are bosons?

1 Answer
Aug 14, 2017

Bosons (such as photons) are quantum mechanical particles with integer spins, that are symmetric upon particle interchange. More than one can exist in a given quantum state, unlike fermions (such as electrons), which have half-integer spins.

In other words...

  • Bosons do not follow the Pauli Exclusion Principle, but fermions do.
  • Bosons have m_s = pm1, but fermions have m_s = pm1/2.
  • Bosons that are interchanged return the same state, so it is easier to double-count the microstates in a bosonic system by accident (there are generally fewer nonredundant configurations possible).

DISOBEYING THE PAULI EXCLUSION PRINCIPLE

The grand canonical partition function for an ensemble of bosons where each state is singly-degenerate is given by:

Xi = prod_(i=1)^(N) 1/(1-e^(-(epsilon_i - mu)//k_BT))

where epsilon_i is the energy of level i and mu is the chemical potential for bosons.

It can be shown that, since

  • PbarV = k_BTln Xi
  • d(PbarV) = barSdT + PdbarV + barNdmu
  • sum_(i=1)^(N) barn_i = barN

the average occupation number barn_i is given by:

barn_i = e^(-(epsilon_i - mu)//k_BT)/(1 - e^(-(epsilon_i - mu)//k_BT))

When barn_i is graphed against the energy above mu (which I called the "Bose Level", analogous to the Fermi Level) at "300 K", this looks like:

This means that you CAN have an infinite number of particles in the ground state of a boson system.

When barn_i is graphed against the temperature instead, it looks like:

This means that at some point (around "1500 K"), the average occupation of the energy level at "0.2 eV" above the ground state increases linearly with temperature, meaning that the bosons no longer see each other.

If epsilon_i - mu was reduced to "0 eV", i.e. all the bosons are in the ground state, we would find barn_i -> oo.

SYMMETRY UPON PARTICLE INTERCHANGE

Define the wave function for two particles (with a specific ordering) as Psi(1,2).

Then we say that

Psi(1,2) = Psi(2,1),

if both particles are bosons (such as photons). That is, the sign of the wave function stays the same upon bosonic particle interchange.

But we say

Psi(1,2) = -Psi(2,1),

If both particles are fermions (such as electrons). That is, the sign of the wave function switches upon fermionic particle interchange.

Or, consider a simple Slater determinant for fermions (such as electrons), which treats the wave function as having a spatial part and a spin part.

Psi(1,2) = 1/sqrt2 |(color(green)(phi_(1s)(1)alpha(1)), color(blue)(phi_(1s)(1)beta(1))),(color(green)(phi_(1s)(2)alpha(2)), color(blue)(phi_(1s)(2)beta(2)))|

= 1/sqrt2 [color(green)(phi_(1s)(1)alpha(1))color(blue)(phi_(1s)(2)beta(2)) - color(blue)(phi_(1s)(1)beta(1))color(green)(phi_(1s)(2)alpha(2))]

(We represent spatial orbitals as phi and spin orbitals as alpha (spin-up) and beta (spin-down). The numbers in parentheses indicate the particle number.)

Another way to depict particle interchange is to switch these two columns (one could choose instead to switch the two rows, but let's not get too complicated). For fermions, we would have:

Psi(1,2) = -Psi(2,1) = -1/sqrt2 |(color(blue)(phi_(1s)(1)beta(1)), color(green)(phi_(1s)(1)alpha(1))),(color(blue)(phi_(1s)(2)beta(2)), color(green)(phi_(1s)(2)alpha(2)))|

= ul(-1/sqrt2 [color(blue)(phi_(1s)(1)beta(1))color(green)(phi_(1s)(2)alpha(2)) - color(green)(phi_(1s)(1)alpha(1))color(blue)(phi_(1s)(2)beta(2))])

On the other hand, for bosons, the analogous determinant uses a + sign instead when evaluating:

Psi(1,2) = Psi(2,1) = 1/sqrt2 |(color(blue)(phi_(1s)(1)beta(1)), color(green)(phi_(1s)(1)alpha(1))),(color(blue)(phi_(1s)(2)beta(2)), color(green)(phi_(1s)(2)alpha(2)))|

= ul(1/sqrt2 [color(blue)(phi_(1s)(1)beta(1))color(green)(phi_(1s)(2)alpha(2)) + color(green)(phi_(1s)(1)alpha(1))color(blue)(phi_(1s)(2)beta(2))])

In the case of bosons, particle interchange is symmetric, and that is seen in that addition is commutative.