# What are bosons?

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 ${m}_{s} = \pm 1$, but fermions have ${m}_{s} = \pm \frac{1}{2}$.
• 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:

$\Xi = {\prod}_{i = 1}^{N} \frac{1}{1 - {e}^{- \left({\epsilon}_{i} - \mu\right) / {k}_{B} T}}$

where ${\epsilon}_{i}$ is the energy of level $i$ and $\mu$ is the chemical potential for bosons.

It can be shown that, since

• $P \overline{V} = {k}_{B} T \ln \Xi$
• $d \left(P \overline{V}\right) = \overline{S} \mathrm{dT} + P \mathrm{db} a r V + \overline{N} \mathrm{dm} u$
• ${\sum}_{i = 1}^{N} {\overline{n}}_{i} = \overline{N}$

the average occupation number ${\overline{n}}_{i}$ is given by:

${\overline{n}}_{i} = {e}^{- \left({\epsilon}_{i} - \mu\right) / {k}_{B} T} / \left(1 - {e}^{- \left({\epsilon}_{i} - \mu\right) / {k}_{B} T}\right)$

When ${\overline{n}}_{i}$ is graphed against the energy above $\mu$ (which I called the "Bose Level", analogous to the Fermi Level) at $\text{300 K}$, this looks like:

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

When ${\overline{n}}_{i}$ is graphed against the temperature instead, it looks like:

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

If ${\epsilon}_{i} - \mu$ was reduced to $\text{0 eV}$, i.e. all the bosons are in the ground state, we would find ${\overline{n}}_{i} \to \infty$.

SYMMETRY UPON PARTICLE INTERCHANGE

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

Then we say that

$\Psi \left(1 , 2\right) = \Psi \left(2 , 1\right)$,

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

$\Psi \left(1 , 2\right) = - \Psi \left(2 , 1\right)$,

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.

$\Psi \left(1 , 2\right) = \frac{1}{\sqrt{2}} | \left(\textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)} , \textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)}\right) , \left(\textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)} , \textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)}\right) |$

$= \frac{1}{\sqrt{2}} \left[\textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)} \textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)} - \textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)} \textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)}\right]$

(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:

$\Psi \left(1 , 2\right) = - \Psi \left(2 , 1\right) = - \frac{1}{\sqrt{2}} | \left(\textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)} , \textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)}\right) , \left(\textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)} , \textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)}\right) |$

$= \underline{- \frac{1}{\sqrt{2}} \left[\textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)} \textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)} - \textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)} \textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)}\right]}$

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

$\Psi \left(1 , 2\right) = \Psi \left(2 , 1\right) = \frac{1}{\sqrt{2}} | \left(\textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)} , \textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)}\right) , \left(\textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)} , \textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)}\right) |$

$= \underline{\frac{1}{\sqrt{2}} \left[\textcolor{b l u e}{{\phi}_{1 s} \left(1\right) \beta \left(1\right)} \textcolor{g r e e n}{{\phi}_{1 s} \left(2\right) \alpha \left(2\right)} + \textcolor{g r e e n}{{\phi}_{1 s} \left(1\right) \alpha \left(1\right)} \textcolor{b l u e}{{\phi}_{1 s} \left(2\right) \beta \left(2\right)}\right]}$

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