# Proving that an alpha particle is a boson

quantum_smile
1. Is there any way to prove that the alpha particle is a boson (its total wave function is symmetric), given that it's made up of two protons (fermions) and two neutrons (fermions)?

## Homework Equations

The total wave function for two identical particles that are
(bosons) ψ_tot = 1/√2 * (ψ_a (particle 1) ψ_b (particle 2) + ψ_b (particle 1) ψ_a ( particle 2))
and
(fermions) ψ_tot = 1/√2 * (ψ_a (particle 1) ψ_b (particle 2) - ψ_b (particle 1) ψ_a ( particle 2)),
where "particle 1" and "particle 2" designate the coordinates of each particle, and
a,b designate states of each of the particles.

## The Attempt at a Solution

For a single alpha particle,
ψ = P*N, where P is the wave function for the two protons and N is the wave function for two fermions.
P=1/√2 * (P_a(Proton 1)P_b(Proton 2) - P_b (Proton 1) P_a (Proton 2))
N= 1/√2 * (N_c(Neutron 1)N_d(Neutron 2) - N_d (Neutron 1) N_c(Neutron 2)),
where a,b describe the states for each of the two protons
and c,d does the same for each of the two neutrons.

For a pair of alpha particles,
ψ_tot = 1/√2 * (ψ_{abcd} (Alpha particle 1) * ψ_{efgh} (Alpha particle 2) \pm ψ_{efgh} (Alpha particle 1) * ψ_{abcd} (Alpha particle 2)},
and our goal is to know whether we should use the plus sign (if the alpha particle is a boson) or the minus sign (if the alpha particle is a fermion).

At this point I'm stuck. How can we find out which sign to use? I appreciate any help!

Last edited:

$$ψ=2^{-0.5}*(ψ_{a}(1)ψ_{b}(2)-ψ_{b}(1)ψ_{a}(2)).$$.