How can quarks, which are fermions, combine to form bosons? Why must an even number combine to form a boson? Give one example by stating the quark substructure of a boson.

The interplay of physical forces, such as electromagnetism and maybe even gravity, is controlled by bosons, which are sometimes referred to as force particles.

Bosons have an integer spin, whereas fermions have a half integer spin. When two particles with spin \({{\rm{S}}_{\rm{1}}}{\rm{/}}{{\rm{S}}_{\rm{2}}}\)and spin projection on one axis \({\rm{S}}_{\rm{1}}^{\rm{z}}{\rm{/S}}_{\rm{2}}^{\rm{z}}\)collide, the final particle's spin is not uniquely determined, i.e., it can take on different values. Simply put, the spin orientation of the component particles determines this. The projection on one axis is determined in a unique way.

\({{\rm{S}}^{\rm{z}}}{\rm{ = S}}_{\rm{1}}^{\rm{z}}{\rm{ + S}}_{\rm{2}}^{\rm{z}}\)

while boundaries are imposed on the total spin

\(\left| {{S^z}} \right| \le S \le \left| {{S_1} + {S_2}} \right|\)

For reasons outside the scope of this response, spin values are quantized so that only \({\rm{1/2}}\) multiples are permitted, and one particle may only differ by multiples of 1, i.e., if \({\rm{S = 1}}\), possible projection values are \({{\rm{S}}_{\rm{z}}}{\rm{ = - 1,0,1}}\) (e.g. \({{\rm{S}}^{\rm{z}}}{\rm{ = 1/2}}\)is not allowed despite satisfying (2)). Quarks, which are fermions, can be combined so that the resulting particle's total spin is an integer. Because fermions have half integer values, their spin projections have half integer values as well, hence \({\rm{S}}_{\rm{1}}^{\rm{z}}{\rm{ = }}\frac{{\rm{m}}}{{\rm{2}}}\)and \({\rm{S}}_{\rm{2}}^{\rm{z}}{\rm{ = }}\frac{{\rm{n}}}{{\rm{2}}}\) may be written, where, \({\rm{m}}\)and\(n\)are odd integers. The spin projection of the final particle would therefore be derived from (2).

\(\begin{array}{c}{S^z} = \frac{m}{2} + \frac{n}{2}\\ = \frac{{m + n}}{2}\end{array}\)

However, because the combination of two odd integers is an even number, the final particle has an integer spin and is thus a boson. The right-hand side of \(\left( {\rm{3}} \right)\)would look like \(\frac{{m + n + p}}{2}\)if there were an odd number of fermion components, where \({\rm{m,n,p}}\)are all odd numbers. Their sum would therefore be an odd number, resulting in a half-integer projection, which would categories the final particle as fermionic. Consider a particle made composed of an up quark (spin \({\rm{1/2}}\)and a down antiquark (spin \({\rm{1/2}}\)). Because they don't have a fixed projection value, any value for the final spin is acceptable, therefore we can have \({\rm{S = 1}}\) (which we get because the spins are aligned, \({\rm{1/2 + 1/2 = 1}}\)) or \({\rm{S = 0}}\) (which we get because the spins are opposite, \({\rm{1/2 - 1/2 = 0}}\)) which corresponds to a pion. Both bosons are particles.