I'm in the process of getting to grips with RF design and am having some trouble understanding why we should always match antennas to \$50\Omega\$.
I understand from the maximum power transfer theorem that if you have a fixed source (e.g. \$50\Omega\$) that you cannot change as a designer/user, then the maximum power transfer occurs if you set your load impedance to be the same as your source impedance. Therefore, if you are using an antenna as a transmitter, then you'd want to make the antenna (load) impedance equal to the impedance of the source for maximum power transfer into the antenna - so you match your antenna to \$50\Omega\$.
However, what if you are only using your antenna to receive? In that case the fixed variable is the load impedance (i.e. your receiver) which will be fixed at \$50\Omega\$ as below (I've also drawn the transmission line for completeness):
simulate this circuit – Schematic created using CircuitLab
If \$Z_L\$ is fixed (\$50\Omega\$), then the maximum received power will be when the voltage across its terminals is maximized. The voltage across \$Z_L\$ is (potential divider): $$ V_{Z_L} = V_s \frac{Z_L}{Z_L+Z_S} $$ \$Z_L\$ is fixed but we can "control" \$Z_S\$. Maximum voltage across \$Z_L\$ (and hence maximum power to the load) occurs when \$Z_S=0\Omega\$
Therefore, following this logic, you'd ideally want your receiving antenna to have a very low impedance (as close to \$0\Omega\$ as possible).
I'm probably missing quite a few fundamental concepts here but at the moment, the only reason I can think to match a receiving antenna to \$50\Omega\$ is in order to avoid re-reflections back to the load. However, if the load itself is already matched well to its \$50\Omega\$ transmission line, there should be minimal reflections already.