How does the Klein-Gordon equation represent a classical field?

Question

This is what I know about the Klein-Gordon equation so far. Suppose we are working with natural units such that $c = 1$. Then we may obtain the Klein-Gordon equation by considering any 4-vector $p^\mu = (E, \textbf{p})$. To enforce Lorentz invariance we require that $(p^\mu)^2 = m^2$. The quantized version of the momentum is given by the operator $P = -i\nabla$ and energy of the system is given by the Hamiltonian, which by the Schrodinger equation we have that the quantized energy of the system is governed by $i\partial_t$. Making these substations in $p^\mu$ we get (using the +--- signature) : $$ p^\mu = (i\partial_t, -i\nabla) \\ \implies (p^\mu)^2 = -\partial_t^2 +\nabla^2 = m^2.$$ Thus we obtain the Klein-Gordon equation: $$(\partial_t^2 - \nabla^2 + m^2)\phi = 0$$ whose interpretation is that it governs the "wave mechanics" of the underlying field $\phi$ and has no classical analogue. This brings me to my question: considering the equation was derived with quantum considerations, how does the solution to the Klein-Gordon equation represent a classical field as opposed to a quantum one? Is it because the solution $\phi(t,x)$ is a real-valued function as opposed to an operator?

Answer 1

Is it because the solution ϕ(t,x) is a real-valued function as opposed to an operator?

Exactly. While the theory is quantum, the field is not. Similarly, solutions to the Schrödinger equation are classical fields, even though they describe quantum particles. To consider a quantum field, one would go through the process of second quantization, essentially promoting the field itself to an operator.

First quantization would be to promote position, momentum, etc to operators.