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?
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Pretty much, yes. I believe the term "classical" in such contexts is an umbrella term for "not quantised." – Song of Physics Dec 21 '22 at 03:23
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@SongofPhysics I see, is that why they call it second quantization? The first quantizes the operators and the second the resulting field? – CBBAM Dec 21 '22 at 03:27
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1You might find this answer useful: https://physics.stackexchange.com/a/734239/27732 – Andrew Dec 21 '22 at 03:33
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1Does this answer your question? Is the Klein-Gordon Equation an equation for a Classical field or a Quantum Field? – Roger V. Dec 21 '22 at 09:22
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Related: https://physics.stackexchange.com/a/738835/247642 – Roger V. Dec 21 '22 at 09:22
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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.
Níckolas Alves
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Awesome thank you! Could you elaborate on how the Schrödinger equation is a classical field? That sounds interesting and I've never seen that interpretation before. – CBBAM Dec 21 '22 at 03:33
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1@CBBAM From the point of view of field theory, you can perfectly well find a Langrangian that will lead to the Schrödinger equation upon imposing the Euler–Lagrange equations. This is an unusual way to think of the Schrödinger equation when you are doing QM, but it is interesting when you are considering the non-relativistic limit of a QFT, for example. I think the Schrödinger field is one of the examples given by Nivaldo Lemos in the field theory chapter of his book Analytical Mechanics – Níckolas Alves Dec 21 '22 at 03:37
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In summary: you have a function of spacetime ruled by some equation of motion (the Schrödinger equation). Regardless of the physical meaning of this function, it is a classical field. Hence, the Klein–Gordon equation, the Schrödinger equation, the Dirac equation, the Maxwell equations, the Navier–Stokes equations all involve classical fields – Níckolas Alves Dec 21 '22 at 03:38
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So whenever the field is real, complex, or vector valued it is classical, and whenever it is operator valued it is quantum? – CBBAM Dec 21 '22 at 03:47
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