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Consider the theory of a complex scalar field $$S[\psi, \psi^\dagger] = -\int d^4x \left(\hbar \partial_\mu \psi^\dagger \partial^\mu\psi + \hbar^{-1} m^2 |\psi|^2\right)$$ giving the Klein-Gordon equation of motion $$ \hbar^2 \square \psi = m^2 \psi $$ The WKB 'semiclassical approximation' consists of subbing in $\psi = e^{iF /\hbar}$ and then taking $\hbar \rightarrow0$. Doing this above gives $$\eta^{\mu\nu} (\partial_\mu F )(\partial_\nu F) = -m^2 .$$ At this point one recognises this as the (relativistic) Hamilton-Jacobi equation for the action of a particle of mass $m$ evaluated along the classical path. One thus ends up identifying $F\equiv S_c$ as the classical action.

Q1) I want to check, what classical action are we actually referring to? Is it indeed not not the original action above $S[\psi,\psi^\dagger]|_{classical \;path}$, but rather the action of a point relativistic particle $S_c = -m\int d\tau |_{classical \;path}$? Or both, in a way?

Next, the general solution to the above differential equation is $$F = a \mp t\sqrt{\vec{b}^2 + m^2} + \vec{x} \vec{b}.$$ Just because two things, $F$ and $S_c$, satisfy the same equation doesn't mean they are the same thing. Pushing the identification nonetheless (?), it is known that the classical action of a point particle (not sure bout field theory) satisfies $$\partial_t S_c = -H \quad \text{and} \quad \partial_i S_c = P_i ,$$ where $H$ and $P_i$ are the Hamiltonian and conjugate momenta respectively, both constant for a free particle. So we should identify $b_i \equiv P_i$.

Notice the $\mp$ sign in the solution for $F$. The Hamiltonian however is actually always positive $H=E>0$, in fact $S_c =-\int E dt + \int p_i dx^i$, $p_i \in R$ in the theory of a free particle.

Q2) So I'm wondering whether the identification $F\equiv S_c$ is only for a particle, whereas for an antiparticle we should have $F\equiv -S_c$? So, technically, for antiparticle the WKB solution should be $\psi = e^{-iS_c/\hbar}$?

I realise the latter doesn't end up making a difference in this case, but in the most general case, it will! e.g. suppose I was interested in changing $S[\psi,\psi^\dagger]$ so that particle and antiparticle behaved differently.

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