Lagrangian density for a single-spin 0-real-bosonic field ($\phi$) is given by, $$\mathcal{L}=-\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{m^2}{2}\phi^2$$
Now if we formulate the Euler Lagrange equation for the above Lagrangian density, we will get, $$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial \phi_{,\mu}}\right)=\frac{\partial \mathcal{L}}{\partial \phi}$$ $$\Rightarrow -\frac{1}{2}\left[\partial_\mu\bigg(\frac{\partial}{\partial\phi_{,\mu}}\bigg)\right]\underbrace{(\partial_\nu\phi g^{\nu\alpha}\partial_\alpha\phi)}_{\text{New dummy indices, $\nu$ and $\alpha$}}=-m^2\phi$$ $$\Rightarrow -\frac{1}{2}\partial_\mu g^{\nu \alpha}(\delta^{\mu}_\nu\partial_\alpha\phi+\delta^\mu_\alpha \partial_\nu\phi)=-m^2\phi$$ $$\Rightarrow -\frac{1}{2}\partial_\mu(\partial^\mu\phi+\partial^\mu\phi)=-m^2\phi$$ $$\Rightarrow \partial_\mu\partial^\mu\phi-m^2\phi=0$$
So, in the end, we get the Klein Gordon Equation, which is a relativistic version of the Schrodinger Equation. Now my question is Euler Lagrange Equation is a classical dynamical equation, which gives a particular type of path in which the corresponding action is an extremum. But according to Feynman Path Integral formalism, in any quantum theory, all possible types of path has some finite probability (Unlike the classical case, where only a particular path, for which action will be extremum, will have a sharp probability, and the probability of other paths tend to 0).$$\\$$ So, my question is,
- how the classical Euler Lagrange equation can formulate a quantum dynamical equation?
- How Euler Lagrange Equation is valid in Quantum Theory
$$$$
N.B.- Same thing I also have noticed in Dirac Field also.
