Books on QFT treats, any quantum field as quantized classical fields. For example, the Klein-Gordon field is first treated as a classical field $\phi(x)$ obeying classical Euler-Lagrange equation $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Big)=0\Rightarrow(\Box+m^2)\phi(x)=0.\tag{1}$$ However, if we restore the units $\hbar$ and $c$, it becomes $$(\Box+\frac{m^2c^2}{\hbar^2})\phi(x)=0.$$

  1. How can the classical equation of motion (1) depend upon $\hbar$?

  2. For electromagnetic field, this problem is not there. The Maxwell's equations do not contain $\hbar$. Is there a deep reason (apart from any accident) why Maxwell's equation do not contain $\hbar$ but the classical KG equation, classical Dirac equation does?


In classical mechanics there is no reason to identify $m$ with a mass. It is just an inverse of a length, in which case there are no factors of $\hbar$ nor $c$ in the KG equation.

  • $\begingroup$ Dear @AFT If you took the Fourier mode expansion of the field $\phi(x)$, each classical mode satisfies relativistic dispersion relation. So I doubt your answer. $\endgroup$ – SRS Jan 24 '17 at 15:26
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    $\begingroup$ @SRS so what? you may have a Fourier variable satisfy $k^2=m^2$ and still there is no reason to identify $m$ with a mass. In classical mechanics $k$ is not the momentum of anything. There are no particles in CM. $\endgroup$ – AccidentalFourierTransform Jan 24 '17 at 15:27
  • $\begingroup$ $m$ in OP's (1) looks like the inverse of a length. $\endgroup$ – yuggib Jan 24 '17 at 15:52
  • $\begingroup$ @AccidentalFourierTransform My idea was that the KG equation was historically obtained in RQM (which contains $\hbar$ and $c$). In field theory, this KG equation is thought of as the Euler-Lagrange equation derived from a Lagrangian. And therefore, this Lagrangian must also have factors of $\hbar$ and $c$. Isn't it? If the classical Lagrangian doesn't have factors of $\hbar$ and $c$, how does it suddenly appear after quantizing the theory? $\endgroup$ – SRS Jan 25 '17 at 10:42
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    $\begingroup$ @SRS $\uparrow$ Yes. $\endgroup$ – AccidentalFourierTransform Feb 7 '18 at 19:15

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