We can take the Klein-Gordon equation describing the evolution of a complex scalar field. Taking the non-relativistic limit yields a classical wave equation that is identical in form to the Schrodinger equation.

We can also quantise the Klein-Gordon equation into a quantum field, and then, in the non-relativistic single particle special case, we end up with the actual free particle non-relativistic Schrodinger equation, complete with a probabilistic interpretation

What's the big picture? Why does doing either of these things produce equations of the same structure? And specifically what about the second approach results in a probabilistic interpretation?

  • $\begingroup$ see physics.stackexchange.com/a/238761/50583 (ignore the last paragraph since it's specific to the RS model in that question, but otherwise I think that's the answer you're looking for) $\endgroup$
    – ACuriousMind
    Mar 25, 2022 at 15:48
  • $\begingroup$ @ACuriousMind I don't understand how it's justified to interpret $\phi (x)|0\rangle$ as a position eigenstate. $\phi$ was just a field operator $\endgroup$
    – Egg Man
    Mar 25, 2022 at 17:10
  • 1
    $\begingroup$ The $a^\dagger_p$ create momentum eigenstates, so $\tilde{\phi}(p)$ acting on the vacuum does, too (the annihilation part of its mode expansion does nothing when acting on the vacuum). Therefore $\phi(x)$ acting on the vacuum creates something that's like a plane wave superposition of momentum eigenstates, i.e. a "position eigenstate" (remember how in ordinary QM a momentum eigenstate is a plane wave in position space and vice versa? this is just like that. Relativistically there isn't a good position operator but in the non-relativistic limit it works out fine) $\endgroup$
    – ACuriousMind
    Mar 25, 2022 at 18:07
  • $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$
    – Qmechanic
    Mar 25, 2022 at 18:54


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