This is what I understood from Klein Gordon equation :

We start from $$E^2=p^2+m^2.$$

We quantize it replacing $E \rightarrow \partial_t$, $p \rightarrow -ih\nabla$, $m \rightarrow m$

Thus, we get the Klein Gordon equation :

$$ (\Box +m^2) \Psi = 0 $$

But we can't interpret directly $\Psi$ as a wavefunction (it leads to some incoherences).

But we can interpret $\Psi$ as an operator acting on an hilbert space.

It is linked to what we call "second quantization" ? I mean, we first quantized the relation $E^2=p^2+m^2$, and we quantized the solution $\Psi$.

  • $\begingroup$ Hi user3183950, I took the liberty of editing out your second question, which would have made your post far too broad, and it would have been closed. $\endgroup$ Mar 29, 2017 at 14:46
  • $\begingroup$ Ok ! But if I ask it in a different post you think it is a too imprecise question ? Or it is just because I asked two differents questions in one post that you prefered to delete it ? $\endgroup$
    – StarBucK
    Mar 29, 2017 at 14:47
  • $\begingroup$ Two different, unrelated questions are always discouraged. You can ask the second question in a different post, but I fear that the question itself is too broad, so it would get closed too. You can try though. $\endgroup$ Mar 29, 2017 at 14:49

1 Answer 1


The way I yielded the Klein-Gordon equation is substituting the Einstein energy and momentum relationship into the Hamiltonian of the schrodinger equation since my first intention was modifying the Schrodinger equation a real wave dynamic equation which requires the derivative with respect to time and coordinates of phi to be second order. As for second quantization, I see the word "second" to mean the second type of quantization. What's more, quantization is just a guessing of the classical theory to expand it to more general use. Hope you find it helpful.


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