# Second quantization and Klein Gordon equation

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 $\Psi$.

• 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. – AccidentalFourierTransform Mar 29 '17 at 14:46
• 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 ? – StarBucK Mar 29 '17 at 14:47
• 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. – AccidentalFourierTransform Mar 29 '17 at 14:49