I Non-relativistic limit of Klein Gordon field

[dRic2]

TL;DR

Non-relativistic limit of Klein Gordon field (not the equation).

From Wikipedia:

The analogous limit of a quantum Klein-Gordon field is complicated by the non-commutativity of the field operator. In the limit v ≪ c, the creation and annihilation operators decouple and behave as independent quantum Schrödinger fields.

Which should be conceptually similar of what happen in the non-relativistic limit of the Dirac equations when you see that the solutions decouple.

Do you have any reference that I can look up where the derivation for the KG field is performed?

Thanks in advance!

 

[HomogenousCow]

That quote from wikipedia doesn't make sense to me. How can the creation and annihilation operators "decouple"? They're hermitian conjugates of each other, not two independent variables linked by dynamics. From what I've seen taking the non-relativisitic limit for the KG field simply involves taking the $\mathbf{k} \rightarrow 0$ limit.

In any case I don't you think you can truly recover the Schrodinger field from the KG field since the latter is quantized as a physical observable right from the beginning while the former is not observable.

 

[dRic2]

That quote from wikipedia doesn't make sense to me. How can the creation and annihilation operators "decouple"?

The KG filed $\phi$ can be decomposed in positive and negative energies field $\phi^+$ and $\phi^-$, and in the KG Lagrangian there are mixed term like $\bar \phi \phi = \phi^+\phi^+ + \phi^+\phi^- + ...$ I think they mean that, in the non relativistic limit, your KG filed will "become" a new field $\psi$ which can not be decomposed in positive/negative energies fields, and so the $\bar \phi \phi$ term will become simply $\bar \psi \psi$. Now $\bar \psi \neq \psi$, as opposed to the KG field that satisfies $\bar \phi = \phi$. In this limit we should also recover particle number conservation. This new field $\psi$ should be what they call the Schrodinger field.

This is how I understood after researching a little bit on the internet, but maybe it's wrong. I don't know QFT... I was just curious about this