# Is the Klein-Gordon equation one evolution equation for states?

I remember when first studying relativistic QM that it was argued that viewing the Klein-Gordon equation as one evolution equation for a state like the Schrodinger equation leads to some issues.

This seems one of the motivations for introducing QFT, on which the Klein-Gordon equation is actually an equation for the field - which is the building block of the observables of the theory, not a state.

Now, when studying QFT in curved spacetimes, and in particular Hawking radiation, one thing that is bothering me is that many sources appear to treat the Klein-Gordon equation as one equation of evolution for states. This seems to conflict with the QFT idea on which it is an equation for the field.

Indeed, a few examples are on the discussion on this question and this question. Both seem to suggest as far as I understood that the Klein-Gordon equation indeed can be seen as the evolution of a single-particle state in QFT.

This seems also to be hinted on Chris Fewster's notes on QFT in curved spacetimes. When discussing he ultrastatic case in section 4.1, page 19, he mentions that one is considering a globally hyperbolic spacetime $$M=\mathbb{R}\times \Sigma$$ and the single-particle state space turns out to be $$L^2(\Sigma)$$.

So combining all this information it seems states are wavepackets defined on Cauchy surfaces and that the time-evolution of a state is governed by the Klein-Gordon equation.

Now is this true? Isn't the Klein-Gordon equation the equation for the field, instead of an equation for states?

Doesn't this picture fall back to the old (non-field) relativistic quantum mechanics which had a bunch of problems?

How this is compatible with QFT?

• If you’re wondering why I deleted my answer — it is because I understood that it’s wrong immediately after posting it :) – Prof. Legolasov Feb 25 at 14:11
• @SolenodonParadoxus I was just about to write essentially the same thing as your deleted answer when I saw your comment! What's wrong with your answer? – knzhou Feb 25 at 15:06
• @knzhou gravity field can produce particles, as described by the Bogolubov coefficients – Prof. Legolasov Feb 25 at 22:55
• @DanYand I think you’re right. That must be what I was missing... so the math is still the same, but there is a major interpretational difference? – Prof. Legolasov Feb 26 at 4:03
• @DanYabd you’re absolutely right, thanks. I got confused :) – Prof. Legolasov Feb 26 at 4:35

As long as your equations of motion are linear in $$\phi$$, your QFT is free and it’s space of states is the Fock space.
Fock space is naturally split into $$n$$-particle subspaces, all of which are symmetrized tensor powers of the $$1$$-particle subspace.
A moment of reflection will convince you that the $$1$$-particle superselection sector is analogous to solutions of Klein-Gordon as evolution equation.