Schrödinger evolution for a Klein-Gordon equation

I have a problem with the transition from quantum relativistic wave equations (specifically Klein-Gordon equation) to QFT, since a lot of assumptions seem implicit. For example I have a problem with the time evolution operator, which is crucial on deriving the perturbative expansion $-$ the main tool in QFT I believe. c So here's what I have a problem with: when we make the leap from Schrödinger equation to a Klein-Gordon equation, we get a second order time derivative, and hence loose the simple concepts from nonrelativistic QM like: the Hamiltonian, time evolution operator etc.

But for a scalar quantum field we can make a Lagrangian density:

$$\mathcal{L}(x) = \hbar^2 c^2 g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi^* - m^2c^4 \phi \phi^*$$

and perform the "second quantization", from which we get a Hamiltonian, canonical commutation relations and the ability to use pictures (Schrödinger's, Heisenberg's...).

So how does this work? Before there was no Hamiltonian in principle, and now there is. Is this the Hamiltonian we pluck into the perturbative expansions' formulas? What changed, when compared to the single solution wave equation in the beginning?

• I suggest you take a look at this presentation of the free Klein-Gordon quantization. It seems sufficiently accurate from the mathematical standpoint to help you clarify your doubts. – yuggib Jul 5 '14 at 9:02
• General remark: 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. – Qmechanic Jul 5 '14 at 9:12

The first thing you should realize is the fact that while $\phi$ has an equation of motion with second time derivatives, it is not the wave function, and therefore there is no problem with QM. The field is just an operator (more or less), not a state. Acting with the fields on the vacuum state you generate the other states which do evolve with an hamiltonian built out of the operators such as $\phi$ itself. And the operators evolve accordingly to the usual Heisenberg equation of motion $[H,\phi(t,x)]=-i\partial_t \phi(t,x)$ (and by Lorentz symmetry, $[P_j,\phi(t,x)]=-i\partial_j \phi(t,x)$ with $P_\mu=(H,P_i)$ as a 4 Lorentz vector). From this Heisenberg picture you can move to the schroedinger picture which it is as in non relativistic QM mechanics, the hamiltonian gives rise to time evolution of the states, $H\rightarrow -i\partial_t$. The fact that the theory is Lorentz invariant just adds other (important) things, but do not change what QM says. QFT implements the principles of QM for a system with infinitely many degrees of freedom that can change the number of particles.
All should become very clear if you realize that the lagrangian for a free scalar boson gives the hamiltonian for a collection of harmonic oscillators, one harmonic oscillator for every momentum $k$ with frequency (aka energy) $\omega^2=k^2+m^2$. In case I find more time, I will add more details to this answer.