I think you are mixing up two different things. Namely:

On the one hand, you can see QM as 0+1 (one temporal dimension) QFT, where the position operators (and their conjugate momenta) in the Heisenberg picture plays the role of the fields (and their conjugate momenta) in the QFT. You can check, for instance, that spatial rotational symmetry in the quantum mechanical theory is translated to an internal symmetry in the QFT.

On the other hand, you can take the non-relativistic limit (by the way, ugly name because Galilean relativity is as relativistic as Special relativity) of the Klein-Gordon or Dirac theory to get the Schrödinger QFT, where $\phi$ (in your notation) is a quantum field instead of a wave function. There is a chapter in Srednicki's book where this issue is raised in a simple and nice way. There, you can also read about spin-stastistic and the wave function of multi-particle states. Let me add some equations that hopefully clarify (I'm using your notation and of course can be wrong factors, units, etc.):

The quantum field is:
$$\phi \sim \int d^3p \, a_p e^{-i(p^2/(2m) \cdot t - p \cdot x)}$$

The Hamiltonian is:

$$H \sim i\int d^3x \left( \phi^{\dagger}\partial_t \phi - (1/2m)\partial _i \phi ^{\dagger} \partial ^i \phi \right) \, \sim \int d^3p \, p^2/(2m) \,a^{\dagger}_p a_p$$

The evolution of the quantum field is given by:

$$i\partial _t \phi \sim [\phi, H] \sim -\nabla ^2 \phi /(2m)$$

1-particle states are given by:

$$|1p> \sim \int d^3p \, \tilde f(t,p) \, a^{\dagger}_p \, |0> $$

(one can analogously define multi-particle states)

This state verifies the Schrödinger equation:

$$H \, |1p>=i\partial _t \, |1p>$$ iff

$$i\partial _t \, f(t,x) \sim -\nabla ^2 f(t,x) /(2m)$$

where $f(t,x)$ is the spatial Fourier transformed of $\tilde f (t,p)$.

$f(t,x)$ is a wave function, while $\phi (t, x)$ is a quantum field.

This is the free theory, one can add interaction in a similar way.

This post imported from StackExchange Physics at 2014-04-01 16:22 (UCT), posted by SE-user drake