What is the difference between a one-particle state in the fock space and single particle wave function (in momentum representation)?

Question

If I consider one single Dirac electron in momentum representation, I use the wavefunction $u(p)e^{-ipx}$, however if I consider an one-particle state in the Fock space I use $|p\rangle$. Should it not be same?

Obviously the Dirac 1-particle wavefunction is a bispinor, and probably $|p\rangle$ is not a spinor. But could it not be spinor?

For a 2-particle wavefunction $|p,k\rangle$, I would use $$\frac{1}{\sqrt2}(u_1(p)u_2(k)e^{ipx_1+ikx_2} - u_2(k)u_1(p)e^{ipx_2 + ikx_1})$$ or something similar. I regret my limited way of expressing correctly. Certainly there is the problem if I consider instead of a half-spin particle a scalar particle, then I would have to build my multi-particles state out scalar wave function instead of spinor wave functions. May be my understanding of the Fock space is incomplete.

Answer 1

Let us first clarify the difference between a state and its wave function representation. A state $|\psi\rangle$ is an element in a Hilbert (or equivalently) Fock space, whereas its wave function $\psi(x)=\langle x |\psi\rangle$ is its representation onto the position (or momentum, respectively) basis.

In quantum field theory things extend a little, although they keep their initial definitions. The Dirac equation is an equation whose solution is the Dirac field $\hat{\psi}$, which in turn can be expressed in spinor notation and in terms of creation and annihilation operators. A state of the Dirac field is given acting with the field upon the vacuum, namely $|p\rangle = \hat{\psi}(p)|0\rangle$. Once so, in order to obtain its wave function you have to take scalar products against whichever basis you choose (position or momentum).