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

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.

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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).
Thank you for your explanation. Just another question:The constructed vector $|p> =\hat{\psi(p)}|0>$ if $\hat{\psi}$ is the solution of the Dirac equation has a priori spinor character (as $\hat{\psi}$ has) ? – Frederic Thomas Dec 4 '15 at 17:36
It depends on what you mean by "spinor character". Its wave function surely inherits the features of $\hat{\psi}$; other than that I do not know what spinor character is. – Gennaro Tedesco Dec 5 '15 at 18:07