# Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,Suppose we want to construct a wave function for 2 free (relativistic) fermions.

Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,
$\mathrm{\Psi }\left({x}_{1},{x}_{2}\right)=-\mathrm{\Psi }\left({x}_{2},{x}_{1}\right)$
If we assume that we can factorize the wave function in terms of single particle wave functions we can write
$\mathrm{\Psi }\left({x}_{1},{x}_{2}\right)={\psi }_{1}\left({x}_{1}\right){\psi }_{2}\left({x}_{2}\right)-{\psi }_{1}\left({x}_{1}\right){\psi }_{2}\left({x}_{2}\right)$
which fulfills the anti-symmetry requirement. The plane wave single particle states are given by,
${\psi }_{\mathbf{k},{m}_{s}}\left(x\right)={u}_{\mathbf{k},{m}_{s}}\left(s\right)\varphi \left(\mathbf{k}\cdot \mathbf{r}\right)$
So expect the total wavefunction to be
$\begin{array}{rl}\mathrm{\Psi }\left({x}_{1},{x}_{2}\right)& ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}\left({s}_{1}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}\right){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}\left({s}_{2}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2}\right)-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}\left({s}_{2}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}\right){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}\left({s}_{1}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1}\right)\\ & ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}\left({s}_{1}\right){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}\left({s}_{2}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2}\right)-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}\left({s}_{2}\right){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}\left({s}_{1}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1}\right)\end{array}$
However
$u\left({\mathbf{k}}_{1},{m}_{{s}_{1}}\right)u\left({\mathbf{k}}_{2},{m}_{{s}_{2}}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2}\right)-u\left({\mathbf{k}}_{2},{m}_{{s}_{2}}\right)u\left({\mathbf{k}}_{1},{m}_{{s}_{1}}\right)\varphi \left({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}\right)\varphi \left({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1}\right)$
If I'm not mistaking one cannot freely change the order of the Dirac spinors $u\left({\mathbf{k}}_{1},{m}_{{s}_{1}}\right)u\left({\mathbf{k}}_{2},{m}_{{s}_{2}}\right)\ne u\left({\mathbf{k}}_{2},{m}_{{s}_{2}}\right)u\left({\mathbf{k}}_{1},{m}_{{s}_{1}}\right)$ so these expressions seem to be uncompatible. What would the correct expression look like?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

ustalovatfog
Your formalism is not correct. Momentum space and polarization space are independent. You have to use tensorial products. For instance, for a one-particle state, you have:
$\psi \left({k}_{1},{s}_{1}\right)=\varphi \left({k}_{1}\right)\otimes \chi \left({s}_{1}\right)$
In this formalism, do not use dirac spinors. If you are talking about electrons, you have 2 possible polarizations, so use simply states $\chi \left({s}_{1}\right)=|0{>}_{1}$ or $\chi \left({s}_{1}\right)=|1{>}_{1}$
For a 2-particle state, momentum state and spin-state of one particle are independent of the states of the other particle, so there is an other tensorial product here.
For instance, an example of an antisymmetric global state could be :
$\left({\psi }_{1}\left({k}_{1}\right){\psi }_{2}\left({k}_{2}\right)+{\psi }_{2}\left({k}_{1}\right){\psi }_{1}\left({k}_{2}\right)\right)\otimes \left(|0{>}_{1}|1{>}_{2}-|1{>}_{1}|0{>}_{2}\right)$
Another example:
$\left({\psi }_{1}\left({k}_{1}\right){\psi }_{2}\left({k}_{2}\right)\otimes |0{>}_{1}|1{>}_{2}-{\psi }_{2}\left({k}_{1}\right){\psi }_{1}\left({k}_{2}\right)\right)\otimes |1{>}_{1}|0{>}_{2}\right)$
Note, that here, the tensorial products into each of the space (momentum space or spin space) are implicit, that is $|0{>}_{1}|1{>}_{2}=|0{>}_{1}\otimes |1{>}_{2}$