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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, $$ \Psi(x_1,x_2) = - \Psi(x_2,x_1) $$ If we assume that we can factorize the wave function in terms of single particle wave functions we can write $$ \Psi(x_1,x_2) = \psi_{1}(x_1) \psi_{2}(x_2) - \psi_{1}(x_1)\psi_2(x_2) $$ which fulfills the anti-symmetry requirement. The plane wave single particle states are given by, $$ \psi_{\mathbf{k},m_s} (x) = u_{\mathbf{k},m_s}(s) \phi( \mathbf{k} \cdot \mathbf{r}) $$ So I would expect the total wavefunction to be \begin{align} \Psi(x_1,x_2) &= u_{\mathbf{k}_1,m_{s_1}}(s_1) \phi( \mathbf{k}_1 \cdot \mathbf{r}_1) u_{\mathbf{k}_2,m_{s_2}}(s_2) \phi( \mathbf{k}_2 \cdot \mathbf{r}_2) - u_{\mathbf{k}_1,m_{s_1}}(s_2) \phi( \mathbf{k}_1 \cdot \mathbf{r}_2) u_{\mathbf{k}_2,m_{s_2}}(s_1) \phi( \mathbf{k}_2 \cdot \mathbf{r}_1) \\ &= u_{\mathbf{k}_1,m_{s_1}}(s_1) u_{\mathbf{k}_2,m_{s_2}}(s_2) \phi( \mathbf{k}_1 \cdot \mathbf{r}_1) \phi( \mathbf{k}_2 \cdot \mathbf{r}_2) - u_{\mathbf{k}_1,m_{s_1}}(s_2) u_{\mathbf{k}_2,m_{s_2}}(s_1) \phi( \mathbf{k}_1 \cdot \mathbf{r}_2) \phi( \mathbf{k}_2 \cdot \mathbf{r}_1) \end{align} However I have seen this written as $$ u(\mathbf{k}_1,m_{s_1}) u(\mathbf{k}_2,m_{s_2}) \phi( \mathbf{k}_1 \cdot \mathbf{r}_1) \phi( \mathbf{k}_2 \cdot \mathbf{r}_2) - u(\mathbf{k}_2,m_{s_2}) u(\mathbf{k}_1,m_{s_1}) \phi( \mathbf{k}_1 \cdot \mathbf{r}_2) \phi( \mathbf{k}_2 \cdot \mathbf{r}_1) $$ If I'm not mistaking one cannot freely change the order of the Dirac spinors ($ u(\mathbf{k}_1,m_{s_1}) u(\mathbf{k}_2,m_{s_2}) \neq u(\mathbf{k}_2,m_{s_2}) u(\mathbf{k}_1,m_{s_1}) $) so these expressions seem to be uncompatible. What would the correct expression look like?

Maybe related to my question is my confusement about the spin coordinate in the Dirac spinor. It is my understanding that the Dirac spinor only depends on the projection of the spin $m_s$ which denotes a quantum number or a quantum state, and is not a coordinate. So is $m_{s_i}$ invariant under a coordinate switch $s_1 \leftrightarrow s_2$? Why explicitly write the spin coordinate?

camelthemammel
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1 Answers1

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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: $$\psi(k_1,s_1) =\phi(k_1)\otimes \chi(s_1) $$.

In this formalism, do not use dirac spinors. If you are talking about electrons, you have 2 possible polarizations, so use simply states $\chi(s_1) = |0>_1$ or $\chi(s_1) = |1>_1$

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 :

$$(\psi_1(k_1)\psi_2(k_2) + \psi_2(k_1)\psi_1(k_2))\otimes(|0>_1|1>_2 -|1>_1|0>_2)$$

Another example :

$$(\psi_1(k_1)\psi_2(k_2)\otimes |0>_1|1>_2 - \psi_2(k_1)\psi_1(k_2)) \otimes|1>_1|0>_2)$$

Note, that here, the tensorial products into each of the space (momentum space or spin space) are implicit, that is $|0>_1|1>_2 = |0>_1 \otimes |1>_2$

Trimok
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