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?


1 Answer 1


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 $|0>_1|1>_2 = |0>_1 \otimes |1>_2$

  • $\begingroup$ Why do you use momentum space ( $\phi(k)$) and not coordinate space ( $\phi(r)$ )? Does the same derivation apply in coordinate space where $k$ denotes a quantum state instead of a coordinate? When I was looking trough related posts I stumbled onto this. In the comments Luboš Motl suggests that coordinate and spin space or not, in general, independent. How does that fit into your argument? It seems it would then be impossible to antisymmeterise the radial or spin parts separately. $\endgroup$ Jun 20, 2013 at 19:28
  • $\begingroup$ Yes, of course, you could use $ϕ(r)$ instead of $ϕ(k)$. It is just a Fourier transform. The point is that the whole function has to be antisymmetric. In the examples I gave, notice that it is the case. Of corse, if you antisymmetrise separately the spatial part and the spin part, the whole function would be symmetric, which is forbidden for a fermion. But you don't need to symmetrise or anti-symmetrise separately the spatial part and the spin part : see my example #2 : $$(\psi_1(k_1)\psi_2(k_2)\otimes |0>_1|1>_2 - \psi_2(k_1)\psi_1(k_2)) \otimes|1>_1|0>_2)$$ $\endgroup$
    – Trimok
    Jun 20, 2013 at 19:50
  • $\begingroup$ The example of Luboš Motl is more complex because it involves nucleons (and not simple electrons), so it involves isospin too. So the whole function (spatial, spin, isospin) has to be antisymmetric. But you stil don't need to symmetrise or antisymmetrise some part. It is only a special case. $\endgroup$
    – Trimok
    Jun 20, 2013 at 19:56
  • $\begingroup$ Ok, in my question I was talking about free fermions and it is quite obvious that the spin and spatial parts must seperate. However am I correct in stating that for the most general hamiltonian the corresponding wave function doesn't necessarily seperate into seperate spin and spatial parts which will either by symmetric or antisymmetric such that the whole wavefunction is antisymmetric? $\endgroup$ Jun 20, 2013 at 21:57
  • $\begingroup$ I am accepting your answer as it more or less answers my original question. However in thinking about this issue another question arose which is very similar to this one. You can find it here if you are interested. $\endgroup$ Jun 20, 2013 at 22:01

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