The Slater determinant takes into account the Pauli principle, but if the fermions have no spin, a degree of freedom is missing. What would the determinant look like then?
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1$\begingroup$ what is the meaning of spinless fermions? $\endgroup$– schris38Commented Jul 8, 2022 at 17:13
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1$\begingroup$ @schris38 See for example this and the comments here. $\endgroup$– Tobias FünkeCommented Jul 8, 2022 at 17:23
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$\begingroup$ Wavefunctions have a spin part and spatial part. If they are effectively spinless, Slater just serves to anti-symmetrize the spatial part. $\endgroup$– Connor BehanCommented Jul 8, 2022 at 17:37
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1 Answer
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A Slater determinant \begin{align} \left\vert \begin{array}{ccc} \psi_1(x_a)&\psi_1(x_b)&\psi_1(x_c)\\ \psi_2(x_a)&\psi_2(x_b)&\psi_2(x_c)\\ \psi_3(x_a)&\psi_3(x_b)&\psi_3(x_c)\end{array} \right\vert \end{align} will be antisymmetric in the spatial part and so can be combined with a symmetric spin states, such as $\vert +\rangle_1\vert +\rangle_2\vert +\rangle_3$, to yield a fully antisymmetric wavefunction.
answered Jul 8, 2022 at 18:20