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How can I prove, that two fermions with a total spin of 1 must have an antisymmetric wave function?

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Note: All products between $\chi$'s are to be understood as tensor products.

Assuming the fermions are both spin 1/2 particles, one recalls that the spin-part of the total wavefunction in the spin 1 triplet is either one or a linear combination of

$$ \chi(j=1,m=1) = \chi(1/2,1/2)\chi(1/2,1/2) $$ $$ \chi(j=1,0) = \frac{1}{\sqrt{2}}\big( \chi(1/2,1/2)\chi(1/2,-1/2) + \chi(1/2,-1/2)\chi(1/2,1/2) \big) $$ $$ \chi(j=1,m=-1) = \chi(1/2,-1/2)\chi(1/2,-1/2) $$

where $ \chi(j,m)$ is to be understood as the eigenvector of say $S^z$ with eigenvalue $m$ in the spin $j$ representation

$$ S^z \chi(j,m) = m\chi(j,m) $$ $$ S^2\chi(j,m) = j(j+1)\chi(j,m) $$

Each of these triplet states is symmetric under particle exchange

$$ P\chi(m_1)\chi(m_2) = \chi(m_2)\chi(m_1) $$

Since for fermions, the total wavefunction1

$$ \Psi(r_1,r_2,m_1,m_2) = \psi(r_1,r_2)\otimes\Sigma(m_1,m_2) $$ is antisymmetric (Pauli exclusion principle), the orbital part $\psi$ is necessarily antisymmetric if the spin part $\Sigma$ is symmetric and vice versa. This is true if $\Sigma$ is in the triplet subspace.

More generally, when combining two fermions of spin $j_1$ and $j_2$, the symmetry of the resulting spin 1 state will depend on $j_1$ and $j_2$. This can be seen from the behavior of the Clebsch-Gordan coefficients under exchange (e.g Wikipedia has the full formulas). In particular, under exchange they pick up a factor $$ (-1)^{j-j_1-j_2} $$ resulting in even $j=1,3,...$ states iff $j_1+j_2$ is odd. For $(3/2,1/2)$ the triplet is in fact antisymmetric. Combining two fermions of same spin indeed always results in a symmetric triplet, since $j_1+j_2=2j_1$ is odd for half-integer $j_1$.


1As was remarked in a comment, the most general form of the wavefunction of two spinnning particles is not a tensor product, but may be a superposition of different angular momenta. Here it is assumed that the two fermions are in a state of definite angular momentum, namely $S=1$

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  • $\begingroup$ Thank you! But there are also other combinations that would lead to a total of spin 1, aren't there? But I guess the argument would look the same, so its just easier to show that on the example that both are spin 1/2 particles? $\endgroup$ – Darius Jan 11 '16 at 9:36
  • $\begingroup$ @Darius There are and I have added a remark about the more general case. $\endgroup$ – Nephente Jan 11 '16 at 10:56
  • $\begingroup$ @EmilioPisanty Thanks for pointing that out. I replaced the erroneous tensor product with a more general spinor part. $\endgroup$ – Nephente Jan 11 '16 at 11:00

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