What is the Slater determinant of a singlet or a triplet?

1. Apr 1, 2015

atat1tata

We have a system of 2 particles, let's say with following hamiltonian:

$$\hat{H} = -\frac{\hbar^2\hat{\nabla}_1^2}{2m} -\frac{\hbar^2\hat{\nabla}_2^2}{2m}$$

The eigenstates are often represented as (spatial wavefunction)*(spin wavefunction), where the spin wavefunction is a singlet or a triplet. However the particle are independent. So the total wavefunction (position*spin) must be written as a Slater determinant of single-particle eigenstates. But I can't find such states!

For example, combining $a(x)|\uparrow\rangle$ and $b(x)|\downarrow\rangle$ I get $a(x_1)b(x_2)|\uparrow\downarrow\rangle - b(x_1)a(x_2)|\downarrow\uparrow\rangle$.

So how can I get the true eigenstates, namely (spatial part)*(spin part), where if one part is symmetric the other is antisymmetric?

2. Apr 1, 2015

DrDu

Singlet and triplet are degenerate when the particles are independent, so you could construct also two degenerate Slater deteminants from them.

3. Apr 2, 2015

vanhees71

You mean that the particles are identical and fermions of spin 1/2. Then the total state must be antisymmetric. Let's take momentum-spin eigenstates (which exist in non-relativistic quantum theory, because the spin operators commute with momentum). Then the two-particle basis can be given in terms of irreducible representations of the rotation group wrt. spin. The Kronecker product of two spin-1/2 representations reduce to spin 0 (singlet) und spin 1 (triplet). The former are the antisymmetrized and symmetrized product states, respectively:
$$|s=0,\sigma=0 \rangle=\frac{1}{\sqrt{2}} |s=1/2,\sigma=1/2 \rangle \otimes |s=1/2,\sigma=-1/2 \rangle - |s=1/2,\sigma=-1/2 \rangle \otimes |s=1/2,\sigma=1/2 \rangle,$$
$$|s=1,\sigma=1 \rangle=|s=1/2,\sigma=1/2 \rangle \otimes |s=1/2,\sigma=1/2 \rangle, \quad |s=1,\sigma=0 \rangle=\frac{1}{\sqrt{2}} |s=1/2,\sigma=1/2 \rangle \otimes |s=1/2,\sigma=-1/2 \rangle +|s=1/2,\sigma=-1/2 \rangle \otimes |s=1/2,\sigma=1/2 \rangle, \quad |s=1,\sigma=-1 \rangle=|s=1/2,\sigma=-1/2 \rangle \otimes |s=1/2,\sigma=-1/2 \rangle.$$
Further you can write the basis as a product of the momentum (orbital) part and the spin part. Since you have fermions, your basis in terms of proper total spin is
$$|\vec{p}_1,\vec{p}_2,s=0,\sigma=0 \rangle=\frac{1}{\sqrt{2}} (|\vec{p}_1 \rangle \otimes \vec{p}_2 \rangle + |\vec{p}_2 \rangle \otimes \vec{p}_1 \rangle) \otimes |s=0,\sigma=0 \rangle,$$
$$|\vec{p}_1,\vec{p}_2,s=1 \rangle=\frac{1}{\sqrt{2}} (|\vec{p}_1 \rangle \otimes \vec{p}_2 \rangle -|\vec{p}_2 \rangle \otimes \vec{p}_1 \rangle) \otimes |s=1,\sigma \rangle, \quad \sigma \in \{-1,0,1 \}.$$

4. Apr 2, 2015

atat1tata

Thanks, but my question is if such a basis can be constructed by Slater determinants of single-particle spin-orbitals

5. Apr 2, 2015

DrDu

Using a somewhat terser notation, each of the two possible slater determinants can be written as a 50-50 superposition of the singlet and triplet state (and vice versa):