# 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):
abdu-baud=((ab-ba)(ud+du)-(ab+ba)(ud-du))/2

As these states are all degenerate, you may use either the Slater determinants or the spin eigenfunctions.