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

  1. Apr 1, 2015 #1
    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 [itex]a(x)|\uparrow\rangle[/itex] and [itex]b(x)|\downarrow\rangle[/itex] I get [itex]a(x_1)b(x_2)|\uparrow\downarrow\rangle - b(x_1)a(x_2)|\downarrow\uparrow\rangle[/itex].

    So how can I get the true eigenstates, namely (spatial part)*(spin part), where if one part is symmetric the other is antisymmetric?
     
  2. jcsd
  3. Apr 1, 2015 #2

    DrDu

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    Singlet and triplet are degenerate when the particles are independent, so you could construct also two degenerate Slater deteminants from them.
     
  4. Apr 2, 2015 #3

    vanhees71

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    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
    $$|\vec{p}_1,\vec{p}_2,s=0,\sigma=0 \rangle=\frac{1}{\sqrt{2}} (|\vec{p}_1 \rangle \otimes \vec{p}_2 \rangle + |\vec{p}_2 \rangle \otimes \vec{p}_1 \rangle) \otimes |s=0,\sigma=0 \rangle,$$
    $$|\vec{p}_1,\vec{p}_2,s=1 \rangle=\frac{1}{\sqrt{2}} (|\vec{p}_1 \rangle \otimes \vec{p}_2 \rangle -|\vec{p}_2 \rangle \otimes \vec{p}_1 \rangle) \otimes |s=1,\sigma \rangle, \quad \sigma \in \{-1,0,1 \}.$$
     
  5. Apr 2, 2015 #4
    Thanks, but my question is if such a basis can be constructed by Slater determinants of single-particle spin-orbitals
     
  6. Apr 2, 2015 #5

    DrDu

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    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):
    abud-badu=((ab-ba)(ud+du)+(ab+ba)(ud-du))/2
    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.
     
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