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Confusion about Slater Determinants

  1. Oct 13, 2015 #1
    Consider a system of 2 identical fermions.
    $$\psi_{k_1,k_2}(x_1,x_2,m_1, m_2) = \langle x_1\,x_2\,m_1\,m_2\mid \psi \rangle$$
    According to what I have read we can construct a state with the right antisymmetry properties by
    $$\psi_{k_1,k_2}(x_1,x_2,m_1, m_2) = \frac{1}{\sqrt{2}}\begin{vmatrix}\phi_{k_1}(x_1, m_1) & \phi_{k_2}(x_1, m_1) \\ \phi_{k_1}(x_2, m_2) & \phi_{k_2}(x_2, m_2)\end{vmatrix} $$
    $$= \frac{1}{\sqrt{2}}\left(\phi_{k_1}(x_1, m_1) \phi_{k_2}(x_2, m_2) - \phi_{k_2}(x_1, m_1) \phi_{k_1}(x_2, m_2) \right) $$
    This vanishes if ##k_1 = k_2## regardless of any of the other variables. I feel like something is amiss and I am either confused or not doing this properly. For ##m_i = \pm \frac{1}{2}##, shouldn't we have 2 particles allowed to have the same value of ##k## as long as they do not have the same spin? What about singlet and triplet states? How do I understand this?
     
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  3. Oct 13, 2015 #2

    blue_leaf77

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    Actually ##k_i## is a collective index representing all 4 quantum numbers, including the spin. The ##m_1## and ##m_2## in your notation only denote to which particle the spin wavefunction specified implicitly in ##k_1## or ##k_2## belong. So, rather than ##m_i = 1/2## or ##-1/2##, I think it should be either ##1## or ##2## denoting particle ##1## or ##2##, respectively.
     
  4. Oct 13, 2015 #3
    Apparently the thing to do is actually
    $$\phi_{k_1, \sigma_1}\left(x_1, m_1 \right) = f_{k_1}(x)\delta_{\sigma_1m_1}$$
    $$\psi_{k_1, k_2, \sigma_1, \sigma_2}(x_1, x_2, m_1, m_2) = \frac{1}{\sqrt{2}}f_{k_1}(x_1)\delta_{\sigma_1m_1}f_{k_2}(x_2)\delta_{\sigma_2m_2} - \frac{1}{\sqrt{2}} f_{k_2}(x_1)\delta_{\sigma_2m_1}f_{k_1}(x_2)\delta_{\sigma_1m_2}$$
    I suppose this can also be expressed
    $$\frac{1}{\sqrt{2}}f_{k_1}(x_1)f_{k_2}(x_2)\mid\sigma_1\rangle\mid\sigma_2\rangle - \frac{1}{\sqrt{2}} f_{k_2}(x_1)f_{k_1}(x_2)\mid\sigma_2\rangle\mid\sigma_1\rangle $$

    So it seems with ##k_1 = k_2, x_1 = x_2##, we get 0 unless ##\sigma_1\sigma_2 = \uparrow\downarrow## or ## \downarrow\uparrow## in which case there is a singlet.

    Also it seems worth noting that when we use the Slater determinant we have less orthogonal ##\psi## than combinations of ##k_1, k_2## (where here I mean I generalized label). There are ##N_k^N## elements in the set of ##\mid k_1\rangle\mid k_2\rangle \cdots \mid k_N \rangle##. It seems the number of orthogonal wave functions from the Slater determinant would be less than this, but I am not certain how many at this time.
     
  5. Oct 13, 2015 #4

    blue_leaf77

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    This notation is different from your original one, in this new notation, #k_i# is reserved only for the spatial quantum numbers. However, remember that in this new notation ##\sigma_i = \pm 1/2## and ##m_i = 1,2##. Anyway, I don't disagree with your new notation.
    There is nothing wrong either with this expression, only that you should realize that you have assigned appearance order to specify the particle number and that ##|\sigma_1\rangle## and ##|\sigma_2\rangle## have fixed values, e.g. ##|\sigma_1\rangle = \uparrow## and ##|\sigma_2\rangle=\downarrow## . So for example, ##|\sigma_2\rangle |\sigma_1\rangle## means first particle has down spin and second particle has up spin.
    I'm not quite sure with your statement there. But given a set of collective indices (those which include the spin as well) ##\{k_1,k_2,...k_N\}## which represent different states, you can only have one ##N\times N## Slater determinant since it contains all possible combinations of those indices.
     
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