Confusion about Slater Determinants

[MisterX]

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?

 

[blue_leaf77]

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.

 

[MisterX]

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.

 

[blue_leaf77]

$$\phi_{k_1, \sigma_1}\left(x_1, m_1 \right) = f_{k_1}(x)\delta_{\sigma_1m_1}$$

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.

$$\frac{1}{\sqrt{2}}f_{k_1}(x_1)f_{k_2}(x_2)\mid\sigma_1\rangle\mid\sigma_2\rangle - \frac{1}{\sqrt{2}} f_{k_2}(x_1)f_{k_1}(x_2)\mid\sigma_2\rangle\mid\sigma_1\rangle $$

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.

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.

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.