Slater Determinants

For two fermions we wrote an antisymmetric combination by hand. For three, four, or $N$ fermions, antisymmetrizing every ordering by hand becomes unwieldy. The Slater determinant packages the whole construction into a single elegant formula.

The determinant idea

Suppose $N$ fermions occupy $N$ distinct single-particle orbitals $\phi_1, \phi_2, \ldots, \phi_N$ . Arrange a matrix whose rows are labelled by particles and whose columns are labelled by orbitals, with entry $\phi_j(x_i)$ in row $i$ , column $j$ . The antisymmetric $N$ -fermion state is the determinant of this matrix, normalized by $1/\sqrt{N!}$ :

\Psi(x_1, \ldots, x_N) = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi_1(x_1) & \phi_2(x_1) & \cdots & \phi_N(x_1) \\ \phi_1(x_2) & \phi_2(x_2) & \cdots & \phi_N(x_2) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_1(x_N) & \phi_2(x_N) & \cdots & \phi_N(x_N) \end{vmatrix}.

Math you'll use here: the determinant. For an $N\times N$ matrix $M$ with entries $M_{ij}$ , the determinant is a signed sum over all $N!$ ways of picking one entry from each row and column,

\det M = \sum_{\sigma} \operatorname{sgn}(\sigma)\, M_{1\sigma(1)} M_{2\sigma(2)} \cdots M_{N\sigma(N)},

where $\sigma$ runs over the permutations of $\{1,\ldots,N\}$ and $\operatorname{sgn}(\sigma)=\pm 1$ counts whether $\sigma$ is built from an even or odd number of swaps. Equivalently you can compute it by cofactor expansion along any row. Two algebraic facts follow directly and are all we need below: (i) swapping two rows (or two columns) multiplies $\det M$ by $-1$ , because each swap flips $\operatorname{sgn}(\sigma)$ ; and (ii) if two columns are identical, $\det M = 0$ , since the swap that leaves $M$ unchanged must also negate the determinant, forcing $\det M = -\det M$ . For the $2\times 2$ case this reduces to $\det\!\begin{pmatrix} p & q \\ r & s \end{pmatrix} = ps - qr$ , the formula the exercise evaluates.

Why a determinant does exactly the right job

A determinant has two properties that match fermionic statistics perfectly:

Swapping two rows flips its sign. Exchanging particles $i$ and $j$ swaps two rows, so $\Psi \to -\Psi$ . The state is automatically antisymmetric.
A determinant with two equal columns is zero. If two fermions occupied the same orbital, $\phi_i = \phi_j$ , two columns would be identical and the determinant would vanish. This is the Pauli exclusion principle, built in for free.

So the Slater determinant enforces both antisymmetry and exclusion without any extra bookkeeping.

The two-particle case

For $N = 2$ the determinant has just two terms:

\Psi(x_1, x_2) = \frac{1}{\sqrt{2}} \begin{vmatrix} \phi_a(x_1) & \phi_b(x_1) \\ \phi_a(x_2) & \phi_b(x_2) \end{vmatrix} = \frac{1}{\sqrt{2}}\big[\phi_a(x_1)\phi_b(x_2) - \phi_b(x_1)\phi_a(x_2)\big].

This is exactly the antisymmetric combination from the previous lessons, now read off a $2\times 2$ determinant. The exercise has you evaluate one such determinant numerically.

Limits of a single determinant

A single Slater determinant captures the statistics of identical fermions exactly, but it does not capture all of their correlation. Real electrons repel each other, and the true ground state is generally a superposition of many determinants. Nonetheless, the single-determinant picture is the indispensable starting point: it guarantees the wavefunction is properly antisymmetric before any correlation corrections are added.

Try it

Evaluate the two-particle Slater determinant $\tfrac{1}{\sqrt{2}}\big(\phi_a(x_1)\phi_b(x_2) - \phi_b(x_1)\phi_a(x_2)\big)$ for the orbital values given in the starter code, and return its numeric value.

Run your code to see the quantum state.