Slater Determinants

Antisymmetry is the law that governs identical fermions: swap any two of them and the total wavefunction must change sign. Permutation Symmetry and the Exchange Operator built the antisymmetrizer $\hat A$ that projects onto this totally antisymmetric sector, and Symmetric and Antisymmetric States wrote out the two-particle members explicitly. But atoms have many electrons, and writing "the antisymmetric combination of $N$ orbitals" by hand means summing $N!$ signed terms. The Slater determinant is the trick that does this bookkeeping for free: pack the orbitals into a matrix, take its determinant, and the algebra of determinants supplies the antisymmetry, the exclusion, and the normalization all at once. The dictionary is exact and it is the whole point. Each column of the matrix is one orbital, each row is one particle's coordinate. Exchanging two particles swaps two rows — and a determinant flips sign when two rows are swapped, so antisymmetry is automatic. Putting two particles into the same orbital makes two columns identical — and a determinant with two equal columns is zero, so the Pauli exclusion principle is automatic too. This article owns the general $N$ -particle construction; Pauli Exclusion Principle used only the $2\times2$ special case, and here we prove that the antisymmetrizer acting on a product of orbitals is exactly the determinant, $\sqrt{N!}\,\hat A[\psi_1\cdots\psi_N]=\det$ , a permutation-sign identity we verify symbolically for $N=2,3,4$ . What the determinant is not, we also state honestly. A single Slater determinant is the best independent-particle wavefunction, not the exact many-body state. It captures the exchange (Fermi) correlation that antisymmetry forces, but not the full Coulomb correlation of electrons dodging each other; the exact ground state is generally an infinite sum of determinants, which is why methods like configuration interaction exist. The energy of a single determinant splits into a Coulomb piece and an exchange piece $J-K$ — but the exchange energy itself is developed by Exchange Forces and Exchange Energy, and the bosonic all-plus analogue, the permanent, is owned by Permanents and Symmetrized Boson States. Here we own the object and its geometry.

Key takeaways

A Slater determinant is the antisymmetric $N$ -fermion wavefunction $\Psi=\frac{1}{\sqrt{N!}}\det[\psi_i(x_j)]$ : columns are orbitals, rows are particles, and the determinant supplies antisymmetry, exclusion, and normalization automatically.
Antisymmetry is exact — swapping two particles swaps two rows, flipping the determinant's sign, so $\hat P_{ij}\Psi=-\Psi$ for every pair.
Exclusion is exact — two particles in the same spin-orbital give two identical columns, and a determinant with two equal columns vanishes; this is the general- $N$ version of the exclusion that Pauli Exclusion Principle wrote for $2\times2$ .
The owned result: the antisymmetrizer $\hat A$ of Permutation Symmetry and the Exchange Operator, applied to an orbital product, equals the determinant up to $1/\sqrt{N!}$ — verified as a permutation-sign identity for $N=2,3,4$ .
For orthonormal orbitals the prefactor $1/\sqrt{N!}$ normalizes the state exactly, and the overlap of two determinants collapses to $\det[\langle\psi'_i|\psi_j\rangle]$ .
A single determinant is a mean-field ansatz — the workhorse of Hartree–Fock and quantum chemistry — but the exact state is generally a sum of many determinants; it captures Fermi correlation, not full Coulomb correlation.
Key idea: pack $N$ orbitals into a matrix and take its determinant — the sign-flip-on-row-swap gives antisymmetry, the zero-on-equal-columns gives Pauli exclusion, and the $1/\sqrt{N!}$ gives normalization, so one algebraic object enforces all of fermionic statistics at once.

The core insight

The Slater determinant $\Psi=\frac{1}{\sqrt{N!}}\det[\psi_i(x_j)]$ turns fermionic antisymmetry into linear algebra: row-swap $\to$ sign flip (exchange), equal columns $\to$ zero (Pauli exclusion), and the determinant is exactly $\sqrt{N!}$ times the antisymmetrizer acting on the orbital product. It is the natural orthonormal basis of the antisymmetric Hilbert space and the starting point of Hartree–Fock, though the exact many-body state is generally a superposition of determinants.

Where this lives today

The Slater determinant is the atom of computational chemistry and condensed-matter physics. Every mean-field electronic-structure calculation — Hartree–Fock, and the Kohn–Sham formulation of density-functional theory that dominates materials modeling — represents its electrons by a single determinant of optimized orbitals. Post-Hartree–Fock methods that chase correlation energy — configuration interaction, coupled cluster, multireference approaches — are, at bottom, prescriptions for combining many determinants. In quantum Monte Carlo, determinants (dressed with correlation factors) are the trial wavefunctions whose statistics are sampled on the largest supercomputers. And in the theory of the fractional quantum Hall effect and of ultracold fermionic atoms, Slater determinants of Landau-level or trap orbitals are the reference states from which exotic correlated physics is built. Ninety years on, when a chemist writes down an $N$ -electron wavefunction, they almost always start by writing a determinant.

The physical picture

Suppose you want the wavefunction of three electrons occupying three orbitals $\psi_1,\psi_2,\psi_3$ . Because electrons are identical fermions, you cannot say "electron 1 is in $\psi_1$ " — the state must be built so that no such labelling survives, and so that swapping any two electrons flips the overall sign. The honest construction sums over all $3!=6$ ways of assigning the three electrons to the three orbitals, each term weighted by $+1$ or $-1$ according to whether the assignment is an even or odd permutation. That is a mouthful. The genius of Slater's 1929 observation is that this signed sum is exactly the expansion of a determinant.

\Psi(x_1,x_2,x_3)=\frac{1}{\sqrt{3!}}\begin{vmatrix}\psi_1(x_1)&\psi_2(x_1)&\psi_3(x_1)\\ \psi_1(x_2)&\psi_2(x_2)&\psi_3(x_2)\\ \psi_1(x_3)&\psi_2(x_3)&\psi_3(x_3)\end{vmatrix} \qquad (1)

Read the matrix as a grid: row $j$ carries particle coordinate $x_j$ , column $i$ carries orbital $\psi_i$ . Now every property of fermions falls out of a property of determinants. Interchange particles $1$ and $2$ — that is, swap $x_1\leftrightarrow x_2$ — and you have swapped rows $1$ and $2$ ; a determinant changes sign under a row swap, so $\Psi\to-\Psi$ . Try to put two electrons in the same orbital, say $\psi_2=\psi_3$ — that makes two columns identical, and a determinant with two equal columns is zero, so the state simply does not exist. Antisymmetry and Pauli exclusion are not extra rules bolted on; they are theorems about determinants.

Figure

The exponential bookkeeping a determinant absorbs: N! signed permutation terms, all packed into one object — the outlier at N=10 is 3,628,800 terms

The figure is the practical case for the determinant. By $N=10$ the explicit signed sum has $3{,}628{,}800$ terms; the determinant packages all of them, with every sign correct, in one symbol. This is why electronic-structure theory speaks in determinants rather than in permutation sums — and why the single-determinant approximation, once written, can be manipulated with the mature machinery of linear algebra.

Three readings of the same object

The determinant has three equivalent faces, and each earns its keep. The first is the determinant itself, Eq. (1). The second is the explicit signed sum over the symmetric group $S_N$ , which makes the permutation content visible:

\Psi=\frac{1}{\sqrt{N!}}\sum_{P\in S_N}(-1)^{P}\,\psi_{P(1)}(x_1)\psi_{P(2)}(x_2)\cdots\psi_{P(N)}(x_N) \qquad (2)

where: $P$ ranges over the $N!$ permutations of $S_N$ ; $(-1)^P$ is the signature, $+1$ for even and $-1$ for odd permutations. This is the definition of a determinant by Leibniz's formula, so Eqs. (1) and (2) are literally the same expression — but Eq. (2) shows why the sign structure is exactly the fermionic one. The third face connects the determinant to the machinery of the previous chapter. Permutation Symmetry and the Exchange Operator built the antisymmetrizer $\hat A=\frac{1}{N!}\sum_P(-1)^P\hat P$ , the projector onto the totally antisymmetric subspace. Applying it to a plain product of orbitals and rescaling gives the determinant:

\Psi=\sqrt{N!}\,\hat A\big[\psi_1(x_1)\psi_2(x_2)\cdots\psi_N(x_N)\big],\qquad \hat A=\frac{1}{N!}\sum_{P}(-1)^P\hat P \qquad (3)

where: $\hat A$ is the antisymmetrization projector ( $\hat A^2=\hat A$ ) whose algebra is owned by Permutation Symmetry and the Exchange Operator; the $\sqrt{N!}$ converts the projector's $1/N!$ into the state's $1/\sqrt{N!}$ . This is the identity we own and prove below — the determinant is not merely like the antisymmetrizer, it is the antisymmetrizer acting on a product, exactly.

Derivation AThe antisymmetrizer applied to an orbital product is exactly the determinant

We prove that $\sqrt{N!}\,\hat A[\psi_1(x_1)\cdots\psi_N(x_N)]$ equals the determinant of the orbital matrix, so that the abstract projector $\hat A$ of Permutation Symmetry and the Exchange Operator and the concrete determinant of Eq. (1) are one object. The goal is to make the equivalence a theorem about permutation signatures, not an analogy: we show the signed permutation sum that defines a determinant (Leibniz's formula) is character-for-character the same as the antisymmetrizer expansion, and we confirm it symbolically for $N=2,3,4$ so no bookkeeping slips through. This is what lets the whole chapter treat "totally antisymmetric $N$ -fermion state" and "Slater determinant" as synonyms.

Step 1. Start from the antisymmetrizer $\hat A=\frac{1}{N!}\sum_{P}(-1)^P\hat P$ . Acting on the product $\Phi=\psi_1(x_1)\psi_2(x_2)\cdots\psi_N(x_N)$ , the permutation $\hat P$ relabels the particle coordinates, sending $x_j\to x_{P(j)}$ . Collecting terms, $\hat A\,\Phi=\frac{1}{N!}\sum_P(-1)^P\prod_j\psi_j(x_{P(j)})$ .

Step 2. Reindex the product. Since $P$ runs over all of $S_N$ , replacing $P$ by its inverse (a bijection of $S_N$ with $(-1)^{P^{-1}}=(-1)^P$ ) rewrites the sum as $\frac{1}{N!}\sum_P(-1)^P\prod_i\psi_{P(i)}(x_i)$ . Multiplying by $\sqrt{N!}$ gives exactly Eq. (2).

Step 3. Recognize Leibniz's formula. The determinant of the $N\times N$ matrix $M$ with entries $M_{ij}=\psi_i(x_j)$ is by definition $\det M=\sum_{P}(-1)^P\prod_i M_{i,P(i)}=\sum_P(-1)^P\prod_i\psi_i(x_{P(i)})$ . This is the same signed sum, so $\sqrt{N!}\,\hat A\,\Phi=\frac{1}{\sqrt{N!}}\det[\psi_i(x_j)]$ — the determinant, prefactor and all.

Step 4. Machine confirmation. Build the symbolic matrix $M_{ij}=\psi_i(x_j)$ and compute $\det M$ with the computer-algebra system; independently sum $\sum_P\mathrm{sgn}(P)\prod_i\psi_{P(i)}(x_i)$ over all permutations using the permutation-group signature. Subtract: the difference simplifies to zero for $N=2$ , $N=3$ , and $N=4$ . The identity holds not approximately but as a polynomial identity in the orbital symbols. Why this matters: every downstream statement — antisymmetry, exclusion, the Slater–Condon energy rules — can now be proved either as a determinant property or as an $\hat A$ property, whichever is easier, because they are the same object. Verified: `verify_aq-qm-slater-determinants.py` — the identity $\det[\psi_i(x_j)]=\sum_P\mathrm{sgn}(P)\prod_i\psi_{P(i)}(x_i)$ symbolically exact for $N=2,3,4$ .

Antisymmetry, exclusion, and normalization for free

With Eq. (1) in hand, the three defining properties of a fermionic state are immediate consequences of standard determinant identities.

Antisymmetry. Exchanging particles $i$ and $j$ swaps rows $i$ and $j$ of the matrix. A single row swap multiplies a determinant by $-1$ , so $\hat P_{ij}\Psi=-\Psi$ for every pair — the state lives in the totally antisymmetric sector, exactly as required.

Exclusion. If two orbitals coincide, $\psi_a=\psi_b$ , two columns of the matrix are identical, and a determinant with two equal columns is zero. So $\Psi\equiv0$ : no two fermions may occupy the same spin-orbital. This is the Pauli exclusion principle, and it is the general- $N$ generalization of the statement Pauli Exclusion Principle proved for the $2\times2$ determinant.

Normalization. For orthonormal orbitals $\langle\psi_i|\psi_j\rangle=\delta_{ij}$ , the prefactor $1/\sqrt{N!}$ makes $\langle\Psi|\Psi\rangle=1$ . The proof is the second derivation card below, and its punchline is a clean counting fact: the double permutation sum collapses to $N!$ surviving unit terms, which the prefactor's $1/N!$ exactly cancels.

Go deeper · Derivation B · 1 minWhy the prefactor is 1/√N! — normalization from the permutation sum

Derivation BWhy the prefactor is 1/√N! — normalization from the permutation sum

We prove that $1/\sqrt{N!}$ is precisely the factor that normalizes a Slater determinant built from orthonormal orbitals, and that the overlap of two different determinants collapses to a single determinant of one-particle overlaps. The goal is to show the normalization is not a fitted constant but the exact residue of a counting argument: when the permutation sum is squared, orthogonality annihilates every cross term between distinct permutations and leaves a diagonal of $N!$ identical unit terms, so the amplitude must carry $1/\sqrt{N!}$ . The overlap result that falls out is the seed of the entire Slater–Condon matrix-element machinery.

Step 1. Square the permutation sum. Using Eq. (2), $\langle\Psi|\Psi\rangle=\frac{1}{N!}\sum_{P,Q\in S_N}(-1)^{P+Q}\prod_j\langle\psi_{P(j)}|\psi_{Q(j)}\rangle$ , a double sum over pairs of permutations.

Step 2. Impose orthonormality. With $\langle\psi_i|\psi_j\rangle=\delta_{ij}$ , a term survives only if $\psi_{P(j)}=\psi_{Q(j)}$ for every $j$ , i.e. only if $P=Q$ . Each such surviving term has sign $(-1)^{2P}=+1$ and value $\prod_j 1=1$ .

Step 3. Count and cancel. There are exactly $N!$ permutations $P$ , so the diagonal contributes $N!$ unit terms: $\langle\Psi|\Psi\rangle=\frac{1}{N!}\cdot N!\cdot 1=1$ . The prefactor $1/\sqrt{N!}$ is therefore forced — its square $1/N!$ cancels the $N!$ surviving terms exactly.

Step 4. Generalize to the overlap of two determinants. Repeat the calculation with two different orbital sets $\{\psi_j\}$ and $\{\psi'_i\}$ . The same permutation algebra reorganizes the double sum into a single determinant of one-particle overlaps, $\langle\Psi'|\Psi\rangle=\det[\langle\psi'_i|\psi_j\rangle]$ . When the two sets coincide and are orthonormal, this is $\det(\mathbb 1)=1$ , recovering Step 3 as a special case — verified numerically as $\det(I_3)=1$ . Why it matters: every matrix element between determinants (energies, transition amplitudes) is built from this overlap identity, which is why the Slater–Condon rules exist in closed form. Verified: `verify_aq-qm-slater-determinants.py` — the $(1/N!)\cdot N!=1$ counting for $N=2,3$ , the prefactor $1/\sqrt{3!}=1/\sqrt6\approx0.4082$ , and $\det(I_3)=1$ .

Figure

Exchange and exclusion as determinant identities: swap two rows and the sign flips; make two columns equal and the state vanishes

What a single determinant misses

A determinant is the best possible independent-particle description, and no better single-configuration wavefunction exists. But "best independent-particle" is not "exact." A single determinant builds in the correlation that antisymmetry forces — same-spin electrons avoid each other, the Fermi hole, exchange — but it does not let electrons of opposite spin dodge each other's instantaneous Coulomb repulsion. That missing physics is called Coulomb correlation, and recovering it requires mixing many determinants. Concretely, the exact ground state of an $N$ -electron system is a (generally infinite) superposition $\Psi_{\text{exact}}=\sum_I c_I\,\Psi_I$ over determinants $\Psi_I$ built from different orbital occupations. The Hartree–Fock method keeps one determinant and optimizes its orbitals variationally; configuration interaction and coupled-cluster methods add more determinants to chase the remaining correlation energy. The single determinant is the foundation stone, not the finished building — a distinction worth holding onto, because a great deal of chemistry lives in the difference.

Limiting cases

Case	Result	Meaning
$N=2$ , distinct orbitals	$\Psi=\frac{1}{\sqrt2}[\psi_a(1)\psi_b(2)-\psi_b(1)\psi_a(2)]$	the two-fermion determinant of Symmetric and Antisymmetric States
Two orbitals coincide, $\psi_a=\psi_b$	$\Psi\equiv0$	Pauli exclusion (equal columns $\to$ zero) — the general- $N$ form of Pauli Exclusion Principle
Non-interacting fermions	one determinant is exact	no Coulomb correlation to capture; the mean field is the truth
Interacting fermions	one determinant is approximate	Coulomb correlation demands a sum of determinants (CI, coupled-cluster)
Bosons instead of fermions	replace $\det$ by the permanent	all-plus signs, owned by Permanents and Symmetrized Boson States
Orbitals not orthonormal	$\langle\Psi\|\Psi\rangle=\det[\langle\psi_i\|\psi_j\rangle]$	general overlap; reduces to $1/\sqrt{N!}$ normalization only in the orthonormal case

Common trapA determinant is one particular antisymmetrized product — not "the" fermion state, and the labels

\psi_i(x_j)

do not mean "particle j is in state i"

Two traps live here. First, a single Slater determinant is the antisymmetrization of one choice of orbitals; the true many-body state is generally a sum of determinants. Reading a lone determinant as the exact wavefunction is the mean-field approximation, and it silently drops Coulomb correlation. Second — and more insidious — do not read the entry $\psi_i(x_j)$ as "particle $j$ occupies state $i$ ." Antisymmetry makes that labelling meaningless: every particle is smeared over every orbital with the signs arranged so that no particle can be assigned to any orbital. The determinant says "these $N$ orbitals are occupied," full stop; which electron is in which orbital is not a question quantum mechanics answers. And when you build a determinant from spin-orbitals, antisymmetry is required in the combined space-and-spin label $x=(\mathbf r,m_s)$ — two electrons may share a spatial orbital only if their spins differ, which is exactly why the $1s$ shell holds two.

Worked examples

Example 1 — the helium ground state as a $2\times2$ determinant. Build the ground-state wavefunction of helium, both electrons in the $1s$ orbital, and show the determinant factorizes into a symmetric spatial part and an antisymmetric spin singlet.

Step 1. Enumerate the spin-orbitals. Both electrons occupy the spatial orbital $1s$ , but they must differ in spin, so the two occupied spin-orbitals are $\chi_1=1s\,\uparrow$ and $\chi_2=1s\,\downarrow$ . Any other choice (two $1s\uparrow$ , say) would repeat a column and vanish — exclusion at work.

Step 2. Write the determinant. With rows = particles and columns = spin-orbitals,

\Psi=\frac{1}{\sqrt2}\begin{vmatrix}1s(1)\uparrow(1)&1s(1)\downarrow(1)\\ 1s(2)\uparrow(2)&1s(2)\downarrow(2)\end{vmatrix}=\frac{1}{\sqrt2}\,1s(1)\,1s(2)\big[\uparrow(1)\downarrow(2)-\downarrow(1)\uparrow(2)\big].

Step 3. Read the factorization — boundary number. The spatial factor $1s(1)1s(2)$ is symmetric under exchange; the spin factor $\frac{1}{\sqrt2}(\uparrow\downarrow-\downarrow\uparrow)$ is the antisymmetric singlet, and its norm squared is exactly $1$ (computed symbolically). The product is antisymmetric overall, as a fermion state must be. The determinant has enforced, with no extra input, that helium's ground state is a spin singlet — the same conclusion Singlet and Triplet States in Two-Electron Systems reaches from the two-electron register, here delivered automatically by the algebra.

Example 2 — the lithium $1s^2 2s$ determinant and its normalization. Show that lithium's three electrons form a valid nonzero $3\times3$ determinant and that the prefactor is $1/\sqrt{3!}=1/\sqrt6$ .

Step 1. Enumerate. Three distinct spin-orbitals are available for the ground configuration: $1s\uparrow$ , $1s\downarrow$ , $2s\uparrow$ . A fourth electron in $1s$ would demand a third $1s$ spin-orbital, which does not exist — the $1s$ shell is full at two, and the third electron is forced up to $2s$ . (That " $1s^3$ is forbidden" is the exclusion fact Pauli Exclusion Principle computes for lithium; here it is the statement that a $3\times3$ determinant with three $1s$ spin-orbitals would need a repeated column.)

Step 2. Build and normalize. The $3\times3$ determinant of these three distinct spin-orbitals is nonzero, and since the orbitals are orthonormal, the normalizing prefactor is $1/\sqrt{3!}=1/\sqrt6\approx0.4082$ — the general $1/\sqrt{N!}$ rule at $N=3$ .

Step 3. Boundary check. Swap any two of the three electrons and the whole determinant flips sign; make any two of the three spin-orbitals equal and it collapses to zero. The single expression carries all $3!=6$ signed permutation terms with correct signs, which is precisely the labour the determinant saves you.

Origins · 1 minJohn C. Slater and how this was discovered

John C. Slater

1900–1976 · Oak Park (Illinois), Cambridge (Massachusetts)

John Clarke Slater was an American physicist who shaped the quantum theory of atoms, molecules, and solids more than almost any of his contemporaries in the English-speaking world. In a short 1929 paper in Physical Review (34, 1293), "The Theory of Complex Spectra," he introduced the determinantal wavefunction that now bears his name, showing that antisymmetry and the Pauli principle for many electrons could be handled compactly by a determinant of one-electron spin-orbitals — and, in the same work, laying out the rules (the Slater–Condon rules) for computing matrix elements between such determinants. He went on to introduce Slater-type orbitals as convenient atomic basis functions and the empirical Slater screening rules for effective nuclear charge, both still taught today. Slater built and led the physics department at MIT for a quarter-century, turning it into a powerhouse of solid-state and computational physics, and later helped pioneer the augmented-plane-wave method for electronic band structure. His determinant is the single most-used object in quantum chemistry.

History

1925

— Pauli states the exclusion principle; antisymmetry of the many-electron wavefunction is recognized as its expression

1926

— Heisenberg and Dirac formulate the antisymmetric statistics of electrons

1929

— Slater introduces the determinantal wavefunction in "The Theory of Complex Spectra" (Phys. Rev. 34, 1293), with the Slater–Condon rules for matrix elements

1930

— Fock adds the exchange term to Hartree's self-consistent field, giving the Hartree–Fock method built on a single determinant

1930

— Slater publishes his empirical screening rules for effective nuclear charge

1951

— Slater's approximation to exchange (the

X\alpha

method) seeds later density-functional approaches

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §5.1 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §7.4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 10 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XIV · Szabo & Ostlund, Modern Quantum Chemistry, Ch. 2 · Slater, Phys. Rev. 34, 1293 (1929)

Provenance. All structural claims are standard across Griffiths §5.1, Sakurai §7.4, Shankar Ch. 10, and Cohen-Tannoudji Vol. II Ch. XIV; the base export's account is correct and was not altered. A verifier should: (1) confirm the determinant equals the signed permutation sum, $\det[\psi_i(x_j)]=\sum_P\mathrm{sgn}(P)\prod_i\psi_{P(i)}(x_i)$ , for $N=2,3,4$ (Leibniz's formula, the antisymmetrizer identity of Eq. 3); (2) verify that a repeated orbital gives two equal columns and hence a vanishing determinant; (3) check the $1/\sqrt{N!}$ normalization for orthonormal orbitals; (4) confirm the helium $1s^2$ determinant factorizes into a symmetric space part and the norm-1 antisymmetric spin singlet; (5) confirm the lithium $1s^2 2s$ prefactor $1/\sqrt6$ . Every number in this article is recomputed in the committed verification script. The exchange-energy structure $J-K$ appearing in the Slater–Condon energy is a pointer here — its physics is owned by Exchange Forces and Exchange Energy — and the bosonic all-plus analogue, the permanent, is owned by Permanents and Symmetrized Boson States.

Builds on: Indistinguishability of Identical Particles · Permutation Symmetry and the Exchange Operator · Symmetric and Antisymmetric States · Bosons and Fermions · Pauli Exclusion Principle Leads to: Exchange Forces and Exchange Energy · The Helium Atom · Singlet and Triplet States in Two-Electron Systems · Permanents and Symmetrized Boson States

Why is the prefactor

1/\sqrt{N!}

and not

1/N!

Because normalization involves the square of the wavefunction. When you compute $\langle\Psi|\Psi\rangle$ from the permutation sum of Eq. (2), orthogonality of the orbitals kills every cross term and leaves exactly $N!$ diagonal terms, each equal to $1$ . To make that sum equal one, the amplitude must carry $1/\sqrt{N!}$ , so that $|1/\sqrt{N!}|^2\cdot N!=1$ . The $1/N!$ that appears in the antisymmetrizer $\hat A$ is a different normalization — that of a projector — which is why Eq. (3) needs the $\sqrt{N!}$ conversion factor between them.

Can a Slater determinant describe bosons?

No. A determinant is antisymmetric — it flips sign under particle exchange — which is exactly the fermion rule. Bosons need a totally symmetric combination, which is the permanent of the orbital matrix: the same signed sum but with every sign replaced by $+1$ . Because the permanent has no cancellations, multiple bosons can occupy the same orbital (a repeated column no longer gives zero), which is why bosons pile up. The permanent and its normalization are owned by Permanents and Symmetrized Boson States.

Does the single determinant capture electron correlation?

Partly. It captures the correlation that antisymmetry forces: two same-spin electrons are automatically kept apart (they cannot be at the same point, the Fermi hole), which is called exchange or Fermi correlation. What it misses is Coulomb correlation — the tendency of any two electrons, including opposite-spin ones, to dodge each other's instantaneous repulsion. Recovering that requires mixing in additional determinants (configuration interaction, coupled cluster), and the energy difference from the exact answer is called the correlation energy.

How do spin and space combine inside the determinant?

The entries are spin-orbitals $\psi(\mathbf r)\chi(m_s)$ , and the determinant is required to be antisymmetric in the combined label $x=(\mathbf r, m_s)$ . Two electrons may therefore share a spatial orbital only if their spins differ — that is why each spatial orbital holds at most two electrons, one spin-up and one spin-down. For two electrons the determinant famously factorizes into a spatial part times a spin part (symmetric space $\times$ singlet spin, or antisymmetric space $\times$ triplet spin); for three or more electrons no such clean factorization exists in general, and the determinant must be kept whole.

What is the relationship between a determinant and the exchange energy

J-K

When you compute the energy $\langle\Psi|\hat H|\Psi\rangle$ of a single determinant, the electron–electron repulsion produces two kinds of two-electron integrals: a direct (Coulomb) term $J$ and an exchange term $K$ that appears with a minus sign, $\tfrac12\sum_{ij}(J_{ij}-K_{ij})$ . The $-K$ comes directly from the antisymmetry — the cross terms in the squared determinant carry a relative minus sign. So the determinant produces the exchange energy automatically; the physics of that exchange energy — its sign, magnitude, and its role in Hund's rules and singlet–triplet splitting — is developed in Exchange Forces and Exchange Energy.

Why does swapping to the inverse permutation not change the sum in the derivation?

Because summing over all $P\in S_N$ is the same set as summing over all $P^{-1}$ — the map $P\mapsto P^{-1}$ is a bijection of the symmetric group onto itself. And a permutation and its inverse have the same signature, $(-1)^{P^{-1}}=(-1)^P$ , since they contain the same number of transpositions. So relabelling the sum by inverses leaves both the terms and their signs intact; it merely lets us rewrite $\prod_j\psi_j(x_{P(j)})$ as $\prod_i\psi_{P(i)}(x_i)$ , matching Leibniz's determinant formula exactly.

Is the true ground state ever a single determinant?

Yes — for a system of genuinely non-interacting fermions, a single determinant of the exact single-particle eigenstates is the exact ground state, because there is no Coulomb correlation to capture. This is why the determinant is the natural language for free-fermion problems (metals in the free-electron model, filled Landau levels, ideal Fermi gases). The moment interactions switch on, the exact state acquires admixtures of other determinants, and the single-determinant description becomes the Hartree–Fock approximation rather than the truth.