Permanents and Symmetrized Boson States

The Slater determinant packages the antisymmetry of $N$ fermions into one compact object; erase every minus sign and you get the permanent, the matching object for $N$ identical bosons. Same sum over all $N!$ ways of assigning particles to orbitals, but with all signs set to $+1$ . That one change flips the physics from exclusion to permission: where the determinant vanishes when two orbitals coincide, the permanent grows, so bosons may pile into a single state without limit. Bosons and Fermions teased this with the "permanent of ones" at $N=2,3$ ; this article owns the general machinery. Two things make the permanent more than a cosmetic edit of the determinant. First, the normalization is subtler: because bosons can share an orbital, the unit-length constant depends on the occupation numbers — we correct a normalization slip carried in the base corpus and prove the right constant $1/\sqrt{N!\,\prod_\alpha n_\alpha!}$ from scratch. Second, the permanent has no efficient shortcut. A determinant falls to Gaussian elimination in $\mathcal{O}(N^3)$ steps; the permanent is $\#\mathrm{P}$ -hard, and Ryser's best-known method is exponential. That asymmetry — signs cancel, plus signs do not — is what boson sampling turns into a claim of quantum computational advantage. We build the symmetric state, prove its normalization, exhibit $\mathrm{perm}(J_N)=N!$ as the bunching enhancement (generalizing the teaser), contrast it term-by-term with the vanishing determinant, and mark the handoffs: the exchange-correlation geometry to Symmetric and Antisymmetric States, the thermal occupation formulas to statistical mechanics, and the operator reformulation to Second Quantization.

Key takeaways

For $N$ identical bosons the total state must be completely symmetric under exchange; the compact object encoding this is the permanent of the amplitude matrix $M_{ij}=\psi_{\alpha_i}(x_j)$ — the determinant's sum over permutations with every sign set to $+1$ .
A permanent does not vanish when two orbitals coincide (as a determinant does); it grows, which is exactly why bosons may multiply occupy one mode while fermions cannot.
The owned structural result: $\mathrm{perm}(J_N)=N!$ for the all-ones matrix, whereas $\det(J_N)=0$ — the sign is the whole difference — verified exactly to $N=5$ .
The unit-normalized state is $\Psi=\frac{1}{\sqrt{N!\,\prod_\alpha n_\alpha!}}\,\mathrm{perm}\,[\psi_{\alpha_i}(x_j)]$ ; the occupation factors $\prod_\alpha n_\alpha!$ correct for permuting identical orbitals among themselves.
The base corpus mis-wrote this prefactor as $\sqrt{\prod_\alpha n_\alpha!/N!}$ , which fails to normalize for repeated orbitals (giving $\langle\Psi|\Psi\rangle=4$ for a doubly-occupied mode, not $1$ ); we correct it and prove the fix.
Permanents are the first-quantized face of bosonic Fock states, but computing them is $\#\mathrm{P}$ -hard — no determinant-like $\mathcal{O}(N^3)$ trick exists — which boson-sampling exploits.
Key idea: the symmetric $N$ -boson state is the all-plus-sign permanent of the single-particle amplitudes, $\Psi=\mathrm{perm}[\psi_{\alpha_i}(x_j)]/\sqrt{N!\prod_\alpha n_\alpha!}$ ; the missing minus signs turn Pauli exclusion into bosonic bunching, and turn a polynomial computation into an exponential one.

The core insight

The totally symmetric $N$ -boson wavefunction is the permanent of the amplitude matrix $M_{ij}=\psi_{\alpha_i}(x_j)$ — the Slater determinant with all its minus signs removed — normalized by $1/\sqrt{N!\,\prod_\alpha n_\alpha!}$ . Removing the signs converts exclusion (determinant vanishes for repeated orbitals) into permission (permanent grows), and converts an $\mathcal{O}(N^3)$ computation into a $\#\mathrm{P}$ -hard one.

Where this lives today

The permanent sits at the meeting point of three active fields. In quantum optics and computing, boson sampling has moved from proposal to hardware: photonic experiments send tens of identical photons through programmable interferometers and measure output distributions governed by permanents of large submatrices, in regimes argued to be beyond classical simulation. In ultracold-atom physics, the same symmetric construction is the wavefunction of a Bose-Einstein condensate in its all-in-one-orbital limit, and Hanbury Brown–Twiss correlation measurements with bosonic atoms directly image the bunching the plus signs encode. In many-body theory the permanent underlies any exact treatment of bosonic modes, though second quantization does the heavy lifting because explicit permanents do not scale. The object Cauchy wrote down beside the determinant, long the poorer cousin for lacking a fast algorithm, turns out to encode the hardest-to-fake signature of quantum indistinguishability.

The physical picture

Run the same algorithm that builds a Slater determinant — sum over every way of matching $N$ particles to $N$ orbital-slots — but strike out each permutation's $\pm1$ parity and weight all by $+1$ . That is the permanent. On the determinant side, the minus signs cancel the terms to zero whenever two orbitals are identical — that cancellation is Pauli exclusion. On the permanent side, those terms add. So bosons do not merely tolerate sharing a state; the amplitude to share is enhanced, growing factorially with how many pile in. The most compressible statement of the boson–fermion split lives in one matrix. Take $J_N$ , the $N\times N$ matrix of all ones — the structure of "every particle in the same orbital." Its determinant is zero: identical rows, the fermion state annihilates itself. Its permanent is $N!$ : every one of the $N!$ orderings survives and adds. One matrix, two answers, and the gap between them is the whole distinction between a Fermi sea and a Bose condensate.

Figure

The permanent–determinant split on one matrix: det collapses to zero, perm explodes as N!, and the computational cost splits with it

Which sign a species carries is fixed by its spin through the spin-statistics theorem — integer spin takes the symmetric $+1$ , half-integer the antisymmetric $-1$ — and that dictionary is owned by Bosons and Fermions and proved in Spin-Statistics Theorem. Here we take "boson $\Rightarrow$ symmetric $\Rightarrow$ permanent" as the working rule and follow its machinery.

From symmetrization to the permanent

Start from single-particle orbitals $\psi_{\alpha_1},\dots,\psi_{\alpha_N}$ (repeats allowed) and the bare product $\psi_{\alpha_1}(x_1)\cdots\psi_{\alpha_N}(x_N)$ . That product is not symmetric — relabelling $1\leftrightarrow2$ changes it. The symmetrizer restores exchange symmetry by summing over all $N!$ relabellings, and the compact name for that sum is the permanent.

\mathrm{perm}\,M=\sum_{\sigma\in S_N}\;\prod_{i=1}^{N}M_{i,\sigma(i)},\qquad \det M=\sum_{\sigma\in S_N}\mathrm{sgn}(\sigma)\prod_{i=1}^{N}M_{i,\sigma(i)} \qquad (1)

where: $M$ — the $N\times N$ amplitude matrix with $M_{ij}=\psi_{\alpha_i}(x_j)$ ; $S_N$ — the symmetric group of all $N!$ permutations; $\sigma$ — a permutation, $\sigma(i)$ its image of $i$ ; $\mathrm{sgn}(\sigma)=\pm1$ its parity. The physical read: the permanent and determinant differ by exactly one thing — the determinant's $\mathrm{sgn}(\sigma)$ weight, which the permanent replaces by $+1$ . Every other structure is shared. The removal of the sign is the removal of Pauli exclusion. Applied to bosons, the permanent gives the totally symmetric state, and its normalization is where care is needed. When several bosons share an orbital, permuting those particles among themselves leaves the term unchanged, so the naive count of $N!$ permutations over-counts by $\prod_\alpha n_\alpha!$ , where $n_\alpha$ is the number of particles in orbital $\alpha$ . The unit-normalized state is:

\Psi_{\{n_\alpha\}}(x_1,\dots,x_N)=\frac{1}{\sqrt{N!\,\prod_\alpha n_\alpha!}}\;\mathrm{perm}\,[\psi_{\alpha_i}(x_j)],\qquad N=\sum_\alpha n_\alpha \qquad (2)

where: $n_\alpha$ — occupation number of orbital $\alpha$ ; the prefactor $1/\sqrt{N!\,\prod_\alpha n_\alpha!}$ makes $\langle\Psi|\Psi\rangle=1$ for orthonormal orbitals. The physical read: for all-distinct orbitals $\prod_\alpha n_\alpha!=1$ and this reduces to $1/\sqrt{N!}$ — the same magnitude as the Slater determinant, only the signs differ. The extra $\prod_\alpha n_\alpha!$ appears only when the boson-specific option — multiple occupation — is used, which is precisely what fermions cannot do. (A slip in the base corpus wrote this prefactor upside down as $\sqrt{\prod_\alpha n_\alpha!/N!}$ ; it fails to normalize repeated-orbital states, and we correct it below.)

The general permanent and its normalization

The owned result of this article is the general permanent machinery: the value $\mathrm{perm}(J_N)=N!$ that generalizes the $N=2,3$ enhancement teased in Bosons and Fermions, and the normalization constant that its worked derivation flagged as "owned downstream."

Derivation AThe permanent of the all-ones matrix equals N!, while its determinant vanishes

Prove $\mathrm{perm}(J_N)=N!$ and $\det(J_N)=0$ for the $N\times N$ all-ones matrix $J_N$ , and read off why this single pair of facts is the boson–fermion split. The goal is to show that the only difference between the two objects — the sign weight $\mathrm{sgn}(\sigma)$ — is exactly what converts "every particle in the same orbital" from a forbidden fermion configuration into the maximally enhanced boson one. This is the general- $N$ statement whose $N=2,3$ instances Bosons and Fermions computed as the "permanent of ones" ( $2!=2$ , $3!=6$ ) and deferred here.

Step 1. Write both objects for $J_N$ , every entry $=1$ . In the sum over $\sigma\in S_N$ , each term $\prod_i (J_N)_{i,\sigma(i)}=\prod_i 1=1$ . So the permanent is $\sum_{\sigma\in S_N}1=|S_N|=N!$ : each of the $N!$ orderings contributes one, and with no sign to cancel them they simply add.

Step 2. The determinant weights the identical terms by $\mathrm{sgn}(\sigma)$ : $\det(J_N)=\sum_\sigma\mathrm{sgn}(\sigma)\cdot1$ . The number of even and odd permutations of $N\ge2$ objects is equal ( $N!/2$ each), so the signed sum is $N!/2-N!/2=0$ . Equivalently, $J_N$ has rank $1$ (every row identical), hence a determinant of zero for $N\ge2$ .

Step 3. Read the physics. The all-ones matrix is the amplitude structure of "all $N$ particles in one orbital." For fermions the determinant kills it — that is Pauli exclusion in its cleanest form. For bosons the permanent returns $N!$ , the maximal bunching enhancement: the amplitude to find all $N$ bosons together is boosted by $\sqrt{N!}$ over distinguishable particles, so the probability is boosted by $N!$ . This is stimulated emission in embryo — a boson is more likely to join an already-occupied mode, by exactly this factor.

Step 4. Machine confirmation. Compute $\mathrm{perm}(J_N)$ and $\det(J_N)$ symbolically: $\mathrm{perm}(J_N)=2,6,24,120$ for $N=2,3,4,5$ (exactly $N!$ ) while $\det(J_N)=0$ throughout. The generic $2\times2$ makes the sign structure explicit: $\mathrm{perm}\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad+bc$ against $\det=ad-bc$ . Why it matters: removing the sign weight removes all cancellation. Every place a determinant vanishes by antisymmetry, the permanent survives — the algebraic root of the entire bosonic phenomenology. Verified: `verify_aq-qm-permanents-symmetrized-boson-states.py` — $\mathrm{perm}(J_N)=N!$ and $\det(J_N)=0$ exact to $N=5$ ; $\mathrm{perm}(2\times2)=ad+bc$ vs $\det=ad-bc$ .

With the enhancement in hand, the remaining owned piece is the normalization — the constant Bosons and Fermions named (" $1/\sqrt{N!\prod_k n_k!}$ ; that machinery is owned downstream") but did not derive. Deriving it correctly is where the base corpus slipped.

Go deeper · Derivation B · 1 minThe unit normalization of the symmetric boson state, and the base-corpus sign of the prefactor corrected

Derivation BThe unit normalization of the symmetric boson state, and the base-corpus sign of the prefactor corrected

Derive the constant $\mathcal{N}$ that makes $\Psi=\mathcal{N}\,\mathrm{perm}[\psi_{\alpha_i}(x_j)]$ a unit-norm state for orthonormal orbitals, and show that the base-corpus form $\sqrt{\prod_\alpha n_\alpha!/N!}$ is inverted — it over-normalizes any state with a multiply-occupied orbital. The goal is to pin the exact occupation dependence and to make the correction a teaching moment: the factor $\prod_\alpha n_\alpha!$ multiplies inside the square root's denominator, not its numerator, and the difference is invisible for distinct orbitals but wrong by a large factor for shared ones.

Step 1. Let $\Phi=\mathrm{perm}[\psi_{\alpha_i}(x_j)]=\sum_{\sigma\in S_N}\prod_i\psi_{\alpha_i}(x_{\sigma(i)})$ be the unnormalized symmetrizer. Its norm is $\langle\Phi|\Phi\rangle=\sum_{\sigma,\tau}\prod_i\langle\psi_{\alpha_i}(x_{\sigma(i)})|\psi_{\alpha_i}(x_{\tau(i)})\rangle$ . With orthonormal orbitals, only permutation pairs $(\sigma,\tau)$ that assign the same orbital to each coordinate survive.

Step 2. Count the survivors. For each of the $N!$ choices of $\sigma$ , the permutations $\tau$ that reproduce the same coordinate-to-orbital assignment are exactly those that permute the identical orbitals among themselves — there are $\prod_\alpha n_\alpha!$ of them. Hence $\langle\Phi|\Phi\rangle=N!\,\prod_\alpha n_\alpha!$ .

Step 3. Normalize: $\mathcal{N}=1/\sqrt{\langle\Phi|\Phi\rangle}=1/\sqrt{N!\,\prod_\alpha n_\alpha!}$ , giving Eq. (2). For all-distinct orbitals $\prod_\alpha n_\alpha!=1$ and $\mathcal{N}=1/\sqrt{N!}$ , matching the Slater magnitude; for all-in-one ( $n_\alpha=N$ ) $\mathcal{N}=1/\sqrt{N!\,N!}=1/N!$ , and since $\mathrm{perm}$ then supplies a factor $N!$ times the bare product, $\Psi$ collapses to the plainly-normalized product $\prod_j\psi_\alpha(x_j)$ .

Step 4. The correction. The base wrote $\mathcal{N}_{\text{base}}=\sqrt{\prod_\alpha n_\alpha!/N!}$ . Test it on a doubly-occupied mode ( $n_a=2$ , $N=2$ ): $\langle\Phi|\Phi\rangle=2!\cdot2!=4$ , so $\langle\Psi|\Psi\rangle=\mathcal{N}_{\text{base}}^2\cdot4=(2!/2!)\cdot4=4\ne1$ . On three-in-one it gives $36$ , not $1$ . The correct $\mathcal{N}=1/\sqrt{N!\prod n_\alpha!}$ returns exactly $1$ in every case. The two forms agree only when all orbitals are distinct — which is why the error hides until a boson actually shares a state, the one thing bosons are for. Why the factor is $\prod_\alpha n_\alpha!$ and not its reciprocal: permuting the $n_\alpha$ particles in orbital $\alpha$ leaves the permanent's term unchanged, so those permutations inflate $\langle\Phi|\Phi\rangle$ ; the normalization divides them back out, placing $\prod_\alpha n_\alpha!$ under the root. Verified: `verify_aq-qm-permanents-symmetrized-boson-states.py` — $\langle\Phi|\Phi\rangle=N!\prod n_\alpha!$ (sympy, five configurations); correct prefactor $\Rightarrow\langle\Psi|\Psi\rangle=1$ always; base prefactor $\Rightarrow4$ ( $n_a=2$ ), $36$ ( $n_a=3$ ), $=1$ only for distinct orbitals.

Figure

Normalization by occupation: distinct orbitals give 1/√N! but a shared mode needs the extra √(∏nₐ!), which the wrong prefactor omits

Bosons versus fermions in one table

The permanent and determinant are twin constructions; reading them side by side is the fastest way to see which physics each enforces.

Feature	Determinant (fermions)	Permanent (bosons)
Sign weight in the sum	$\mathrm{sgn}(\sigma)=\pm1$	all $+1$
Two equal orbitals	vanishes (Pauli exclusion)	grows (multiple occupation allowed)
$J_N$ (all-ones matrix)	$\det=0$	$\mathrm{perm}=N!$
Normalization (orthonormal)	$1/\sqrt{N!}$	$1/\sqrt{N!\,\prod_\alpha n_\alpha!}$
Exchange correlation	repulsion (Fermi hole)	attraction (bunching)
Computational cost	$\mathcal{O}(N^3)$ (Gaussian elimination)	$\#\mathrm{P}$ -hard (Ryser $\mathcal{O}(2^N N)$ )
Second-quantized image	$(a_\alpha^\dagger)^2=0$	$(a_\alpha^\dagger)^{n_\alpha}$ , no cap

The last row previews Second Quantization: the entire content of the permanent — the all-plus symmetry and the $\sqrt{n_\alpha!}$ factors — reappears automatically from the bosonic operators $[a_\alpha,a_\beta^\dagger]=\delta_{\alpha\beta}$ , which is why large- $N$ calculations abandon explicit permanents for operator algebra. The commutators are owned by Creation and Annihilation Operators.

Limiting cases

Case	Result	Meaning
All orbitals distinct ( $n_\alpha=1$ )	$\Psi=\mathrm{perm}[\psi_{\alpha_i}(x_j)]/\sqrt{N!}$	same magnitude as a Slater determinant — only the signs differ
All particles in one orbital ( $n_\alpha=N$ )	$\Psi=\prod_{j}\psi_\alpha(x_j)$	the bare symmetric product; impossible for fermions (would vanish) — this is the condensate limit
Two particles, two orbitals	$\Psi=\tfrac1{\sqrt2}[\psi_a(x_1)\psi_b(x_2)+\psi_a(x_2)\psi_b(x_1)]$	the familiar symmetric two-boson state, the $+$ partner of the antisymmetric fermion pair
Large $N$ , mixed occupation	$N!$ terms, $\#\mathrm{P}$ -hard	no polynomial shortcut; pass to Fock space and operators — the basis of boson sampling
Two dimensions	permanent $\to$ anyonic braiding phases	the $\pm$ dichotomy is a 3D fact; Anyons and Fractional Statistics owns the 2D generalization

Common trapThe normalization occupation factor divides — and the permanent is not a determinant with the signs "fixed later"

Two traps sit here. First, the occupation correction $\prod_\alpha n_\alpha!$ belongs under the square root in the denominator: $\mathcal{N}=1/\sqrt{N!\,\prod_\alpha n_\alpha!}$ . Writing it as $\sqrt{\prod_\alpha n_\alpha!/N!}$ (numerator) — a form carried in the base corpus — leaves multiply-occupied states over-normalized: a doubly-occupied mode then has $\langle\Psi|\Psi\rangle=4$ instead of $1$ , and a triply-occupied one gives $36$ . The two agree only when every orbital is distinct, so the error hides until a boson actually shares a state. Second, do not treat the permanent as "a determinant whose signs you patch up afterward." There is no Gaussian-elimination-style reduction for a permanent; its $\#\mathrm{P}$ -hardness is not a failure to find the trick but a proof (modulo standard complexity assumptions) that no efficient trick exists. The missing minus signs are exactly what forbids the fast algorithm.

Worked examples

Example 1 — normalize three bosons with a shared orbital. Two bosons occupy orbital $a$ and one occupies orbital $b$ (orthonormal). Find the normalization constant and compare it to both the all-distinct and all-shared cases.

Step 1. Enumerate the occupation. Here $N=3$ , occupations $n_a=2,\,n_b=1$ , so $\prod_\alpha n_\alpha!=2!\cdot1!=2$ .

Step 2. Apply the constant. By Eq. (2), $\mathcal{N}=1/\sqrt{N!\,\prod_\alpha n_\alpha!}=1/\sqrt{3!\cdot2}=1/\sqrt{12}\approx0.2887$ . Contrast: three distinct orbitals give $\prod n_\alpha!=1$ , so $\mathcal{N}=1/\sqrt{6}\approx0.4082$ ; all three in one orbital give $\prod n_\alpha!=3!=6$ , so $\mathcal{N}=1/\sqrt{6\cdot6}=1/6\approx0.1667$ .

Step 3. Boundary check. The constant shrinks as particles crowd into fewer orbitals ( $0.4082\to0.2887\to0.1667$ ), compensating the permanent's growing value so the state stays unit-length. The wrong base prefactor on the $n_a=2$ case returns $\langle\Psi|\Psi\rangle=4$ — a state four times too long. The occupation factor is what keeps probability equal to one when bosons share.

Example 2 — the bunching enhancement across $N$ . Compute the amplitude enhancement for $N$ identical bosons converging on a single mode, for $N=2$ through $6$ , and connect it to stimulated emission.

Step 1. Use the owned result. The enhancement is $\mathrm{perm}(J_N)=N!$ : the joint-detection amplitude is boosted by $\sqrt{N!}$ over distinguishable particles, the probability by $N!$ . For $N=2,3,4,5,6$ this is $2,6,24,120,720$ .

Step 2. Read the growth. The factor is not linear but factorial — a sixfold probability boost at $N=3$ , but $720$ -fold at $N=6$ . Bosons and Fermions computed the $N=2$ ( $2!=2$ ) and $N=3$ ( $3!=6$ ) entries as the "permanent of ones" and deferred the general law here; this is that law.

Step 3. Physical tie. In stimulated emission the rate to add a photon to a mode holding $n$ carries the factor $n+1$ — the incremental face of this permanent-of-ones bunching. Building a mode from empty to $N$ multiplies these to $1\cdot2\cdots N=N!=\mathrm{perm}(J_N)$ . The permanent of an all-ones matrix is the gain of a laser.

Origins · 1 minpermanent and how this was discovered

permanent

coined 1812 · named ~1882 · linear algebra → quantum optics

The permanent was introduced by Augustin-Louis Cauchy in 1812, in the same memoir that systematized the determinant: the identical expansion over all $N!$ permutations, with the alternating sign removed. The name was fixed by Thomas Muir around 1882, because the object is permanent — invariant — under the row and column permutations that flip a determinant's sign. For most of a century it was a curiosity with no efficient evaluation, in contrast to the determinant's rapid Gaussian elimination. Its physical destiny arrived with quantum statistics: the totally symmetric boson wavefunction is exactly a permanent of single-particle amplitudes, so bosonic interference is permanent arithmetic. Leslie Valiant proved in 1979 that computing the permanent is $\#\mathrm{P}$ -complete — as hard as counting solutions of any $\mathrm{NP}$ problem — and in 2011 Scott Aaronson and Alex Arkhipov turned that hardness into "boson sampling," a proposal for quantum advantage using identical photons through a linear-optical network whose output amplitudes are permanents of submatrices of the unitary. The nineteenth-century curiosity became a twenty-first-century benchmark for quantum machines.

History

1812

— Cauchy introduces both the determinant and the permanent in a single memoir on symmetric functions

1882

— Muir's theory of determinants fixes the name "permanent" (invariant under row/column permutations)

1926

— Dirac and Fermi formulate the symmetric (bosonic) and antisymmetric (fermionic) statistics

1927

— Dirac's radiation-field quantization recasts the symmetric states via creation operators

1963

— Marcus and Minc survey the theory of permanents and its open computational questions

1979

— Valiant proves computing the permanent is

\#\mathrm{P}

-complete

2011

— Aaronson and Arkhipov propose boson sampling, making photonic permanents a quantum-advantage benchmark

Going further

References. Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XIV (Identical Particles) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 7 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 10 · Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., Ch. 5 · Aaronson & Arkhipov, "The Computational Complexity of Linear Optics" (2011) · Valiant, "The complexity of computing the permanent," Theor. Comput. Sci. 8, 189 (1979).

Provenance. The permanent/determinant contrast and the symmetric-state construction are standard across Cohen-Tannoudji Vol. II Ch. XIV, Sakurai Ch. 7, and Shankar Ch. 10. A verifier should: (1) confirm $\mathrm{perm}(J_N)=N!$ and $\det(J_N)=0$ ; (2) re-derive $\mathcal{N}=1/\sqrt{N!\prod_\alpha n_\alpha!}$ via $\langle\Phi|\Phi\rangle=N!\prod n_\alpha!$ and check the all-distinct and all-same limits; (3) verify $\#\mathrm{P}$ -hardness against Valiant (1979) and boson sampling against Aaronson & Arkhipov (2011). Base-error note (this rebuild): the AllQFT base wrote the prefactor as $\sqrt{\prod_\alpha n_\alpha!/N!}$ , which is inverted — it fails to normalize any multiply-occupied state ( $\langle\Psi|\Psi\rangle=4$ for a doubly-occupied mode, $36$ for a triply-occupied one). The corrected $1/\sqrt{N!\prod_\alpha n_\alpha!}$ agrees with the served canon of Bosons and Fermions and Cohen-Tannoudji; the proof is in the verifier. The $\langle(x_1-x_2)^2\rangle$ exchange-correlation geometry the base sketched here is owned by Symmetric and Antisymmetric States (recalled by name); the Bose-Einstein/Fermi-Dirac distributions are owned by statistical mechanics. Every value — the permanents and the constants $1/\sqrt6$ , $1/\sqrt{12}$ , $1/6$ — is recomputed in `verify_aq-qm-permanents-symmetrized-boson-states.py`.

Builds on: Indistinguishability of Identical Particles · Permutation Symmetry and the Exchange Operator · Symmetric and Antisymmetric States · Bosons and Fermions Leads to: Slater Determinants · Exchange Forces and Exchange Energy · Second Quantization · Spin-Statistics Theorem

Why does the normalization depend on occupation numbers for bosons but not for fermions?

Because only bosons can multiply occupy an orbital. For fermions every occupation is $0$ or $1$ , so $\prod_\alpha n_\alpha!=1$ always and the normalization is uniformly $1/\sqrt{N!}$ . For bosons, when $n_\alpha$ particles share orbital $\alpha$ , permuting those $n_\alpha$ particles leaves the permanent's term unchanged, inflating $\langle\Phi|\Phi\rangle$ by $\prod_\alpha n_\alpha!$ . The normalization divides that back out, so the constant is $1/\sqrt{N!\prod_\alpha n_\alpha!}$ . The occupation dependence is the mathematical fingerprint of the option — multiple occupation — that distinguishes bosons from fermions.

Is the permanent really just the determinant with the minus signs removed?

Algebraically, yes: both are sums over the same $N!$ permutations of the same products $\prod_i M_{i,\sigma(i)}$ ; the determinant weights each by $\mathrm{sgn}(\sigma)=\pm1$ , the permanent by $+1$ . But that single change has enormous consequences. Physically it flips exclusion into permission. Computationally it flips a polynomial-time object into a $\#\mathrm{P}$ -hard one, because the sign structure is exactly what makes Gaussian elimination work — you can add a multiple of one row to another without changing a determinant, but this operation does change a permanent. So "just remove the signs" is true as a definition and profoundly misleading as an intuition about difficulty.

What does the permanent have to do with photon bunching and stimulated emission?

The amplitude for $N$ identical bosons to occupy one mode is the permanent of the all-ones matrix, $\mathrm{perm}(J_N)=N!$ , so the joint-detection probability is enhanced by $N!$ over distinguishable particles. The incremental version of this — the rate to add one more photon to a mode already holding $n$ — carries the factor $n+1$ , which is stimulated emission. Building a mode from empty to $N$ multiplies these increments to $1\cdot2\cdots N=N!=\mathrm{perm}(J_N)$ . The Hanbury Brown–Twiss experiment measures the two-photon face of the same enhancement as a coincidence peak. So permanent arithmetic is the quantitative content of "light likes to travel with light."

Why do large-N boson calculations use second quantization instead of permanents?

Because a permanent of an $N\times N$ matrix has $N!$ terms and no efficient evaluation, so explicit symmetrized wavefunctions become intractable well before $N$ is large. Second quantization repackages the identical information — the all-plus symmetry and the $\sqrt{n_\alpha!}$ factors — into bosonic operators satisfying $[a_\alpha,a_\beta^\dagger]=\delta_{\alpha\beta}$ . A Fock state $\prod_\alpha (a_\alpha^\dagger)^{n_\alpha}/\sqrt{n_\alpha!}\,|0\rangle$ carries the same normalization automatically. Second Quantization and Occupation-Number Representation and Fock Space develop this; the operator algebra is owned by Creation and Annihilation Operators.

If the permanent can't be computed efficiently, how does boson sampling "compute" it?

It doesn't compute the permanent — that is the point. A boson-sampling device samples from the output distribution of identical photons through a linear-optical network, and each output probability equals $|\mathrm{perm}(U_{\text{sub}})|^2$ for a submatrix of the network's unitary. The device produces samples drawn from those permanent-valued probabilities without ever evaluating a permanent. The claim of quantum advantage is that no efficient classical algorithm can reproduce the same samples, since doing so would require estimating permanents. The physics performs the sampling; the hardness is what makes it meaningful.

Does the permanent picture survive in two dimensions, like anyons?

No — the strict $\pm1$ dichotomy between permanents (bosons) and determinants (fermions) is a three-dimensional fact, resting on the exchange operator squaring to the identity so its eigenvalues are $\pm1$ . In two dimensions the topology of exchange is richer: worldlines braid, and the state can pick up an arbitrary phase $e^{i\theta}$ or a non-abelian factor. The permanent and determinant are then the $\theta=0$ and $\theta=\pi$ endpoints of a continuum owned by Anyons and Fractional Statistics; the permanent here is the boson endpoint.