Permanents

A permanent is the bosonic analogue of a determinant: it sums over all permutations of orbitals assigned to particle slots, but with no alternating signs. For an $N\times N$ matrix $M$,

$$\operatorname{perm} M = \sum_{\pi\in S_N} \prod_{i=1}^{N} M_{i,\pi(i)}.$$

For bosonic wavefunctions, the matrix entries are one-particle orbitals evaluated at particle slots. Permanents build completely symmetric many-boson wavefunctions in the same way that Slater determinants build completely antisymmetric many-fermion wavefunctions.

Permanents are conceptually important because they make bosonic exchange symmetry explicit. They are less convenient than occupation-number notation for most calculations, because writing all $N!$ permutation terms quickly becomes impractical.

Permanent Definition

Compare the determinant and permanent of an $N\times N$ matrix $M$:

$$\det M = \sum_{\pi\in S_N} \operatorname{sgn}(\pi) \prod_{i=1}^{N}M_{i,\pi(i)},$$

while

$$\operatorname{perm} M = \sum_{\pi\in S_N} \prod_{i=1}^{N}M_{i,\pi(i)}.$$

The only formal difference is the factor $\operatorname{sgn}(\pi)$. That difference is physically decisive.

For fermions, the signs make the wavefunction change sign when two particle slots are exchanged. For bosons, all signs are positive, so the wavefunction remains unchanged when particle slots are exchanged.

For a $2\times2$ matrix,

$$M= \begin{pmatrix} a & b\\ c & d \end{pmatrix},$$

one has

$$\det M=ad-bc, \qquad \operatorname{perm}M=ad+bc.$$

The permanent is not a determinant with a harmless sign convention changed. It has different algebraic properties, different normalization consequences, and very different computational behavior.

Two-Boson Permanent

Let $\varphi_a(q)$ and $\varphi_b(q)$ be one-particle wavefunctions, where $q$ denotes all one-particle labels. Form the matrix

$$\begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) \end{pmatrix}.$$

Its permanent is

$$\varphi_a(q_1)\varphi_b(q_2) + \varphi_b(q_1)\varphi_a(q_2).$$

If the two one-particle states are distinct and orthonormal, the normalized two-boson wavefunction is

$$\Psi_B(q_1,q_2) = \frac{1}{\sqrt2} \operatorname{perm} \begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) \end{pmatrix}.$$

Equivalently,

$$\Psi_B(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ \varphi_a(q_1)\varphi_b(q_2) + \varphi_b(q_1)\varphi_a(q_2) \bigr].$$

Exchanging $q_1$ and $q_2$ swaps the two rows of the matrix. The permanent is unchanged by a row swap, so the wavefunction is symmetric.

Same-Mode Case

If both bosons occupy the same normalized mode $\varphi_a$, the permanent matrix has two identical columns:

$$\begin{pmatrix} \varphi_a(q_1) & \varphi_a(q_1)\\ \varphi_a(q_2) & \varphi_a(q_2) \end{pmatrix}.$$

Its permanent is

$$2\,\varphi_a(q_1)\varphi_a(q_2).$$

The normalized two-boson same-mode wavefunction is not obtained by multiplying this permanent by $1/\sqrt2$. The correct state is

$$\Psi_B(q_1,q_2) = \varphi_a(q_1)\varphi_a(q_2).$$

The extra factor of $2$ in the permanent is canceled by the repeated-occupation normalization. This is the simplest place where bosonic permanents differ from the naive rule “sum over permutations and divide by $\sqrt{N!}$.”

N-Boson Wavefunctions with Distinct Orbitals

For $N$ bosons occupying $N$ distinct orthonormal one-particle orbitals $\varphi_1,\ldots,\varphi_N$, the normalized first-quantized wavefunction is

$$\Psi_B(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \operatorname{perm} \bigl[ \varphi_j(q_i) \bigr]_{i,j=1}^{N}.$$

Expanding the permanent gives

$$\Psi_B(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} \prod_{i=1}^{N} \varphi_{\pi(i)}(q_i).$$

Each term assigns the same set of occupied orbitals to the $N$ particle slots in a different order. The sum over all assignments removes any observable meaning from slot labels.

If two rows are exchanged, the terms in the permanent are merely reordered. Hence

$$\Psi_B(\ldots,q_s,\ldots,q_r,\ldots) = \Psi_B(\ldots,q_r,\ldots,q_s,\ldots).$$

This is exactly the bosonic exchange rule.

Repeated Occupations

Bosons may occupy the same one-particle mode more than once. Suppose an orthonormal mode basis contains modes labeled by $\mu$, and the occupation numbers are

$$n_\mu=0,1,2,\ldots, \qquad N=\sum_\mu n_\mu.$$

Choose a list

$$a_1,\ldots,a_N$$

in which each mode label $\mu$ appears $n_\mu$ times. Then the normalized bosonic wavefunction is

$$\Psi_{\{n_\mu\}}(q_1,\ldots,q_N) = \frac{ \operatorname{perm} \bigl[ \varphi_{a_j}(q_i) \bigr]_{i,j=1}^{N} }{ \sqrt{ N!\prod_\mu n_\mu! } }.$$

The product of factorials is the repeated-occupation correction. It appears because permuting identical columns does not produce a new distinct assignment.

For example, with two bosons in mode $a$ and one in mode $b$, the formula reduces to

$$\begin{aligned} \Psi_{2_a,1_b} &= \frac{1}{\sqrt3} \bigl[ \varphi_b(q_1)\varphi_a(q_2)\varphi_a(q_3)\\ &\qquad + \varphi_a(q_1)\varphi_b(q_2)\varphi_a(q_3)\\ &\qquad + \varphi_a(q_1)\varphi_a(q_2)\varphi_b(q_3) \bigr]. \end{aligned}$$

The coefficient $1/\sqrt3$ reflects the three distinct places where the single $b$ occupation can appear among the three slots.

For $N$ bosons all in the same mode $a$,

$$n_a=N,$$

and the normalized wavefunction becomes

$$\Psi(q_1,\ldots,q_N) = \prod_{i=1}^{N}\varphi_a(q_i).$$

This state is symmetric, normalized, and nonzero.

Contrast with Slater Determinants

The determinant and permanent constructions are parallel, but their consequences are almost opposite.

For fermions, a determinant changes sign under a row exchange:

$$\det(\text{row-exchanged matrix}) = - \det(\text{original matrix}).$$

For bosons, a permanent is unchanged:

$$\operatorname{perm}(\text{row-exchanged matrix}) = \operatorname{perm}(\text{original matrix}).$$

If two columns of a Slater determinant are identical, the determinant vanishes. That is the determinant form of Pauli exclusion. If two columns of a permanent are identical, the permanent usually does not vanish. Instead, it counts repeated bosonic assignments with positive multiplicity.

Thus:

• determinants encode antisymmetry and no repeated fermionic occupation;
• permanents encode symmetry and allow repeated bosonic occupation;
• determinants are usually compact and algebraically tractable;
• permanents become cumbersome quickly and are usually replaced by occupation-number notation.

Occupation-Number Notation Is Usually Better

The permanent formula is valuable because it displays exchange symmetry directly. But it is rarely the most efficient representation for many-boson calculations.

The occupation vector

$$\lvert n_1,n_2,\ldots\rangle_B$$

already says how many bosons occupy each mode. It avoids listing all slot permutations. In creation-operator notation,

$$\lvert n_1,n_2,\ldots\rangle_B = \prod_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle.$$

The factorials here are the same repeated-occupation normalization factors that appear in the permanent wavefunction. The operator algebra keeps the symmetrization implicit.

For this reason, explicit permanents are mostly used when the first-quantized wavefunction matters, when comparing bosons and fermions, or when studying linear-optical transition amplitudes. General many-boson theory is usually written in Fock space.

Boson Sampling Preview

Permanents also appear in noninteracting bosonic interference. In a passive linear-optical network, a single-particle unitary matrix $U$ maps input modes to output modes. If identical bosons enter specified input modes and are detected in specified output modes, the transition amplitude is proportional to a permanent of a submatrix of $U$.

For input occupations $\{s_i\}$ and output occupations $\{t_j\}$ with the same total particle number $N$, one standard form is

$$\langle \mathbf t\vert \widehat U\vert\mathbf s\rangle = \frac{ \operatorname{perm} U_{\mathbf t,\mathbf s} }{ \sqrt{ \prod_i s_i! \prod_j t_j! } }.$$

Here $U_{\mathbf t,\mathbf s}$ is the $N\times N$ matrix obtained by repeating input columns according to the input occupations and repeating output rows according to the output occupations.

This formula is the mathematical core of boson sampling: many-boson interference amplitudes involve matrix permanents, and permanents are computationally difficult in general. This page only uses boson sampling as a preview of why permanents matter beyond notation. The detailed computational-complexity and experimental story belongs elsewhere.

Nonorthogonal Orbitals

The clean occupation formula above assumes an orthonormal one-particle mode basis. If the one-particle orbitals are not orthonormal, the normalization is no longer controlled only by repeated occupation numbers.

Let

$$S_{ij} = \langle\varphi_i\vert\varphi_j\rangle$$

be the overlap matrix for the orbitals used in a symmetrized product. The norm of the bosonic symmetrized state is controlled by a permanent of this overlap matrix. This is the bosonic counterpart of the determinant overlap formula for fermionic Slater determinants.

In most many-body applications, one chooses an orthonormal mode basis precisely so that occupation-number states are orthonormal and these overlap permanents do not appear in every calculation.

Common Mistakes

• Using the determinant formula but changing the minus signs without checking normalization for repeated bosonic occupations.
• Dividing every permanent wavefunction by $\sqrt{N!}$ even when two or more occupied orbitals are the same.
• Thinking repeated columns make a bosonic permanent vanish; that is a determinant property, not a permanent property.
• Treating slot labels in a permanent as physical particle identities.
• Assuming that bosonic symmetry means all bosons must occupy the same mode.
• Treating boson sampling as a statement about all bosonic systems rather than a special linear-optical interference problem.
• Forgetting that occupation-number notation is basis-dependent even though it avoids explicit slot labels.

Cross-Links

• Bosons
• Exchange Operators
• Symmetric and Antisymmetric Wavefunctions
• Slater Determinants
• Symmetrization Postulate
• Occupation-Number Basis
• Bosonic Fock Space
• Number States
• Creation and Annihilation Operators
• Bosonic Commutation Relations
• Identical Particle Exercises
• Formula Sheet

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Wiley, 1977.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• S. Aaronson and A. Arkhipov, “The Computational Complexity of Linear Optics,” Theory of Computing 9, 143-252, 2013.
• S. Scheel, “Permanents in linear optical networks,” arXiv:quant-ph/0406127, 2004.

Exercises

1. Two-boson permanent. Compute the permanent of

$$\begin{pmatrix} \varphi_a(q_1) & \varphi_b(q_1)\\ \varphi_a(q_2) & \varphi_b(q_2) \end{pmatrix}$$

and show that it gives a symmetric two-boson wavefunction after normalization for distinct orthonormal modes.

Solution

The permanent is

$$\varphi_a(q_1)\varphi_b(q_2) + \varphi_b(q_1)\varphi_a(q_2).$$

For distinct orthonormal modes, the normalized state is

$$\frac{1}{\sqrt2} \bigl[ \varphi_a(q_1)\varphi_b(q_2) + \varphi_b(q_1)\varphi_a(q_2) \bigr].$$

Exchanging $q_1$ and $q_2$ interchanges the two terms, so the state is unchanged.

2. Same-mode normalization. Explain why

$$\frac{1}{\sqrt2} \operatorname{perm} \begin{pmatrix} \varphi_a(q_1) & \varphi_a(q_1)\\ \varphi_a(q_2) & \varphi_a(q_2) \end{pmatrix}$$

is not normalized when $\varphi_a$ is normalized.

Solution

The permanent is

$$2\varphi_a(q_1)\varphi_a(q_2).$$

Multiplying by $1/\sqrt2$ gives

$$\sqrt2\,\varphi_a(q_1)\varphi_a(q_2),$$

whose norm is $\sqrt2$, not $1$. The normalized same-mode state is

$$\varphi_a(q_1)\varphi_a(q_2).$$

3. Row exchange. Show that exchanging two rows of a permanent leaves it unchanged.

Solution

The permanent sums over all ways of choosing one entry from each row and each column:

$$\operatorname{perm} M = \sum_{\pi\in S_N} \prod_i M_{i,\pi(i)}.$$

Exchanging two rows only relabels which row index appears in the product. Since all permutations are still included and there is no sign factor, the sum is the same. For determinants, the same row exchange would introduce a minus sign.

4. Identical columns. Compare a determinant and a permanent when two columns are identical.

Solution

A determinant with two identical columns is zero because swapping those columns changes the determinant by a minus sign but leaves the matrix unchanged. Therefore the determinant must equal its negative.

A permanent has no sign change under a column swap. Identical columns do not force it to vanish. For bosons, identical columns describe repeated occupation of the same one-particle mode and require factorial normalization rather than exclusion.

5. Occupation formula. Starting from the general repeated-occupation formula, derive the normalized wavefunction for two bosons in mode $a$ and one in mode $b$.

Solution

Here $N=3$, $n_a=2$, and $n_b=1$. The prefactor is

$$\frac{1}{\sqrt{3!\,2!\,1!}} = \frac{1}{\sqrt{12}}.$$

The permanent sum has each distinct placement of the single $b$ mode repeated twice, because the two $a$ columns are identical. Thus the coefficient of each distinct term is

$$\frac{2}{\sqrt{12}} = \frac{1}{\sqrt3}.$$

Therefore

$$\begin{aligned} \Psi_{2_a,1_b} &= \frac{1}{\sqrt3} \bigl[ \varphi_b(q_1)\varphi_a(q_2)\varphi_a(q_3)\\ &\qquad + \varphi_a(q_1)\varphi_b(q_2)\varphi_a(q_3)\\ &\qquad + \varphi_a(q_1)\varphi_a(q_2)\varphi_b(q_3) \bigr]. \end{aligned}$$

6. Linear-optical preview. Why do permanents, rather than determinants, appear in ideal bosonic linear-optical transition amplitudes?

Solution

Each indistinguishable boson can take one of several single-particle paths through the network, and the total amplitude sums over assignments of input particles to output detections. Because the particles are bosons, exchanging two particle slots does not introduce a minus sign. The sum over assignments is therefore a permanent of the relevant single-particle transition submatrix. Fermions would instead produce determinants because odd exchanges carry a minus sign.