$\DeclareMathOperator\perm{perm}\DeclareMathOperator\diag{diag}$Using MacMahon's master theorem, the properties of complex gaussian integrals, and Cauchy's integral theorem one can show that the permanent of a matrix $A$ satisfies \begin{equation} \perm(A) = \frac{1}{\pi^N} \int_{\mathbb{R}^{2N}} d^N\mathbf{u} \, \exp\Bigl( -\sum_{i=1}^N \mathbf{u}_i^2 \Bigr) \prod_{i=1}^{N} \Bigl(\mathbf{u}_i\cdot\sum_{j=1}^NA_{ij} \mathbf{u}_j\Bigr), \end{equation} where $\mathbf{u}_i \equiv (u_{i1}, u_{i2})$ and $d^{N}\mathbf{u} \equiv \prod^{N}_{i =1}du_{i1} \wedge du_{i2}$.

My guess is that this result is not new, but I'm not familiar with the literature on such identities. Is this identity known? Alternatively, where would be a good place to check for similar such identities?

Proof of Identity

The truth of the identity could be inferred from noting that we need our gaussian to integrate a function where $\mathbf{u}_i$ for each $i$ appears exactly twice.

To prove it directly, we note that MacMahon's master theorem states that the permanent of a matrix $A$ is the $x_1x_2\dotsm x_N$ coefficient of the quantity \begin{equation} \frac{1}{\det(I - XA)}, \end{equation} where $I$ is the $N\times N$ identity matrix and $X \equiv \diag(x_1, x_2, \dotsc, x_N)$. Thus, we have \begin{equation} \perm(A) = \frac{1}{(2\pi i)^N} \oint \left[\prod_{i=1}^{N} \frac{dq_i}{q_i^2} \right] \frac{1}{\det(I- QA)}, \label{eq:perm_def} \end{equation} where $Q = \diag(q_1, q_2, \ldots, q_N)$, $q_i$ is a complex variable, and we are performing $N$ contour integrations in sequence. Defining a set of $N$ two-dimensional vectors $\{\mathbf{u}_i\}$ as $\textbf{u}_i \equiv (u_{i1}, u_{i2})$, we have \begin{equation} \frac{1}{\det(I-QA)} = \frac{1}{\pi^N} \int_{\mathbb{R}^{2N}} d^{N}\mathbf{u}\,\exp\Bigl(- \sum_{i, j=1}^N \mathbf{u}_i \cdot \left(\delta_{ij} - q_{i} A_{ij}\right) \mathbf{u}_{j}\Bigr). \label{eq:gauss_result} \end{equation} Using this determinant expression in MacMahon's theorem and performing the contour integrations yields the stated result.

  • $\begingroup$ How is it related to MacMahon master theorem? This identity looks pretty straightforward: if we expand the brackets, any term not contributing to the permanent is odd with respect to at least one of variables, thus it disappears after integration. $\endgroup$ Commented Nov 1, 2020 at 18:43
  • $\begingroup$ @Fedor I agree that checking the identity is straightforward, but personally I wouldn't have thought to write it down without the prior derivation. Perhaps, the straightforwardness is a reason it is not in the literature? Though with the well known connection between determinants and gaussian integrals, I would have expected to see some reciprocal connection between permanents and gaussians $\endgroup$ Commented Nov 1, 2020 at 19:45
  • $\begingroup$ Also @ Fedor, I'll write up the explicit connection to MacMahon's master theorem as an addendum to this question. $\endgroup$ Commented Nov 1, 2020 at 19:58

1 Answer 1


In case anyone from the future comes to this post:

I wasn't able to find relevant literature that answered my question so I wrote a preprint on it (https://arxiv.org/abs/2106.11861). It turns out that the gaussian above is more complicated than necessary. The main result of the paper is a generalization of the identity above:

Let $p_X: \Omega_X \to \mathbb{R}$ be a probability distribution defined over the domain $\Omega_X$ with zero mean and unit variance. Let $A$ be an $n\times n$ matrix with elements $a_{i, j}$. Then the permanent of $A$ is \begin{equation} \text{perm}(A) = \int_{\Omega^n_{X}} d^n\textbf{x}\, \prod_{i=1}^n p_{X}(x_i) \, x_{i} \sum_{j=1}^n a_{i, j} x_j, \label{eq:fund_thm} \end{equation} where $\Omega^n_X = \Omega_X \otimes \cdots \otimes \Omega_X$ is the $n$-factor product over the single-variable domain of integration. In condensed notation, we can write this result as the expectation value \begin{equation} \text{perm}(A) = \left\langle \prod_{i=1}^nx_{i} \sum_{j=1}^n a_{i, j} x_j \right\rangle_{x_i \sim p_X}, \end{equation} where the average is over $\{x_i\}$, a set of independent identically distributed random variables each of which is drawn from $p_X$.

When $p_X = N(0, 1)$, we get something like the equation in the question.


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