Integrability of $\Phi ^4$ matrix model as N-body harmonic oscillator system

Abstract

We study a Hermitian matrix model with a kinetic term given by $Tr (H \Phi ^2 )$, where H is a positive definite Hermitian matrix, similar as in the Kontsevich Matrix model, but with its potential $\Phi ^3$ replaced by $\Phi ^4$. We show that its partition function solves an integrable Schrödinger-type equation for a non-interacting N-body Harmonic oscillator system.

1 Introduction

Let $\Phi$ be a Hermitian $N\times N$ matrix, E be a positive diagonal $N\times N$ matrix $E := diag (E_1, E_2 , \ldots ,E_N )$ without degenerate eigenvalues, and $\eta$ be a positive real number as a coupling constant. We deal in this paper with the following one Hermitian matrix model defined using this E:

$$\begin{aligned} S'&= N~ Tr \{ E \Phi ^2 + \frac{\eta }{4} \Phi ^4 \} \nonumber \\&= N \left( \sum _{i,j}^N E_{i}\Phi _{ij}\Phi _{ji} + \frac{\eta }{4} \sum _{i,j,k,l}^N \Phi _{ij}\Phi _{jk}\Phi _{kl}\Phi _{li} \right) . \end{aligned}$$

(1.1)

This matrix model is obtained by changing the potential of the Kontsevich model [15] from $\Phi ^3$ to $\Phi ^4$. It was introduced while studying a scalar field defined on a deformed four-dimensional space-time and studied over years [7, 8]. An additional oscillator term was added in order to resolve the IR-UV mixing problem. This term leads to an external matrix E with equally spaced eigenvalues. Recent developments are summarized in Branahl et al. [2].

The main theorem of this paper is expressed as follows.

Theorem 1.1

Let $Z(E, \eta )$ be the partition function defined by

$$\begin{aligned} Z(E, \eta )= \int d \Phi ~e^{-S'} . \end{aligned}$$

Let $\Delta (E)$ be the Vandermonde determinant $\Delta (E) := \prod _{k<l} (E_l -E_k)$. Then the function

$$\begin{aligned} \Psi (E, \eta ) := e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E) Z(E, \eta ) \end{aligned}$$

(1.2)

is a zero-energy solution of a Schrödinger-type differential equation being 2-nd order in each of its variables,

$$\begin{aligned} {\mathcal {H}}_{HO} \Psi (E, \eta ) = 0, \end{aligned}$$

where $\mathcal {H}_{HO}$ is the Hamiltonian of the N-body harmonic oscillator without interaction:

$$\begin{aligned} \mathcal {H}_{HO}:= - \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 + \frac{N}{\eta }\sum _{i=1}^N (E_i)^2 . \end{aligned}$$

In this sense, this matrix model is a solvable system.

2 Schwinger–Dyson equation

Let $\Phi$ be a Hermitian $N\times N$ matrix. Let H be a positive Hermitian $N\times N$ matrix with nondegenerate eigenvalues $\{E_1, E_2 , \ldots ,E_N ~ | ~ E_i \ne E_j ~\text{ for }~ i \ne j \}$. $\eta$ is a real positive number. We consider the following action

$$\begin{aligned} S&= N~ Tr \{ H \Phi ^2 + \frac{\eta }{4} \Phi ^4 \} \nonumber \\&= N \left( \sum _{i,j,k}^N H_{ij}\Phi _{jk}\Phi _{ki} + \frac{\eta }{4} \sum _{i,j,k,l}^N \Phi _{ij}\Phi _{jk}\Phi _{kl}\Phi _{li} \right) . \end{aligned}$$

(2.1)

The partition function is defined by

$$\begin{aligned} Z(E, \eta ) := \int _{h_N} d \Phi ~e^{-S} , \end{aligned}$$

(2.2)

and we denote the expectation value with this action S by $\displaystyle \langle O \rangle := \int \nolimits _{h_N} d \Phi ~ O e^{-S}$. Note that we do not normalize it here, i.e., $\langle 1 \rangle = Z(E, \eta ) \ne 1$. Here, the integral measure is the ordinary Haar measure. Using the real variables defined by $\Phi _{ij}= \Phi _{ij}^{Re} + i \Phi _{ij}^{Im}$, the measure is given as $\displaystyle \int \nolimits _{h_N} d \Phi := \prod \nolimits _{i}^N \int _{-\infty }^{\infty } d\Phi _{ii} \prod \nolimits _{k<l} \int _{-\infty }^{\infty } d\Phi _{kl}^{Re} \int _{-\infty }^{\infty } d\Phi _{kl}^{Im}$. Note that the partition function $Z(E, \eta )$ depends only on the eigenvalues of H because the integral measure is U(N) invariant. Indeed $Z(E, \eta )$ is equal to the partition function obtained from the action defined by $S'$ in (1.1).

In the following, we use the notation:

$$\begin{aligned} \frac{\partial }{\partial \Phi _{ij}}= \frac{1}{2} \left( \frac{\partial }{\partial \Phi _{ij}^{Re}} - i \frac{\partial }{\partial \Phi _{ij}^{Im}} \right) \quad \text{ for } \ (i \ne j ) . \end{aligned}$$

(2.3)

For the diagonal elements $\Phi _{ii} (i= 1,2, \ldots ,N)$, the corresponding partial derivatives are the usual ones. The Schwinger–Dyson equation is derived from

$$\begin{aligned} \int _{h_N} \frac{\partial }{\partial \Phi _{ij}} \left( \Phi _{ij} e^{-S} \right) = 0, \end{aligned}$$

(2.4)

which is expressed as

$$\begin{aligned} Z(E, \eta ) - N \sum _k (\langle H_{ki}\Phi _{ij}\Phi _{jk} \rangle +\langle H_{jk}\Phi _{ki}\Phi _{ij} \rangle ) -N\eta \sum _{k,l} \langle \Phi _{jk}\Phi _{kl}\Phi _{li}\Phi _{ij} \rangle =0 . \end{aligned}$$

(2.5)

Taking sum over the indices i, j and using

$$\begin{aligned} \frac{\partial Z(E, \eta ) }{\partial H_{ij}} = -N \sum _k \langle \Phi _{jk}\Phi _{ki} \rangle , \quad \frac{\partial ^2 Z(E, \eta ) }{\partial H_{ij}\partial H_{mn}} = N^2 \sum _{k,l} \langle \Phi _{jk}\Phi _{ki} \Phi _{nl}\Phi _{lm} \rangle , \end{aligned}$$

(2.6)

a partial differential equation is obtained:

$$\begin{aligned} \mathcal {L}_{SD}^H Z(E, \eta ) = 0 . \end{aligned}$$

(2.7)

Here, $\mathcal {L}_{SD}^H$ is a second-order differential operator defined by

$$\begin{aligned} \mathcal {L}_{SD}^H:= N^2 + 2 \sum _{i,k} H_{ki} \frac{\partial }{\partial H_{ki}} -\frac{\eta }{N} \sum _{i,k} \left( \frac{\partial }{\partial H_{ki}}\frac{\partial }{\partial H_{ik}} \right) . \end{aligned}$$

(2.8)

Next we rewrite this Schwinger–Dyson equation by using eigenvalues of H, i.e., $E_n (n= 1,2, \ldots , N)$. References [13, 14] are helpful in the following calculations. Let P(x) be the characteristic polynomial:

$$\begin{aligned}P(x): = \det (x~ Id_N - H) = \prod _{i=1}^N (x-E_i).\end{aligned}$$

Using this P(x),

$$\begin{aligned} \frac{\partial E_j}{\partial H_{ki}} = \frac{(-1)^{k+i} | E_j~ Id_N - H |_{ki} }{P'(E_j)} \end{aligned}$$

(2.9)

is obtained. Here, $| M |_{kj}$ denote the minors of the matrix M defined by the determinant of the smaller matrix obtained by removing the k-th row and j-th column from M. Using the formula (2.9),

$$\begin{aligned} \sum _{i,j} H_{ij} \frac{\partial Z(E, \eta )}{\partial H_{ij}} =&\sum _{i,j,k} H_{ij} \frac{\partial E_k}{\partial H_{ij}} \frac{\partial Z(E, \eta )}{\partial E_{k}} \nonumber \\ =&\sum _{i,j,k} H_{ij} \frac{(-1)^{i+j} | E_k~ Id_N - H |_{ij} }{P'(E_k)} \frac{\partial Z(E, \eta )}{\partial E_{k}} \nonumber \\ =&-\sum _{i,j,k} (\delta _{ij}E_k - H_{ij} ) \frac{(-1)^{i+j} | E_k~ Id_N - H |_{ij} }{P'(E_k)} \frac{\partial Z(E, \eta )}{\partial E_{k}} \nonumber \\&+ \sum _{i,j,k} \delta _{ij}E_k \frac{(-1)^{i+j} | E_k~ Id_N - H |_{ij} }{P'(E_k)} \frac{\partial Z(E, \eta )}{\partial E_{k}}. \end{aligned}$$

(2.10)

The first term in the last line is equal to $\displaystyle -\sum _{k,i} \frac{P(E_k)}{P'(E_k)} \frac{\partial Z(E, \eta )}{\partial E_{k}} = 0$, because P(x) is the characteristic polynomial and $E_k$ is one of eigenvalues of H. $\sum _{i,j} \delta _{ij} (-1)^{i+j} | E_k~ Id_N - H |_{ij}$ in the second term is $P'(E_k)$. Then we find

$$\begin{aligned} \sum _{i,j} H_{ij} \frac{\partial Z(E, \eta )}{\partial H_{ij}} = \sum _k E_k \frac{\partial Z(E, \eta )}{\partial E_{k}} . \end{aligned}$$

(2.11)

Next step, we rewrite the Laplacian $\displaystyle \sum \nolimits _{i,k} \left( \frac{\partial }{\partial H_{ki}}\frac{\partial }{\partial H_{ik}} \right) Z(E, \eta )$ by $E_k$. It is a well-known fact that by using the Vandermonde determinant $\Delta (E) := \prod _{k<l} (E_l -E_k)$ the Jacobian for the change of variables is obtained as follows:

$$\begin{aligned} d H := \prod _{i}^N dH_{ii} \prod _{k<l} dH_{kl}^{Re} dH_{kl}^{Im} = \Delta ^2 (E) \left( \prod _{i}^N dE_{i}\right) \left( \prod _{k<l} \left( U^{-1}dU\right) _{kl}^{Re} \left( U^{-1}dU\right) _{kl}^{Im}\right) . \end{aligned}$$

Then, the Laplacian is rewritten as

$$\begin{aligned} \sum _{i,k} \left( \frac{\partial }{\partial H_{ki}}\frac{\partial }{\partial H_{ik}} \right) ~Z(E, \eta )= & {} \frac{1}{\Delta ^2 (E)} \sum _i^N \frac{\partial }{\partial E_i} \left( \Delta ^2 (E) \frac{\partial }{\partial E_i} \right) ~ Z(E, \eta ) \nonumber \\= & {} \left\{ \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 +\sum _{i \ne j} \frac{1}{E_i - E_j} \left( \frac{\partial }{\partial E_i} - \frac{\partial }{\partial E_j} \right) \right\} ~ Z(E, \eta ) . \nonumber \\ \end{aligned}$$

(2.12)

Here, $\displaystyle \sum \nolimits _{i \ne j}$ means $\displaystyle \sum \nolimits _{i,j=1 , i \ne j}^N$. From (2.7), (2.11), and (2.12), we obtain the following.

Theorem 2.1

The partition function defined by (2.2) satisfies

$$\begin{aligned} \mathcal {L}_{SD} Z(E, \eta ) = 0 , \end{aligned}$$

(2.13)

where

$$\begin{aligned} \mathcal {L}_{SD} := \left\{ \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 + \frac{\eta }{N} \sum _{i \ne j} \frac{1}{E_i - E_j} \left( \frac{\partial }{\partial E_i} - \frac{\partial }{\partial E_j} \right) -2 \sum _k E_k \frac{\partial }{\partial E_{k}} -N^2 \right\} ~. \end{aligned}$$

(2.14)

In Appendix A, this Schwinger–Dyson equation is checked by using perturbative calculations, as a cross-check.

3 Diagonalization of $\mathcal {L}_{SD}$

In this section, we prove the main theorem (Theorem 1.1).

As the first step, we prove the following proposition.

Proposition 3.1

The differential operator $\mathcal {L}_{SD}$ defined in (2.14) is transformed into the Hamiltonian of the N-body harmonic oscillator as

$$\begin{aligned} e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E) \mathcal {L}_{SD} \Delta ^{-1}(E) e^{\frac{N}{2\eta } \sum _i E_i^2} = \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 - \frac{N}{\eta }\sum _{i=1}^N (E_i)^2 . \end{aligned}$$

(3.1)

We denote this Hamiltonian by $\mathcal {H}_{HO}$:

$$\begin{aligned} -\mathcal {H}_{HO}:= \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 - \frac{N}{\eta }\sum _{i=1}^N (E_i)^2 . \end{aligned}$$

(3.2)

Proof

The proof is done by direct calculations.

We calculate $\Delta ^{-1}(E) e^{\frac{N}{2\eta } \sum _i E_i^2} \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E)$ at first.

$$\begin{aligned}&\Delta ^{-1}(E) e^{\frac{N}{2\eta } \sum _i E_i^2} \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E) \nonumber \\ =&\frac{\eta }{N} \sum _{i=1}^N \Bigg \{ \sum _{j,k=1, j\ne i, k\ne i}^N \frac{1}{(E_i -E_j )(E_i -E_k)} -\sum _{j=1, j\ne i}^N \frac{1}{(E_i - E_j)^2} \Bigg \} \end{aligned}$$

(3.3)

$$\begin{aligned}&-\sum _{i=1}^N \Bigg \{ \Bigg (\sum _{j=1, j\ne i}^N \frac{ 2 E_i}{E_i -E_j} \Bigg ) + 1 \Bigg \} \end{aligned}$$

(3.4)

$$\begin{aligned}&+ \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_i} \right) ^2 + \frac{\eta }{N} \sum _{i \ne j} \frac{1}{E_i - E_j} \left( \frac{\partial }{\partial E_i} - \frac{\partial }{\partial E_j} \right) -2 \sum _{k=1}^N E_k \frac{\partial }{\partial E_{k}} + \frac{N}{\eta } \sum _{i=1}^N (E_i)^2. \end{aligned}$$

(3.5)

(3.3) is equal to 0 since

$$\begin{aligned}&\sum _{j\ne i, k\ne i, j\ne k} \frac{1}{(E_i -E_j )(E_i -E_k)} \\&\qquad = \frac{1}{3} \sum _{j\ne i, k\ne i, j\ne k} \Big ( \frac{1}{(E_i -E_j )(E_i -E_k)}+ \frac{1}{(E_j -E_i )(E_j -E_k)}\\&\qquad \quad \qquad \quad \qquad \quad \qquad \quad + \frac{1}{(E_k -E_i )(E_k -E_j)} \Big )=0. \end{aligned}$$

(3.4) is written as follows.

$$\begin{aligned} -\sum _{i=1}^N \left\{ \left( \sum _{j=1, j\ne i}^N \frac{ 2 E_i}{E_i -E_j} \right) + 1 \right\}&= - \sum _{i\ne j} \Big ( \frac{ E_i}{E_i -E_j} - \frac{ E_j}{E_i -E_j} \Big ) - N \nonumber \\&= - (N^2 -N )-N = -N^2 . \end{aligned}$$

(3.6)

Then, we obtain

$$\begin{aligned} \Delta ^{-1}(E) e^{\frac{N}{2\eta } \sum _i E_i^2} ~\mathcal {H}_{HO} ~e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E) = - \mathcal {L}_{SD} . \end{aligned}$$

(3.7)

$\square$

We introduce a transformed partition function $\Psi (E, \eta )$ by

$$\begin{aligned} \Psi (E, \eta ) := e^{-\frac{N}{2\eta } \sum _i E_i^2} \Delta (E) Z(E, \eta ) . \end{aligned}$$

(3.8)

Note that this transformation is invertible. Then, the following theorem follows from Proposition 3.1 immediately.

Theorem 3.2

The transformed partition function $\Psi (E, \eta )$ is a zero-energy solution of the Schrödinger-type differential equation:

$$\begin{aligned} \mathcal {H}_{HO} \Psi (E, \eta ) = 0. \end{aligned}$$

(3.9)

Here, $\mathcal {H}_{HO}$ is the N-body harmonic oscillator Hamiltonian (3.2).

This N-body harmonic oscillator system has no interaction terms between the oscillators, so it is a trivial quantum integrable system.

Theorem 1.1 is proved as above. In the next section, we calculate the solution $\Psi (E, \eta )$ more concretely and give another proof using it.

4 From partition function to zero-energy solution

A new expression of the zero-energy solution of the N-body harmonic oscillator system is constructed by a direct calculation of the partition function.

Let us carry out the integration of the off-diagonal components of $\Phi$ in the definition of the partition function after the change of variables to $U(N)\times {\mathbb R}^N$. We denote the eigenvalues of $\Phi$ by $x_1 , x_2 , \ldots , x_N$. By using a unitary matrix U, $\Phi$ is diagonalized as $X=U \Phi U^\dagger$, where $X= diag(x_1, x_2 , \ldots , x_N )$. Then,

$$\begin{aligned} Z(E, \eta )&= \int _{h_N} d \Phi ~e^{-S'} \nonumber \\&= \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i~ e^{-N V(x_i)}\right) \left( \prod _{l < k}(x_k -x_l)^2 \right) \int _{U(N)} dU e^{-N Tr U E U^{\dagger } X^2 }, \end{aligned}$$

(4.1)

where $V(x):= \frac{\eta }{4}x^4$.

Let us use the Harish–Chandra–Itzykson–Zuber integral [12, 19] for the unitary group U(N) :

$$\begin{aligned} \int _{U(N)}\exp \left( t\textrm{tr}\left( AUBU^{\dagger }\right) \right) dU=&\tilde{c}_{N}\frac{\displaystyle \det _{1\le i,j\le N}\left( \exp \left( t\lambda _{i}(A)\lambda _{j}(B)\right) \right) }{t^{\frac{(N^{2}-N)}{2}} \displaystyle \Delta (\lambda (A))\displaystyle \Delta (\lambda (B))}. \end{aligned}$$

(4.2)

Here, A and B are Hermitian matrices whose eigenvalues are denoted by $\lambda _{i}(A)$ and $\lambda _{i}(B)$ $(i=1,\ldots ,N)$, respectively. t is a nonzero complex parameter, $\displaystyle \Delta (\lambda (A)):=\prod \nolimits _{1\le i<j\le N}(\lambda _{j}(A)-\lambda _{i}(A))$ is the Vandermonde determinant, and $\displaystyle \tilde{c}_{N}:=\left( \prod \nolimits _{i=1}^{N-1}i!\right) \times \pi ^{\frac{N(N-1)}{2}}$. $\left( \exp \left( t\lambda _{i}(A)\lambda _{j}(B)\right) \right)$ is the $N\times N$ matrix with the i-th row and the j-th column being $\exp \left( t\lambda _{i}(A)\lambda _{j}(B)\right)$. After adapting this formula, the partition function is described by

$$\begin{aligned} Z(E, \eta )&= \frac{c_N}{\Delta (E)} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N V(x_i)}\right) \left( \prod _{l< k}\frac{x_k -x_l}{x_k +x_l} \right) \det _{1\le i,j\le N} \left( e^{-NE_i x_j^2 } \right) \nonumber \\&=\sum _{\sigma \in S_N} \frac{c_N}{\Delta (E)} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N V(x_i)}\right) \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) (-1)^\sigma \prod _{j=1}^N e^{-N E_j x_{\sigma (j)}^2} ,\nonumber \\ \end{aligned}$$

(4.3)

where $c_N = \tilde{c}_N (-1/ N)^{\frac{N^2-N}{2}}$ and $S_N$ denotes the symmetric group. This integral representation (4.3) should be regarded as a Cauchy principal value. Consider the change of variables $x_i \mapsto x_{\sigma ^{-1}(i)}$. Note that the sign of $\prod _{l < k}(x_k -x_l)$ changes as $(-1)^\sigma \prod _{l < k}(x_k -x_l)$ , and the following formula is obtained by this change of variables.

$$\begin{aligned} Z(E, \eta )&= \sum _{\sigma \in S_N} \frac{c_N}{\Delta (E)} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N V(x_i)}\right) \left( \prod _{l< k}\frac{x_k -x_l}{x_k +x_l} \right) \prod _{j=1}^N e^{-N E_j x_{j}^2} \nonumber \\&=\frac{N! c_N}{\Delta (E)} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N (V(x_i) + E_i x_i^2)} \right) \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) . \end{aligned}$$

(4.4)

Then, the zero-energy solution of (3.9) is obtained by (1.2).

Theorem 4.1

The function

$$\begin{aligned} \Psi (E, \eta ) =N! c_N \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N (\frac{\eta }{4} x_i^4 + E_i x_i^2 + \frac{1}{2\eta } E_i^2)} \right) \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \end{aligned}$$

(4.5)

satisfies the Schrödinger-type differential equation (3.9).

Since this fact follows from Theorem 3.2, there is no need to prove it, but it would be worthwhile to show the differential equation (3.9) directly from expression (4.5) as a confirmation.

At first, we prove the following Lemma:

Lemma 4.2

$$\begin{aligned} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i ~ e^{-N (\frac{\eta }{4} x_i^4 + E_i x_i^2 + \frac{1}{2\eta } E_i^2)} \right) \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \sum _{j=1}^N (\eta N x_j^4 +2N E_j x_j^2 -1 ) = 0 . \end{aligned}$$

(4.6)

Proof

For simplicity, $\displaystyle \sum \nolimits _i^N N \left( \frac{\eta }{4} x_i^4 + E_i x_i^2 + \frac{1}{2\eta } E_i^2 \right)$ will be abbreviated as f(x, E). From the following identity:

$$\begin{aligned}&\int _{{\mathbb R}^N} (\prod _{i=1}^N dx_i) ~ \frac{d}{dx_j} \Big \{ x_j e^{-f(x,E)} (\prod _{l < k}(x_k -x_l)^2) \int _{U(N)} dU e^{-N Tr U E U^{\dagger } X^2 } \Big \} =0, \end{aligned}$$

(4.7)

we obtain the formal expression:

$$\begin{aligned} \sum _{j=1}^N \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i \right) \frac{d}{d x_j}\left\{ x_j e^{-f(x,E)} \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \right\} =0. \end{aligned}$$

Then, we get

$$\begin{aligned} \int _{{\mathbb R}^N} \left( \prod _{i=1}^N dx_i \right)&e^{-f(x,E)} \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \nonumber \\&\times \left\{ \sum _{j=1}^N (\eta N x_j^4 +2N E_j x_j^2 -1 ) -\sum _{m \ne n} \frac{2x_m x_n}{(x_m -x_n)(x_m +x_n)} \right\} = 0. \end{aligned}$$

From the identity

$$\begin{aligned} \sum _{m \ne n} \frac{2x_m x_n}{(x_m -x_n)(x_m +x_n)} =\sum _{m \ne n} \Big ( \frac{x_m x_n}{(x_m -x_n)(x_m +x_n)} +\frac{x_n x_m}{(x_n -x_m)(x_n +x_m)} \Big ) =0 , \end{aligned}$$

(4.6) is obtained. $\square$

Using Lemma 4.2, let us prove Theorem 4.1 by direct calculations.

Proof

From (4.5),

$$\begin{aligned}&\frac{\eta }{N} \sum _i^N \left( \frac{\partial }{\partial E_i} \right) ^2 \Psi (E, \eta ) \nonumber \\&=N! c_N \int _{{\mathbb R}^N} \left( \prod _{j=1}^N dx_j \right) e^{-f(x,E)} \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \left\{ \sum _{i=1}^N (\eta N x_i^4 +2N E_i x_i^2 -1 + \frac{N}{\eta }E_i^2) \right\} \nonumber \\&= \frac{N}{\eta } \sum _i^N E_i^2 \Psi (E, \eta ) . \end{aligned}$$

(4.8)

For the last equality, we used Lemma 4.2. $\square$

$\Psi (E, \eta )$ can also be expressed using Pfaffian. It is described in Appendix B.

As described above, we have also directly proved that the function $\Psi (E, \eta )$ obtained from the partition function of the matrix model satisfies the Schrödinger-type differential equation for the N-body harmonic oscillator system without interactions.

5 Discussions and remarks for $N=1$

The matrix model studied in this paper is related to a renormalizable scalar $\Phi ^4$ theory on Moyal space [8] in the large N limit. There are mainly two approaches to study the question of integrability of this matrix model: One relies on the model, where one replaces the $\Phi ^4$ interaction by a constant times $\Phi ^3$. This gives the Kontsevich model, for which it is known, that the logarithm of the partition function is the $\tau$ function for the KdV hierarchy and fulfills a Hirota bilinear equation [10, 13, 15, 21]. Another approach follows topological recursion. While the Kontsevich model follows topological recursion, it turned out that the $\Phi ^4$ model follows the more sophisticated blobbed topological recursion, (proven for genus one and two) [2, 3, 11]. Due to these complications, it was unexpected, to obtain such a simple answer. This N-body harmonic oscillator system is known as an integrable system and this system has been studied for a long time. See, for example, [17, 18, 20] and references therein. Note that the solution required by the Schrödinger-type equation (3.9) is a zero-energy solution, which is different from the well-known harmonic oscillator solutions by using Hermite polynomials for nonzero-energy solutions. In particular, the case $N=1$ corresponds to what is called the Weber equation. In the following, we will consider the case $N=1$ as a particularly simplest case and see how $\Psi (E, \eta )$ corresponds to a solution to the Weber equation.

Introducing new variables $\displaystyle u_i := \sqrt{\frac{N}{\eta }}E_i$, the Schrödinger-type equation (3.9) is deformed into

$$\begin{aligned} \sum _{i=1}^N \left( \frac{\partial }{\partial u_i} \right) ^2 y(u) = \sum _{i=1}^N u_i^2 y(u) . \end{aligned}$$

(5.1)

So the $N=1$ case, this is a kind of the Weber equation [6]:

$$\begin{aligned} y'' (u) = u^2 y(u) . \end{aligned}$$

(5.2)

The series solution of this Weber equation is given as follows. For $y(u)= \sum _{n=0}^{\infty } a_n u^n$, (5.2) requires

$$\begin{aligned}{} & {} a_{4n+2}=a_{4n+3}=0, \nonumber \\{} & {} a_{4n} = \frac{1}{4n(4n-1)}\cdot \frac{1}{(4n-4)(4n-5)} \cdots \frac{1}{4\cdot 3} a_0 \nonumber \\{} & {} a_{4n+1} = \frac{1}{(4n+1)4n}\cdot \frac{1}{(4n-3)(4n-4)} \cdots \frac{1}{5\cdot 4} a_1 . \end{aligned}$$

(5.3)

So the boundary conditions are given by $y(0)=a_0$ and $y'(0) =a_1$. For $N=1$ case, the partition function is

$$\begin{aligned} Z(E, \eta ):=&\int _{-\infty }^{\infty } dx ~e^{-Ex^2 - \frac{\eta }{4}x^4} =\int _{-\infty }^{\infty } dx ~e^{-\sqrt{\eta } u x^2 - \frac{\eta }{4} x^4}=: Z(u,\eta ), \end{aligned}$$

(5.4)

and using this $Z(u,\eta ),$ (3.8) implies that the solution of (5.2) is given as

$$\begin{aligned} \Psi (u) := e^{-\frac{E^2}{2\eta }} Z(u,\eta ) = e^{-\frac{u^2}{2}} \int _{-\infty }^{\infty } dx~ e^{-\sqrt{\eta } u x^2 - \frac{\eta }{4} x^4} . \end{aligned}$$

(5.5)

Indeed, we can prove that $\Psi (u)$ satisfies (5.2) as follows. As similar to the proof for Lemma 4.2,

$$\begin{aligned}0= \int _{-\infty }^{\infty } dx \frac{d}{dx}( x e^{-f(u)}) = \int _{-\infty }^{\infty } dx (1- 2\sqrt{\eta } u x^2 - \eta x^4 ) e^{-f(u)},\end{aligned}$$

where $-f(u)= -\sqrt{\eta } u x^2 - \frac{\eta }{4} x^4 -\frac{u^2}{2}$. Using this formula, (5.2) is derived:

$$\begin{aligned} \left( \frac{d}{d u} \right) ^2 \Psi (u) = \int _{-\infty }^{\infty } dx (u^2 + 2\sqrt{\eta } u x^2 + \eta x^4 -1 ) e^{-f(u)} = u^2 \Psi (u) . \end{aligned}$$

Furthermore, for $u>0$ , by using modified Bessel function of the second kind $K_{\frac{1}{4}}(u)$, $\Psi (u)$ can also be written as

$$\begin{aligned} \Psi (u) = \frac{1}{\eta ^{\frac{1}{4}}} \sqrt{u} K_{\frac{1}{4}} \big ( \frac{u^2}{2} \big ). \end{aligned}$$

We find that the boundary condition for this solution is required as

$$\begin{aligned} a_0&= \Psi (0) =Z(0,\eta ) = \int _{-\infty }^{\infty } dx e^{- \frac{\eta }{4} x^4} = \frac{\Gamma \left( \frac{1}{4}\right) }{\sqrt{2}\eta ^{\frac{1}{4}} } ,\\ a_1&= \Psi ' (0) =-\sqrt{\eta } \int _{-\infty }^{\infty } dx~ x^2 e^{- \frac{\eta }{4} x^4} =- \frac{\sqrt{2} \Gamma \left( \frac{3}{4}\right) }{\eta ^{\frac{1}{4}} }. \end{aligned}$$

Thus, in the case of $N=1$, the results are derived using known special functions.

Additional comments

During the peer-review process of this paper, a follow-up paper [9] was submitted by the authors, Kanomata and Wulkenhaar, which was published first. Paper [9] corresponds to the application of the technology of this paper. For the benefit of the readers of this journal, we comment on it below.

In this paper, we have seen that the Schwinger–Dyson equation satisfied by the partition function of the Hermitian matrix model (1.1) derives the Schrödinger equation for the Hamiltonian of N-body harmonic oscillator system. The N-body harmonic oscillator system can be extended to the integrable Calogero–Moser model [5, 16]. It is thus natural to think that there should be matrix models whose partition functions satisfy the Schrödinger equation for the Calogero–Moser model. The calculations with Hermitian matrices in this paper are performed by replacing them with real symmetric matrices and showing that the Schwinger–Dyson equation satisfied by the partition function corresponds to the Schrödinger equation of the Calogero–Moser model in Grosse et al. [9]. Furthermore, since the Calogero–Moser model admits a Virasoro algebra representation, it gives rise to a family of differential equations satisfied by the partition function ${Z}(E,\eta )$ in Grosse et al. [9]. It is also possible to construct Virasoro algebra representations for N-body harmonic oscillator systems. Therefore, a similar family of differential equations satisfied by the partition function ${Z}(E,\eta )$ can also be constructed for the Hermitian matrix model of this paper.

Next, we also comment on the generalization of this model. As mentioned above, the matrix model of the Hermitian matrix and the matrix model of the real symmetric matrix correspond to the harmonic oscillator system and the Calogero–Moser model, respectively. As a similar extension, it is natural to extend the matrix $\Phi$ from a Hermitian to a quaternion self-dual matrix without changing the form of the action (1.1). In this case, it is still expected to correspond to the Schrödinger equation in the Calogero–Moser model.

A more non-trivial generalization is to consider higher-order potentials like $\Phi ^6 , \Phi ^8 ,$ etc. In the case of this paper, the $\Phi ^4$ interaction term leads to the Laplacian in the Schrödinger equation, as seen from its derivation process. Equations in (2.6) and the subsequent calculations show this. By similar considerations, it can be expected that higher-order differential terms will appear, as the order of interactions increases. In such cases, it remains a challenging problem for the future what kind of theory can be developed. We intend to return to such models and to treat also different observables beyond partition functions.

Appendices

Appendix: Perturbative check for Schwinger–Dyson equation

In Appendix A, we check that the partition function given in Sect. 4 satisfies the Schwinger–Dyson equation (2.13) by using perturbation theory.

For simplicity, we use the action (1.1) in the following perturbative calculations. We use the fact that the theory itself is equivalent even if the action is changed from (2.1) to (1.1). Then, the partition function is written as

$$\begin{aligned} Z(E, \eta )= \int _{h_N} d \Phi ~e^{-S'} . \end{aligned}$$

Note that

$$\begin{aligned} \frac{\eta }{N} \sum _{i=1}^N \left( \frac{\partial }{\partial E_{i}} \right) ^2 Z(E, \eta )&= \eta N \sum _{i,j,k}^N \langle \Phi _{ij}\Phi _{ji} \Phi _{ik}\Phi _{ki} \rangle , \end{aligned}$$

(A.1)

$$\begin{aligned} \frac{\eta }{N} \sum _{i\ne j} \frac{1}{E_i -E_j}\left( \frac{\partial }{\partial E_{i}} -\frac{\partial }{\partial E_{j}} \right) Z(E, \eta )&= -\eta \sum _{i\ne j} \frac{1}{E_i -E_j} \sum _{k=1}^N \left( \langle \Phi _{ik}\Phi _{ki} \rangle - \langle \Phi _{jk}\Phi _{kj} \rangle \right) , \end{aligned}$$

(A.2)

$$\begin{aligned} -2 \sum _k E_k \frac{\partial }{\partial E_{k}} Z(E, \eta )&= 2N \sum _{k,j} E_k \langle \Phi _{kj}\Phi _{jk} \rangle . \end{aligned}$$

(A.3)

The Schwinger–Dyson equation (2.13) is expressed as

$$\begin{aligned} (A.1)+(A.2)+(A.3) = N^2 Z(E, \eta ) . \end{aligned}$$

(A.4)

So, we verify this equation (A.4) by perturbation expansion up to first order in $\eta$. We calculate the three expectation values (A.1), (A.2), and (A.3) by using Feynman rules.

We shall now summarize the Feynman rules. We denote the free model action by

$$\begin{aligned} S_f&= N~ Tr E \Phi ^2 = N \left( \sum _{i,j}^N E_{i}\Phi _{ij}\Phi _{ji} \right) . \end{aligned}$$

(A.5)

Introducing an $N\times N$ Hermitian matrix J as an external field, the partition function for the free theory is defined as

$$\begin{aligned} Z_f [J] := \int _{h_N} d\Phi ~ e^{-N Tr E \Phi ^2 +Tr J\Phi } . \end{aligned}$$

(A.6)

After Gaussian integral, this is expressed as

$$\begin{aligned} Z_f [J] := Z_f e^{\frac{1}{2N} \sum _{i,j} J_{ij} J_{ji} \frac{1}{E_i + E_j}}, \end{aligned}$$

(A.7)

where

$$\begin{aligned} Z_f := c \frac{1}{\prod _i (E_i)^{\frac{1}{2}} \cdot \prod _{l>k}(E_l +E_k)} , \quad \left( c= (\sqrt{2\pi /N})^{N^2} \right) . \end{aligned}$$

(A.8)

We denote the vacuum expectation value of the free action $S_f$ by $\langle O \rangle _f := \int _{h_N} d\Phi ~ O e^{-S_f}$. Note that we do not normalize this vacuum expectation value, i.e., $\langle 1 \rangle _f = Z_f$. From (A.7), the propagator is given by

(A.9)

the interaction is given as

(A.10)

and each loop corresponds to a sum $\sum _n$. The Feynman diagrams contributing to the 2-point function $\langle \Phi _{ij}\Phi _{kl} \rangle$ to zeroth and first order in $\eta$ are shown in Fig. 1. If the terms are written in the same order as each graph in Fig. 1, they contribute as follows.

$$\begin{aligned}&\langle \Phi _{ij}\Phi _{kl} \rangle = Z_f\Big \{ \frac{1}{N}\frac{\delta _{kj} \delta _{il}}{E_i + E_j} \nonumber \\&+\frac{1}{N^3}\frac{\delta _{kj} \delta _{il}}{(E_i + E_j)^2} \sum _n \frac{1}{E_i + E_n} \frac{-\eta N}{4}\times 4 +\frac{1}{N^3}\frac{\delta _{kj} \delta _{il}}{(E_i + E_j)^2} \sum _n \frac{1}{E_j + E_n} \frac{-\eta N}{4}\times 4 \nonumber \\&+\frac{1}{N^3}\frac{\delta _{ij} \delta _{kl}}{2E_i 2E_k(E_i + E_k)} \frac{-\eta N}{4}\times 4 +\frac{1}{N}\frac{\delta _{kj} \delta _{il}}{E_i + E_j} \times B \Big \} +O(\eta ^2), \end{aligned}$$

(A.11)

where B represents the terms from bubble graphs in Fig. 1:

$$\begin{aligned} B := -\frac{1}{N}\eta \Big ( \frac{1}{2}\sum _{m,n,s} \frac{1}{(E_m + E_n)(E_m + E_s)} +\frac{1}{16} \sum _m \frac{1}{E_m^2} \Big ). \end{aligned}$$

(A.12)

(See also (A.19).) This equation can be rearranged as follows.

$$\begin{aligned} \langle \Phi _{ij}\Phi _{kl} \rangle =&\frac{Z_f}{N}\frac{\delta _{kj} \delta _{il}}{E_i + E_j}\Big \{ (1+B) - \frac{\eta }{N}\frac{1}{(E_i + E_j)} \sum _n \left( \frac{1}{E_i + E_n} + \frac{1}{E_j + E_n} \right) \Big \} \nonumber \\&+\frac{Z_f}{N^2} \frac{ -\eta \delta _{ij} \delta _{kl}}{4E_i E_k(E_i + E_k)} +O(\eta ^2) . \end{aligned}$$

(A.13)

Fig. 1

Feynman diagrams for 2-point function

Now that we are ready to calculate contributuions by using Feynman rules, we shall immediately carry out the calculations of the three expectation values (A.1), (A.2), and (A.3). At first, we estimate (A.1). By using Wick contraction, we get

$$\begin{aligned}&\eta N \sum _{i,j,k}^N \langle \Phi _{ij}\Phi _{ji} \Phi _{ik}\Phi _{ki} \rangle \nonumber \\&= \eta \frac{N}{Z_f} \sum _{i,j,k}^N \{ \langle \Phi _{ij}\Phi _{ji}\rangle \langle \Phi _{ik}\Phi _{ki}\rangle + \langle \Phi _{ij}\Phi _{ik}\rangle \langle \Phi _{ji}\Phi _{ki}\rangle +\langle \Phi _{ij}\Phi _{ki}\rangle \langle \Phi _{ji}\Phi _{ik}\rangle \}. \end{aligned}$$

(A.14)

Substituting (A.13) for (A.14), finally we obtain

$$\begin{aligned} \eta N \sum _{i,j,k}^N \langle \Phi _{ij}\Phi _{ji} \Phi _{ik}\Phi _{ki} \rangle&= \frac{\eta Z_f }{N} \Bigg \{ \sum _{i,j,k}^N \frac{1}{ (E_i + E_j)(E_i + E_k)} \nonumber \\&\quad + \sum _{i=1}^N \frac{1}{4 E_i^2 } + \sum _{i, j}^N \frac{1}{ (E_i +E_j)^2 } \Bigg \} +O(\eta ^2 ) . \end{aligned}$$

(A.15)

Next, after substituting (A.13), (A.2) is given as

$$\begin{aligned}&-\eta \sum _{i\ne j} \frac{1}{E_i -E_j} \sum _{k=1}^N \left( \langle \Phi _{ik}\Phi _{ki} \rangle - \langle \Phi _{jk}\Phi _{kj} \rangle \right) \nonumber \\&\quad = \frac{\eta Z_f }{N} \sum _{i \ne j} \sum _{k=1}^N \frac{1}{ (E_i + E_k)(E_j + E_k)} +O(\eta ^2 ) . \end{aligned}$$

(A.16)

Similarly, (A.3) is given as

$$\begin{aligned} 2N \sum _{k,j} E_k \langle \Phi _{kj}\Phi _{jk} \rangle&= 2Z_f \sum _{k, j} \frac{E_k}{ E_k + E_j} \Bigg \{ 1+B - \frac{\eta }{N} \frac{1}{ E_k + E_j}\nonumber \\&\qquad \sum _{n=1}^N \Big ( \frac{1}{ E_k + E_n} + \frac{1}{ E_j + E_n} \Big ) \Bigg \} \nonumber \\&\qquad - \frac{\eta Z_f}{4 N} \sum _{k=1}^N \frac{1}{ E_k^2} +O(\eta ^2 ) . \end{aligned}$$

(A.17)

For the right hand side of (A.4), next we calculate

$$\begin{aligned} N^2 Z(E, \eta ) = N^2 \sum _{k=0}^{\infty } \frac{1}{k!} \Big ( -\frac{ N \eta }{4} \Big )^k \langle (Tr \Phi ^4 )^k \rangle _f . \end{aligned}$$

(A.18)

$\eta ^0$-term is $N^2 \langle 1 \rangle _f = N^2 Z_f$. $\eta ^1$-term is given by $-\frac{ N^3 \eta }{4} \langle Tr \Phi ^4 \rangle _f$. This is the same calculation for B:

$$\begin{aligned}&-\frac{ N^3 \eta }{4} \langle Tr \Phi ^4 \rangle _f \nonumber \\&= -\frac{ N^3 \eta }{4 Z_f} \sum _{i,j,k,l}^N \Bigg \{ \langle \Phi _{ij}\Phi _{jk}\rangle _f \langle \Phi _{kl}\Phi _{li}\rangle _f\nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad + \langle \Phi _{ij}\Phi _{kl}\rangle _f \langle \Phi _{jk}\Phi _{li}\rangle _f +\langle \Phi _{ij}\Phi _{li}\rangle _f \langle \Phi _{jk}\Phi _{kl}\rangle _f \Bigg \}. \nonumber \\&=-\frac{ N \eta Z_f}{4} \sum _{i,j,k,l}^N \Bigg ( \frac{\delta _{ik}}{ (E_i + E_j)(E_k + E_l)} +\frac{\delta _{jk} \delta _{il} \delta _{kl} \delta _{ji} }{ (E_i + E_j)(E_j + E_k)}\nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + \frac{\delta _{jl}}{ (E_i + E_j)(E_j + E_k)} \Bigg ) \nonumber \\&= N^2 Z_f B \end{aligned}$$

(A.19)

Now is the time to check the Schwinger–Dyson equation (A.4). (A.1), (A.2), and (A.3) are calculated as (A.15), (A.16), and (A.17), respectively. Note that B is the term proportional to $\eta$. The $\eta ^0$-term in (A.1)+ (A.2)+ (A.3) exists only in (A.17):

$$\begin{aligned} 2Z_f \sum _{k, j} \frac{E_k}{ E_k + E_j}= Z_f \sum _{k, j}\left( \frac{E_k}{ E_k + E_j} +\frac{E_j}{ E_k + E_j} \right) = N^2 Z_f. \end{aligned}$$

(A.20)

On the other hand, $\eta ^0$-term from $N^2 Z(E, \eta )$, that is the right hand side of (A.4), is $N^2 Z_f$. Thus, it is confirmed that Eq. (A.4) holds in zeroth order of $\eta$.

Next order terms in (A.15) + (A.16) + (A.17) are summarized as

$$\begin{aligned}&\frac{\eta Z_f }{N} \left\{ 2 \sum _{i,j,k}^N \frac{1}{ (E_i + E_j)(E_i + E_k)} -2 \sum _{k, j} \frac{E_k}{ (E_k + E_j)^2} \sum _{n=1}^N \Big ( \frac{1}{ E_k + E_n} + \frac{1}{ E_j + E_n} \Big ) \right\} \nonumber \\&\quad \qquad \quad + Z_f N^2 B . \end{aligned}$$

(A.21)

Here, we use the same way in (A.20) to obtain the last term. From the following observation,

$$\begin{aligned} 2 \sum _{k, j} \frac{E_k}{ (E_k + E_j)^2} \sum _{n=1}^N \Big ( \frac{1}{ E_k + E_n} + \frac{1}{ E_j + E_n} \Big )&= 2 \sum _{i,j,k} \frac{1}{(E_i + E_j) (E_i + E_k)} , \end{aligned}$$

(A.22)

we find (A.21) is given by

$$\begin{aligned} Z_f N^2 B . \end{aligned}$$

(A.23)

On the other hand, $\eta$-linear terms from $N^2 Z(E, \eta )$ are given by (A.19), then we checked that Eq. (A.4) holds in first order of $\eta$.

Appendix: Pfaffian expression

We can rewrite $Z(E, \eta )$ and $\Psi (E, \eta )$ by using Pfaffian. For simplicity, we consider only 2N case of (4.4).

$$\begin{aligned} Z(E, \eta )&=\frac{(2N)! c_{2N}}{\Delta (E)} \int _{{\mathbb R}^{2N}} \left( \prod _{i=1}^{2N} dx_i e^{-2N (V(x_i) + E_i x_i^2)} \right) \left( \prod _{l < k}\frac{x_k -x_l}{x_k +x_l} \right) \\&= \frac{(2N)! c_{2N}}{\Delta (E)} \int _{{\mathbb R}^{2N}} \left( \prod _{k=1}^{2N} dx_k \right) \det _{i,j} \left( \delta _{ij} e^{-2N (V(x_i) + E_i x_i^2)} \right) \text {pf}_{k,l} \left( \frac{x_k -x_l}{x_k +x_l} \right) . \nonumber \end{aligned}$$

(B.1)

As noted in Sect. 4, the integral shall be treated as a principal value integral. We use $C= ((2N)!)^2 c_{2N}$ below. After applying Bruijn’s formula [4], this is obtained by

$$\begin{aligned} Z(E, \eta )&= \frac{C}{\Delta (E)} \text {pf}_{i,j} M_{ij} \end{aligned}$$

(B.2)

where

$$\begin{aligned} M_{ij}&:= \int _{{\mathbb R}^{2}} dx dy \left( \frac{x -y}{x +y}\right) e^{-2N(V(x)+E_i x^2 + V(y)+E_j y^2)} . \end{aligned}$$

(B.3)

The justification for this process is discussed in Borot and Wulkenhaar [1]. Therefore, the zero-energy solution is given by

$$\begin{aligned} \Psi (E, \eta )&= {C} e^{-\frac{N}{2\eta } \sum _i E_i^2 } \text {pf}_{i,j} M_{ij}. \end{aligned}$$

(B.4)