Bose-Einstein-Like condensation of deformed random matrix: A replica approach

Here we use the method developed by Kabashima and Takahashi [30]. To investigate the minimal eigenvalue $\lambda_{\rm min}$ and corresponding vector $\bm{x}_{\rm min}$ of $W$, we consider a system interacting with the following potential:

where the $N$ dimensional vector $\bm{x}=\{x_{1},\dots,x_{N}\}$ denotes the state variable. We also introduced the sub-vectors:

We impose that the state vector $\bm{x}$ satisfies the spherical constraint:

When $h_{i}=0$, the model Eq. (4) can be identified with the $p=2$ spin spherical model, which has been fully investigated before [31, 32, 33].

Under the above setup, it is easy to show that when $\bm{x}=\bm{x}_{\rm min}$, we get the ground state energy [30, 34]:

Therefore, the minimal eigenvalue is calculated as $\lambda_{\rm min}=2H_{\rm GS}/N$.

To investigate the ground state, we introduce the partition function [30]:

and the free-energy

where $\beta=1/T$ denotes the inverse temperature, and the overline denotes the average for the quenched randomness $J$. The ground state energy per particle is given by taking the zero temperature limit of the free-energy

Below we omit the subscript ${\rm GS}$ unless it causes confusion. To perform the disordered average in Eq. (9), we use the replica trick [21]:

where we have introduced the replicated partition function as follows:

Since the distribution of $J_{ij}$ is a Gaussian Eq. (2), the quenched average can be taken analytically [18] ¹¹1 Here and in subsequent calculations, we omit constants and sub-leading terms that are not relevant to the final result.:

where we have defined the overlaps as follows:

When we change the variable from $\{\bm{x}_{k}^{a}\}_{a=1,\dots,n}$ to $\{Q_{k}^{ab}\}_{a,b=1,\dots n}$, the following Jacobian apepars [35]:

Summarizing the above results, we get

where

We should minimize $S(Q)$ with the spherical constraint

To proceed the calculation, we assume the replica symmetric Ansatz [21]:

Then, we get

where

Finally, by taking the $n\to 0$ limit, we get the free-energy

To get the ground state energy, we should take the zero temperature limit $T\to 0$. This is possible by using the harmonic approximation:

which is validated at sufficiently low $T$ [35]. Substituting Eq. (23) into Eq. (22) and taking $T\to 0$ limit, we get

where

Now we minimize it for $\chi_{k}$ and $q_{k}$. We first consider the saddle point condition for $\chi_{k}$:

Using this equation, one can eliminate $\chi_{k}$ from Eq. (24):

Next, we should minimize $e$ w.r.t $q_{k}$ with the spherical constraint $q=\sum_{k=1}^{M}q_{k}/M=1$. To this purpose, we introduce the Lagrange multiplier $\mu$:

The saddle point condition for $q_{k}$ leads to

Since $\sqrt{q_{k}}\geq 0$, $\mu$ should satisfy

The Lagrange multiplier $\mu$ should be determined by the following condition:

where we have introduced the distribution of $h_{k}$

and the self-overlap of spins subjected to the external field $h$

Similar equations as Eq. (31) have been previously obtained for a sparse random matrix [30] and deformed random matrices [8, 9, 11]. Substituting the above results into Eq. (7), one can calculate $\lambda_{\rm min}$ as follows:

We first check the result for a single delta peak:

which is tantamount to consider the matrix:

where $I$ is the $N\times N$ identity matrix. The minimal eigenvalue of this matrix is $\lambda_{\rm min}=-2+\Delta$ [6]. Below, we check if our method can correctly reproduce this result.

The spherical constraint Eq. (31) in this case is

Solving this equation for $\mu$, we get

The minimal eigenvalue is calculated as

The known result has been correctly reproduced.

Here we consider a simple binary distribution:

where $c\in[0,1]$ and $\Delta$ is a positive constant. Assuming the distribution Eq. (40) is tantamount to set the external field in Eq. (3) as

Now the spherical constraint Eq. (31) is written as follows

where

The Lagrange multiplier $\mu$ should be determined so as to satisfy Eq. (42). In Fig. 1, we plot $\mu$ for several $c$. For later comparison with the result of the continuous distribution, we are in particular interested in the limit $c\to 0$. A naive expectation is that Eq. (42) in this limit reduces to

Solving this equation, we get

Eq. (45) however implies that $\mu$ becomes negative when $\Delta>1$, which is prohibited by Eq. (30). What was wrong? What we missed is that when $\mu\sim 0$, the first term on the right-hand side of Eq. (42), $cq(0)=c/\mu^{2}$ can no longer be ignored. Let we assume that this term takes a finite value for $\Delta>1$, then from Eq. (42), we get

From Eqs. (43), (45), and (46), we can deduce the behavior of $\mu$, $q(0)$ and $q(\Delta)$ in the limit $c\to 0$ as follows:

In Figs. 1 and 2, we plot $\mu$, $q(0)$, and $cq(0)$ for several $c$ to show how these results converge to Eqs. (47) in the limit $c\to 0$.

From Eq. (34), ground state energy is calculated as

Substituting Eqs. (47) into the above equation, we get in the limit $c\to 0$

In Fig. 3, we plot this equation with the results of finite $c$’s.

Now we discuss the degree of the localization. For this purpose, we define the participation ratio:

where

The partition ratio takes ${\rm PR}=O(1)$ when $\bm{x}$ is extended, while ${\rm PR}=0$ when $\bm{x}$ is localized. To calculate the forth moment of $x_{i}$, we assume that $x_{i}$ follows the normal distribution of zero mean and variance $q(0)$ for $i\leq cN$ and variance $q(\Delta)$ for $i>cN$ [8]. Then, we get

In the limit $c\to 0$, Eq. (50) reduces to

Therefore, the eigenvector of the minimal eigenvalue is localized for $\Delta>1$.

In Fig. 4, we plot ${\rm PR}$ for several $c$ to see how the results converge to Eq. (53) in the limit $c\to 0$.

In the limit $M\to\infty$, one expects that $P(h)$ is approximated by a continuous function. To simplify the calculation, here we only consider the following function:

where $\Delta$ is a positive constant. The pre-factor has been chosen so that $\int_{-\infty}^{\infty}dhP(h)=1$. The Lagrange multiplier is determined by the spherical constraint:

In Fig. 5, we plot the results for several $n$.

The integral in Eq. (55) takes a maximum at $\mu=0$ ($\mu$ can not be negative due to Eq. (30)). If $n>1$, the integral at $\mu=0$ converges to a finite value :

When $1>\Delta^{-2}(n+1)/(n-1)$ or equivalently

Eq. (55) has no solution. This is similar to the situation of the previous section, and the term corresponding to $h_{k}=0$ should be carefully treated. For this purpose, let we explicitly write down the summation in Eq. (31) as

where

Eq. (59) guarantees that the distribution of $h_{k}$ converges to Eq. (54) in the limit $M\to\infty$. A necessary condition for the sum to be rewritten as an integral is that each term of the sum goes to zero in the limit of $M\to\infty$. Below we will check this condition. The terms for $k>1$ are evaluated as

where we used $h_{k}=O(M^{-\frac{1}{n+1}})$, see Eq. (59). Therefore, $q_{k}/M\to 0$ if $n>1$. This is not true for the first term

when $\mu\sim 0$. From the above consideration, one realizes that the first and other terms should be treated separately to rewrite the sum to an integral for $\Delta>\Delta_{c}$. In the limit $M\to\infty$, we obtain

Substituting back it into Eq. (58), we get for $\Delta>\Delta_{c}$

which is the essentially the same equation as Eq. (46). Eq. (63) implies that above $\Delta_{c}$, the eigenvector tends to condensate to unperturbed sites for which $h_{k}=0$. The mathematical structure that causes the condensation is very similar to that of the Bose-Einstein condensation, as mentioned in Refs. [36, 8, 9].

In Fig. 6, we plot $\mu$ calculated by Eq. (58) for $n=2$ and several $M$. For $\Delta\leq\Delta_{c}=\sqrt{3}\approx 1.73$, the results nicely converge to that of the continuum limit $\mu_{\infty}$ calculated by Eq. (55), while for $\Delta>\Delta_{c}$, the results converge to $\mu_{\infty}=0$ in the limit $M\to\infty$.

In Fig. 7, we plot $q_{1}$ and $q_{1}/M$ for several $M$. For $\Delta\leq\Delta_{c}$, $q_{1}$ converges to $1/\mu_{\infty}^{2}$ in the limit $M\to\infty$, see Fig. 7 (a). On the contrary, for $\Delta>\Delta_{c}$, $q_{1}/M$ converges to Eq. (63), see Fig. 7 (b).

As in Eq. (53), we use a Gaussian approximation to calculate the participation ratio [8]:

For $\Delta\leq\Delta_{c}$, the summation is expressed by an integral, and we get

At the transition point, the denominate is evaluated as

Therefore, at the transition point, Eq (65) vanishes for $n\in(1,3]$ and has a finite value for $n>3$. On the contrary, for $\Delta>\Delta_{c}$, the condensation $q_{1}=O(M)$ leads to ${\rm PR}\to 0$ in the limit $M\to\infty$. Those arguments suggest that on approaching the transition point, ${\rm PR}$ continuously goes to zero for $n\in(1,3]$, while it changes discontinuously from a finite value to zero for $n>3$. In Fig. 8, we plot ${\rm PR}$ for finite $M$ calculated by Eq. (64) and for $M\to\infty$ calculated by Eq. (65) for $n=2$. One can see that${\rm PR}$ changes continuously at $\Delta_{c}$, in contrast with the binary distribution where ${\rm PR}$ changes discontinuously at the transition point, see Fig. 4.

Finally, In Figs. 9, 10, and 11, we compare the theoretical prediction and numerical results obtained by direct diagonalization of $W$ for $M=10$ and $100$. We fond good agreement for $M=10$, while there are small but visible finite size effects for $M=100$. This is a natural result because our theory requires $N\gg M$. So we expect larger finite size effects for larger $M$.