Spectral Density of 1D Ising Model in n-Vicinity Method

Abstract

For a 1D-Ising system we obtain an exact combinatorial expression for its spectral density in the n-vicinity of the ground state. We show that the energies of the states from each n-vicinity take n different values and obtain a formula for a degeneracy of each energy level. We find out that in any n-vicinity there is N states whose energies differ from the energy of the ground state by an infinitesimal small value, where N is the number of spins. The obtained expressions are generalized to the case $N \to \infty$ when the variables of the problem become continuous. We compare the obtained expression with the normal approximation of the spectral density that is usually used in the framework of the n-vicinity method and discuss the reasons why it does not work in the case of the 1D-Ising system. The normal distribution provides an accurate approximation of the spectral density at the center of the energy interval but at it boundaries the behavior of the functions differs significantly. It is reasonable to say that this discrepancy leads to incorrect results when we apply the n-vicinity method to the 1D-Ising system.

Appendix

1. Let us write the connection matrix ${\mathbf{J}}$ of the 1D-Ising model as a sum of a one-step shift matrix ${\mathbf{T}}$ and the transpose of this matrix ${\mathbf{T}}^{ + }$:

$$ {\mathbf{J}} = \left( {\begin{array}{*{20}c} 0 & 1 & 0 & 1 \\ 1 & 0 & \ddots & 0 \\ 0 & \ddots & \ddots & 1 \\ 1 & 0 & 1 & 0 \\ \end{array} } \right) = {\mathbf{T}} + {\mathbf{T}}^{ + } ,\,{\text{where}}\,{\mathbf{T}} = \left( {\begin{array}{*{20}c} 0 & 0 & 0 & 1 \\ 1 & 0 & \ddots & 0 \\ 0 & \ddots & \ddots & 0 \\ 0 & 0 & 1 & 0 \\ \end{array} } \right). $$

We set the interaction constant equal to one: $J_{ii + 1} = 1$.

Let ${\mathbf{e}}_{i}$ be the i-th unit vector in the space ${\mathbf{R}}^{{\mathbf{N}}}$: $\left( {{\mathbf{e}}_{i} ,{\mathbf{e}}_{j} } \right) = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker symbol and $i = 1, \ldots ,N$. The matrix ${\mathbf{T}}$ transforms the i-th unit vector into the $(i + 1)$-th unit vector: ${\mathbf{Te}}_{i} = {\mathbf{e}}_{i + 1}$. Since we imposed the cyclic boundary conditions, we have ${\mathbf{Te}}_{N} = {\mathbf{e}}_{1}$. By $F({\mathbf{s}})$ we denote the quadratic form $F({\mathbf{s}}) = \left( {{\mathbf{Ts}},{\mathbf{s}}} \right)$. Then the energy of the state ${\mathbf{s}}$ is equal to

$$ E({\mathbf{s}}) = - \frac{{\left( {{\mathbf{Js}},{\mathbf{s}}} \right)}}{2N} = - \frac{{F({\mathbf{s}})}}{N}. $$

(A1)

It is more convenient to work with the function $F({\mathbf{s}})$ than with the energy $E({\mathbf{s}})$.

As the initial configuration we choose the ground state ${\mathbf{s}}_{0} = (1,1, \ldots ,1) = \sum\nolimits_{i = 1}^{N} {{\mathbf{e}}_{i} }$. By ${\mathbf{s}}_{{i_{1} i_{2} ..i_{n} }}$ we denote a configuration from the n-vicinity of ${\mathbf{s}}_{0}$ in which at $i_{1}$,$i_{2}$,.., and $i_{n}$ the values of the spins are equal to $- 1$: ${\mathbf{s}}_{{i_{1} i_{2} ..i_{n} }} = {\mathbf{s}}_{0} - 2\left( {{\mathbf{e}}_{{i_{1} }} + {\mathbf{e}}_{{i_{2} }} \ldots + {\mathbf{e}}_{{i_{n} }} } \right)$. It is easy to see that

$$ F({\mathbf{s}}_{{i_{1} i_{2} ..i_{n} }} ) = N - 4 \cdot n + 4 \cdot \left( {{\mathbf{e}}_{{i_{1} + 1}} + {\mathbf{e}}_{{i_{2} + 1}} + \ldots + {\mathbf{e}}_{{i_{n} + 1}} ,{\mathbf{e}}_{{i_{1} }} + {\mathbf{e}}_{{i_{2} }} + \ldots + {\mathbf{e}}_{{i_{n} }} } \right). $$

(A2)

By $\Delta_{{i_{1} i_{2} ..i_{n} }}$ we denote the scalar product

$$ \Delta_{{i_{1} i_{2} ..i_{n} }} = \left( {{\mathbf{e}}_{{i_{1} + 1}} + {\mathbf{e}}_{{i_{2} + 1}} + \ldots + {\mathbf{e}}_{{i_{n} + 1}} ,{\mathbf{e}}_{{i_{1} }} + {\mathbf{e}}_{{i_{2} }} + \ldots + {\mathbf{e}}_{{i_{n} }} } \right). $$

(A3)

The energy spectrum of the states from the n-vicinity is defined by a set of different $\Delta_{{i_{1} i_{2} ..i_{n} }}$. It is easy to understand the structure of these scalar products. At first, we suppose that the indices come in sequence: $i_{2} = i_{1} + 1,\;i_{3} = i_{2} + 1,..,i_{n} = i_{n - 1} + 1$. It is straightforward to see that in this case the value of $\Delta_{{i_{1} i_{2} ..i_{n} }}$ is equal to $n - 1$.

Next, let a set of indices falls into two isolated groups and inside each of the group the indices come in sequence. That is the first group the indices are $i,i + 1,..,i + a$ and in the second group they are $k,k + 1,..,k + b$. Evidently, the equality $a + b + 2 = n$ is fulfilled and there is a gap between the last index of the first group and the first index of the second group: $k > i + a + 1$. It is easy to see that independent of the content of the groups the value of $\Delta_{{i_{1} i_{2} ..i_{n} }}$ is equal to $n - 2$. Similarly, when the set of the indices $i_{1} ,i_{2} ,..,i_{n}$ falls into three isolated groups inside each of which the indices $i_{j}$ come in sequence, independent of the content of the groups the value of $\Delta_{{i_{1} i_{2} ..i_{n} }}$ is equal to $n - 3$ and so on. When the set of the indices $i_{1} ,i_{2} ,..,i_{n}$ falls into $k$ isolated groups inside each of which the indices come in sequence, independent of the content and the size of the groups the value of $\Delta_{{i_{1} i_{2} ..i_{n} }}$ is equal to $n - k$. When $n \le N/2$ the maximal number of the isolated groups is $k_{\hbox{max} } = n$. Using Eqs. (A1), (A2), (A3) we obtain the energies of the states from the n-vicinity of the configuration ${\mathbf{s}}_{0}$. These energies take on exactly $n$ values

$$ E_{k} = - \left( {1 - 4k/N} \right),\;k = 1,2, \ldots n;\;n = 1,2,..,N/2. $$

2. Now, let us find out the degeneration of an energy $E_{k}$ for the states from the n-vicinity. (As a rule, in different n-vicinities the degenerations of the same energy $E_{k}$ are different.) By $D(n,k)$ we denote the number of states in the n-vicinity whose energy is equal to $E_{k}$.

The 0-vicinity consists of the configuration ${\mathbf{s}}_{0}$. Here $F({\mathbf{s}}_{0} ) = \left( {{\mathbf{s}}_{0} ,{\mathbf{s}}_{0} } \right) = N$ and we can write that $E_{0} = - 1$ and $D(0,0) = 1$.

The 1-vicinity of ${\mathbf{s}}_{0}$ consists of N configurations that differ from ${\mathbf{s}}_{0}$ in the opposite value of one spin only. Consequently, all N configurations from the 1-vicinity are characterized by the same energy $E_{1}$ and $D(1,1) = N$.

It is easy to see that $\left( {_{2}^{N} } \right)$ configurations of the 2-vicinity of the ${\mathbf{s}}_{0}$ fall into two groups. They are, first, N different configurations where two “−1” are in succession. They are characterized by the energy $E_{1}$. The second group consists of all the other configurations from the 2-vicinity, which are characterized by the energy $E_{2}$. Consequently,

$$ D(2,1) = N,\;D(2,2) = \frac{N(N - 3)}{2}. $$

Next, in the 3-vicinity there are exactly $N$ configurations characterized by the energy $E_{1}$, $N(N - 4)$ configurations, which are characterized by the energy $E_{2}$, and the energy of all other configurations is $E_{3}$. Then we obtain

$$ D(3,1) = N,\;D(3,2) = N(N - 4),\;D(3,3) = \frac{N(N - 4)(N - 5)}{3!}. $$

When n increases, the analysis is somewhat more difficult, but with the aid of standard methods it is not a problem to obtain the following expressions for the 4-th and 5-th vicinities

$$ \begin{array}{*{20}c} {D(4,1) = N,\;\;D(4,2) = \frac{3}{2}N(N - 5),\;\;D(4,3) = \frac{N(N - 5)(N - 6)}{2}} \\ {D(4,4) = \frac{N(N - 5)(N - 6)(N - 7)}{4!},\quad } \\ \end{array} $$

and

$$ \begin{array}{*{20}c} {D(5,1) = N,\;\quad D(5,2) = 2N(N - 6),\;\quad D(5,3) = N(N - 6)(N - 7),\quad \quad } \\ {D(5,4) = \frac{N(N - 6)(N - 7)(N - 8)}{3!},\;D(5,5) = \frac{N(N - 6)(N - 7)(N - 8)(N - 9)}{5!},} \\ \end{array} $$

respectively. The analysis of the obtained expressions for the frequencies of occurrence of the energy $E_{k}$ for the five n-vicinities allowed us to write down the following combinatorial formula,

$$ D(n,k) = \frac{Nk}{(N - n)n}C_{N - n}^{k} C_{n}^{k} ,\;\;k = 1,2, \ldots n;\;\;n = 1,2, \ldots N/2. $$

We confirmed these formulae by means of a lot computer simulations performed for different values of $N,n$ and k.