Cubic Interaction Vertices and One-loop Self-energy in the Stable String Bit Model

TL;DR

This paper develops a formalism for calculating cubic interaction vertices and one-loop self-energy corrections in a stable string bit model with spin degrees of freedom, revealing how the energy scales with the number of bits and spin.

Contribution

It introduces a new formalism for cubic interaction vertices in a string bit model with no spatial movement, enabling calculation of one-loop energy corrections.

Findings

Ground state one-loop self-energy scales as M^{5-s/4} for even s and M^{4-s/4} for odd s.

For s=24, the self-energy scales as 1/M, resembling Poincaré invariance in 1+1 dimensions.

Analytical calculations for M=3, s=1,2 and numerical confirmation for s≤4.

Abstract

We provide a formalism to calculate the cubic interaction vertices of the stable string bit model, in which string bits have $s$ spin degrees of freedom but no space to move. With the vertices, we obtain a formula for one-loop self-energy, i.e., the $\mathcal{O}\left(1/N^{2}\right)$ correction to the energy spectrum. A rough analysis shows that, when the bit number $M$ is large, the ground state one-loop self-energy $\Delta E_{G}$ scale as $M^{5-s/4}$ for even $s$ and $M^{4-s/4}$ for odd $s$. Particularly, in $s=24$, we have $\Delta E_{G}\sim 1/M$, which resembles the Poincar\'e invariant relation $P^{-}\sim 1/P^{+}$ in $(1+1)$ dimensions. We calculate analytically the one-loop correction for the ground energies with $M=3$ and $s=1,\,2$. We then numerically confirm that the large $M$ behavior holds for $s\leq4$ cases.

Figures

Tables

Table 1. Table 1: Gradings of functions and operators.

$T_r$	${\psi }_r$	${\overline{\psi }}_r$	${\psi }_G$	${\overline{\psi }}_G$	$F_{\left\{r\right\}}$
$g_r$	$g_r-sM$	$g_r$	$0$	$sM$	$g_r-sM$

Table 2. Table 2: 1-bit and 2-bit energy eigenstates of s = 1 𝑠 1 s=1 that do not contain zero modes.

$M$	$\psi$	Conjugate	Energy	Grading of $\psi$
1	${\psi }_G^{\left(1\right)}=1$	${\overline{\psi }}_G^{\left(1\right)}={\theta }_1$	$E_G^{\left(1\right)}=0$	even
2	${\psi }_1^{\left(2\right)}=F_1^{\left(2\right)†}{\psi }_G^{\left(2\right)}$	${\overline{\psi }}_1^{\left(2\right)}={\tilde{F}}_1^{\left(2\right)}{\overline{\psi }}_G^{\left(2\right)}$	$E_1^{\left(2\right)}=4$	odd

Table 3. Table 3: 1-bit and 2-bit energy eigenstates of s = 2 𝑠 2 s=2 that do not contain zero modes.

$M$	$\psi$	Conjugate	Energy	Grading of $\psi$
1	${\psi }_G^{\left(1\right)}=1$	${\overline{\psi }}_G^{\left(1\right)}={\theta }_1$	$E_G^{\left(1\right)}=0$	even
2	${\psi }_G^{\left(2\right)}$	${\overline{\psi }}_G^{\left(2\right)}$	$E_G^{\left(2\right)}=-8$	even
2	${\psi }_1^{\left(2\right)}=F_{1,a=1}^{\left(2\right)†}{F}_{1,a=2}^{\left(2\right)†}{\psi }_G^{\left(2\right)}$	${\overline{\psi }}_1^{\left(2\right)}={\tilde{F}}_{1,a=2}^{\left(2\right)}{\tilde{F}}_{1,a=1}^{\left(2\right)}{\overline{\psi }}_G^{\left(2\right)}$	$E_1^{\left(2\right)}=8$	even

Equations

$${\left({\bar{\varphi }}_{a_1\cdots a_n}\right)}_{\alpha }^{\beta },a_i=1,\cdots ,s,n=0,\cdots ,s,\alpha ,\beta =1,\cdots ,N.$$

$$H=\sum _{i=1}^5{H}_i+s\xi \Delta H,$$

$$\psi \left(\theta \right)=\sum _{k=0}^s\frac{1}{k!}{\bar{\varphi }}_{c_1\cdots c_k}{\theta }^{c_1}\cdots {\theta }^{c_k}.$$

$$T\left({\theta }_1,\cdots ,{\theta }_k\right)=\mathrm{Tr}\psi \left({\theta }_1\right)\psi \left({\theta }_2\right)\cdots \psi \left({\theta }_k\right),$$

$$T\left({\theta }_1,\cdots ,{\theta }_K\right)T\left({\eta }_1,\cdots ,{\eta }_L\right)\cdots |0⟩,$$

$${\bar{h}}_{kl}$$

$$\bar{h}$$

$$HT\left({\theta }_1,\cdots ,{\theta }_M\right)|0⟩=\bar{h}T\left({\theta }_1\cdots {\theta }_M\right)|0⟩+\frac{1}{N}\sum _{k=1}^M\sum _{l\ne k+1}{\bar{h}}_{kl}T\left({\theta }_l\cdots {\theta }_k\right)T\left({\theta }_{k+1}\cdots {\theta }_{l-1}\right)|0⟩$$

$$\begin{gathered} HT\left({\theta }_1\cdots {\theta }_K\right)T\left({\eta }_1\cdots {\eta }_L\right)|0⟩=\left({\bar{h}}_{\theta }+{\bar{h}}_{\eta }\right)T\left({\theta }_1\cdots {\theta }_K\right)T\left({\eta }_1\cdots {\eta }_L\right) \\ +\frac{1}{N}\sum _{k=1}^K\sum _{l=1}^L{\bar{h}}_{kl}T\left({\theta }_{k+1}\cdots {\theta }_k{\eta }_l\cdots {\eta }_{l-1}\right)|0⟩ \\ +\frac{1}{N}\sum _{k=1}^K\sum _{l=1}^L{\bar{h}}_{lk}T\left({\theta }_k\cdots {\theta }_{k-1}{\eta }_{l+1}\cdots {\eta }_l\right)|0⟩+\frac{1}{N}\text{Fission Terms}. \end{gathered}$$

$$T_r|0⟩=\int d^s{\theta }_1\cdots d^s{\theta }_{M}T\left({\theta }_1,\cdots ,{\theta }_M\right){\psi }_r\left({\theta }_1,\dots ,{\theta }_M\right)|0⟩,$$

$${\psi }_r\left({\theta }_1,\cdots ,{\theta }_M\right)={\left(-\right)}^{s\left(M-1\right)}{\psi }_r\left({\theta }_2,\cdots ,{\theta }_M,{\theta }_1\right)$$

$${\bar{h}}_{kl}{T}_r|0⟩=\int d\theta {\bar{h}}_{kl}T\left(\theta \right){\psi }_r\left(\theta \right)|0⟩=\int d\theta T\left(\theta \right)h_{kl}{\psi }_r\left(\theta \right)|0⟩$$

$$h_{kl}=-2\left(s-2{\theta }_k^a\frac{d}{d{\theta }_k^a}\right)-2{\theta }_k^a\frac{d}{d{\theta }_l^a}-2{\theta }_l^a\frac{d}{d{\theta }_k^a}-2i{\theta }_k^a{\theta }_l^a-2i\frac{d}{d{\theta }_k^a}\frac{d}{d{\theta }_l^a}+2s\xi +2s{\delta }_{k,l}.$$

$$h=\sum _{k=1}^M\left(h_{k,k+1}-2s\xi \right).$$

$$h{\psi }_r\left({\theta }_1,\dots ,{\theta }_M\right)=E_r{\psi }_r\left({\theta }_1,\dots ,{\theta }_M\right).$$

$${\alpha }_n$$

$${\theta }_k$$

$$n$$

$$\left\{{\alpha }_n,{\beta }_m\right\}={\delta }_{m+n,M}+{\delta }_{m,0}{\delta }_{n,0}.$$

$${\alpha }_n^{†}={\beta }_{M-n},{\beta }_n^{†}={\alpha }_{M-n},\left\{{\alpha }_n,{\alpha }_m^{†}\right\}=\left\{{\beta }_n,{\beta }_m^{†}\right\}={\delta }_{m,n},0\le n,m\le M-1.$$

$$h=2\sum _{n=1}^{M-1}\left[\left({\alpha }_n{\alpha }_{M-n}+{\beta }_n{\beta }_{M-n}\right)\sin \frac{2n\pi }{M}+2\left(1-\cos \frac{2n\pi }{M}\right)\left({\alpha }_n{\beta }_{M-n}+{\alpha }_{M-n}{\beta }_n\right)\right]-2M,$$

$$F_k=r_k{\alpha }_k+{\beta }_k,\left[h,F_k\right]={ϵ}_k{F}_k,$$

$${ϵ}_k^{\pm }=\pm 8\sin \frac{k\pi }{M},r_k^{\pm }=\tan \frac{k\pi }{M}\pm \sec \frac{k\pi }{M}.$$

$$F_k=s_k{\alpha }_k+c_k{\beta }_k,{\bar{F}}_k=c_k{\alpha }_k-s_k{\beta }_k,k=1,\cdots ,M-1,$$

$$F_k^{†}={\bar{F}}_{M-k}=c_k{\alpha }_{M-k}+s_k{\beta }_{M-k},1\le k\le M-1,$$

$$F_0=F_M=e^{i\pi /4}{\beta }_0,F_0^{†}={\bar{F}}_0={\bar{F}}_M=e^{-i\pi /4}{\alpha }_0.$$

$$\left\{F_k,F_l\right\}=\left\{F_k^{†},F_l^{†}\right\}=0,\left\{F_k,F_l^{†}\right\}={\delta }_{kl},0\le k,l\le M-1.$$

$${\psi }_G^{s=1}=\prod _{k=1}^{\lfloor \left(M-1\right)/2\rfloor }\left(c_k-s_k{\alpha }_{M-k}{\alpha }_k\right),$$

$$F_k\left(c_k-s_k{\alpha }_{M-k}{\alpha }_k\right)$$

$$\left[F_k,c_l-s_l{\alpha }_{M-l}{\alpha }_l\right]$$

$$E_G^{s=1}=-4\sum _{k=1}^{M-1}\sin \frac{k\pi }{M}=-4\cot \frac{\pi }{2M}=-\frac{8M}{\pi }+\frac{2\pi }{3M}+\mathcal{O}\left(M^{-3}\right).$$

$${\psi }_G={\psi }_G^{\left(1\right)}{\psi }_G^{\left(2\right)}\cdots {\psi }_G^{\left(s\right)},E_G=-4s\cot \frac{\pi }{2M},$$

$${\psi }_r$$

Full text

1 Introduction

In the string bit model [1], a string is a chain comprised of pointlike entities called string bits. While the chain is discretized, it behaves like a continuous string when the bit number $M$ is large enough.

The string bit model is an implementation of ’t Hooft’s idea of holography [2, 3, 4]. In Lorentz invariant theory, spacetime can be described by lightcone coordinates with transverse dimensions $\mathbf{x}=\left(x^{2},\cdots,x^{D-1}\right)$ and the ‘$\pm$’ dimensions $x^{\pm}=\left(x^{0}\pm x^{1}\right)/\sqrt{2}$. In the string bit model, the $x^{-}$ coordinate of string bits is missing, and hence, the Lorentz invariance is not present a priori. String bits enjoy the dynamic of Galilean symmetry, under which the $+$-component momentum $P^{+}=\left(P^{0}+P^{1}\right)/\sqrt{2}$ is identified as $mM$, where $m$ is the mass of one string bit. When $M$ is large enough and $P^{+}$ is fixed, $P^{+}$ can be considered as a continuous variable and its conjugate $x^{-}$ can be interpreted as the missing coordinate. The Lorentz invariance can be therefore regained and string theory emerges.

With ’t Hooft’s large $N$ limit [5, 6], the type II-B superstring was formulated in ref. [7] as a string bit model. In the model, a superstring bit creation operator, which was an adjoint representation of $U\left(N\right)$ color group, has up to $s$ spin indices and moves in transverse space. A more drastic form of holography was studied in recent papers [8, 9, 10, 11], where string bits have no transverse coordinate and hence no space to move. However, new compactified bosonic coordinates can be generated from spin degrees of freedom of string bits. If suitable dynamics is chosen, these spin degrees of freedom are converted to one-dimensional spin waves, which then act as compactified bosonic coordinates. The $1/N$ perturbation of the latter model was studied in ref. [11], where the cubic interaction vertices and their application to the calculation of the one-loop self-energy were discussed.

Following the main idea of ref. [11], we continue the work in the following way.

• •

A more detailed study of the cubic interaction vertices is performed. We present a systematic way to build conjugates of energy eigenfunctions, determine the sign factors of the vertices, and (anti)symmetrize the vertices, which are denoted as $V_{qpr}$ and $W_{rpq}$ and shown as Figure 1, over the indices $p$ and $q$. We then show that the interaction vertices can be calculated by finding the vacuum expectation values of ladder operators. These are necessary for the use of interaction vertices in our calculation of observables.

• •

The calculation of the one-loop self-energy is improved, and its large $M$ behavior for the ground states is analyzed. We assemble the ingredients necessary to calculate the one-loop self-energy. The one-loop self-energies of ground states, $\Delta E_{G}$, are studied, and their large $M$ behavior is analyzed. We calculate $\Delta E_{G}$ analytically for the $M=3,\,s=1$ and $M=3,\,s=2$ cases. A qualitative analysis shows that $\Delta E_{G}$ scales as $M^{5-s/4}$ for even $s$ and $M^{4-s/4}$ for odd $s$. The scaling behavior is consistent with Lorentz invariance in $1+1$ dimensions when $s=24$, the critical Grassmann dimension, and the protostring model [11] emerges.

• •

$\Delta E_{G}$ is determined numerically for higher $M$ and $s$. We confirm the large $M$ behavior of $\Delta E_{G}$ for $s\leq 4$. We also verify that $\Delta E_{G}$ increases exponentially with respect to $s$ when $M$ is fixed. We generalize the Hamiltonian of the model by adding $\mathcal{O}\left(1/N\right)$ terms $s\xi\Delta H$ and numerically show that, for the $s=2$ case, the Hamiltonian is bounded from below with respect to $M$ only when $\xi\geq 1$. Our analysis suggests that this is true for all the even $s$ cases. The result shows that the $s\xi\Delta H$ generalization is necessary for building a physical string bit model.

The rest of this paper is organized as follows. In section 2, we review some results of stable string bit models obtained by [11]. Specifically, we introduce the Hamiltonian of the model, solve for the energy spectrum of the model at $N=\infty$, and summarize the three chains overlap calculation. In section 3, we provide a systematic approach to build conjugate eigenfunctions, which will be used in the calculation of the $1/N$ expansion. In section 4, the cubic interaction vertices are studied by $1/N$ perturbation. In section 5, we use the cubic interaction vertices to calculate one-loop self-energies. Numerical results for the one-loop self-energy are analyzed in section 6. The main text is closed with a conclusion section. Finally, several Appendixes are included for technical details.

2 Stable string bit model

The purpose of this section is to review some results of stable string bit models obtained in ref. [11] and introduce useful notations. These results are necessary for setting up the $1/N$ expansion of the model. Meanwhile some modifications specific to this paper are incorporated. To be clear, the modifications are as follows. In Sec. 2.1, we add an $\mathcal{O}\left(1/N\right)$ term $\xi\Delta H$ to the Hamiltonian of the model. In Sec. 2.2, the diagonalization of the Hamiltonian at $N=\infty$ is done via different intermediate variables.

2.1 Hamiltonian

The superstring bit creation operator is

[TABLE]

where $a_{i}$ are totally antisymmetric spin indices and $\alpha$, $\beta$ color indices of $U\left(N\right)$. $\bar{\phi}$ is bosonic when $n$ is even and fermionic when $n$ is odd. In Fock space, a closed string is represented by a color singlet trace operator acting on the vacuum state, that is of the form $\mathrm{Tr}\,\bar{\phi}\cdots\bar{\phi}\left|0\right\rangle$. The number of $\bar{\phi}$ in the trace operator is the eigenvalue of the bit number operator $M=\sum_{n}\frac{1}{n!}\mathrm{Tr}\,\bar{\phi}_{a_{1}\cdots a_{n}}\phi_{a_{1}\cdots a_{n}}$.

The Hamiltonian $H$ to be studied in this paper reads

[TABLE]

where expressions of $H_{i}$ and $\Delta H$ are given in eqs. (A.3) and (A.6). The $H_{i}$s make an $\mathcal{O}\left(1\right)$ contribution to $H$, while $\Delta H$ makes only $\mathcal{O}\left(1/N\right)$ contribution and hence does not affect the large $N$ limit. We note that $H$ is a generalization of the $s=1$ Hamiltonian in refs. [8, 10]. The $H_{i}$ parts have been proposed in refs. [9, 11]; $\Delta H$ is the new term added by this paper and its derivation is given in Appendix A.1.

Let us now consider the action of $H$ on trace states space, which is defined as follows. We introduce $s$ Grassmann coordinates $\theta^{a}$, $a=1,\cdots,s$ and then define a superbit creation operator

[TABLE]

and a single trace operator

[TABLE]

where $\theta_{i}$ are $s$-component Grassmann variables. The trace states space, i.e., color singlet subspace of Fock space, is then spanned by states like

[TABLE]

where $\left|0\right\rangle$ is the vacuum state. The action of each $H_{i}$ and $\Delta H$ on trace states is given in Appendix A. To summarize the results, let us define

[TABLE]

Then the actions of $H$ on single and double trace states can be written as111The actions of each $H_{i}$ on single and double trace states are shown in Appendix A.

[TABLE]

[TABLE]

Note that in eqs. (6), the $-2s\delta_{kl}$ term of $\bar{h}_{kl}$ should be zero even if $k=l$, as they label different variables.

While $\bar{h}_{kl}$ acts on the trace states, to solve for energy eigenstates, it is helpful to convert $\bar{h}_{kl}$ to an equivalent form acting on the wave function of an energy eigenstate at $N=\infty$. The wave function $\psi_{r}$ is defined as follows. It follows from eq. (5) that, at $N=\infty$, $H$ evolves single trace states to single trace states. Therefore, we can express a single trace energy state as

[TABLE]

where $\psi_{r}$ is the wave function. Since $T\left(\theta_{1},\cdots,\theta_{M}\right)$ is invariant under the cyclic permutation $\theta_{i}\to\theta_{i+1}$, we can constrain $\psi_{r}$ by

[TABLE]

without loss of generality. The sign factor follows from the fact that the measure $d^{s}\theta_{1}\cdots d^{s}\theta_{M}$ is changed by a factor $\left(-\right)^{s\left(M-1\right)}$ under the cyclic transformation $\theta_{i}\to\theta_{i+1}$. Now, the action of $\bar{h}_{kl}$ on $T_{r}\left|0\right\rangle$ is

[TABLE]

where we have performed an integration by parts in the last step and

[TABLE]

Note that, in the derivation of $h_{kl}$, the $k=l$ case needs special treatment. Likewise, the action of $\bar{h}$ on $T\left(\theta\right)$ is equivalent to the action on $\psi_{r}\left(\theta\right)$ by

[TABLE]

2.2 Diagonalizing Hamiltonian

at $N=\infty$

Now let us solve for the energy spectrum of the model at $N=\infty$. A single trace energy eigenstate is determined by an eigenfunction $\psi_{r}$ satisfying the equation

[TABLE]

To solve the eigenvalue problem eq. (12), we need to find the lowering and raising eigenoperators of $h$. This has been done by ref. [11]. Here, we repeat the procedure with different sets of intermediate variables.

From (10), we see that each term of $h$ contains only variables or derivatives of the same $\theta^{a}$. It implies the variables can be separated and we only need to solve the equation of one variable. We therefore drop the spin index $a$ in the following calculation.

We introduce Fourier transforms [10, 8]

[TABLE]

which satisfy

[TABLE]

In ref. [11], instead of $\theta_{k}$ and $\frac{d}{d\theta_{k}}$, the diagonalization was done via the Grassmann variables $S_{k}=\theta_{k}+\frac{d}{d\theta_{k}}$, $\tilde{S}_{k}=i\left(\theta_{k}+\frac{d}{d\theta_{k}}\right)$, and their Fourier transforms. Such different choices should not affect the eigenoperators and the energy spectrum.

The Hermiticity of the Hamiltonian implies that $\theta_{k}^{\dagger}=\frac{d}{d\theta_{k}},\,\frac{d}{d\theta_{k}}^{\dagger}=\theta_{k},$ from which it follows that

[TABLE]

We now express $h$ in terms of $\alpha_{n}$ and $\beta_{n}$ as

[TABLE]

and seek for eigenoperators of $h$,

[TABLE]

where $r_{k}$ and $\epsilon_{k}$ are constants. Substituting (15) into (16) yields

[TABLE]

We then normalize the coefficients of $F_{k}$ to obtain the lowering and raising operators for $k\geq 1$,

[TABLE]

The phase factors are chosen so that the expression of $h_{kl}$ in terms of eigenoperators will have a simple form; see eq. (53). A direct calculation shows that the eigenoperators satisfy the following anticommutation relations

[TABLE]

To obtain the energy spectrum, we need to find the ground energy $E_{G}$ and the ground eigenfunction $\psi_{G}$, which is annihilated by all the lowering operators. Since the zero mode does not change energy eigenvalues, there are degeneracies in ground state. To eliminate the ambiguity, we require the ground eigenfunction to be annihilated by the zero mode $F_{0}$ as well. The ground eigenfunction can be [10]

[TABLE]

where $\left\lfloor\left(M-1\right)/2\right\rfloor$ indicates the integral part of $\left(M-1\right)/2$. To verify $F_{m}\psi_{G}^{s=1}=0$, one only needs to check that

[TABLE]

Acting $h$ on the ground eigenfunction, we obtain the ground energy

[TABLE]

We can now build general eigenfunctions for arbitrary $s$ case. The ground eigenfunction and energy are

[TABLE]

where each $\psi_{G}^{\left(a\right)}$ has the form of (19). A general energy eigenfunction $\psi_{r}$ and its corresponding energy can be written as

[TABLE]

where we have defined $F_{\left\{r\right\}}$ as a string of eigenoperators and we choose $0\leq r_{a,1}<r_{a,2}<\cdots\leq M-1$ as a convention. To build a physical state, the modes $r_{a,i}$ (24a) need to satisfy the cyclic constraint (8). Under the cyclic permutation $\theta_{k}^{a}\to\theta_{k+1}^{a}$, $F_{k}^{a\dagger}$ transforms as $F_{k}^{a\dagger}\to e^{-2ik\pi/M}F_{k}^{a\dagger}$. It then follows from eq. (8) that the modes must satisfy

[TABLE]

Since the zero modes do not change the energy, the ground energy eigenstate has at least $2^{s}$ degeneracies. This is the consequence of $H$ commuting with supersymmetry operators $Q^{a}$, as defined in eq. (A.8). The constraint (25) has a profound impact on the energy spectrum of the model. When $s$ is even, all the ground states are allowed by (25) and are hence physical. But when $s$ is odd, the ground state is allowed only when $M$ is odd. It then follows that the lowest single trace state for even $M$ is the one corresponding to $F_{M/2}^{a\dagger}\psi_{G}$.

2.3 Three chains overlap

We have constructed the energy eigenfunctions for $N=\infty$. To obtain the $1/N$ expansion results, we also need to calculate the overlap among three chains: one large chain of $M$ bits and two small chains of $K$ bits and $L=M-K$ bits. The calculation can be done by establishing the relation among the eigenoperators of large chain and two small chains. Here, we recap the results of ref. [11].

Let us only consider the $s=1$ case. Let $F_{m}^{\left(K\right)}$ and $F_{n}^{\left(L\right)}$ be lowering operators of $L$-bit and $K$-bit chains. Define a set of operators

[TABLE]

which satisfy the anticommutation relationship $\left\{f_{n},f_{m}\right\}=\left\{f_{n}^{\dagger},f_{m}^{\dagger}\right\}=0$ and $\left\{f_{n},f_{m}^{\dagger}\right\}=\delta_{nm}$. Note that $f_{0}$ equals $F_{0}$ of the large chain [11]. We then express the large chain operators in terms of $f$ and $f^{\dagger}$ as

[TABLE]

The anticommutation relation among $F_{m}$ and $F_{m}^{\dagger}$ requires

[TABLE]

The matrix elements of $C$ and $S$ are given by

[TABLE]

and [11]

[TABLE]

where $1\leq m\leq M-1$ in eqs. (29). When $M$ is large, the determinate of $C$ can be approximated as [11]

[TABLE]

We then express the ground eigenfunction of the large chain as

[TABLE]

where $\psi_{G}^{\left(K\right)}$ and $\psi_{G}^{\left(L\right)}$ are ground eigenfunction for two small chains. The constraints $F_{m}\psi_{G}=0$ imply

[TABLE]

From the above construction, it is clear that the first rows and columns of the matrices $C$, $S$, and $D$ are trivial. One can therefore write them as $C=\begin{pmatrix}1\end{pmatrix}\oplus C^{\prime}$, $S=\begin{pmatrix}0\end{pmatrix}\oplus S^{\prime}$, and $D=\begin{pmatrix}0\end{pmatrix}\oplus D^{\prime}$ where $C^{\prime}$, $S^{\prime}$, and $D^{\prime}$ are nontrivial matrices of dimension $\left(M-1\right)\times\left(M-1\right)$.

With (28) and (32), we can simplify (31):

[TABLE]

[TABLE]

3 Conjugate eigenfunction

We have built energy eigenfunctions of the model at $N=\infty$ in Sec. 2.2. To calculate $1/N$ expansion results, we also need to find functions that conjugate to the energy eigenfunctions. For convenience, we call these functions conjugate eigenfunctions. In this section, we will construct conjugate eigenfunctions systematically.

A conjugate eigenfunction $\bar{\psi}_{r}$ is a function of $\theta_{i}$ that satisfies the normalization condition [11]

[TABLE]

and the completeness relation

[TABLE]

where the delta function $\tilde{\delta}\left(\theta-\eta\right)$ is understood to be symmetrized under cyclic constraint like (8).222To be specific, it means that

$\int d^{M}\theta\,f\left(\theta_{1},\cdots,\theta_{M}\right)\tilde{\delta}\left(\theta-\eta\right)=\frac{1}{M}\sum_{k=0}^{M-1}\left(-\right)^{ks\left(M-1\right)}f\left(\eta_{k+1},\cdots,\eta_{k+M}\right).$

We stress that, once there is a complete set of $\bar{\psi}_{r}$ and $\psi_{r}$ fulfilling the normalization condition, the completeness relation is satisfied automatically.

To construct $\bar{\psi}_{r}$ explicitly, it is convenient to define operators $\tilde{F}_{k}^{\pm}$ as conjugate to $F_{k}$ under integration by parts,

[TABLE]

where the $+$ superscript is chosen if $\psi\left(\theta\right)$ is Grassmann even and $-$ is chosen otherwise. It then follows from eqs. (17) that

[TABLE]

[TABLE]

In the remainder of this paper, we may suppress the superscript $\pm$ if there is no danger of ambiguity.

In the $s=1$ case, we claim that the conjugate to the ground eigenfunction $\psi_{G}^{s=1}$ is

[TABLE]

In Appendix B, we verify that $\bar{\psi}_{G}^{s=1}$ satisfies the normalization condition (34). The function conjugate to the general eigenfunction (24a) can be built by acting on $\bar{\psi}_{G}$ with a string of $\tilde{F}_{k}^{\left(a\right)}$ as

[TABLE]

where all the $\tilde{F}$s pick $\tilde{F}^{+}$ if $\bar{\psi}_{G}$ is Grassmann even and $\tilde{F}^{-}$ otherwise. The normalization condition (34) can be easily verified,

[TABLE]

where we used (36) in the second equality and (18) in the third equality. In the last equality, the sign factor of $\bar{\psi}_{G}$ cancels the sign introduced by the rearrangement of the measure from $d^{s}\theta_{1}\cdots d^{s}\theta_{M}$ to $\left(\prod_{i=1}^{M}d\theta_{i}^{\left(1\right)}\right)\cdots\left(\prod_{i=1}^{M}d\theta_{i}^{\left(s\right)}\right)$.

By analogy with (31), for the $s=1$ case, the overlap of conjugate eigenfunctions among the large chain and two small chains is given by

[TABLE]

where $\tilde{f}$ picks $\tilde{f}^{+}$ if $M$ is even and $\tilde{f}^{-}$ if $M$ is odd and all the notations follow the ones of Sec. 2.3.

Let us conclude this section by discussing the grading of energy eigenstates and eigenfunctions. We define

[TABLE]

Now, we can write the trace operator $T\left(\theta\right)$ as a linear combination of $\bar{\psi}_{r}$. Let $T\left(\theta\right)=\sum_{t}X_{t}\bar{\psi}_{t}\left(\theta\right)$, where $X_{t}$ is independent of $\theta$; then

[TABLE]

where the sign factor comes from the commutation of the measure and $X_{t}$. It implies that $X_{r}$ differs from $T_{r}$ only by a sign factor. So, we have $\mathrm{g}\left(X_{r}\right)=\mathrm{g}\left(T_{r}\right)$ and

[TABLE]

Finally, from eqs. (42), (34), and (23), we obtain the gradings (modulo 2) of functions and operators as Table 1. These results will be used in the next section.

4 Cubic interaction vertices

Let $T_{p}\left|0\right\rangle$, $T_{q}\left|0\right\rangle$, and $T_{r}\left|0\right\rangle$ be energy eigenstates of strings with $K$, $L$, and $M=K+L$ bits respectively; then the interaction vertices $V_{qpr}$ and $W_{rpq}$ are defined as[11]

[TABLE]

The vertex $V_{qpr}$ represents the amplitude of breaking one large string into two small strings and the vertex $W_{rpq}$ represents the amplitude of joining two small strings into one large string. Without loss of generality, we can (anti)symmetrize the vertices over indices $p$ and $q$ as

[TABLE]

In this section, we shall find that

[TABLE]

Several notations are used in (45) for convenience. $F_{\left\{qp\right\}}^{a}\equiv F_{\left\{q\right\}}^{a}F_{\left\{p\right\}}^{a}$ and the superscript $a$ indicates that only operators of spin index $a$ are involved. The brackets $\left\langle\cdot,\cdot\right\rangle_{V,W}$ stand for vacuum expectation values of operators

[TABLE]

where the matrix $D$ and operators $f_{k}$ are defined as (32) and (26) and $h_{kl}$ is given by (10). The vacuum of (46a) is the state annihilated by all lowering operators of $L$-bit and $K$-bit systems, i.e., $F_{i}^{\left(K\right)}\left|0\right\rangle=F_{i}^{\left(L\right)}\left|0\right\rangle=0$. In the following, we first mark remarks on the interaction vertices in the Sec. 4.1 and then give all the techenical details of the derivation of (45) in Sec. 4.2.

4.1 Remarks on vertices

The form of vertices in (45) can be interpreted as follows. The prefactor $M$ of $V_{qpr}$ shows that, when a large chain splits into two small chains of $K$ and $L$ bits, there are $M$ ways to choose the break points, and each way contributes equally to $V_{qpr}$. Likewise, the prefactor $KL$ of $W_{rpq}$ shows that, when two small chains join into a large chain, there are $K\times L$ ways to choose the joint points, and each way contributes equally to $W_{rpq}$. The operator $h_{V}^{a}=\frac{1}{2}\left(h_{K,1}^{a}+h_{M,K+1}^{a}\right)$ reflects the fact that, to break one $M$-bit string into $K$-bit and $L$-bit strings, one needs to connect bit $1$ to bit $K$ and bit $\left(K+1\right)$ to bit $M$. Similarly, the operator $h_{W}=h_{K,K+1}^{a}+h_{M,1}^{a}$ reflects the fact that, to join back the above two small strings into one, one needs to connect bit $K$ to bit $\left(K+1\right)$ and bit $M$ to bit $1$. The difference of factor $2$ between $h_{V}^{a}$ and $h_{W}^{a}$ is because that, when joining two strings, one can inverse the labels of the first small string as $1+i\leftrightarrow K-i$ to obtain a different large string.

4.2 Derivation of $V_{qpr}$ and $W_{rpq}$

Now, let us derive the formula (45). Acting the Hamiltonian to the zeroth order energy eigenstate $T_{r}\left|0\right\rangle$ and using (7) and (5), we have

[TABLE]

where in the second equality we renamed the indices as $j\to i+K+1$ and in the last equality we used (42). Comparing (47) with (43a), we arrive at

[TABLE]

The vertex is decorated with a tilde because we have not yet applied the constraint (44) to it. Note that the sign factor is changed due to the reorder of $T_{p}$ and $T_{q}$.

The action of $H$ on the double trace produces both fusion and fission terms:

[TABLE]

Comparing the above with (43b), we have $\tilde{W}_{rpq}=W_{rpq}^{\left(1\right)}+W_{rpq}^{\left(2\right)}$, where

[TABLE]

[TABLE]

Note that so far the derivation of $V$ and $W$ follows the one of ref. [11] except that we changed the notation slightly and determined the sign factors of the vertices, which are overlooked by ref. [11] in eqs. (21) and (27).

Now, let us simplify $\tilde{V}$ and $\tilde{W}$. We denote the integral with index $i$ in (48) as $\tilde{V}_{qpr}^{\left(i\right)}$. It can be shown as follows that all the $M$ integrals $\tilde{V}_{qpr}^{\left(i\right)}$ are the same. For the integral with index $i$, we can rename all integration variables as $\theta_{j}\to\theta_{j+1}$ and then use the cyclic constraint (8) to bring $\psi_{r}$ and the measure to their original form. The value of the integral is invariant under both changes but $\tilde{V}_{qpr}^{\left(i\right)}$ is changed to $\tilde{V}_{qpr}^{\left(i+1\right)}$. It implies that $\tilde{V}_{qpr}^{\left(i\right)}$ is independent of $i$ and we can choose $i=M$ for every integral to give

[TABLE]

To find the vertex satisfying the constraint (44), we let $V_{qpr}=\frac{1}{2}\left(\tilde{V}_{qpr}+\left(-\right)^{g_{p}g_{q}}\tilde{V}_{pqr}\right)$, where $\tilde{V}_{pqr}$ can be obtained by exchanging $p\leftrightarrow q$, $K\leftrightarrow L$:

[TABLE]

We therefore have

[TABLE]

where $h_{V}$ is given by (46b).

We perform a similar calculation for the $\tilde{W}$ vertex. All the integrals of $W^{\left(1\right)}$ and $W^{\left(2\right)}$ are independent of the indices $k$ and $l$. So we can simply replace the sums over $k$ and $l$ with the factor $K\times L$. We then rename $\eta_{1},\dots,\eta_{L}$ to $\theta_{K+1},\dots,\theta_{M}$ and fix the indices as $k=K,\,l=K+1$ for $W^{\left(1\right)}$ and $k=1,\,l=M$ for $W^{\left(2\right)}$ to give

[TABLE]

Exchanging $p\leftrightarrow q$ and $K\leftrightarrow L$, we have

[TABLE]

Renaming the integral variables as $\left\{\theta_{1},\cdots,\theta_{L}\right\}\to\left\{\theta_{K+1},\cdots,\theta_{M}\right\}$, $\left\{\theta_{L+1},\cdots,\theta_{M}\right\}\to\left\{\theta_{1},\cdots,\theta_{K}\right\}$, under which $h_{L,L+1}+h_{M,1}$ becomes $h_{M,1}+h_{K,K+1}$, and then applying the property that $\bar{\psi}_{r}\left(\theta_{1}\cdots\theta_{M}\right)$ is invariant under the cyclic permutation $\theta_{k}\to\theta_{k+1}$333One can show that $\bar{\psi}_{r}\left(\theta_{1}\cdots\theta_{M}\right)$ is invariant under the cyclic permutation $\theta_{k}\to\theta_{k+1}$ as follows. From eq. (39a) and (39b), we see that $\bar{\psi}_{G}^{s=1}\to\left(-\right)^{M-1}\bar{\psi}_{G}^{s=1}$ as $\theta_{k}\to\theta_{k+1}$. It then follows that $\bar{\psi}_{G}$ transforms as $\bar{\psi}_{G}\to\left(-\right)^{s\left(M-1\right)}\bar{\psi}_{G}$. From the cyclic constraint (25), we see that $\tilde{F}_{\left\{r\right\}}$ transforms in the same way as $\bar{\psi}_{G}$. Therefore, $\bar{\psi}_{r}=\tilde{F}_{\left\{r\right\}}\bar{\psi}_{G}$ is invariant., we obtain that $\tilde{W}_{rqp}=\left(-\right)^{g_{p}g_{q}}\tilde{W}_{rpq}$, which implies that $W_{rpq}=\left(\tilde{W}_{rpq}+\left(-\right)^{g_{p}g_{q}}\tilde{W}_{rqp}\right)=\tilde{W}_{rpq}$.

Let us now get rid of the integral in the expression of $V$. For simplicity, we consider the $s=1$ case. We use (24a) and (40a) to write $\psi_{r}=F_{\left\{r\right\}}^{\dagger}\psi_{G}^{\left(M\right)}$, $\bar{\psi}_{r}=\tilde{F}_{\left\{r\right\}}\bar{\psi}_{G}^{\left(M\right)}$ and similarly for states $p$ and $q$. We then use (33) to express $\psi_{G}^{\left(M\right)}$ in terms of $\psi_{G}^{\left(L\right)}$ and $\psi_{G}^{\left(K\right)}$. By a little algebra, we arrive at

[TABLE]

The ground eigenfunctions $\psi_{G}^{\left(L\right)}$ and $\psi_{G}^{\left(K\right)}$ are annihilated by any lowering eigenoperators of the small chains. Their conjugates $\bar{\psi}_{G}^{\left(L\right)}$ and $\bar{\psi}_{G}^{\left(K\right)}$ can be annihilated by any raising eigenoperators of the small chains, as eq. (B.3) shows. Therefore, the rhs of (52) can be interpreted as a vacuum expectation value of the operator $F_{\left\{q\right\}}F_{\left\{p\right\}}h_{V}F_{\left\{r\right\}}^{\dagger}\exp\left(\frac{1}{2}\sum_{kl}f_{k}^{\dagger}D_{kl}f_{l}^{\dagger}\right)$. We therefore have

[TABLE]

where the vacuum is understood to be the state annihilated by all $F_{i}^{\left(K\right)}$ and $F_{i}^{\left(L\right)}$. We perform a similar calculation for $W_{rpq}$ and find

[TABLE]

Note that $V_{qpr}^{s=1}$ and $W_{rpq}^{s=1}$ have the same sign factor $\left(-\right)^{L\left(g_{p}-K\right)}$. We shall see that physical observables, like one-loop self-energies, only depend on products like $W_{rpq}V_{qpr}$. It implies that the sign factors are unphysical and can be dropped in the calculation of physical observables. So, for arbitrary $s$, up to a common unphysical sign factor, we can express $V$ and $W$ as products of vacuum expectation values over spin index $a$. We therefore obtain the formula (45).

To calculate the vacuum expectation values, we need to express $h_{V}$ and $h_{W}$ in terms of eigenoperators. From eqs. (C.4), (C.3), and (C.5), we have

[TABLE]

where

[TABLE]

5 One-loop self-energy

One application of the interaction vertices is to calculate the one-loop self-energy, i.e., the $\mathcal{O}\left(1/N^{2}\right)$ correction to energy spectrum. In this section, we will first express the one-loop self-energy in terms of cubic interaction vertices [11]. We then apply the results of previous sections and obtain a formula for analytic and numerical computation.

For a finite $N$ energy eigenstate, we use the ansatz

[TABLE]

where the coefficients $C_{pq}=\left(-\right)^{g_{p}g_{q}}C_{qp}$ are c-numbers of order $1/N$. Imposing the eigenvalue equation $\left(H-E\right)\left|E\right\rangle=0$ and using perturbation theory, we obtain [11]

[TABLE]

where $\Delta E_{r}$ is the leading order correction to $E_{r}$, i.e., $E=E_{r}+\Delta E_{r}+\mathcal{O}\left(1/N^{3}\right)$. We stress that the vertices in (57) should be the ones satisfying the constraint (44); otherwise, it would lead to an incorrect $\Delta E_{r}$.

We now apply the formulas of $V$ and $W$ to (57). Let us first consider the $s=1$ case. The zero modes require special treatment. Substitute (45a) and (45b) into (57) and write the sum over zero modes explicitly,

[TABLE]

where we wrote $F_{0}^{\left(K\right)}$ as $F_{0,K}$ for convenience and ${\sum_{p,q}}^{\prime}$ indicates the sum over states without zero modes. We can replace $F_{0}^{\left(K\right)}$ and $F_{0}^{\left(L\right)}$ with $f_{0}$ and $f_{M-1}$ given the following reasoning. The sum over $\lambda$ and $\kappa$ produces four terms. For the term with $\lambda=\kappa=1$, we find $F_{0}^{\left(K\right)}F_{0}^{\left(L\right)}=e^{i\pi/4}f_{0}f_{M-1}$ by eqs. (26a) and (26d). The phase is irrelevant. The $\lambda=1,\,\kappa=0$ and $\lambda=0,\,\kappa=1$ terms are quadratic forms of $F_{0}^{\left(K\right)}$ and $F_{0}^{\left(L\right)}$. One can easily verify that $F_{0}^{\left(K\right)*}F_{0}^{\left(K\right)}+F_{0}^{\left(L\right)*}F_{0}^{\left(L\right)}=f_{0}^{*}f_{0}+f_{M-1}^{*}f_{M-1}$. So, the sum over $F_{0}^{\left(K\right)}$ and $F_{0}^{\left(L\right)}$ can be replaced by the one over $f_{0}$ and $f_{M-1}$. We then have

[TABLE]

For arbitrary $s$, $\left|\det C\right|$ is replaced by $\left|\det C\right|^{s}$, and each term inside the summation becomes a product over $a$. So, we have

[TABLE]

Note that the sum over $\lambda_{i,j}$ can be performed for each $a$ independently. So, we can move the sum over $\lambda_{i,j}$ inside the product over $a$ to give

[TABLE]

where $Z^{a}=\left(1,\,f_{0}^{a},\,f_{M-1}^{a},\,f_{0}^{a}f_{M-1}^{a}\right)$.

5.1 Ground energy correction

In principle, we can now calculate one-loop self-energy for any single trace energy state with eq. (58). But in general, the calculation is tedious. Let us consider the simplest case that $\psi_{r}$ is the ground state, i.e., $F_{\left\{r\right\}}=1$. For convenience, we denote $\left\langle O,1\right\rangle_{V,W}$ as $\left\langle O\right\rangle_{V,W}$. We only consider the $s=1$ case here, since $s>1$ cases are simply products of the $s=1$ case.

We need to calculate the vacuum expectation value $\left\langle\cdots h\exp\left(\frac{1}{2}f_{k}^{\dagger}D_{kl}f_{l}^{\dagger}\right)\right\rangle$. In terms of eigenoperators, $h_{kl}$ contains quadratic terms of the form $A_{nm}^{\dagger}F_{m}^{\dagger}F_{n}^{\dagger}$, $A_{-n,m}F_{m}^{\dagger}F_{n}$, and $A_{nm}F_{n}F_{m}$ and a constant term $\mu$, as eq. (53) shows. Since $F_{m}\exp\left(\frac{1}{2}f_{k}^{\dagger}D_{kl}f_{l}^{\dagger}\right)\psi_{G}^{\left(K\right)}\psi_{G}^{\left(L\right)}=F_{m}\psi_{G}^{\left(M\right)}=0$, only the $F^{\dagger}F^{\dagger}$ and the constant terms make a nonzero contribution:

[TABLE]

To calculate the result of $A_{nm}^{\dagger}F_{n}^{\dagger}F_{m}^{\dagger}$ term, we need to express $F_{m}$ in terms of a linear combination of $f_{k}$ and $f_{k}^{\dagger}$, as (27) shows, and commute $f_{k}$ through the exponential. This is done in Appendix D. Using eq. (D.1), we have

[TABLE]

where

[TABLE]

with $\mu_{V,W}$ and $A_{V,W}$ defined in (54). Finally, the vacuum expectation values on the rhs of (59) can be calculated using

[TABLE]

where $S_{2n}$ is the set of all permutations of $2n$ integers, $\left(-\right)^{P}$ is the signature of permutation $P$, and ${\sum}^{\prime}$ indicates the sum over permutations satisfying

[TABLE]

Combining the above together, we can calculate the one-loop self-energy of the ground state. As the complete formula is very complicated, we do not bother writing it here. In Appendix E, we show examples of using formula (58) to calculate the one-loop self-energies of the $M=3,\,s=1$ and $M=3,\,s=2$ cases. For $M=3,\,s=1$, we have

[TABLE]

and for $M=3,\,s=2$, we have

[TABLE]

In general, $\Delta E_{G}$ is a polynomial of $\xi$ of degree $2s$.

5.2 Large $M$ behavior

We conclude this section by considering the large $M$ behavior of $\Delta E_{G}$444The large $M$ discussion is mainly based on comments by Charles Thorn.. The vacuum expectation values in (58) only depends on the ratio $K/M$ and therefore can be considered as $\mathcal{O}\left(1\right)$. So, when $M$ is large,

[TABLE]

In (60), the factor $KLM$ scales as $M^{3}$, $\left|\det C\right|$ scales as $M^{-s/4}$ by eq. (30), and the sum over $K$ gives another factor of $M$. These three parts produce a factor scale as $M^{4-s/4}$.

We then consider the large $M$ behavior of $1/\left(E_{G}-E_{p}-E_{q}\right)$. When $s$ is even, both $p$ and $q$ can be ground states, and hence $1/\left(E_{G}-E_{p}-E_{q}\right)\sim\mathcal{O}\left(M\right)$ by eq. (22). When $s$ is odd, $M$ has to be odd in order to have the physical $M$-bit ground state, and one of the small strings must have an even bit number. It implies that the ground state of one small chain is forbidden by the cyclic constraint (25). Therefore, $1/\left(E_{G}-E_{p}-E_{q}\right)\sim\mathcal{O}\left(1\right)$ for odd $s$.

Combining the above together, we have

[TABLE]

In analogy with the standard string theory, we can infer from eq. (61) the critical Grassmann dimension of the model, where Lorentz invariance in $1+1$ dimensions is regained. In the lightcone coordinates, $P^{+}$ is identified as $mM$, and $P^{-}$ is identified as $E$. So, the Poincar invariant dispersion relation $P^{-}\sim 1/P^{+}$ implies $E\sim 1/M$. Therefore, the Lorentz invariance requires $s=24$. The model in the special $s=24$ case is called the protostring model[11].

6 Numerical results

We have derived a formula for the one-loop correction to the ground energy. As Appendix E shows, however, the calculation is tedious even for the simplest case. We therefore turn to numerical computation555The source code for the numerical computation can be found in ref. [12].. As the complexity of the calculation grows dramatically, the highest $M$ for which we performed numerical computation is 27 for $s=1$ and 16 for $s=2$ and continues decreasing as $s$ increases. Since only the ground energy is considered, we will simply write the ground energy as $E$ and its correction as $\Delta E$ and also suppress the $1/N^{2}$ factor.

We first compare the perturbation results with the exact numerical results, which are obtained by the method of ref. [10]. Figure 2 plots the change of ground energy with respect to the $1/N$ for $M=3$ and $5$ in the $s=1$ case. The solid lines are exact numerical results, and the dashed lines are $\mathcal{O}\left(1/N^{2}\right)$ perturbation results. We see that the two types of results match very well for $N$ large enough. One interesting observation is that, when $N$ is small, the perturbation results of $M=3$ are lower than the exact results, while the perturbation results of $M=5$ are above the exact results. It implies that the $\mathcal{O}\left(1/N^{4}\right)$ correction is positive for $M=3$ and negative for $M=5$.

We then verify the large $M$ behavior of $\Delta E$. Instead of plotting $\Delta E$ with respect to $M$, we study its “inner structure”, that is the contribution of each $K$ to $\Delta E$, denoted by $\Delta E_{i}$ and defined as

[TABLE]

Since the power of $M$ in the large $M$ behavior of $\Delta E_{i}$ is 1 lower than that of $\Delta E$, we introduce the normalized $\Delta E_{i}$ to remove the $M$ dependence:

[TABLE]

We expect that, for fixed $s$ and $\xi$, $\Delta\hat{E}_{i}$ only depends on the ratio $K/M$.

The plots of $\Delta\hat{E}_{i}$ as a function of $i=K/M$ are shown in Fig. 3, where $\xi=0$ for all four plots. When $s$ is odd, only odd values of $M$ are allowed and each $M$ has two curves, one for odd $K$ points and the other one for even $K$ points, for a reason will be clear shortly. For $s=2,\,3,\,4$ cases, the curves of different $M$ values are very close to each other, so the asymptotic behavior is evident. For the $s=1$ case, the gaps between consecutive curves become smaller as $M$ increases, which is consistent with the expected asymptotic behavior. It is therefore fair to conclude that the large $M$ behavior is confirmed.

The fact that there are two curves for each $M$ in odd $s$ cases can be understood as follows. Let us consider the $s=1$ case and take examples of $K=1$ and $K=2$, where the former has a much lower contribution to $\Delta E$ than the latter according to the plots. Assuming that $M$ is large enough, we have the other small chains with bit number $L\gg K$. Since $M$ is odd, $L$ is even for $K=1$ and odd for $K=2$. The lowest energies of these two cases, which are equal to $-4\cot\frac{\pi}{2L}-4\cos\frac{\pi}{2K}+8$ according to (24b) and the cyclic constraint (25), differ only by $\mathcal{O}\left(1\right)$. Now, we compare these two cases in the low energy regime, in which the gap between energy levels and the lowest energies are at most of order $1/M$. Consider the numbers of states in the low energy regime. Because of the cyclic constraint, only chains with an even bit number have excited states with energy gaps of order $1/M$ above the lowest energy. For $K=1$, the number of states in the low energy regime roughly equals $P\left(L/2\right)$, the partition number of $L/2$; for $K=2$, it equals $P\left(2/2\right)=1$. It implies that the low energy regime of $K=1$ is much denser than the one of $K=2$. Therefore, for large enough $M$, the $K=1$ case has much lower average energy than the $K=2$ case. This reasoning holds when $K$ is small. Hence, small odd $K$ cases have a lower contribution to $\Delta E$ than small even $K$.

We next consider the effect of the $\xi$ parameter. Figure 4 shows the plots of $\Delta\hat{E}_{i}$ with respect to $i=K/M$ for $s=2$ with different values of $\xi$. From the plots, the $\xi=0.5$ and $\xi=1.5$ cases show a smooth asymptotic behavior as the cases in Fig. 3. But when $\xi$ is close to $1$, curves are not smooth and intersect each other. When $\xi<1$, the curve moves downward as $M$ increases, which implies that $\Delta E$ decreases as $M$ increases. So, $\Delta E$ is not bounded from below, and the system is not stable. In contrast, when $\xi>1$, the curve moves upward as $M$ increases, which implies a stable system. This is related to a special feature of the $\xi=1$ case. Recall that the Hamiltonian has an $H_{1}$ part shown as (A.3a). This part produces a term like $-s\mathrm{Tr}\,\bar{\phi}_{12\cdots s}\bar{\phi}_{12\cdots s}\phi_{12\cdots s}\phi_{12\cdots s}$. When $s$ is even, $\phi_{12\cdots s}$ is a scalar and this term behaves like a scalar potential with a negative coefficient, which leads to a dangerous instability. But when $\xi=1$, this term is canceled exactly by $s\xi\Delta H$. That being said, for even $s$, $\xi=1$ is the minimal value for the potential to be bounded from below. To build a physical string bit model for even $s$, we should require $\xi\geq 1$.

We next study the dependence of $\Delta\hat{E}=\sum_{i}\Delta\hat{E}_{i}$ on $s$. Figure 5 plots the change of $\ln\left|\Delta\hat{E}\right|$ with respect to $s$ for chains of $M=5$ and $M=6$. For $M=5$, we sampled $s$ from 1 to $10$; for $M=6$, only even $s$ points are sampled as its ground states only survive in even $s$ cases. For each $M$, we choose $\xi=0,\,1,\,2,\,3$. For $M=6$, all the curves almost rise linearly. Of all four curves, $\xi=3$ is the steepest one, and $\xi=1$ is the flattest one. $\xi=0$ and $\xi=2$ almost coincide with each other. For the $M=5$ case, the overall trends of the curves are the same as $M=6$ except for slight oscillations between even and odd $s$ points. For $\xi=0,\,1$, the oscillation is relatively noticeable, and for $\xi=3$, it is negligible. Actually, if only even $s$ points of $M=5$ are sampled, the plots are almost the same as $M=6$. The exponential dependence of $\Delta\hat{E}$ on $s$ stems from the fact that each ground state has $2^{s}$ degeneracies. The fact that $\xi=1$ has a lower slope than others is also related to the fact that $\xi=1$ is the boundary for $\Delta E$ to be bounded from below.

7 Conclusion

We have presented a formalism to calculate the cubic interaction vertices for the stable string bit model. With the vertices, we calculated the one-loop self-energies of the model in both analytical and numerical ways.

From the large $M$ behavior of one-loop self-energies, we found that the Lorentz invariance requires the critical dimension of the model to be $s=24$, which then leads to the protostring model. One interesting interpretation of $s=24$ is as follows[13]. Out of the 24 dimensions, 16 of them are paired to form 8 compactified bosonic dimensions, and the rest 8 remain as fermionic dimensions. Thus, it has the same degrees of freedom as the superstring model. The large $M$ behavior of $\Delta E_{G}$ is determined by the ground states contribution of the small chains. Notwithstanding that the number of excited states grows exponentially with respect to $M$ [10], the excited states contributions are canceled out due to the fermionic nature of string bits. These results support the idea of formulating string theory by string bit models.

The future research of this work can be done in several ways. One can improve the numerical computation to study higher $M$ or $s$ cases. One can also apply the formalism to other calculations, e.g., four strings interaction, or to study higher-loop corrections and find the Feynman rules of the model.

8 Acknowledgments

We thank Charles Thorn, Pierre Ramond, and Sourav Raha for helpful discussions and comments on this work. This research was supported in part by the Department of Energy under Grant No. DE-SC0010296.

Appendix A Hamiltonian and its action on color singlets

The (anti)communication relations among string bit creation and annihilation operators is

[TABLE]

where the sum runs over all permutations of $1,2,\dots,n$.

The Hamiltonian of the model consists of $\mathcal{O}\left(1\right)$ terms and $\mathcal{O}\left(1/N\right)$ terms. The $\mathcal{O}\left(1\right)$ terms are the generalization of the Hamiltonian of the $s=1$ string bit model [8, 10]

[TABLE]

where $\bar{a}=\bar{\phi}$ and $\bar{b}=\bar{\phi}_{1}$. $H^{s=1}$ produces the Green-Schwarz Hamiltonian [14, 15] at $N=\infty$.

$H^{s=1}$ is generalized to $\sum_{i=1}^{5}H_{i}$, where[9, 11]

[TABLE]

One can check that for $s=1$ eq. (A.3) is reduced to eq. (A.2) if one identifies $\bar{\phi}$ as $\bar{a}$ and $\bar{\phi}_{1}$ as $\bar{b}$.

We now add $\mathcal{O}\left(1/N\right)$ terms to the Hamiltonian. As refs. [10, 8] show, the $N=\infty$ behavior is not affected by the $\mathcal{O}\left(1/N\right)$ terms

[TABLE]

By analogy with $H^{s=1}$, $\Delta H^{s=1}$ can be generalized to the arbitrary $s$ case as

[TABLE]

Combining the two parts together, we have the complete form of the Hamiltonian for arbitrary $s$,

[TABLE]

where $\xi$ is a real constant.

$H$ commutes with the supersymmetry operators

[TABLE]

[TABLE]

which will guarantee equal numbers of bosonic and fermionic eigenstates at each energy level.

Using the commutation relations (A.1), we obtain the action of $H_{i}$ on single trace states [11]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Similarly, the action of $\Delta H$ on a single trace state is

[TABLE]

The actions of $H_{i}$ on double traces are [11]

[TABLE]

[TABLE]

Similarly, the action of $\Delta H$ on double traces is

[TABLE]

A.1 Derivation of $\Delta H$

It is not obvious how to generalize $\Delta H^{s=1}$ to arbitrary $s$ cases. We actually obtain the generalization from the relation

[TABLE]

which has been proven in Appendix E of ref. [10] for $s=1$. Here, the color operator $G$ is defined as [7]

[TABLE]

and both $\Delta H$ and $H^{\prime}$ are supersymmetric and of $\mathcal{O}\left(1/N\right)$. The notation${:}\!\mathrel{\phi_{a_{1}\cdots a_{n}}\bar{\phi}_{a_{1}\cdots a_{n}}}\!{:}$ indicates the normal ordering of $\phi_{a_{1}\cdots a_{n}}\bar{\phi}_{a_{1}\cdots a_{n}}$. In $s=1$, we have[10]

[TABLE]

One can verify that the action of $G_{\alpha}^{\beta}$ on any color singlet vanishes: $G_{\alpha}^{\beta}\left|\text{any color singlet}\right\rangle=0$. We therefore have $\left(\Delta H-H^{\prime}\right)=0$ in the color singlet space.

To find $\Delta H$, we expand $\mathrm{Tr}\,G^{2}$ and match its terms with $H^{\prime s=1}$ and $\Delta H^{s=1}$. By direct calculation, we have

[TABLE]

We calculate each term on the rhs of $\mathrm{Tr}\,G^{2}$ and obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Combining the above together, we have

[TABLE]

Comparing the terms of $\mathrm{Tr}\,G^{2}$ with $H^{\prime s=1}$ and $\Delta H^{s=1}$, we can identify

[TABLE]

and $\Delta H$ as (A.6). One can verify that both $H^{\prime}$ and $\Delta H$ commute with the supersymmetry operators $Q^{a}$ (A.8).

Appendix B Verifying the normalization condition

for $\bar{\psi}_{G}$

In this Appendix we show that the conjugate eigenfunction of $s=1$, defined as eq. (39), satisfies the normalization condition (34). We first show that $\int d^{M}\theta\,\bar{\psi}_{G}\psi_{G}=1$. For odd $M$,

[TABLE]

where in the last step we used666We do not prove the formula (B.1) here. But we have verified it by the Mathematica program.

[TABLE]

Similarly, we can show $\int d^{M}\theta\,\bar{\psi}_{G}\psi_{G}=1$ for even $M$.

To show$\int d^{M}\theta\,\bar{\psi}_{G}\psi_{r}=0$ for $r\neq G$, it suffices to show that $\int d^{M}\theta\,\bar{\psi}_{G}F_{k}^{\dagger}\psi^{\prime}$ vanishes for all $0\leq k\leq M-1$ and any eigenfunction $\psi^{\prime}$. If $k=0$, it clearly vanishes because both $F_{0}^{\dagger}$ and $\bar{\psi}_{G}$ contain the Grassmann odd operator $\alpha_{0}$. If $0<k<M/2$,

[TABLE]

The rhs of (B.2) vanishes because of

[TABLE]

which can be verified by checking that

[TABLE]

[TABLE]

Similarly, we can show that $\tilde{F}_{k}^{\dagger\pm}\bar{\psi}_{G}=0$ for $M/2\leq k\leq M-1$. Therefore, the normalization condition $\int d^{M}\theta\,\bar{\psi}_{G}\psi_{r}=\delta_{Gr}$ is proved.

Appendix C Calculation of $h_{kl}$

In this Appendix, we will find the expression of $h_{kl}^{a}$ in terms of lowering and raising operators. The $h_{kl}^{a}$ in the language of $\theta_{k}$ is

[TABLE]

We now temporarily drop the last two constant terms and will add them back in the end of the calculation.

Using (13b), we express $\theta_{k}$ and $\frac{d}{d\theta_{k}}$ in terms of $\alpha_{n}$ and $\beta_{n}$:

[TABLE]

Substituting the above into $h_{kl}$ and rearranging, we obtain

[TABLE]

where $h_{kl}^{\left(0\right)}$ are the terms with zero modes and $h_{kl}^{\left(1\right)}$ are the terms without,

[TABLE]

[TABLE]

Let us first consider $h_{kl}^{\left(1\right)}$. We express nonzero modes $\alpha_{m}$ and $\beta_{m}$ in terms of raising and lowering operators. Using

[TABLE]

we have

[TABLE]

We then apply eqs. (C.1) to $h_{kl}^{\left(1\right)}$, collect like terms, and antisymmetrize $F^{\dagger}F^{\dagger}$ and $FF$ terms to give

[TABLE]

where

[TABLE]

Similarly, applying

[TABLE]

[TABLE]

[TABLE]

[TABLE]

to $h_{kl}^{\left(0\right)}$ yields

[TABLE]

where

[TABLE]

Now from eq. (C.3), we see that

[TABLE]

Hence, to add $h_{kl}^{\left(0\right)}$ terms to $h_{kl}^{\left(1\right)}$, we can simply change the $m,n$ index of (C.2) to start from [math]. Finally, adding back the constant terms, we have

[TABLE]

where $A_{mn}$ is given by (C.3) and

[TABLE]

Appendix D Calculation of $\left\langle\cdots\bar{A}_{nm}F_{n}^{\dagger}F_{m}^{\dagger}\exp\left(\frac{1}{2}f_{k}^{\dagger}D_{kl}f_{l}^{\dagger}\right)\right\rangle$

In this Appendix we will derive the formula

[TABLE]

where $\bar{A}\equiv A^{\dagger}$ and $B=C^{-1}\bar{A}\left(C^{-1}\right)^{T}$, and the relations among $F_{n}^{\dagger}$, $f_{k}$, and $f_{k}^{\dagger}$ are given by

[TABLE]

with $C_{mn}D_{nl}+S_{ml}=0$.

Let $X=\frac{1}{2}f_{k}^{\dagger}D_{kl}f_{l}^{\dagger}$, $\left|G\right\rangle=\exp\left(X\right)\left|0\right\rangle$; then

[TABLE]

Now let us calculate each term in the parentheses of the rhs of eq. (D.3). For the first term

[TABLE]

It follows that the higher order commutations all vanish. Substituting eqs. (D.4) into eq. (D.3), we have

[TABLE]

where in the third equality we antisymmetrized the $2DS^{\dagger}\bar{A}C^{*}$ term to be $DS^{\dagger}\bar{A}C^{*}-\left(DS^{\dagger}\bar{A}C^{*}\right)^{T}$ and then used the fact that $\bar{A}$ and $D$ are antisymmetric matrices. Now,

[TABLE]

where in the second-to-last equality $I+DD^{\dagger}=C^{-1}C^{-1\dagger}$ follows from eqs. (28) and (32). We therefore have

[TABLE]

which implies (D.1).

Appendix E Examples of $M=3$

In this Appendix, as a demo of using (58) to calculate one-loop self-energy, let us consider the one-loop self-energy for the ground state of the $M=3,\,s=1$ and $M=3,\,s=2$ cases. For $M=3$, we only need to calculate the $K=1$ case since the contribution of $K=2$ is the same as $K=1$.

The $C$, $S$, and $D$ matrices are

[TABLE]

and matrices $A$, $B$, and constant $\mu^{\prime}$ are

[TABLE]

The operators $f_{n}$ are

[TABLE]

For $s=1$, the eigenfunctions and their conjugates of $1$-bit and $2$-bit chains are shown in Table 2.

The contribution of $K=1$ to the energy correction is

[TABLE]

So, we need to calculate $\left\langle F_{1}^{\left(2\right)}f_{0}\right\rangle_{V,W}$ and $\left\langle F_{1}^{\left(2\right)}f_{1}\right\rangle_{V,W}$:

[TABLE]

[TABLE]

Likewise,

[TABLE]

Substituting above results and $E_{G}=-4\sqrt{3}$ into (E.1) yields $\Delta E_{G}^{K=1}=-\frac{3}{2}\left(3\sqrt{3}-5\right)\xi^{2}+\left(12-7\sqrt{3}\right)\xi-\frac{3}{4}\left(3\sqrt{3}-5\right)$. We then have

[TABLE]

For $s=2$, the matrices and constants are the same as the $s=1$ case. But as Table 3 shows, the energy eigenstates of small chains are different. The energy correction now is given by

[TABLE]

Since we have calculated the $\left\langle F_{1}^{\left(2\right)}f_{0}\right\rangle_{V,W}$ and $\left\langle F_{1}^{\left(2\right)}f_{2}\right\rangle_{V,W}$ in the $s=1$ case, we only need to find $\left\langle 1\right\rangle_{V,W}$ and $\left\langle f_{2}f_{0}\right\rangle_{V,W}$.

For $K=1$,

[TABLE]

[TABLE]

Likewise

[TABLE]

Substituting the above into eq. (E.2), we obtain

[TABLE]

From the results and the formula (58), we see that $\Delta E_{G}$ is a polynomial of $\xi$ of degree $2s$.