Addition of SU(3) generators and its Singlet Hilbert space

TL;DR

This paper constructs the singlet Hilbert space for SU(3) generators in lattice QCD, explicitly building basis states with Schwinger Bosons and analyzing operator actions to facilitate understanding of gauge invariance.

Contribution

It introduces an explicit construction of the singlet Hilbert space for SU(3) using Schwinger Bosons, advancing the mathematical framework for lattice QCD.

Findings

Explicit basis states for the singlet Hilbert space are constructed.

The action of basic singlet operators on these states is computed.

Provides a foundation for solving Gauss law in lattice QCD.

Abstract

We construct the singlet Hilbert space associated with addition of SU(3) generators. This corresponds to the solution of Gauss law in lattice QCD. The normalized basis states are explicitly constructed using Schwinger Bosons. Further, we compute the action of basic singlet operators on these basis states.

Figures

Tables

Table 1. Table 1: Action of various SU(2) invariant operators on the | n 1 , n 2 , n 3 ⟩ ket subscript 𝑛 1 subscript 𝑛 2 subscript 𝑛 3 |n_{1},n_{2},n_{3}\rangle basis.

$N_i|n_1,n_2,n_3⟩=n_i|n_1,n_2,n_3⟩$

$\left(a_3^{†}\cdot {\tilde{a}}_i^{†}\right)|n_i,n_{\overline{i}},n_3⟩=\frac{1}{2}\sqrt{\left(m+n_i+n_{\overline{i}}+4\right)\left(m+n_i-n_{\overline{i}}+2\right)}|n_i+1,n_{\overline{i}},n_3+1⟩$

$\left(a_3^{†}\cdot a_1\right)|n_1,n_2,n_3⟩=\frac{1}{2}\sqrt{\left(n_3-n_1+n_2+2\right)\left(n_1-n_3+n_2\right)}|n_1-1,n_2,n_3+1⟩$

$\left(a_3\cdot {\tilde{a}}_1\right)|n_1,n_2,n_3⟩=\frac{1}{2}\sqrt{\left(n_3+n_1+n_2+2\right)\left(n_3+n_1-n_2\right)}|n_1-1,n_2,n_3-1⟩$

Equations

$$E^a$$

$$|j,m⟩\equiv |n_1,n_2⟩=\frac{(a_1^{†}{)}^{n_1}(a_2^{†}{)}^{n_2}}{\sqrt{n_1!n_2!}}|0⟩$$

$$|n_1,n_2,n_3⟩=|l_{12},l_{23},l_{31}⟩=\frac{(a_1^{†}\cdot {\tilde{a}}_2^{†}{)}^{l_{12}}(a_2^{†}\cdot {\tilde{a}}_3^{†}{)}^{l_{23}}(a_3^{†}\cdot {\tilde{a}}_1^{†}{)}^{l_{31}}}{\sqrt{l_{12}!l_{23}!l_{31}!(l_{12}+l_{23}+l_{31}+1)!}}|0⟩$$

$$k_1$$

$$k_2$$

$$k_0$$

$$a_{\alpha }^{†}={\bar{a}}_{\alpha }^{†}-\frac{1}{n+m+1}{k}_{+}{\bar{b}}_{\alpha }{b}_{\alpha }^{†}={\bar{b}}_{\alpha }^{†}-\frac{1}{n+m+1}{k}_{+}{\bar{a}}_{\alpha }$$

$$\displaystyle\big{[}a_{\alpha}~{},~{}a^{\dagger}_{\beta}\big{]}$$

$$\displaystyle\big{[}b_{\alpha}~{},~{}b^{\dagger}_{\beta}\big{]}$$

$$\displaystyle\big{[}a_{\alpha}~{},~{}b^{\dagger}_{\beta}\big{]}$$

$$E^a=a^{†}\frac{{\lambda }^a}{2}a-b^{†}\frac{{\lambda }^{\ast a}}{2}b$$

$$|l,p⟩=\frac{1}{\sqrt{S(l,p)}}|l,p{⟩}_u=\{\begin{array}{ll}\frac{1}{\sqrt{S(l,p)}}\prod _{\stackrel{i=1,2,3}{j=1,2,3}}(a_i^{†}\cdot b_j{)}^{l_{ij}}(ϵa_1^{†}{a}_2^{†}{a}_3^{†}{)}^{|p|}|0⟩& p\ge 0\\ \frac{1}{\sqrt{S(l,p)}}\prod _{\stackrel{i=1,2,3}{j=1,2,3}}(a_i^{†}\cdot b_j{)}^{l_{ij}}(ϵb_1^{†}{b}_2^{†}{b}_3^{†}{)}^{|p|}|0⟩& p\le 0\end{array}$$

$$⟨l^{\prime },p^{\prime }|l,p⟩=\frac{1}{S(l,p)}⟨0|(ϵa_1{a}_2{a}_3{)}^{p^{\prime }}(a_i\cdot b_j{)}^{l_{ij}^{\prime }}(a_i^{†}\cdot b_j^{†}{)}^{l_{ij}}(ϵa_1^{†}{a}_2^{†}{a}_3^{†}{)}^p|0⟩={\delta }_{l{l}^{\prime }}\delta p{p}^{\prime }$$

$$O|l,p⟩$$

$$\displaystyle\equiv\sum\limits_{l^{\prime},p^{\prime}}~{}\Bigg{[}\sqrt{\frac{S(l^{\prime},p^{\prime})}{S(l,p)}}~{}\Bigg{]}~{}C_{l^{\prime},p^{\prime}}~{}|l^{\prime},p^{\prime}\rangle$$

$$(a_1^{†}\cdot b_2^{†})|l,p⟩$$

$$c_1^{12}$$

$$\displaystyle(a^{\dagger}_{1}\cdot a_{2})~{}|l,p\rangle=d_{1}^{12}~{}|l_{13}+1,l_{23}-1\rangle~{}+~{}d_{2}^{12}~{}\Bigg{|}{\scriptsize\begin{matrix}l_{12}+1,l_{21}-1\\ l_{31}+1,l_{32}-1\\ \end{matrix}}\Bigg{\rangle}$$

$$d_1^{12}$$

$$d_2^{12}$$

$$\displaystyle(b^{\dagger}_{1}\cdot b_{2})|l,p\rangle=e_{1}^{12}~{}|l_{31}+1,l_{32}-1\rangle~{}+~{}e_{2}^{12}~{}\Bigg{|}{\scriptsize\begin{matrix}l_{21}+1,l_{12}-1\\ l_{13}+1,l_{23}-1\end{matrix}}\Bigg{\rangle}$$

$$e_1^{12}$$

$$e_2^{12}$$

$$\displaystyle(a_{1}\cdot b_{2})|l,p\rangle=f_{1}^{12}|l_{12}-1\rangle+f_{2}^{12}~{}\Bigg{|}{\scriptsize\begin{matrix}l_{21}-1,\\ l_{23}+1,l_{32}-1\\ l_{31}+1,l_{13}-1,\end{matrix}}\Bigg{\rangle}+f_{3}^{12}~{}\Bigg{|}{\scriptsize\begin{matrix}l_{21}+1,~{}l_{12}-2\\ l_{13}+1,~{}l_{31}-1\\ l_{32}+1,l_{23}-1\end{matrix}}\Bigg{\rangle}$$

$$f_1^{12}$$

$$f_2^{12}$$

$$\displaystyle(l_{32})(l_{13})l_{21}\bigg{(}\frac{n_{1}+m_{1}+n_{2}+m_{2}+2|p|+3-l_{21}}{(n_{1}+m_{1}+|p|+1)(n_{2}+m_{2}+|p|+1)}\bigg{)}$$

$$f_3^{12}$$

$$\displaystyle~{}~{}~{}\bigg{(}\frac{(n_{1}+m_{1}+|p|+2)(l_{23})(l_{31})}{(n_{1}+m_{1}+|p|+1)}\bigg{)}\frac{l_{12}(l_{12}-1)}{(n_{2}+m_{2}+|p|+1)(n_{1}+m_{1}+|p|+2)}$$

$${\bar{f}}_1^{12}(l_{12})$$

$$\displaystyle-\frac{l_{23}(l_{13}+1)(l_{12})}{n_{1}+m_{1}+|p|+1}\bigg{(}\frac{n_{1}+m_{1}+|p|+1-l_{21}}{n_{2}+m_{2}+|p|+1}\bigg{)}$$

$$\displaystyle(\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2})|l,p\rangle=\begin{cases}g_{1}^{322}|l_{12}-1,p+1\rangle&p\geq 0\\ h_{1}^{322}\Bigg{|}{\scriptsize\begin{matrix}p+1\\ l_{31}+1,l_{23}+1\end{matrix}}\Bigg{\rangle}+h_{2}^{322}\Bigg{|}{\scriptsize\begin{matrix}p+1\\ l_{12}-1,l_{21}+1\\ l_{13}+1,l_{32}+1\end{matrix}}\Bigg{\rangle}&p<0\end{cases}$$

$$\displaystyle g_{1}^{322}=\Bigg{[}\sqrt{\frac{(|p|+3)^{3}}{\bar{f}_{1}^{12}(l_{12})}}\Bigg{]}l_{12}$$

$$\displaystyle h_{1}^{322}=\Bigg{[}\sqrt{\frac{(\bar{f}_{1}^{31}(l_{31}+1))\Big{|}_{l_{12}=l_{21}=l_{13}=0}\bar{f}_{1}^{23}(l_{23}+1)\Big{|}_{l_{12}=l_{21}=l_{13}=l_{31}=0}}{(|p|+2)^{3}}}\Bigg{]}(|p|+l_{12})$$

$$\displaystyle h_{2}^{322}=\Bigg{[}\sqrt{\frac{\bar{f}_{1}^{21}(l_{21}+1))\Big{|}_{l_{12}=0}(\bar{f}_{1}^{13}(l_{13}+1))\Big{|}_{l_{12}=l_{21}=0}(\bar{f}_{1}^{32}(l_{32}+1))\Big{|}_{l_{12}=l_{21}=l_{13}=l_{31}=l_{23}=0}}{(|p|+2)^{3}\bar{f}_{1}^{12}(l_{12})\Big{|}_{l_{12}=l_{21}=l_{13}=l_{31}=0}}}\Bigg{]}l_{12}$$

$$\displaystyle(\epsilon b_{3}b_{2}a^{\dagger}_{2})|l,p\rangle=\begin{cases}i_{1}^{322}\Bigg{|}{\scriptsize\begin{matrix}l_{12}-1,l_{21}+1\\ l_{23}-1,l_{31}-1\\ p+1\end{matrix}}\Bigg{\rangle}+i_{2}^{322}\Bigg{|}{\scriptsize\begin{matrix}l_{13}-1,l_{32}-1\\ p+1\end{matrix}}\Bigg{\rangle}&p\geq 0\\ j_{1}^{322}\Bigg{|}l_{21}+1,p+1\Bigg{\rangle}+j_{2}^{322}\Bigg{|}{\scriptsize\begin{matrix}l_{13}-1,l_{31}+1\\ l_{32}-1,l_{23}+1\\ l_{12}+1,p+1\end{matrix}}\Bigg{\rangle}+j_{3}^{322}\Bigg{|}{\scriptsize\begin{matrix}l_{12}-1,l_{21}+2\\ l_{23}-1,l_{32}+1,p+1\\ l_{13}+1,l_{31}-1\end{matrix}}\Bigg{\rangle}&p<0\end{cases}$$

$$i_1^{322}=\sqrt{\frac{{\bar{f}}_1^{21}(l_{21}+1){|}_{l_{12}=0}(|p|+3{)}^3}{{\bar{f}}_1^{12}(l_{12}){\bar{f}}_1^{23}(l_{23}){|}_{l_{12}=l_{21}=l_{13}=l_{31}=0}{\bar{f}}_1^{31}(l_{31}){|}_{l_{12}=l_{21}=l_{13}=0}}}{\bar{i}}_1^{322},$$

$$i_2^{322}=\sqrt{\frac{(|p|+3{)}^3}{{\bar{f}}_1^{13}(l_{13}){|}_{l_{12}=l_{21}=0}{\bar{f}}_1^{32}(l_{32}){|}_{l_{12}=l_{21}=l_{13}=l_{31}=l_{23}=0}}}{\bar{i}}_2^{322}$$

$$j_1^{322}=\sqrt{\frac{{\bar{f}}_1^{21}(l_{21}+1){|}_{l_{12}=0}}{(|p|+2{)}^3}}{\bar{j}}_1^{322},$$

Full text

I Introduction

In the Hamiltonian formulation of lattice gauge theory it was envisagedrs1 to construct explicit gauge invariant Hilbert space by a simple technique of solving the Gauss law called ’splitting the point’ . In this method, the most primitive element of this construction is the addition of two generators in the adjoint representation $E_{1}^{a},E_{2}^{a}$ to give a third $E_{3}^{a}$. The Gauss law amounts to looking at only those states which satisfy the condition $(E_{1}^{a}+E_{2}^{a}+E_{3}^{a})|\rangle=0$. Such states are indeed singlets and it is important to construct these explicitly at each vertex. These states have been explicitly constructedrs1 for SU(2). Here we pursue the same endeavor for SU(3) which is directly relevant to latticelgt QCD. Furthermore, we construct various basic singlet operators and their action on the above basis states.

We would like to remark about Clebsch- Gordon coefficients vis a vis the present problem. In the classic addition of generators of algebra we have the celebrated Wigner-Eckart theorem which states that the matrix elements of any arbitrary operator with definite irreducible representation (irrep) can be factorized into a Clebsch Gordon term which depends on the azimuthal quantum numbers and a ’reduced matrix element’ which only depends upon the casimir quantum numbers of the bra and ket states along with the casimirs of the operator irrep. Clebsch- Gordon coefficients for SU(3) have been explicitly calculated by Prakash et. al sarat . Here we are looking at a class of reduced matrix elements which occur in lattice gauge theory.

The plan of the paper is as follows. In section II, following a short description of the representation of SU(2) in terms of Schwinger bosons, we use them to construct an orthonormal basis for the most general singlet representation from three arbitrary SU(2) irreps. We then calculate the action of basic invariant operators construct out of three SU(2) Schwinger bosons on this basis. In section III, we repeat the same for SU(3).

II SU(2)

In this section, we will discuss the problem within the simpler context of SU(2). Consider the direct product space of three arbitrary SU(2) irrep with a basis $|j_{1},m_{1}\rangle|j_{2},m_{2}\rangle|j_{3},m_{3}\rangle$. These basis states are eigenstates of $E_{1}^{2},E_{2}^{2},E_{3}^{2},E_{1}^{(3)},E_{2}^{(3)},E_{3}^{(3)}$ with eigenvalues $j_{1}(j_{1}+1),j_{2}(j_{2}+1),j_{3}(j_{3}+1),m_{1},m_{2},m_{3}$ respectively, where $E_{1}^{a},E_{2}^{a},E_{3}^{a}$ are the generators of the three SU(2). In order to add these three angular momenta one writes the direct product space as a direct sum of irreducible representation spaces of SU(2) by changing the basis to simultaneous eigenvectors of $(E_{1}+E_{2}+E_{3})^{2},(E_{1}+E_{2}+E_{3})^{(3)},(E_{1}+E_{2})^{2},E_{1}^{2},E_{2}^{2},E_{3}^{2}$ with eigenvalues $j_{123},m_{123},j_{12},j_{1},j_{2},j_{3}$ (There are two other choices where $(E_{1}+E_{2})^{2}$ is replaced by $(E_{1}+E_{3})^{2}$ or $(E_{2}+E_{3})^{2}$). We are interested in extracting a subspace which satisfies the constraint $E_{1}^{a}+E_{2}^{a}+E_{3}^{a}=0$ i.e, a subspace which transform as a singlet under SU(2). As $j_{123}=m_{123}=0$ and $j_{12}=j_{3}$, such a space is characterised by three quantum numbers $j_{1},j_{2},j_{3}$ only. In order to construct such a singlet subspace it is convenient to write down prep ; schwinger each SU(2) generators in terms of harmonic oscillator doublets called Schwinger bosons:

[TABLE]

where $a^{\dagger}_{\alpha},a_{\beta}$ are the creation, annihilation operator doublets of harmonic oscillators satisfying $[a_{\alpha},a^{\dagger}_{\beta}]=\delta_{\alpha\beta}$. The basis of a general, irreducible representation space of SU(2) can be created as

[TABLE]

Above, $j=\frac{n_{1}+n_{2}}{2},m=\frac{n_{1}-n_{2}}{2}$ are the eigenvalues of $E^{2},E^{(3)}$ and $|0\rangle$ is the harmonic oscillator vacuum $a_{\alpha}|0\rangle=0$. $a^{\dagger}_{\alpha}$ transform under the fundamental representation of SU(2). Therefore, one can directly write down a basis of such a singlet space using Schwinger bosons rs1 as follows:

[TABLE]

where $\tilde{a}^{\dagger}_{i,\alpha}=\epsilon_{\alpha\beta}a^{\dagger}_{i,\beta}$ and $n_{1}=l_{12}+l_{31},n_{2}=l_{12}+l_{23},n_{3}=l_{23}+l_{31}$ are the eigenvalues of the number operators of the three schwinger bosons under consideration. These quantum numbers are related to the original $j_{i}$ quantum numbers as $n_{i}=2j_{i}$. The set of basic invariant operators can be constructed by contracting the operators in the set $\{a^{\dagger}_{i},\tilde{a}_{i}\}$ with those in $\{a_{j},\tilde{a}^{\dagger}_{j}\}$. Action of various invariant operators so constructed on this basis are tabulated in table 1.

III SU(3)

A general unitary irreducible representation of SU(3) is labelled by two nonnegative integers (p,q) and is of dimension $\frac{1}{2}(p+1)(q+1)(p+q+2)$. A basis of such a representation space can be created by diagonalizing a complete set of commuting operators given by $C_{1}=E^{a}E^{a},C_{2}=d_{abc}E^{a}E^{b}E^{c},I^{2}(E)=E^{(1)}E^{(1)}+E^{(2)}E^{(2)}+E^{(3)}E^{(3)},E^{(3)},Y$; where $C_{1},C_{2}$ are the two casimirs and $E^{a},I^{a},Y$ are the generators of SU(3), Isospin (SU(2) subgroup) and hypercharge (U(1) subgroup) respectively. $d_{abc}$ are the symmetric coefficients defined by $\{\frac{\lambda_{a}}{2},\frac{\lambda_{b}}{2}\}=\frac{1}{3}\delta_{ab}+d_{abc}\frac{\lambda_{c}}{2}$, where $\lambda_{a}$ are the generators of SU(3) in the 3 representation. Such an eigenbasis is denoted as $|p,q,I,M,Y\rangle$ with the eigenvalues of $C_{1},C_{2},I^{2},E^{(3)},Y$ given by $\frac{1}{3}(p^{2}+q^{2}+3p+3q+pq),\frac{1}{18}(p-q)(3+p+2q)(3+q+2p),I(I+1),M,Y$ respectively. We are interested in the subspace of the direct product space of three SU(3) irreps which satisfies the 8 constraints $E^{a}_{1}+E^{a}_{2}+E^{a}_{3}=0$ and therefore is parametrized by seven ($5\times 3-8$) quantum numbers. In order to construct such a singlet Hilbert space, it is convenient to go to the Schwinger boson representation of SU(3).

SU(3) has two inequivalent fundamental representations $3$ and $3^{*}$ which in the above notation is (1,0) and (0,1) respectively. However, a naive generalization of SU(2) schwinger bosons using a harmonic oscillator triplet and anti-triplet leads to multiplicity problemmc ; su3rmi In SU(2), the states (2) constructed by taking an arbitrary power of harmonic oscillator creation operators covers every SU(2) irrep basis state uniquely and any irrep basis state can be constructed this way. This is no longer true for SU(3). i.e, all positive integer powers of the creation operators $\bar{a}^{\dagger},\bar{b}^{\dagger}$ of $3$ and $3^{*}$ representation acting on the Schwinger boson vacuum gives the basis of a reducible space su3rmi in general. This problem arises due to existence of an SU(3) invariant operator $(\bar{a}^{\dagger}\cdot\bar{b}^{\dagger})$ which acting on any irrep state gives another state which transforms the same way (in contrast, in SU(2), $2$ representation is equivalent to $2^{*}$ representation and therefore no such invariants exist, for instance $a^{\dagger}\cdot\tilde{a}^{\dagger}=0$). One can use the following Sp(2,R) algebra to label mc such multiplicities.

[TABLE]

where $\hat{n}=a^{\dagger}\cdot a,\hat{m}=b^{\dagger}\cdot b$. One can define $k_{\pm}=k_{1}\pm ik_{2}$ which raises/ lowers $k_{0}$ eigenvalue as $[k_{0},k_{\pm}]=\pm k_{\pm}$. The above Sp(2,R) generators commutes with the SU(3) generators. The eigenvalue of $k_{0}$ distinguishes different multiplicities. States which satisfies the constraint $k_{-}|\rangle=0$, where $|\rangle$ is an arbitrary state, are SU(3) irrep states without multiplicites . One can construct irreducible Schwinger bosons su3rmi which directly creates SU(3) irrep states without multiplicities by solving the above constraints and is given by :

[TABLE]

These irreducible SU(3) prepotentials have the following modified commutation relations (Dirac brackets):

[TABLE]

Above, $\tilde{N}=\frac{1}{\hat{N}+2}$ where $\hat{N}=\hat{n}+\hat{m}$ is the number operator for the total number of oscillators. All other possible commutators vanish. Further, they automatically satisfy the following relation: $(a^{\dagger}\cdot b^{\dagger})|\rangle=0$.

SU(3) generators can be written in terms of irreducible SU(3) Schwinger bosons as follows

[TABLE]

where $\lambda^{a}$ are the eight generators of SU(3) in the 3 representation.

Consider three SU(3) irreps constructed using three pairs of irreducible Schwinger bosons $(a^{\dagger}_{i},b^{\dagger}_{i});i=1,2,3$. These three irreps are denoted pictorially as in figure 1. A basis of the SU(3) singlet representation can be constructed by taking all the singlet operators constructed from the Schwinger boson creation operators and acting on the Schwinger boson vacuum state as follows:

[TABLE]

where, $l\equiv\{l_{12},l_{21},l_{13},l_{31},l_{23},l_{32}\};(\epsilon a^{\dagger}_{1}a^{\dagger}_{2}a^{\dagger}_{3})\equiv(\epsilon_{\alpha\beta\gamma}a^{\dagger}_{1,\alpha}a^{\dagger}_{2,\beta}a^{\dagger}_{3,\gamma})$ (clockwise phase convention used); $|l,p\rangle_{u}$ is the unnormalized basis and the normalization $S(l,p)={}_{u}\langle l,p|l,p\rangle_{u}$ is calculated in appendix A. Orthogonality of the basis can be shown as follows. When $p\geq 0$

[TABLE]

Last equality is because when $l\neq l^{\prime}$ or $p\neq p^{\prime}$, there are extra annihilation or creation operators which hits $|0\rangle$ or $\langle 0|$ to give 0. This is because in the commutation relations (6), the difference between the total number of annihilation operators and creation operators are preserved in each of the terms. Similarly, orthogonality can be shown for $p<0$ case also.

The action of a gauge invariant operator $O$ on a general normalised basis state is

[TABLE]

Therefore, in order to compute the action of various operators on normalized states, one only need the action of those operators on the unnormalized states and ratio of norms. These are calculated in appendix A. Complete set of basic invariant operators are given by (a^{\dagger}_{i}\cdot b^{\dagger}_{j}),(a^{\dagger}_{i}\cdot a_{j}),(b^{\dagger}_{i}\cdot b_{j}),\Big{(}\epsilon(a/b^{\dagger})_{i}(a/b^{\dagger})_{j}(a/b^{\dagger})_{k}\Big{)},\Big{(}\epsilon(a^{\dagger}/b)_{i}(a^{\dagger}/b)_{j}(a^{\dagger}/b)_{k}\Big{)}. Action of various invariant operators on the normalized basis states are as follows (also pictorially shown in fig. 2). We define $n_{1}=l_{12}+l_{13},m_{1}=l_{21}+l_{31},n_{2}=l_{21}+l_{23},m_{2}=l_{12}+l_{32}$.

$a^{\dagger}_{1}\cdot b^{\dagger}_{2}$

[TABLE] 2. 2.

$(a^{\dagger}_{1}\cdot a_{2})$

[TABLE]

[TABLE] 3. 3.

$(b^{\dagger}_{1}\cdot b_{2})$

[TABLE]

[TABLE] 4. 4.

$(a_{1}\cdot b_{2})$

[TABLE]

[TABLE]

[TABLE] 5. 5.

[TABLE] 6. 6.

$(\epsilon b_{3}b_{2}a^{\dagger}_{2})$

[TABLE]

where

[TABLE]

where, $\bar{i}_{1}^{322},\bar{i}_{2}^{322},\bar{j}_{1}^{322},\bar{j}_{2}^{322},\bar{j}_{3}^{322}$ are given in (52). 7. 7.

($\epsilon a^{\dagger}_{1}a^{\dagger}_{2}a^{\dagger}_{3}$)

[TABLE]

where

[TABLE]

In the above expressions,

[TABLE]

In the above expressions all $l_{ij}$ are numbers labelling the L.H.S ket vector. The ket vectors in R.H.S are the same as the L.H.S, only the changed labels are mentioned in the notation. $\bar{f}_{1}^{21},\bar{f}_{1}^{13},\bar{f}_{1}^{31},\bar{f}_{1}^{23},\bar{f}_{1}^{32}$ can be obtained from $\bar{f}_{1}^{12}$ by the two symmetry operations described below.

Since $a$ and $b$ are in the same footing in the basis states and three SU(3) irreps are also in the same footing, one can make the following operations on the equation giving the action of various operators and arrive at another valid equation:

simultaneously perform the following operations:

• •

$a\leftrightarrow b$

• •

$l_{ij}\leftrightarrow l_{ji}$

• •

$p>0$ case and $p<0$ case interchanged.

• •

where ever p increases by some value in the state, replace it by p decreasing by the same value and vice versa. 2. 2.

cyclic permutation of $(1,2,3)$

For example, in Figure 2, (3) is obtained from (2) by $a\leftrightarrow b,l_{ij}\leftrightarrow l_{ji}$.

Action of all other invariant operators can be easily deduced either by using the above symmetries or by reducing those operators in terms of the basic operators given in fig.2. We will illustrate this for the operators $a_{1}\cdot b_{2},\epsilon b_{3}b_{2}a^{\dagger}_{2},\epsilon a_{3}a_{2}b^{\dagger}_{2}$ and $\epsilon a_{1}a_{2}a_{3}$ in appendix A.

IV Concluding remarks

We have studied the problem of constructing a singlet Hilbert space from the direct product of three arbitrary SU(3) irreps and calculating the action of invariant operators on this space. Singlet spaces are of importance when the Hilbert space has to satisfy gauge invariance constraints. In order to study the dynamics of such systems one requires the action of Hamiltonian on such a Hilbert space. Hamiltonian can be written in terms of the invariant operators. Therefore, one needs to evaluate the action of such basic invariant operators on the basis states. For SU(3), we have evaluated the action of these operators. In particular, the construction described in this paper is useful in defining the physical Hilbert space of lattice gauge theories and studying its dynamics.

One can consider product of Hilbert spaces defined here. For example, doing a product of two such Hilbert space with one common casimir will generate a diagram $>-<$ where the intermediate casimir is identified to be the same. This kind of product can be done in three different ways. All the three Hilbert spaces are unitary transforms of each other. Indeed the transformation coefficients are the 6-j symbols in the case of SU(2). These 6-j symbols depend only on the invariant casimir quantum numbers. We comment that the most primitive Hilbert space which only depend on casimirs is the singlet Hilbert space considered here.

One can generalize these results to SU(N) again by using SU(N) irreducible sunb Schwinger bosons. Further, another generalization which is of use, is to replace one or more of Schwinger boson pairs by fermionic oscillator pairs. This corresponds to having Gauss law in the presence of charged currents such as quarks. It will be considered in a future publication.

Appendix A SU(3):Calculation of action of various invariant operators on the unnormalized basis

The action of a gauge invariant operator $O$ on the $|l,p\rangle_{u}$ can be computed by using the commutation relations (6) to shift $O$ across $(a^{\dagger}_{i}\cdot b^{\dagger}_{j})$ and $(\epsilon_{\alpha\beta\gamma}a^{\dagger}_{i,\alpha}a^{\dagger}_{j,\beta}a^{\dagger}_{k,\gamma})$ or $(\epsilon_{\alpha\beta\gamma}b^{\dagger}_{i,\alpha}b^{\dagger}_{j,\beta}b^{\dagger}_{k,\gamma})$ until it hits $|0\rangle$. In the following, whenever there is no confusion, only the $l_{ij},p$ values which are different from the $l,p$ are written. Also, it is convenient to introduce the following notation $(i\cdot j)\equiv a^{\dagger}_{i}\cdot b^{\dagger}_{j},N_{i}\equiv n_{i}+m_{i}+|p|$.

$(a^{\dagger}_{1}\cdot a_{2})$

[TABLE]

Using (6),

[TABLE]

Repeating this $l_{12}$ times we get,

[TABLE]

Using the relation $(a^{\dagger}_{1}\cdot a_{2})(a^{\dagger}_{2}\cdot b^{\dagger}_{1})=(a^{\dagger}_{2}\cdot b^{\dagger}_{1})(a^{\dagger}_{1}\cdot a_{2})-\tilde{N}_{2}(a^{\dagger}_{1}\cdot b^{\dagger}_{2})(b^{\dagger}_{1}\cdot b_{2})$ repeatedly, we get

[TABLE]

Using, $(b^{\dagger}_{1}\cdot b_{2})(2\cdot 1)=(1-\tilde{N}_{2})(2\cdot 1)(b^{\dagger}_{1}\cdot b_{2})$, we get

[TABLE]

We now calculate $(a^{\dagger}_{1}\cdot a_{2})~{}|l_{12}=0,l_{21}=0\rangle_{u}$. Using $(a^{\dagger}_{1}\cdot a_{2})(a^{\dagger}_{3}\cdot b^{\dagger}_{2})=(a^{\dagger}_{3}\cdot b^{\dagger}_{2})(a^{\dagger}_{1}\cdot a_{2})-\tilde{N}_{2}(a^{\dagger}_{1}\cdot b^{\dagger}_{2})(a^{\dagger}_{3}\cdot a_{2})$, we get:

[TABLE]

Since, $(a^{\dagger}_{3}\cdot a_{2})|l_{12}=0,l_{21}=0,l_{32}=0\rangle_{u}=0$ and $a^{\dagger}_{1}\cdot a_{2})(a^{\dagger}_{2}\cdot b^{\dagger}_{3})=(1\cdot 3)+(2\cdot 3)(a^{\dagger}_{1}\cdot a_{2})-\tilde{N}_{2}(1\cdot 2)(b^{\dagger}_{3}\cdot b_{2})$, we get

[TABLE]

Putting this back into (26) gives:

[TABLE]

We now calculate $(b^{\dagger}_{1}\cdot b_{2})~{}|l_{12}=0,l_{21}=0\rangle_{u}$: Since,

[TABLE]

we have,

[TABLE]

Since, (b^{\dagger}_{3}\cdot b_{2})(2\cdot 3)=\Big{(}\frac{N_{2}+1}{N_{2}+2})(2\cdot 3)(b^{\dagger}_{3}\cdot b_{2}) and

[TABLE]

we have

[TABLE]

Now, $(b^{\dagger}_{1}\cdot b_{2})|l_{12}=0,l_{21}=0,l_{23}=0\rangle_{u}=l_{32}(3\cdot 1)|l_{12}=0,l_{21}=0,l_{23}=0,l_{32}-1\rangle_{u}$. Therefore,

[TABLE]

Putting (27) and (33) back in (24), we get

[TABLE]

Summing over $r$ and simplifying gives:

[TABLE]

[TABLE] 2. 2.

$(b^{\dagger}_{1}\cdot b_{2})$:

After a similar calculation,

[TABLE]

[TABLE] 3. 3.

$(a_{1}\cdot b_{2})$

Using the following relation:

[TABLE]

repeatedly, we get

[TABLE]

Using,

[TABLE]

repeatedly we get,

[TABLE]

Since, $(a_{1}\cdot b_{2})|l_{12}=0,l_{21}=0\rangle=0$, putting back (38) into (37) gives

[TABLE]

Using (35) and (36) and summing over r, we get

[TABLE]

[TABLE] 4. 4.

$\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2}$

• •

When $p\geq 0$

Since \Big{[}\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2},\epsilon a^{\dagger}_{1}a^{\dagger}_{2}a^{\dagger}_{3}\Big{]}=0, we have

[TABLE]

where $l_{\bar{i}\bar{j}}$ denote $l_{21},l_{13},l_{31},l_{23},l_{32}$. Now, we use $(\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2})(a^{\dagger}_{1}\cdot b^{\dagger}_{2})=(\epsilon a^{\dagger}_{3}a^{\dagger}_{1}a^{\dagger}_{2})+(a^{\dagger}_{1}\cdot b^{\dagger}_{2})(\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2})$ to get

[TABLE]

Therefore,

[TABLE]

• •

When $p<0$

Since $(\epsilon a^{\dagger}_{3}b_{2}a^{\dagger}_{2})(\epsilon b^{\dagger}_{1}b^{\dagger}_{2}b^{\dagger}_{3})$, we have

[TABLE]

Since $(\epsilon a^{\dagger}_{1}a^{\dagger}_{2}a^{\dagger}_{3})(\epsilon b^{\dagger}_{1}b^{\dagger}_{2}b^{\dagger}_{3})=(1\cdot 2)(2\cdot 3)(3\cdot 1)+(2\cdot 1)(1\cdot 3)(3\cdot 2)$, we have

[TABLE]

Therefore,

[TABLE]

where,

[TABLE]

Action of all other invariant operators can be calculated from the action of the invariant operators (1), (2) and (4) above. We will illustrate this for the operators $a_{1}\cdot b_{2}$, $\epsilon b_{3}b_{2}a^{\dagger}_{2}$ and $\epsilon a_{1}a_{2}a_{3}$.

$a_{1}\cdot b_{2}$

[TABLE]

From eqns (35) and (36), we know

[TABLE]

Substituting values for $d_{1}^{21},d_{2}^{21},e_{1}^{32},e_{2}^{32}$ and simplifying gives (39). 2. 2.

$\epsilon b_{3}b_{2}a^{\dagger}_{2}$

[TABLE]

Since we know the action of the left hand side of above equation on $|l,p\rangle$ from (36), (43) and (45), we get,

[TABLE]

where,

[TABLE] 3. 3.

$\epsilon a_{3}a_{2}b^{\dagger}_{2}$

Since, this operator can be obtained from $\epsilon b_{3}b_{2}a^{\dagger}_{2}$ by $a\leftrightarrow b$, the action of this operator can be obtained by the symmetry operation 1 discussed in section (III).

[TABLE]

[TABLE] 4. 4.

$(\epsilon a_{1}a_{2}a_{3})$

[TABLE]

The action of the LHS of the above equation on $|l,p\rangle_{u}$ is easily obtained using equations (53), (39) thereby giving $(\epsilon a_{3}a_{2}a_{1})|l,p\rangle_{u}$.

Appendix B SU(3): norm

Since we know the action of $a_{i}\cdot b_{j}$ on the $|l,p\rangle_{u}$, the norm $S(l,p)$ of the unnormalised state can be calculated using the following recursion relation:

[TABLE]

To calculate $S(l,p)$, we further need $(\epsilon^{~{}}_{\alpha\beta\gamma}a^{~{}}_{1,\alpha}a^{~{}}_{2,\beta}a^{~{}}_{3,\gamma})|l=0,p\rangle$ which is computed as :

[TABLE]

When $p<0$, a similar calculation goes through and gives the same result. The norm is given by :

[TABLE]

where

[TABLE]

and similar expressions for $F_{21}(l,p),F_{13}(l,p),F_{31}(l,p),F_{23}(l,p),F_{32}(l,p)$.