Formalism

Our purpose is to build a Hilbert space equipped with full $\mathrm{SO}(3)$ rotation symmetry and part of the non-Abelian flavour symmetry, in addition to the $\mathrm{U}(1)$ quantum numbers of which FuzzifiED is capable, and find the eigen-states and the energy of the Hamiltonian on that basis. Although it is impractical to fully diagonalize the total angular momentum $L^2$ on the full system, such diagonalization is practical on a segment (e. g. one of the two flavours in the model for the Ising CFT), and the segment states can assemble into states on the full system through angular-momentum composition

\[|(l_1l_2)lm\rangle=\sum_{m_1m_2}|l_1m_1\rangle|l_2m_2\rangle\langle l_1m_1,l_2m_2|lm\rangle\]

where $\langle l_1m_1,l_2m_2|lm\rangle$ is the Clebsch-Gordan (CG) coefficient.

Our strategy is therefore to

• split the system into segments,
• construct the Hilbert space resolved by $\mathrm{SO}(3)$ spatial rotation and flavour symmetry and the operators on each segment,
• assemble the segment Hilbert spaces and the segment operators into composite Hilbert space and composite operators,
• find the eigen-system of the Hamiltonian on the composite Hilbert space, and make measurements.

Composing the Segment Hilbert Spaces

For simplicity, we first consider bi-partition of the full system. E. g., for the Ising model, we split it into two segment Hilbert spaces of spin-up and spin-down. We will build a Hilbert space

\[ \mathscr{H}(Q,l,C_{2,\{p\}})\]

with an assigned total angular momentum $l$, a set of total $\mathrm{U}(1)$ quantum number $Q$, and representations under flavour symmetry within each segment specified by the values of the quadratic Casimir $C_{2,p}$ ($p=1,2$). For that purpose, on each segment of the system we build the basis with a set of $\mathrm{U}(1)$ quantum numbers $Q_p$ and diagonalize simultaneously the total angular momentum $L^2$ and the quadratic Casimir $C_{2,p}$ of the sub-group of the flavour symmetry on that segment.

\[ \mathscr{H}_p(C_{2,p})=\Big\{|Q_pC_{2,p},l_pm_p,\alpha_p\rangle\Big\}=\bigoplus_{Q_p,l_p}\mathscr{H}_p(Q_p,l_p,C_{2,p})\]

Here, we only need to keep states with a specific value of $C_{2,p}$. On the contrary, although we only need a single value of $Q$ and $l$ on the composite space, we need to keep all values of $Q_p$ and $l$ in the segment space, as different sets of $Q_p$ and $l_p$ can all compose into the same total $Q$ and $l$. The index $\alpha_p$ denotes the multiplicity.

Now we build the composite Hilbert space from the segment Hilbert spaces. Taking one state from each segment, the composite state is

\[ \left|Q_{\{12\}}C_{2,{\{12\}}},(l_1l_2)lm,\alpha_{\{12\}}\right\rangle=\sum_{m_1m_2}\left|Q_1C_{2,1},l_1m_1,\alpha_1\right\rangle\left|Q_2C_{2,p},l_2m_2,\alpha_2\right\rangle\langle l_1m_1,l_2m_2|lm\rangle\]

Here $Q_{\{12\}}$ is a short-hand notation for $\left\{Q_1,Q_2\right\}$, and the same notation applies to the multiplicity $\alpha_{\{12\}}$ and $C_{2,{\{12\}}}$. The segment quantum numbers and angular momenta constraints that $Q_1+Q_2=Q$ and $|l_1-l_2|\leq l\leq l_1+l_2$.

The composite Hilbert space is then the collection of these states

\[ \mathscr{H}(Q,l,C_{2,\{12\}})=\Big\{\left|Q_{\{12\}}C_{2,{\{12\}}},(l_1l_2)lm,\alpha_{\{12\}}\right\rangle\Big\}=\bigoplus_{\substack{Q_1+Q_2=Q\\|l_1-l_2|\leq l\leq l_1+l_2}}\mathscr{H}_1(Q_1,l_1,C_{2,1})\otimes\mathscr{H}_2(Q_2,l_2,C_{2,2})\]

In practice, we pick a state with a representative $m$ within each $\mathrm{SO}(3)$ multiplet, and work in a component-free notation. The composition can therefore be denoted by

\[ \left\|Q_{\{12\}}C_{2,{\{12\}}},(l_1l_2)l,\alpha_{\{12\}}\right\rangle=\big[\left\|Q_1C_{2,1},l_1,\alpha_1\right\rangle\otimes\left\|Q_2C_{2,2},l_2,\alpha_2\right\rangle\big]_l.\]

Decomposing the Operator

Having contructed the Hilbert space, we now calculate the matrix element of the Hamiltonian. For that purpose, we decompose the Hamiltonian into direct product of $\mathrm{SO}(3)$-covariant (i. e. carrying definite spin) operators acting on the segments that carry definite $\mathrm{SO}(3)$ spin.

\[ H=\sum_d g_d\Big[[\Phi^{(d)}_1]_{L_1^{(d)}}\otimes[\Phi^{(d)}_2]_{L^{(d)}_2}\Big]_{LM=00}\]

where $d$ is the index of decomposition. As above, this is a short-hand notation for

\[ \Big[[\Phi_1]_{L_1}\otimes[\Phi_2]_{L_2}\Big]_{LM}=\sum_{M_{\{12\}}}[\Phi_1]_{L_1M_1}[\Phi_2]_{L_2M_2}\langle L_1M_1,L_2M_2|LM\rangle\]

We take the Ising model as an example and explain how this decomposition is accomplished.

Recoupling the Pseudo-Potentials

We consider a two-body interaction written in terms of pseudo-potentials

\[ H=\sum_l\tilde{V}_l\Big[[c^\dagger_1 c^\dagger_2]_l\otimes [c_3c_4]_l\Big]_0\]

E. g., in the Ising model, the fermions 1 and 4 are spin-up, the fermions 2 and 3 are spin-down, and $\tilde{V}_{0,1}$ are non-zero. Here, we use the brackets to indicate objects composite by the rule of CG-coefficient, and we adopt a notation where $[\Phi]_{lm}$ increases the $L^z$ by $m$. In this notation,

\[ [c^\dagger]_{sm}=c^\dagger_m,\qquad[c]_{sm}=(-1)^{m}c_{-m}.\]

In this convension, the $\tilde{V}_l$ is connected with the ordinary pseudo-potential by $\tilde{V}_l=(-1)^lV_l/\sqrt{2l+1}$. We now want to rewrite the Hamiltonian in a different channel

\[ H=\sum_l\tilde{W}_j\Big[[c^\dagger_1 c_4]_j\otimes [c^\dagger_2c_3]_j\Big]_0.\]

E. g., in the Ising model, now $c^\dagger_1 c_4$ is completely in the spin-up segment, and $c^\dagger_2c_3$ is completely in the spin-down segment, so the re-coupling accomplishes the decomposition. The relation of $\tilde{V}$ and $\tilde{W}$ can be expressed in a Dirac bracket notation, which corresponds to a $9j$-symbol that could be further reduced to a $6j$-symbol.

\[\begin{aligned} \tilde{W}_j&=\sum_l\tilde{V}_l\langle((s_1s_2)l,(s_3s_4)l)0|((s_1s_4)j,(s_2s_3)j)0\rangle\\ &=\sum_l\tilde{V}_l\,(-1)^{s_3+s_4-l}(2l+1)(2j+1)\begin{Bmatrix}s_1&s_2&l\\s_4&s_3&l\\j&j&0\end{Bmatrix}. \end{aligned}\]

For an operator with a total spin $L$, one only needs to replace the nineth element $0$ by $L$.

Translate from a Contact Coupling

We now discuss a second case where the Hamiltonian can be written in terms of a contact coupling

\[ H=\int\mathrm{d}^2\mathbf{r}\,\Phi_1(\mathbf{r})\Phi_2(\mathbf{r}).\]

The local operators $\Phi_{1,2}$ can be decomposed into spherical components

\[ \Phi^{(s)}(\mathbf{r})=\sum_{lm} (\Phi)_{lm}^{(s)}Y_{lm}^{(s)}.\]

Here we use the parentheses to indicate that these objects compose by the rule of monopole harmonics instead of CG-coefficients.

\[ (\Phi_1\Phi_2)^{(s)}_{lm}=\sum_{m_1m_2}(\Phi_1)^{(s_1)}_{l_1m_1}(\Phi_2)^{(s_2)}_{l_2m_2}\langle l_1m_1,l_2m_2|lm\rangle\langle l_1(-s_1),l_2(-s_2)|l(-s)\rangle\sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}}.\]

E. g., for the electron and density operators

\[ (c^\dagger)^{(s)}_{sm}=c^\dagger_m,\qquad(c)^{(-s)}_{sm}=(-1)^{m-s}c_{-m}\\ (n)_{lm}=(c^\dagger c)_{lm}=\sum_{m_1m_2}c^\dagger_{m_1} c_{-m_2}\frac{(-1)^{m_2-s}(2s+1)}{\sqrt{4\pi(2l+1)}}\langle sm_1,s(-m_2)|lm\rangle\langle s(-s),ss|l0\rangle.\]

The Hamiltonian can therefore be expressed as

\[ H=\sum_l\tilde{V}_l\Big[[\Phi_1]_{l}\otimes[\Phi_2]_{l}\Big]_0\]

where

\[ [\Phi_p]_{lm}=(\Phi_p)_{lm}\\ \tilde{V}_l=\langle l(-s_1),l(-s_2)|00\rangle\sqrt{\frac{(2l_1+1)(2l_2+1)}{(2l+1)}}.\]

E. g., for the Ising interaction, we take $\Phi_1(\mathbf{r})=n_\uparrow(\mathbf{r})$ and $\Phi_2(\mathbf{r})=n_\downarrow(\mathbf{r})$ or $\nabla^2n_\downarrow(\mathbf{r})$ ; for the transverse field, we take $\Phi_{1,2}(\mathbf{r})=c^{(\dagger)}_{\uparrow,\downarrow}(\mathbf{r})$. They completely lie in one of the two segments ; hence, this expression accomplishes the decomposition.

Composing the Segment Operators

Having obtained the decomposed Hamiltonian, we first calculate the matrix elements of the segment operator $[H_{d,p}]_{l_d}$. By Wigner-Eckart theorem, the matrix elements with different $L^z$-numbers are connected through a $3j$-symbol. Omitting the irrelevant indices,

\[ \langle l'm'|[\Phi]_{LM}|lm\rangle =\langle lm, LM|l'm'\rangle\langle l'\|[\Phi]_L\|l\rangle.\]

Here, $\langle l'\|[\Phi]_L\|l\rangle$ is independent of the choice of states $m,m'$ in the multiplet and is called the reduced matrix element. To calculate the reduced matrix of the segment operator, we need only pick one representative state from each multiplet in the segment Hilber space. We typically pick the representative state as $m'=M=m=0$ except when the $3j$-symbols vanish, i. e. when $l'+L+l\in 2\mathbb{Z}+1$, in which cases we pick $m=1$ instead.

We now assemble the segment operator into the composite operator acting on the full Hilbert space. This involves re-coupling through the $9j$-symbol.

\[ \langle(l_1'l_2')l'\|[\Phi_{1,L_1}\otimes\Phi_{2,L_2}]_L\|(l_1 l_2)l\rangle =\sqrt{(2l+1)(2L+1)(2l'_1+1)(2l'_2+1)}\begin{Bmatrix}l_1&l_2&l\\L_1&L_2&L\\l_1'&l_2'&l'\end{Bmatrix} \langle l_1'\|\Phi_{1,L_1}\|l_1\rangle\,\langle l_2'\|\Phi_{2,L_2}\|l_2\rangle .\]

Note that when exchanging $|l_1\rangle$ and $\Phi_2$, a factor related to fermion parity may arise.

Generalization to Multiple Parts

For multiple parts, the angular-momentum composes successively along a chain, so a state is specified not only by the total angular momentum $l$, but also by the cumulative angular momentum of parts $1\cdots p$ for each $p$

\[ \|(\cdots((l_1 l_2)l_{12}\,l_3)l_{123}\cdots l_p)l_{1\cdots p}\cdots l_{N_p})l\rangle,\]

Each channel is specified by $2N_p-1$ angular momenta in total — $l_p$ ($p=1,\dots,N_p$), $l_{1\cdots p}$ ($p=2,\dots,N_p-1$) and $l$. The coupling of segment operators are specified in a similar way. E. g., for the contact coupling

\[ H=\int\mathrm{d}^2\mathbf{r}\,\Phi_1(\mathbf{r})\Phi_2(\mathbf{r})\cdots\Phi_p(\mathbf{r})\cdots\Phi_{N_p}(\mathbf{r}),\]

it can be rewritten in terms of channels as

\[ H=\sum V_{\{L\}}\Big[\dots[[\Phi_1]_{L_1}[\Phi_2]_{L_2}]_{L_{12}}\dots [\Phi_p]_{L_p}\big]_{L_{1\cdots p}} \cdots [\Phi_{N_p}]_{L_{N_p}}\Big]_{L=0}\]

Here we use $\{L\}$ as a short-hand notation for the coupling channel. The coefficient reads

\[ V_{\{L\}}=\sqrt{\frac{\prod_p(2L_p+1)}{(4\pi)^{N_p-2}(2L+1)}}\prod_p\langle L_{1\dots (p-1)}(-s_{L\dots(p-1)}),L_p(-s_p)|L_{1\cdots p}(-s_{1\cdots p})\rangle\]

where $s_{1\cdots p}=s_1+\cdots+s_p$. To calculate the matrix element, each re-coupling produces a $9j$-symbol, so the final

\[ \langle \{l'\}\|\Phi_{\{L\}}\|\{l\}\rangle =\prod_{p=2}^{N_p}\left[\sqrt{(2l_{1\cdots p}+1)(2k_{1\cdots p}+1)(2l'_{1\cdots (p-1)}+1)(2l'_p+1)} \begin{Bmatrix}l_{1\cdots(p-1)}&l_p&l_{1\cdots p}\\L_{1\cdots(p-1)}&L_p&L_{1\cdots p}\\l'_{1\cdots(p-1)}&l'_p&l'_{1\cdots p}\end{Bmatrix}\right] \prod_{p=1}^{N_p}\langle l'_p\|\Phi_{p,L_p}\|l_p\rangle ,\]