From the Jaynes–Cummings model to non-abelian gauge theories: a guided tour for the quantum engineer

In this section we investigate the JCM focusing on its global and local symmetries. The discussion will allow us to formulate and study extended systems with abelian and non-abelian local symmetries. First we define the Hilbert space. The two level atom can be represented by one fermionic particle in two fermionic modes (fermionic Schwinger representation). The ground state |1⟩ is identified with ${\hat{\psi }}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ and the excited state |0⟩ with ${\hat{\psi }}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$, where ${\hat{\psi }}^{{\dagger}}\left(0\right)$ and ${\hat{\psi }}^{{\dagger}}\left(1\right)$ are the creation operators of the fermionic modes and |vac⟩_F is the fermionic Fock vacuum. The bosonic and fermionic nature of the operators $\hat{b}$ and $\hat{\psi }\left(x\right)$ is manifested in the commutation and anti-commutation relations

$$\begin{equation}\left[\hat{b},{\hat{b}}^{{\dagger}}\right]=1, \left\{\hat{\psi }\left(x\right),{\hat{\psi }}^{{\dagger}}\left(y\right)\right\}=\delta \left(x,y\right),\end{equation} \tag{ 2 }$$

with x, y ∈ {0, 1}, and $\left[\hat{A},\hat{B}\right]=\hat{A}\hat{B}-\hat{B}\hat{A}$ denotes a commutator, whereas $\left\{\hat{A},\hat{B}\right\}=\hat{A}\hat{B}+\hat{B}\hat{A}$ is the anti-commutator. The label x of the fermionic operator $\hat{\psi }\left(x\right)$ denotes the ground or excited state of the two state atom, but x can also be interpreted as the position of a fermionic atom on a lattice site. Using the fermionic Schwinger representation the JCM can be written as

$$\begin{equation}{\hat{H}}_{\text{JC}}={\omega }_{\text{c}}{\hat{b}}^{{\dagger}}\hat{b}+\frac{{\omega }_{\text{a}}}{2}\left[{\hat{\psi }}^{{\dagger}}\left(0\right)\hat{\psi }\left(0\right)-{\hat{\psi }}^{{\dagger}}\left(1\right)\hat{\psi }\left(1\right)\right]+\frac{{{\Omega}}_{0}}{2}\left[{\hat{\psi }}^{{\dagger}}\left(1\right){\hat{b}}^{{\dagger}}\hat{\psi }\left(0\right)+\text{h.c.}\right].\end{equation} \tag{ 3 }$$

Next we define operators, which measure the charge at the lattice site x i.e.

$$\begin{equation}\hat{Q}\left(0\right)={\hat{\psi }}^{{\dagger}}\left(0\right)\hat{\psi }\left(0\right),\end{equation} \tag{ 4a }$$

$$\begin{equation}\hat{Q}\left(1\right)=-\hat{\psi }\left(1\right){\hat{\psi }}^{{\dagger}}\left(1\right)={\hat{\psi }}^{{\dagger}}\left(1\right)\hat{\psi }\left(1\right)-1.\end{equation} \tag{ 4b }$$

We defined $\hat{Q}\left(x\right)$ such that the state $\vert 1\rangle ={\hat{\psi }}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ is associated with the absence of any charges at both sites: $\hat{Q}\left(0\right)\vert 1\rangle =\hat{Q}\left(1\right)\vert 1\rangle =0\vert 1\rangle$. On the other hand, the state $\vert 0\rangle ={\hat{\psi }}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ has charge +1 at site 0 and −1 at site 1, i.e. $\hat{Q}\left(0\right)\vert 0\rangle =+1\vert 0\rangle$ and $\hat{Q}\left(1\right)\vert 0\rangle =-1\vert 0\rangle$. In particular, we will use the notion of such a local charge to discuss local gauge symmetries, which we will introduce in the following. We call a unitary transformation a symmetry, if the Hamiltonian remains invariant with respect to this transformation. These can be space transformations, such as translations and rotations, or transformations between internal degrees of freedom such as flavor and color. A transformation which is applied to all degrees of freedom of a system is called a global transformation, while a local transformation applies only to a finite range of space. In particular the JCM is invariant under the following two local transformations

$$\begin{equation}\hat{\psi }\left(0\right)\to {\text{e}}^{\text{i}\theta \left(0\right)}\hat{\psi }\left(0\right), \hat{b}\to {\text{e}}^{+\text{i}\theta \left(0\right)}\hat{b} \quad \left(\text{transformation}\;\text{at}\;x=0\right),\end{equation} \tag{ 5a }$$

$$\begin{equation}\hat{\psi }\left(1\right)\to {\text{e}}^{i\theta \left(1\right)}\hat{\psi }\left(1\right), \hat{b}\to {\text{e}}^{-\text{i}\theta \left(1\right)}\hat{b} \quad \left(\text{transformation}\;\text{at}\;x=1\right),\end{equation} \tag{ 5b }$$

which are parametrized by the angle θ(0), θ(1) ∈ [0, 2π). These transformations are local since they only involve the space point 0 or 1 respectively, and are realized by the unitary operators

$$\begin{equation}\hat{S}\left(0\right)={\text{e}}^{\text{i}\theta \left(0\right){\hat{G}}_{\text{JC}}\left(0\right)},\end{equation} \tag{ 6a }$$

$$\begin{equation}\hat{S}\left(1\right)={\text{e}}^{\text{i}\theta \left(1\right){\hat{G}}_{\text{JC}}\left(1\right)},\end{equation} \tag{ 6b }$$

with the generators

$$\begin{equation}{\hat{G}}_{\text{JC}}\left(0\right)=-{\hat{b}}^{{\dagger}}\hat{b}-\hat{Q}\left(0\right),\end{equation} \tag{ 7a }$$

$$\begin{equation}{\hat{G}}_{\text{JC}}\left(1\right)={\hat{b}}^{{\dagger}}\hat{b}-\hat{Q}\left(1\right).\end{equation} \tag{ 7b }$$

The unitary operations S(x) leave the Hamiltonian invariant,

$$\begin{equation}\hat{S}\left(x\right){\hat{H}}_{\text{JC}}{\hat{S}}^{{\dagger}}\left(x\right)={\hat{H}}_{\text{JC}}\quad \forall \enspace x\end{equation} \tag{ 8 }$$

or, equivalently, their generators commute with the Hamiltonian

$$\begin{equation}\left[{\hat{G}}_{\text{JC}}\left(x\right),{\hat{H}}_{\text{JC}}\right]=0\quad \forall \enspace x.\end{equation} \tag{ 9 }$$

The local symmetry transformation $\hat{S}\left(x\right)$ is called a local gauge transformation, and the commutation relation with the Hamiltonian implies that ${\hat{H}}_{\text{JC}}$ and the symmetry generators ${\hat{G}}_{\text{JC}}\left(x\right)$ can be diagonalized simultaneously. Choosing a basis which diagonalizes ${\hat{G}}_{\text{JC}}\left(x\right)$ leads to a block-diagonal Hamiltonian. The blocks of the Hamiltonian act on subspaces of the Hilbert space. These separate sectors are characterized by the eigenvalues of the symmetry generators, and are not mixed by the dynamics generated by ${\hat{H}}_{\text{JC}}$. A physical state belongs to a particular sector and does not involve any superposition of different sectors. We formalize this by considering the eigenvalue problem of all generators

$$\begin{equation}{\hat{G}}_{\text{JC}}\left(x\right)\vert \phi \rangle =q\left(x\right)\vert \phi \rangle \quad \forall \enspace x\end{equation} \tag{ 10 }$$

with the eigenvalues q(x), which we call static charges. These equations split the Hilbert spaces into different sectors, labeled by sets of static charges. We call the eigenstates of the ${\hat{G}}_{\text{JC}}\left(x\right)$ operators physical states and the eigenvalue equation (10) are denoted as Gauss's laws. The Gauss's law enforces a charge superselection rule, which enforces physical states to reside within one respective sector. Furthermore, the physical states |ϕ⟩ are gauge invariant i.e. they are invariant up to a phase after performing a local gauge transformation, and this phase depends on the static charges

$$\begin{equation}\hat{S}\left(x\right)\vert \phi \rangle ={\text{e}}^{\text{i}\theta \left(x\right)q\left(x\right)}\vert \phi \rangle \quad \forall \enspace x.\end{equation} \tag{ 11 }$$

The local gauge invariance guarantees the block diagonal structure of the Hamiltonian, which is a direct sum over the different static charge sectors,

$$\begin{equation}H=\underset{q}{\bigoplus}H{\vert }_{q}.\end{equation} \tag{ 12 }$$

Besides the local symmetry, the JCM is invariant with respect to the global gauge transformation for fermionic operators $\hat{\psi }\left(x\right)\to {\text{e}}^{\text{i}\theta }\hat{\psi }\left(x\right)$ leading to the conservation of the total number of fermions

$$\begin{equation}{\hat{N}}_{\text{F}}=\sum _{x}{\hat{\psi }}^{{\dagger}}\left(x\right)\hat{\psi }\left(x\right),\end{equation} \tag{ 13 }$$

i.e. $\left[{\hat{N}}_{\text{F}},{\hat{H}}_{\text{JC}}\right]=0$. Introducing also the operator of the total number of excitations (the number of photons and the number of atomic excitations)

$$\begin{equation}{\hat{N}}_{\text{ex}}={\hat{b}}^{{\dagger}}\hat{b}+{\hat{\psi }}^{{\dagger}}\left(0\right)\hat{\psi }\left(0\right),\end{equation} \tag{ 14 }$$

with $\left[{\hat{N}}_{\text{ex}},{\hat{H}}_{\text{JC}}\right]=0$, the generators ${\hat{G}}_{\text{JC}}\left(0\right)$ and ${\hat{G}}_{\text{JC}}\left(1\right)$ can be represented in terms of the conserving operators ${\hat{N}}_{\text{ex}}$ and ${\hat{N}}_{\text{F}}$ as:

$$\begin{equation}{\hat{G}}_{\text{JC}}\left(0\right)=-{\hat{N}}_{\text{ex}},\end{equation} \tag{ 15a }$$

$$\begin{equation}{\hat{G}}_{\text{JC}}\left(1\right)={\hat{N}}_{\text{F}}-1+{\hat{N}}_{\text{ex}}.\end{equation} \tag{ 15b }$$

For all physical states one has ${\hat{N}}_{\text{F}}\vert \phi \rangle ={n}_{\text{F}}\vert \phi \rangle$ with n_F = 1. In fact, the atom can be only in the internal states $\vert x\rangle ={\hat{\psi }}^{{\dagger}}\left(x\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ containing one fermion in the Schwinger representation. Consequently the total charge operator

$$\begin{equation}\hat{Q}=\sum _{x}\hat{Q}\left(x\right)={\hat{N}}_{\text{F}}-1\end{equation} \tag{ 16 }$$

has zero eigenvalues for physical states with n_F = 1, giving

$$\begin{equation}\sum _{x}q\left(x\right)=0.\end{equation} \tag{ 17 }$$

In order to determine the spectrum of the Jaynes–Cummings Hamiltonian we use the property q(0) = −q(1), which implies that the physical sectors can be denoted by one positive integer q ≡ q(1). The eigenstates of the two Gauss's laws (10) are

$$\begin{equation}\vert 1,q\rangle ={\hat{\psi }}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert {q\rangle }_{\text{B}},\end{equation} \tag{ 18a }$$

$$\begin{equation}\vert 0,q\rangle ={\hat{\psi }}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert q-{1\rangle }_{\text{B}},\end{equation} \tag{ 18b }$$

where |vac⟩_F is the empty fermionic Fock state and |q⟩_B is a bosonic state with q particles. Note that the static charge q ≡ q(1) represents the number of excitations (photons or excited atoms) in the system. Each state in the sector q can be expanded as

$$\begin{equation}\vert {\phi }_{q}\rangle ={\alpha }_{q}\vert 1,q\rangle +{\beta }_{q}\vert 0,q\rangle .\end{equation} \tag{ 19 }$$

Within the sector with q ⩾ 1 the Hamiltonian matrix using the basis {|1, q⟩, |0, q⟩} will take the form

$$\begin{equation}H{\vert }_{q}=\left(\begin{pmatrix}\hfill {\omega }_{\text{c}}q-\frac{{\omega }_{\text{a}}}{2}\hfill & \hfill \frac{{{\Omega}}_{0}}{2}\sqrt{q}\hfill \\ \hfill \frac{{{\Omega}}_{0}}{2}\sqrt{q}\hfill & \hfill {\omega }_{\text{c}}\left(q-1\right)+\frac{{\omega }_{\text{a}}}{2}\hfill \end{pmatrix}\right).\end{equation} \tag{ 20 }$$

The eigenvalues of the Hamiltonian (20)

$$\begin{equation}{E}_{{\pm},q}={\omega }_{c}\left(q-\frac{1}{2}\right){\pm}\frac{1}{2}\sqrt{{{\Delta}}^{2}+{{\Omega}}_{0}^{2}q},\end{equation} \tag{ 21 }$$

make the Jaynes–Cummings ladder by taking different sectors q, where we introduced the detuning Δ = ω_c − ω_a. The corresponding eigenvectors are

$$\begin{equation}\vert q,+\rangle =\mathrm{cos}\left({\alpha }_{q}/2\right)\vert 1,q\rangle +\mathrm{sin}\left({\alpha }_{q}/2\right)\vert 0,q\rangle ,\end{equation} \tag{ 22a }$$

$$\begin{equation}\vert q,-\rangle =\mathrm{sin}\left({\alpha }_{q}/2\right)\vert 1,q\rangle -\mathrm{cos}\left({\alpha }_{q}/2\right)\vert 0,q\rangle \end{equation} \tag{ 22b }$$

with $\mathrm{tan}\left(\alpha \right)={{\Omega}}_{0}\sqrt{q}/{\Delta}$. For q = 0 there is only one eigenstate of the Gauss's law $\vert 1,q\rangle ={\hat{\psi }}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert {q\rangle }_{\text{B}}$ characterized by zero eigenvalue q. The state |1, q⟩ is an eigenvector of the Hamiltonian with an eigenenergy −ω_c/2.

In this subsection we will discuss a model consisting out of two sites and one link connecting these sites. This model will be a minimal instance of a U(1) lattice gauge theory. As in the previous section the fermions reside on the sites. We consider states which are dynamically connected to the Dirac sea $\vert D\rangle ={\hat{\psi }}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ with zero charge. The occupation of the site 0 can be understood as a particle anti-particle state $\vert p\bar{p}\rangle ={\hat{\psi }}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}$ with fermionic charges Q(1) = −1 and Q(0) = 1.

In the JCM considered in section 2, the Hilbert space of the link between the two lattice sites comprises a boson mode described by raising and lowering operators ${\hat{b}}^{{\dagger}}$ and $\hat{b}$. In the U(1) gauge theory the Hilbert space of the link is different and represents the Hilbert space of a particle on a ring. The position coordinate φ of a particle on a ring is an angle, residing in a compact configuration space. On the link, the angular position operator $\hat{\varphi }$ defines a continuous basis through its eigenstates |φ⟩_G, which span the Hilbert space of the link. The Hilbert space of the line will play the role of a gauge degree of freedom, hence the subscript G. Moreover we use the angle operator $\hat{\varphi }$ to define the unitary operator

$$\begin{equation}\hat{U}={\text{e}}^{\text{i}\hat{\varphi }},\end{equation} \tag{ 25 }$$

which is invariant with respect to shifting the phase operator $\hat{\varphi }$ by 2π. The operator $\hat{U}$ induces a representation of the U(1) group on the Hilbert space of a particle i.e.

$$\begin{equation}\hat{U}\vert {\varphi \rangle }_{\text{G}}={\text{e}}^{\text{i}\varphi }\vert {\varphi \rangle }_{\text{G}}.\end{equation} \tag{ 26 }$$

The operator $\hat{U}$ is called the gauge connection or group element operator. The conjugate momentum of the angle operator will be the angular momentum operator $\hat{E}$, which we call the electric field. It is given by −i∂_φ in the position representation of the angular motion. The angular momentum operator has a non-bounded integer spectrum

$$\begin{equation}\hat{E}\vert {j\rangle }_{\text{G}}=j\vert {j\rangle }_{\text{G}},\end{equation} \tag{ 27 }$$

where $j\in \mathbb{Z}$. The canonical commutation relation

$$\begin{equation}\left[\hat{\varphi },\hat{E}\right]\equiv \left[\varphi ,-\mathrm{i}{\partial }_{\varphi }\right]=i\end{equation} \tag{ 28 }$$

leads to

$$\begin{equation}\left[\hat{E},\hat{U}\right]=\hat{U}\end{equation} \tag{ 29 }$$

— that is, $\hat{U}$ raises the electric field by one unit:

$$\begin{equation}\hat{U}\vert {j\rangle }_{\text{G}}=\vert j+{1\rangle }_{\text{G}}\end{equation} \tag{ 30 }$$

and the conjugate operator ${\hat{U}}^{{\dagger}}$ lowers the electric field.

After introducing link variables we are able to formulate a one link gauge theory in analogy to the JCM. The Hamiltonian takes the following form:

$$\begin{equation}{\hat{H}}_{1}=\frac{{g}^{2}}{2}{\hat{E}}^{2}+M\left[{\hat{\psi }}^{{\dagger}}\left(0\right)\hat{\psi }\left(0\right)-{\hat{\psi }}^{{\dagger}}\left(1\right)\hat{\psi }\left(1\right)\right]+{\epsilon}\left[{\hat{\psi }}^{{\dagger}}\left(0\right)\hat{U}\hat{\psi }\left(1\right)+\text{h.c}\right].\end{equation} \tag{ 31 }$$

In this model, the cavity energy ${\omega }_{\text{c}}{\hat{b}}^{{\dagger}}\hat{b}$ featured in the Jaynes–Cummings Hamiltonian (3) is substituted by the operator ${\hat{E}}^{2}$ corresponding to the kinetic energy of a rotating particle. The second term in equation (31) is the staggered mass term and the last term describes the creation of the particle-antiparticle pair out of the Dirac vacuum associated with the increase of the electric field on the connecting link and vice versa.

The Hamiltonian is invariant under to the following local transformations:

$$\begin{equation}\hat{\psi }\left(0\right)\to {\text{e}}^{\text{i}\theta \left(0\right)}\hat{\psi }\left(0\right), \hat{U}\left(0\right)\to {\text{e}}^{+\text{i}\theta \left(0\right)}\hat{U}\left(0\right) \quad \left(\text{transformation}\;\text{at}\;x=0\right),\end{equation} \tag{ 32a }$$

$$\begin{equation}\hat{\psi }\left(1\right)\to {\text{e}}^{\text{i}\theta \left(1\right)}\hat{\psi }\left(1\right), \hat{U}\left(0\right)\to {\text{e}}^{-\text{i}\theta \left(1\right)}\hat{U}\left(0\right) \quad \left(\text{transformation}\;\text{at}\;x=1\right),\end{equation} \tag{ 32b }$$

where the phase operator transformed as

$$\begin{equation}\hat{\varphi }\to \hat{\varphi }+\theta \left(0\right)\quad \left(\text{transformation}\;\text{at}\;x=0\right),\end{equation} \tag{ 33a }$$

$$\begin{equation}\hat{\varphi }\to \hat{\varphi }-\theta \left(1\right)\quad \left(\text{transformation}\;\text{at}\;x=1\right).\end{equation} \tag{ 33b }$$

The transformation of the phase resembles the form of the gauge transformation of the vector potential in the continuum. Thus one identifies the phase operator $\hat{\varphi }$ with a compact vector potential $\hat{A}=\hat{\varphi }$ obeying the canonical commutation relation (28), i.e. $\left[\hat{A},\hat{E}\right]=\mathrm{i}$. This is a very natural choice of the vector potential, as, using the solid state terminology, the unitary operator $\hat{U}={\text{e}}^{\text{i}\hat{A}}$ entering the Hamiltonian (31) can be interpreted as a Peierls's phase factor [66–68] for fermion transition between the lattice sites 1 and 0. The vector potential $\hat{A}$ acts then in the Hilbert space of the quantum gauge field living on the link between the two lattice sites. The local transformations are implemented by the operators

$$\begin{equation}\hat{S}\left(0\right)={\text{e}}^{\text{i}\theta \left(0\right)\hat{G}\left(0\right)},\end{equation} \tag{ 34a }$$

$$\begin{equation}\hat{S}\left(1\right)={\text{e}}^{\text{i}\theta \left(1\right)\hat{G}\left(1\right)},\end{equation} \tag{ 34b }$$

with the generators

$$\begin{equation}\hat{G}\left(0\right)=\hat{E}-\hat{Q}\left(0\right),\end{equation} \tag{ 35a }$$

$$\begin{equation}\hat{G}\left(1\right)=-\hat{E}-\hat{Q}\left(1\right),\end{equation} \tag{ 35b }$$

where $\hat{Q}\left(1\right)$ and $\hat{Q}\left(0\right)$ are local charge operators given by equation (24). The unitary transformation $\hat{S}\left(x\right)$ leaves the Hamiltonian invariant

$$\begin{equation}\hat{S}\left(x\right){\hat{H}}_{1}{\hat{S}}^{{\dagger}}\left(x\right)={\hat{H}}_{1}\quad \forall \enspace x\end{equation} \tag{ 36 }$$

or, in terms of the generators,

$$\begin{equation}\left[\hat{G}\left(x\right),{\hat{H}}_{1}\right]=0\quad \forall \enspace x.\end{equation} \tag{ 37 }$$

Again we define physical states as the eigenstates of the Gauss's law operators

$$\begin{equation}\hat{G}\left(x\right)\vert \phi \rangle =q\left(x\right)\vert \phi \rangle \quad \forall \enspace x.\end{equation} \tag{ 38 }$$

Where the last equation immediately implies the splitting of the Hilbert space into charge sectors. As in the Jaynes–Cummings case, half-filling of fermions (n_F = 1) guarantees that ${\sum }_{x}q\left(x\right)=0$. Thus one can write $q\equiv q\left(0\right)$ and $q\left(1\right)=-q$. The charge sector q may be spanned by two states, as in equations (18a) and (18b) for the JCM:

$$\begin{equation}\vert D,q\rangle ={\hat{\psi }}^{{\dagger}}\left(1\right)\;\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert {q\rangle }_{\text{G}},\end{equation} \tag{ 39a }$$

$$\begin{equation}\vert p\bar{p},q\rangle ={\hat{\psi }}^{{\dagger}}\left(0\right)\;\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert q+{1\rangle }_{\text{G}},\end{equation} \tag{ 39b }$$

where |D, q⟩ corresponds to a state with no particles and anti-particles (Dirac sea), in which the static charges are the only source of electric field, while $\vert p\bar{p},q\rangle$ is a meson-like state with a particle anti-particle pair, serving as the source of an electric flux tube on top of the one originating from the static charges.

A general state of the q sector is then spanned by

$$\begin{equation}\vert {\phi }_{q}\rangle ={\alpha }_{q}\vert D,q\rangle +{\beta }_{q}\vert p\bar{p},q\rangle \end{equation} \tag{ 40 }$$

and the matrix block of the Hamiltonian, which corresponds to the q sector, takes the form

$$\begin{equation}{H}_{1}{\vert }_{q}=\left(\begin{pmatrix}\hfill \frac{{g}^{2}}{2}{q}^{2}-M\hfill & \hfill {\epsilon}\hfill \\ \hfill {\epsilon}\hfill & \hfill \frac{{g}^{2}}{2}{\left(q+1\right)}^{2}+M\hfill \end{pmatrix}\right).\end{equation} \tag{ 41 }$$

The off-diagonal elements of the previous matrix are independent of q, whereas the off-diagonal elements of the Jaynes–Cummings Hamiltonian (20) depend on the sector.

In the limit of small interaction (negligible off diagonal terms), the self energy is approximately given by the diagonal terms. If the static charges have negative values, q < 0, they imply an external electric field which is of opposite sign to the one created by a particle-antiparticle pair. Hence the creation of the state $\vert p\bar{p},q\rangle$ actually lowers the electric field amplitude in the system. For strong enough negative values of the static charge q, this effect even makes the state $\vert p\bar{p},q\rangle$ of lower energy than the state |D, q⟩. From equating the diagonal terms, we find that this happens for

$$\begin{equation}q\approx -\left(\frac{2M}{{g}^{2}}+\frac{1}{2}\right),\end{equation} \tag{ 42 }$$

which can be understood in terms of a two level system. For a single spin in an external magnetic field with the Hamiltonian ${H}_{1}{\vert }_{q}=\frac{{g}^{2}}{2}\left[{\left(q+\frac{1}{2}\right)}^{2}+\frac{1}{2}\right]\mathbb{1}+\left(2M+2q+1\right){\hat{s}}_{z}+2{\epsilon}{\hat{s}}_{x}$, where ${\hat{s}}_{x,z}$ are spin-half operators. The z-component of the magnetic field inverts sign at (42), hence inverting the state of minimal energy. The one-link cQED is the minimal model allowing to demonstrate the instability of the fermionic vacuum in the presence of strong electric fields, see also recent work [69, 70]. The effect is reminiscent of the widely studied Schwinger pair production in extended systems [71–73].

Next, we extend the one-link model of the previous section and include several vertices and links, while staying one-dimensional. The resulting model is a one dimensional lattice gauge theory: compact quantum electrodynamics (cQED) in one space dimension with open boundary conditions. The lattice has an even number of sites $\mathcal{N}=2\mathcal{M}+2$, which are labeled by $x=0,\dots ,\mathcal{N}-1$. On each link we have the Hilbert space of a particle on a ring, with the gauge connection operators $\hat{U}\left(x\right)={\text{e}}^{\text{i}\hat{\varphi }\left(x\right)}$ and electric fields $\hat{E}\left(x\right)$. The Hamiltonian for (1 + 1) cQED is

$$\begin{equation}{\hat{H}}_{\text{cQED}}=\frac{{g}^{2}}{2}\sum _{x}{\hat{E}}^{2}\left(x\right)+M\sum _{x}{\left(-1\right)}^{x}{\hat{\psi }}^{{\dagger}}\left(x\right)\hat{\psi }\left(x\right)+{\epsilon}\sum _{x}\left[{\hat{\psi }}^{{\dagger}}\left(x\right)\hat{U}\left(x\right)\hat{\psi }\left(x+1\right)+\text{h.c.}\right],\end{equation} \tag{ 43 }$$

where the summation over x covers all the lattice sites ($x=0,\dots ,\mathcal{N}-1$) in the second term, whereas the sum is over all links ($x=0,\dots ,\mathcal{N}-2$) in the first and the third terms. For a graphical illustration of the Hamiltonian equation (43) see figure 3. The first term is the electric field energy with g the coupling constant. The second term is the staggered mass term and the last term is the coupling between the gauge field and the matter degrees of freedom. In the single link case ($\mathcal{N}=2$) equation (43) reduces to the Hamiltonian given by equation (31) subject to replacement ${\hat{E}}^{2}\left(0\right)\to {\hat{E}}^{2}$ and $\hat{U}\left(0\right)\to \hat{U}$. H_cQED is symmetric under the local (gauge) transformations

$$\begin{equation}\hat{\psi }\left(x\right)\to {\text{e}}^{\text{i}\theta \left(x\right)}\hat{\psi }\left(x\right), \hat{U}\left(x\right)\to {\text{e}}^{\text{i}\theta \left(x\right)}\hat{U}\left(x\right), \hat{U}\left(x-1\right)\to {\text{e}}^{-\text{i}\theta \left(x\right)}\hat{U}\left(x-1\right),\end{equation} \tag{ 44 }$$

for $0{< }x{< }\mathcal{N}-1$, and at the boundaries

$$\begin{equation}\hat{\psi }\left(0\right)\to {\text{e}}^{\text{i}\theta \left(0\right)}\hat{\psi }\left(0\right), \hat{U}\left(0\right)\to {\text{e}}^{\text{i}\theta \left(0\right)}\hat{U}\left(0\right),\end{equation} \tag{ 45a }$$

$$\begin{equation}\hat{\psi }\left(\mathcal{N}-1\right)\to {\text{e}}^{\text{i}\theta \left(\mathcal{N}-1\right)}\hat{\psi }\left(\mathcal{N}-1\right), \hat{U}\left(\mathcal{N}-2\right)\to {\text{e}}^{-\text{i}\theta \left(\mathcal{N}-1\right)}\hat{U}\left(\mathcal{N}-2\right).\end{equation} \tag{ 45b }$$

The local gauge transformations of the matter field $\hat{\psi }\left(x\right)$ and the link $\hat{U}\left(x\right)$ are different in the bulk (44) and at the boundaries equation (45). In particular, the gauge transformations are different at the boundaries since matter cannot move across the boundary and there is no electric flux out of the system. Further, matter not moving out of the system ensures that the total charge is conserved. This local gauge transformation is represented by the unitary transformation

$$\begin{equation}\hat{S}\left(x\right)={\text{e}}^{\text{i}\theta \left(x\right)\hat{G}\left(x\right)}\end{equation} \tag{ 46 }$$

with the generators

$$\begin{equation}\hat{G}\left(x\right)=\hat{E}\left(x\right)-\hat{E}\left(x-1\right)-\hat{Q}\left(x\right)\end{equation} \tag{ 47 }$$

for $0{< }x{< }\mathcal{N}-1$, and at the boundaries

$$\begin{equation}\hat{G}\left(0\right)=\hat{E}\left(0\right)-\hat{Q}\left(0\right),\end{equation} \tag{ 48a }$$

$$\begin{equation}\hat{G}\left(\mathcal{N}-1\right)=-\hat{E}\left(\mathcal{N}-2\right)-\hat{Q}\left(\mathcal{N}-1\right),\end{equation} \tag{ 48b }$$

using the local charges Q(x) defined in (24). The symmetry generators fulfill

$$\begin{equation}\left[\hat{G}\left(x\right),{\hat{H}}_{\text{cQED}}\right]=0\quad \forall \enspace x\end{equation} \tag{ 49 }$$

and in a complete analogy to the two previous cases, we define the gauge invariant, or physical states |ϕ⟩, as

$$\begin{equation}\hat{G}\left(x\right)\vert \phi \rangle =q\left(x\right)\vert \phi \rangle .\end{equation} \tag{ 50 }$$

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The Gauss's laws give rise to superselection sectors labeled by the configuration of static charges $\left\{q\left(x\right)\right\}$. In particular the Gauss's laws of the bulk are

$$\begin{equation}\left[\hat{E}\left(x\right)-\hat{E}\left(x-1\right)\right]\vert \phi \rangle =\left[\hat{Q}\left(x\right)+q\left(x\right)\right]\vert \phi \rangle \end{equation} \tag{ 51 }$$

— the divergence of electric fields at a vertex equals the sum of dynamical and static charges there, as expected in a Gauss's law. Since the global and local U(1) symmetry are related, the conservation of the total number of fermions is connected to the global charge of the system:

$$\begin{equation}\hat{Q}=\sum _{x}\hat{Q}\left(x\right)={\hat{N}}_{\text{F}}-\left(\mathcal{M}+1\right).\end{equation} \tag{ 52 }$$

Thus the operators of the total particle number ${\hat{N}}_{\text{F}}$ and the global charge $\hat{Q}$ differ only by a constant shift of $\mathcal{M}+1=\mathcal{N}/2$ representing the number of odd sites in the lattice. The latter $\mathcal{M}+1$ equals to the unity in equation (16) for the JCM. As in the Jaynes–Cummings case, adding all the Gauss's laws leads to a constraint for the static charges:

$$\begin{equation}\sum _{x}q\left(x\right)\vert \phi \rangle =\sum _{x}\hat{G}\left(x\right)\vert \phi \rangle =-\sum _{x}\hat{Q}\left(x\right)\vert \phi \rangle =0,\end{equation} \tag{ 53 }$$

where the last equality (... = 0) holds since we consider fermionic half-filling (${n}_{\text{F}}=\mathcal{M}+1$) configurations in the lattice, which includes also the Dirac sea state.

The Hilbert space of the (1 + 1) dimensional lattice gauge theory is a tensor product of the Fock space of the matter sites, which is finite dimensional for a finite system, and the Hilbert spaces on the links, which are infinite dimensional. However, the gauge symmetry restricts the physical Hilbert space into a subspace of the infinite many-body Hilbert space. It is a union of all possible static charge sectors which are labeled by all possible integers $\left\{q\left(x\right)\right\}$ satisfying the condition ∑_xq(x) = 0. A sector denoted by a specific static charge configuration $\left\{q\left(x\right)\right\}$ cannot be left with the gauge invariant dynamics governed by the Hamiltonian ${\hat{H}}_{\text{cQED}}$ given by (43). For a finite chain, the dimension of one sector will be finite, like in the U(1) single link model considered in section 3.1. In the following we derive an upper bound for the dimension of one sector for the lattice of $\mathcal{N}=2\mathcal{M}+2$ sites.

The Gauss's law on the first vertex is

$$\begin{equation}\hat{E}\left(0\right)\vert \phi \left\{q\left(x\right)\right\}\rangle =\left[q\left(0\right)+\hat{Q}\left(0\right)\right]\left\vert \phi \left\{q\left(x\right)\right\}\right\rangle .\end{equation} \tag{ 54 }$$

Because the eigenvalues of $\hat{Q}\left(0\right)$ can only be 0 or 1, we will only have two possibilities for $\hat{E}\left(0\right)$, viz. $\hat{E}\left(0\right)=q\left(0\right)$ and $\hat{E}\left(0\right)=q\left(0\right)+1$. The Gauss's law at the next link is

$$\begin{equation}\hat{E}\left(1\right)\vert \phi \left\{q\left(x\right)\right\}\rangle =\left[\hat{E}\left(0\right)+q\left(1\right)+\hat{Q}\left(1\right)\right]\vert \phi \left\{q\left(x\right)\right\}\rangle .\end{equation} \tag{ 55 }$$

Since Q(1) = 0, −1 we deduce that we have three options for E(1): q(0) + q(1) − 1, q(0) + q(1) and q(0) + q(1) + 1. Similarly, the next link will require a four dimensional Hilbert space, spanned by the electric field values E(2) equal to q(0) + q(1) + q(1) − 1, q(0) + q(1) + q(1), q(0) + q(1) + q(1) + 1 and q(0) + q(1) + q(1) + 2. Moving on recursively, we obtain that the allowed values of E(n) are ${\sum }_{x=0}^{n}q\left(x\right)-\frac{n}{2},\dots ,{\sum }_{x=0}^{n}q\left(x\right)+\frac{n}{2}+1$ if the site n is even, and ${\sum }_{x=0}^{n}q\left(x\right)-\frac{n+1}{2},\dots ,{\sum }_{x=0}^{n}q\left(x\right)+\frac{n+1}{2}$ if n is odd.

Similarly we can start from the Gauss's law at position $\mathcal{N}-1$ and move toward x = 0 repeating the recursive argument. Hence the Hilbert space dimension of the link at position n and at position $2\mathcal{M}-n$ is the same i.e. $\mathrm{dim}\left(n\right)=\mathrm{dim}\left(2\mathcal{M}-n\right)$ and, for $0{\leqslant}n{\leqslant}\mathcal{M}$ the dimension is given by

$$\begin{equation}\mathrm{dim}\left(n\right)=n+2.\end{equation} \tag{ 56 }$$

Therefore the maximal Hilbert space dimension of all the links is given by

$$\begin{equation}{\mathcal{D}}_{\text{Gauge}}\left(\mathcal{M}\right)\equiv {\left(\prod _{n=0}^{\mathcal{M}-1}\left(n+2\right)\right)}^{2}\left(\mathcal{M}+2\right)=\left(\mathcal{M}+1\right)!\left(\mathcal{M}+2\right)!,\end{equation} \tag{ 57 }$$

which is bounded for a finite system size.

The fermionic Fock space dimension reduces in the half filling case to $\left(\genfrac{}{}{0.0pt}{}{2\mathcal{M}+2}{\mathcal{M}+1}\right)$-number of possibilities to distribute $\mathcal{M}+1$ on $2\mathcal{M}+2$ sites. Hence,

$$\begin{equation}\mathcal{D}\left(\mathcal{M}\right){< }\left(\mathcal{M}+2\right)\left(2\mathcal{M}+2\right)!=\mathcal{N}!\left(\mathcal{N}/2+1\right)\end{equation} \tag{ 58 }$$

is an upper bound for Hilbert space describing the physics of (1 + 1) cQED with open boundary conditions. This bound is independent of the charge sector. In particular it is not a strict bound since gauge invariance has not been fully exploited. In one space dimension with open boundaries one can solve the Gauss's law explicitly, but this gives rise to non-local interactions [53, 74–77]. The above upper bound denotes dimensions which will be required for building a quantum simulator which does not eliminate the Gauss's law and involves only local interaction terms.

We begin by reviewing relevant properties of SU(2) and use this to formulate a non-Abelian one link model with local gauge invariance. We first introduce the relevant Hilbert spaces on the links and on the vertices. The Hilbert space, which will be used on the link, is the same as the quantum rigid rotor, whereas the Hilbert space of vertices will be built by fermionic particles. Most importantly the link Hilbert space as well as the fermionic Hilbert space on the vertices will support representations of SU(2).

We start by introducing the group SU(2) as the rotation of an object parametrized by two angles α and β e.g. a particle on a sphere, a spin or a diatomic molecule. However, the group SU(2) also appears by studying the rotation of a rigid object. In classical mechanics [79] as well as in quantum mechanics the rigid body is determined by three angles—the Euler angles-α, β, γ. While we shall use Euler angles to characterize the rigid body there are other ways to parametrize any group element g ∈ SU(2), which will involve three independent parameters ${\theta }_{a}\left(g\right)$.

A general spin rotation of a state is implemented by the unitary operator

$$\begin{equation}{\hat{S}}_{\text{g}}={\text{e}}^{\text{i}{\theta }_{a}\left(g\right){\hat{J}}_{a}},\end{equation} \tag{ 59 }$$

where a summation is implied over a Cartesian index a repeated twice. Here ${\theta }_{a}\left(g\right)$ are the group parameters, uniquely identifying one particular rotation g ∈ SU(2), and the operators J_a (with a = x, y, z) are the three rotation generators, also known as the angular momentum operators. These operators do not commute, but rather satisfy the SU(2) lie algebra

$$\begin{equation}\left[{\hat{J}}_{a},{\hat{J}}_{b}\right]=\mathrm{i}{{\epsilon}}_{abc}{\hat{J}}_{c},\end{equation} \tag{ 60 }$$

where _abc is the Levi-Civita symbol. From these three operators one can build the total angular momentum or quadratic Casimir operator

$$\begin{equation}{\hat{\mathbf{J}}}^{2}={\hat{J}}_{a}{\hat{J}}_{a}={\hat{J}}_{x}^{2}+{\hat{J}}_{y}^{2}+{\hat{J}}_{z}^{2}.\end{equation} \tag{ 61 }$$

It commutes with all the generators

$$\begin{equation}\left[{\hat{\mathbf{J}}}^{2},{\hat{J}}_{a}\right]=0\end{equation} \tag{ 62 }$$

and thus one may choose one particular generator—usually J_z, and combine it with J² to a maximally mutually commuting set of operators, which can be simultaneously diagonalized. The quantum numbers given by the eigenvalues of these operators are used as quantum numbers, labeling quantum states with respect to their transformation properties. Those two operators define the angular momentum states |jm⟩ as

$$\begin{equation}{\hat{\mathbf{J}}}^{2}\vert jm\rangle =j\left(j+1\right)\vert jm\rangle ,\end{equation} \tag{ 63 }$$

$$\begin{equation}{\hat{J}}_{z}\vert jm\rangle =m\vert jm\rangle \end{equation} \tag{ 64 }$$

with $j=\frac{1}{2}n$, $n\in \mathbb{N}$ and m = −j, −j + 1 ⋯ , j. The quantum number j denotes the irreducible representation of SU(2), which contains 2j + 1 states with different m values.

General rotations mix the different states within a representation, but not different representations: the rotation operators are block diagonal in the representations. Therefore the generators may be expressed as a direct sum in the representations,

$$\begin{equation}{\hat{J}}_{a}=\underset{j}{\bigoplus}{\left({T}_{a}^{j}\right)}_{mn}\vert jm\rangle \langle jn\vert ,\end{equation} \tag{ 65 }$$

where summations are implied over the quantum numbers m and n repeated twice. For each irreducible representation j, ${T}_{a}^{j}$ is a set of matrices with dimension $\mathrm{dim}\left(j\right)\equiv 2j+1$ satisfying the algebra of the group

$$\begin{equation}\left[{T}_{a}^{j},{T}_{b}^{j}\right]=\mathrm{i}{{\epsilon}}_{abc}{T}_{c}^{j}.\end{equation} \tag{ 66 }$$

In particular for j = 1/2, the fundamental representation, the matrices ${T}_{a}^{j}$ are given by

$$\begin{equation}{\left({T}_{a}^{1/2}\right)}_{mn}=\frac{1}{2}{\sigma }_{a},\end{equation} \tag{ 67 }$$

where σ_a denotes the three Pauli matrices. The j irreducible representation of a group element g is given by the unitary matrix

$$\begin{equation}{D}^{\enspace j}\left(g\right)={\text{e}}^{\text{i}{\theta }_{a}\left(g\right){T}_{a}^{j}},\end{equation} \tag{ 68 }$$

which are called Wigner matrices. In terms of the Euler angles α, β and γ the Wigner matrices read [80]

$$\begin{equation}{D}^{\enspace j}\left(\alpha ,\beta ,\gamma \right)={\text{e}}^{-\text{i}\alpha {T}_{z}^{j}}{\text{e}}^{-\text{i}\beta {T}_{y}^{j}}{\text{e}}^{-\text{i}\gamma {T}_{z}^{j}}.\end{equation} \tag{ 69 }$$

A general rotation operator may therefore be written in the block diagonal form

$$\begin{equation}\hat{S}\left(\alpha ,\beta ,\gamma \right)=\underset{j}{\bigoplus}{D}_{{m}^{\prime }m}^{\enspace j}\left(\alpha ,\beta ,\gamma \right)\left\vert j{m}^{\prime }\right\rangle \left\langle \enspace jm\right\vert \end{equation} \tag{ 70 }$$

implying that

$$\begin{equation}\hat{S}\left(\alpha ,\beta ,\gamma \right)\vert jm\rangle ={D}_{{m}^{\prime }m}^{\enspace j}\left(\alpha ,\beta ,\gamma \right)\vert j{m}^{\prime }\rangle .\end{equation} \tag{ 71 }$$

In the case of U(1) we considered a Hilbert space parametrized by one angle—the coordinate for a particle on a ring. In the previous section we discussed rotations of a particle parametrized by two angles. Next, we consider a Hilbert space parametrized by three Euler-angles |α, β, γ⟩—the configuration (coordinate) space of the quantum mechanical rigid rotor [81, 82]. Let us consider the rotation of an isotropic rigid rotor, whose Hamiltonian is

$$\begin{equation}\hat{H}=\frac{{\hat{\mathbf{J}}}^{2}}{2I},\end{equation} \tag{ 72 }$$

where ${\hat{\mathbf{J}}}^{2}$ is the total angular momentum. Since we consider an isotropic rigid body, the moment of inertia is the same for all axis and denoted by I.

The rotation may be described in the inertial laboratory or space frame, but as we are dealing with a rigid body, we may also choose another frame of reference fixed to the body—for example, with its origin at its center of mass—the body frame. The two frames are related by a rotation [81, 82]. Measuring the components of (pseudo) vector operators in two frames will give rise to different results. One thus has to introduce different operators in both frames, which are related by a unitary operation [82]. The two sets of angular momentum operators fulfill in the space frame,

$$\begin{equation}\left[{\hat{R}}_{a},{\hat{R}}_{b}\right]=\mathrm{i}{{\epsilon}}_{abc}{\hat{R}}_{c},\end{equation} \tag{ 73 }$$

the SU(2) algebra, and in the body frame

$$\begin{equation}\left[{\hat{L}}_{a},{\hat{L}}_{b}\right]=-\mathrm{i}{{\epsilon}}_{abc}{\hat{L}}_{c}\end{equation} \tag{ 74 }$$

with an opposite sign in the algebra which results from rotating the angular momentum operator from the space frame to the body frame [81, 82]. Since one may measure the angular momentum in one frame without disturbing the measuring angular momentum in the other frame the generators of space and body rotations must commute

$$\begin{equation}\left[{\hat{R}}_{a},{\hat{L}}_{b}\right]=0.\end{equation} \tag{ 75 }$$

On the other hand, the two sets of rotations are not fully independent: the space and body frames are related by a rotation, and hence a measurement of the total angular momentum will give rise to the same result in both non-independent frames [82]. We define a total angular momentum operator, also called Casimir operator, as

$$\begin{equation}{\hat{\mathbf{J}}}^{2}\equiv {\hat{\mathbf{R}}}^{2}={\hat{\mathbf{L}}}^{2},\end{equation} \tag{ 76 }$$

which is the same in both frames.

Again we choose a maximal set of mutually commuting operators: the total angular momentum ${\hat{\mathbf{J}}}^{2}$ shared by both frames and the z components i.e. ${\hat{L}}_{z}$ and ${\hat{R}}_{z}$. This allows to span the Hilbert space of the rigid rotor by angular momentum or representation basis state |jmn⟩, defined by

$$\begin{equation}{\hat{\mathbf{J}}}^{2}\vert jmn\rangle =j\left(j+1\right)\vert jmn\rangle ,\end{equation} \tag{ 77a }$$

$$\begin{equation}{\hat{L}}_{z}\vert jmn\rangle =m\vert jmn\rangle ,\end{equation} \tag{ 77b }$$

$$\begin{equation}{\hat{R}}_{z}\vert jmn\rangle =n\vert jmn\rangle .\end{equation} \tag{ 77c }$$

The eigenstates of the rigid body rotor |jmn⟩ are characterized by three quantum numbers j, m and n, where m and n are associated with two projections of the total angular momentum j. In contrast the angular motion of a quantum particle in a spherical potential is characterized just by two quantum numbers j and m, where m is the projection of the total angular momentum j for a fixed radial quantum number [81]. There are two quantum numbers, because the angular motion of the particle is described by two spherical angles θ and φ. On the other hand, the rigid body rotation is represented by three Euler angles α, β and γ and the additional angular degree of freedom is reflected by an extra quantum number n resulting in |jmn⟩.

For fixed j the ${\left(2j+1\right)}^{2}$ eigenstates |jmn⟩ form a multiplet, so the generators may be expressed in the block diagonal form

$$\begin{equation}\hat{\mathbf{L}}=\sum _{j}\vert jmn\rangle {\mathbf{T}}_{{m}^{\prime }m}^{j}\langle \enspace j{m}^{\prime }n\vert ,\end{equation} \tag{ 78 }$$

$$\begin{equation}\hat{\mathbf{R}}=\sum _{j}\vert jmn\rangle {\mathbf{T}}_{n{n}^{\prime }}^{j}\langle \enspace jm{n}^{\prime }\vert .\end{equation} \tag{ 79 }$$

In terms of state vectors |jmn⟩ one can define the body and space operators of the rigid rotor ${{\Theta}}_{g}^{R}$ and ${{\Theta}}_{g}^{L}$, respectively, satisfying

$$\begin{equation}{{\Theta}}_{\text{g}}^{\text{R}}\vert jmn\rangle =\vert jm{n}^{\prime }\rangle {D}_{{n}^{\prime }n}^{j}\left(g\right)=\vert jm{n}^{\prime }\rangle {D}_{{n}^{\prime }n}^{j}\left(\alpha ,\beta ,\gamma \right),\end{equation} \tag{ 80a }$$

$$\begin{equation}{{\Theta}}_{\text{g}}^{\text{L}}\vert jmn\rangle ={D}_{m{m}^{\prime }}^{j}\left(g\right)\vert j{m}^{\prime }n\rangle ={D}_{m{m}^{\prime }}^{j}\left(\alpha ,\beta ,\gamma \right)\vert j{m}^{\prime }n\rangle .\end{equation} \tag{ 80b }$$

— out of which we can rename the R_a and L_a operators right and left generators, respectively, generating right and left group transformations.

The state vectors |jmn⟩ form a full set of eigenstates of the total angular momentum operator (76). In order to obtain their wavefunctions in coordinate space, we introduce the dual basis of group elements |g⟩ or parameters |α, β, γ⟩. The transition between the bases gives the wavefunctions of the free isotropic rigid rotor

$$\begin{equation}\left\langle \alpha \beta \gamma \vert jmn\right\rangle =\sqrt{\frac{2j+1}{8{\pi }^{2}}}{D}_{mn}^{j}\left(\alpha ,\beta ,\gamma \right).\end{equation} \tag{ 81 }$$

as solved by Wigner [80].

As in the U(1) case we introduce the group element operator as the operator inducing a representation on the states |α, β, γ⟩. These operators, which will later be used as the gauge connections, are defined by

$$\begin{equation}{\hat{U}}_{mn}^{j}\vert \alpha ,\beta ,\gamma \rangle ={D}_{mn}^{\enspace j}\left(\alpha ,\beta ,\gamma \right)\vert \alpha ,\beta ,\gamma \rangle .\end{equation} \tag{ 82 }$$

Here ${\hat{U}}_{mn}^{j}$ is a $\left(2j+1\right){\times}\left(2j+1\right)$ matrix of operators acting on the rigid rotor Hilbert space. From the definition of the gauge connection we obtain immediately

$$\begin{equation}\left[{\hat{U}}_{mn}^{j},{\hat{U}}_{{m}^{\prime }{n}^{\prime }}^{{j}^{\prime }}\right]=0\end{equation} \tag{ 83 }$$

and

$$\begin{equation}{\hat{U}}_{mn}^{j}{\left({\hat{U}}^{j}\right)}_{nk}^{{\dagger}}={\delta }_{mk}.\end{equation} \tag{ 84 }$$

Using the definition of the unitary operators (80a) and (80b) the gauge connection ${\hat{U}}_{mn}^{j}$ transforms as

$$\begin{equation}{{\Theta}}^{\text{R}}\left(\alpha ,\beta ,\gamma \right){U}_{mn}^{j}{{\Theta}}^{\text{R}{\dagger}}\left(\alpha ,\beta ,\gamma \right)={U}_{m{n}^{\prime }}^{j}{D}_{{n}^{\prime }n}^{j}\left(\alpha ,\beta ,\gamma \right),\end{equation} \tag{ 85a }$$

$$\begin{equation}{{\Theta}}^{\text{L}}\left(\alpha ,\beta ,\gamma \right){U}_{mn}^{j}{{\Theta}}^{\text{L}{\dagger}}\left(\alpha ,\beta ,\gamma \right)={D}_{m{m}^{\prime }}^{j}\left(\alpha ,\beta ,\gamma \right){U}_{{m}^{\prime }n}^{j},\end{equation} \tag{ 85b }$$

leading, by considering infinitesimal transformations, to

$$\begin{equation}\left[{\hat{R}}_{a},{\hat{U}}_{mn}^{j}\right]={\hat{U}}_{m{n}^{\prime }}^{j}{\left({T}_{a}^{j}\right)}_{{n}^{\prime }n},\end{equation} \tag{ 86a }$$

$$\begin{equation}\left[{\hat{L}}_{a},{\hat{U}}_{mn}^{j}\right]={\left({T}_{a}^{j}\right)}_{m{m}^{\prime }}{\hat{U}}_{{m}^{\prime }n}^{j}.\end{equation} \tag{ 86b }$$

The transformation laws and commutation relations for the non-abelian case appear as an analogy to the U(1) case. In the abelian case the $\hat{U}$ operator raises the electric field, which is manifested by $\left[\hat{E},\hat{U}\right]=\hat{U}$. The previous equations are the non-abelian generalization of the transformation laws and commutation relations. U(1) is an abelian group whose irreducible representations are one dimensional, and therefore the representation states are labeled only by the quantum number j (no m and n are needed). Raising |j⟩ to |j + 1⟩ by $\hat{U}$ could be seen as combining the representations j and 1 of U(1) to j + 1.

The definition of the gauge connection in equation (82) allows to derive the action of the operator ${\hat{U}}^{j}$ on the states |jmm'⟩, where the states |jmm'⟩ are the eigenstates defined by equation (77). The explicit expression of ${\hat{U}}^{j}$ is given by

$$\begin{equation}{\hat{U}}_{m{m}^{\prime }}^{j}=\sum _{J,K}\sqrt{\frac{\mathrm{dim}\left(J\right)}{\mathrm{dim}\left(K\right)}}\left\langle JMjm\vert KN\right\rangle \left\langle K{N}^{\prime }\vert J{M}^{\prime }j{m}^{\prime }\right\rangle \vert KN{N}^{\prime }\rangle \langle JM{M}^{\prime }\vert .\end{equation} \tag{ 87 }$$

A proof of this relation is given in [78]. If ${\hat{U}}_{m{m}^{\prime }}^{j}$ acts on a |JMM'⟩ state, the resulting states |KNN'⟩ are those reachable by adding angular momentum j to the state |JMM'⟩. The addition of angular momentum manifests in the Clebsch–Gordan coefficients ⟨JMjm|KN⟩ and ⟨KN'|JM'jm'⟩. In particular, the structure of ${\hat{U}}_{m{m}^{\prime }}^{j}$ makes sure that both the initial and final states of the body and space frame remain in the same representation. The state with j = 0, |000⟩, is called the singlet state, as it is invariant under all the group transformations—${\hat{{\Theta}}}^{\text{R}}\left(\alpha ,\beta ,\gamma \right)\vert 000\rangle ={\hat{{\Theta}}}^{\text{L}}\left(\alpha ,\beta ,\gamma \right)\vert 000\rangle =\vert 000\rangle$. The states |jmn⟩ can be created by acting on it with the group element operator [78]

$$\begin{equation}\vert jmn\rangle =\sqrt{\mathrm{dim}\left(j\right)}{\hat{U}}_{mn}^{j}\left\vert 000\right\rangle .\end{equation} \tag{ 88 }$$

The last ingredient in order to discuss the SU(2) lattice gauge theory is the matter—fermions. Therefore we introduce fermionic creation operators ${\hat{\psi }}_{m}^{j{\dagger}}$, that form the elements of a spinor in the j representation. The number of components is the dimension of the representation-2j + 1 in the case of SU(2). In the important case of the fundamental representation (j = 1/2) such a spinor has two components, which we label with m = 1, 2. Again we define unitary transformations of the spinor such that

$$\begin{equation}{\hat{\psi }}_{m}\to {D}_{m{m}^{\prime }}\left(\alpha ,\beta ,\gamma \right){\hat{\psi }}_{{m}^{\prime }}\end{equation} \tag{ 89 }$$

— this transformation mixes the elements of the spinor, just like a conventional j multiplet. The transformation of the fermions is generated by the set of generators

$$\begin{equation}{\hat{Q}}_{a}={\hat{\psi }}_{m}^{j{\dagger}}{\left({T}_{a}\right)}_{mn}{\hat{\psi }}_{n},\end{equation} \tag{ 90 }$$

which fulfill the lie algebra of the group:

$$\begin{equation}\left[{\hat{Q}}_{a},{\hat{Q}}_{b}\right]=\mathrm{i}{{\epsilon}}_{abc}{\hat{Q}}_{c}.\end{equation} \tag{ 91 }$$

Since the operators ${\hat{Q}}_{a}$ fulfill the SU(2) algebra, one can use ${\hat{\mathbf{Q}}}^{2}={\hat{Q}}_{x}^{2}+{\hat{Q}}_{y}^{2}+{\hat{Q}}_{z}^{2}$ and ${\hat{Q}}_{z}$ as maximal set of commuting operators to characterize fermionic states. For example, consider the fundamental representation j = 1/2, where the charges are defined via the Pauli matrices

$$\begin{equation}{\hat{Q}}_{a}=\frac{1}{2}{\hat{\psi }}_{m}^{{\dagger}}{\left({\sigma }_{a}\right)}_{mn}{\hat{\psi }}_{n}.\end{equation} \tag{ 92 }$$

Note that we omitted the index j, which we will suppress whenever the fundamental representation is discussed. In the following we discuss the transformation properties of the fermionic states created from the vacuum |vac⟩_F. The four possible states are:

• The vacuum-|vac⟩_F—a singlet state:
  $$\begin{equation}{\hat{\mathbf{Q}}}^{2}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=0,\end{equation} \tag{ 93a }$$
  $$\begin{equation}{\hat{Q}}_{z}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=0.\end{equation} \tag{ 93b }$$
• The two single fermion states—forming a fundamental, j = 1/2 multiplet,
  $$\begin{equation}{\hat{\mathbf{Q}}}^{2}{\hat{\psi }}_{m}^{{\dagger}}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=\frac{1}{2}\left(\frac{1}{2}+1\right){\hat{\psi }}_{m}^{{\dagger}}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}},\end{equation} \tag{ 94a }$$
  $$\begin{equation}{\hat{Q}}_{z}{\hat{\psi }}_{m}^{{\dagger}}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=m{\hat{\psi }}_{m}^{{\dagger}}\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}.\end{equation} \tag{ 94b }$$
• The doubly occupied state is a singlet:
  $$\begin{equation}{\hat{\mathbf{Q}}}^{2}\left(\frac{1}{2}{{\epsilon}}_{mn}{\hat{\psi }}_{m}^{{\dagger}}{\hat{\psi }}_{n}^{{\dagger}}\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=0,\end{equation} \tag{ 95a }$$
  $$\begin{equation}{\hat{Q}}_{z}\left(\frac{1}{2}{{\epsilon}}_{mn}{\hat{\psi }}_{m}^{{\dagger}}{\hat{\psi }}_{n}^{{\dagger}}\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}=0,\end{equation} \tag{ 95b }$$
  where _ij is the 2 × 2 antisymmetric tensor. The doubly occupied state is created by adding two fermions with j = 1/2, each with an opposite spin which can add up to j = 0 and j = 1. However, the fermionic statistics only allow for the antisymmetric j = 0 representation as can be checked by direct inspection.

In this subsection we formulate a single link theory with a local non-abelian gauge invariance. The Hamiltonian takes the form

$$\begin{equation}H=\frac{{g}^{2}}{2}{\hat{\mathbf{J}}}^{2}+M\left[{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{\psi }}_{m}\left(0\right)-{\hat{\psi }}_{m}^{{\dagger}}\left(1\right){\hat{\psi }}_{m}\left(1\right)\right]+{\epsilon}\left[{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{U}}_{mn}{\hat{\psi }}_{n}\left(1\right)+\text{h.c.}\right]\end{equation} \tag{ 96 }$$

with ${\hat{U}}_{mn}\equiv {\hat{U}}_{mn}^{{j}^{\prime }=1/2}$. A graphical illustration of the Hamiltonian equation (96) can be found in figure 4. As indicated in the paragraph above equation (87), the SU(2) operator ${\hat{U}}^{{j}^{\prime }}$ adds j' to the existing representation j, thus changing the rotational energy of the system. Hence the operator ${\hat{U}}_{mn}$ featured in the Hamiltonian (96) describes changes of the rotation quantum number j by 1/2 after the fermion undergoes a transition from the site 1 to the site 0. In this way ${\hat{U}}_{mn}$ plays a similar role as the operator $\hat{U}$ featured in the Hamiltonian (31) for the U(1) link model.

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The Hamiltonian (96) is invariant under the local (gauge) transformations

$$\begin{equation}{\hat{\psi }}_{m}\left(0\right)\to {D}_{m{m}^{\prime }}\left(g\left(0\right)\right){\hat{\psi }}_{{m}^{\prime }}\left(0\right),{\hat{U}}_{mn}\to {D}_{m{m}^{\prime }}\left(g\left(0\right)\right){\hat{U}}_{{m}^{\prime }n},\end{equation} \tag{ 97a }$$

$$\begin{equation}{\hat{\psi }}_{m}\left(1\right)\to {D}_{m{m}^{\prime }}\left(g\left(1\right)\right){\hat{\psi }}_{{m}^{\prime }}\left(1\right),{\hat{U}}_{mn}\to {\hat{U}}_{m{n}^{\prime }}{D}_{{n}^{\prime }n}^{{\dagger}}\left(g\left(1\right)\right),\end{equation} \tag{ 97b }$$

which are generated by

$$\begin{equation}{\hat{G}}_{a}\left(0\right)={\hat{L}}_{a}-{\hat{Q}}_{a}\left(0\right),\end{equation} \tag{ 98 }$$

$$\begin{equation}{\hat{G}}_{a}\left(1\right)=-{\hat{R}}_{a}-{\hat{Q}}_{a}\left(1\right)\end{equation} \tag{ 99 }$$

—at each site x exists a set of non-commuting generators—three for SU(2) satisfying the left algebra of the group

$$\begin{equation}\left[{\hat{G}}_{a}\left(x\right),{\hat{G}}_{b}\left(x\right)\right]=-i{{\epsilon}}_{abc}{\hat{G}}_{c}\left(x\right)\quad \forall \enspace x,\end{equation} \tag{ 100 }$$

and hence it is impossible to diagonalize all three generators with a common basis. Moreover, the generators of the symmetry commute with the Hamiltonian

$$\begin{equation}\left[{\hat{G}}_{a}\left(x\right),\hat{H}\right]=0\quad \forall \enspace x,a.\end{equation} \tag{ 101 }$$

First, we consider the sector with no static charges, that is, the sector whose states |ϕ({0})⟩ are all strictly invariant under gauge transformations,

$$\begin{equation}{S}_{\text{g}}\left(x\right)\vert \phi \left(\left\{0\right\}\right)\rangle =\vert \phi \left(\left\{0\right\}\right)\rangle \quad \forall \enspace g,x.\end{equation} \tag{ 102 }$$

In this sector, and only in this sector, all generators commute, because of all Gauss's laws

$$\begin{equation}{\hat{G}}_{a}\left(x\right)\vert \phi \left(\left\{0\right\}\right)\rangle =0,\end{equation} \tag{ 103 }$$

which implies that the divergence of non-abelian electric fields—left from one side and right from the other—equals exactly the dynamic charge at the vertex.

Further, we introduce static charge localized on the vertex x, which implies that the action of ${\hat{G}}_{a}\left(x\right)$ on such a charged state is not zero. However, it is not possible to diagonalize all the generators simultaneously, which makes it necessary to label the states by nonzero quantum numbers determined by a maximal set of commuting operators at the vertex-${\hat{\mathbf{G}}}^{2}\left(x\right)$ and ${\hat{G}}_{z}\left(x\right)$. This results in the usual multiplet structure: static charges form multiplets, associated with a representation j of SU(2). Each site of our lattice therefore is associated with labels ${j}_{q}\left(x\right),{m}_{q}\left(x\right)$ specifying the multiplet of local static charges, and the position within the multiplet. We therefore label our static charge sectors by a collection $\left\{{j}_{q}\left(x\right),{m}_{q}\left(x\right)\right\}$, and the Gauss's laws in a given sector takes the form

$$\begin{equation}{\hat{G}}_{a}\left(x\right)\left\vert \hat{\phi }\left(\left\{{j}_{q}\left(x\right),{m}_{q}\left(x\right)\right\}\right)\right\rangle ={\left({T}_{a}^{{j}_{q}\left(x\right)}\right)}_{{m}_{q}^{\prime }\left(x\right),{m}_{q}\left(x\right)}\vert \psi \left(\left\{{j}_{q}\left(x\right),{m}_{q}^{\prime }\left(x\right)\right\}\right)\rangle \end{equation} \tag{ 104 }$$

—the generators of gauge transformations do not leave the states invariant in general, but rather mix the elements of the so-called static charge multiplet.

If a vertex contains no static charge (j_q = 0), the action of ${\hat{G}}_{a}$ on this vertex will result in ${T}_{a}^{0}=0$ indicating invariance as expected. If an additional vertex possesses a fundamental static charge j_q = 1/2, two different m_q values will be necessary. Acting with the operators ${\hat{G}}_{a}$ on such a states will rotate the states by ${T}_{a}=\frac{1}{2}{\sigma }_{a}$. Consequently the action of ${\hat{G}}_{z}$ results in two eigenstate equations (for the two different m_q states) with eigenvalues ± 1/2. The ${\hat{G}}_{x},{\hat{G}}_{y}$ equations, on the other hand, will mix these two ${\hat{G}}_{z}$ eigenstates with respect to the matrix elements of σ_x/2 and σ_y/2, respectively.

To formalize this procedure one defines a Hilbert space sector as the set of all states which share the same ${\hat{\mathbf{G}}}^{2}\left(x\right)$ eigenvalues everywhere, and label it by the set of the representation indices $\left\{{j}_{q}\left(x\right)\right\}$. A sector is closed under the dynamics and under all possible gauge transformations. We define a Hilbert space subsector as the subset of states within a sector, that also share the same ${\hat{G}}_{z}\left(x\right)$ eigenvalues everywhere—labeled by $\left\{{j}_{q}\left(x\right),{m}_{q}\left(x\right)\right\}$. A subsector is closed by the dynamics, but is invariant only under gauge transformations generated by ${\hat{G}}_{z}$: arbitrary gauge transformations mix the subsectors within a sector. Moreover, the Hamiltonian or more generally every gauge invariant operator is block diagonal in the subsectors—since such operators commute with all the ${\hat{\mathbf{G}}}^{2},{\hat{G}}_{z}$ operators. Furthermore, the matrix elements of any gauge invariant operator will be identical for all the subsectors within a given sector

In order to understand the spectrum of the single link Hamiltonian we exploit the Gauss's law at the vertices i.e.

$$\begin{equation}{\hat{L}}_{a}\vert \phi \rangle ={\hat{Q}}_{a}\left(0\right)\vert \phi \rangle ,\end{equation} \tag{ 105 }$$

$$\begin{equation}{\hat{R}}_{a}\vert \phi \rangle =-{\hat{Q}}_{a}\left(1\right)\vert \phi \rangle ,\end{equation} \tag{ 106 }$$

which yield

$$\begin{equation}{\hat{\mathbf{L}}}^{2}\vert \phi \rangle ={\hat{\mathbf{Q}}}^{2}\left(0\right)\vert \phi \rangle ,\end{equation} \tag{ 107 }$$

$$\begin{equation}{\hat{\mathbf{R}}}^{2}\vert \phi \rangle ={\hat{\mathbf{Q}}}^{2}\left(0\right)\vert \phi \rangle .\end{equation} \tag{ 108 }$$

Since ${\hat{\mathbf{L}}}^{2}={\hat{\mathbf{R}}}^{2}={\hat{\mathbf{J}}}^{2}$, we obtain that the representation of the link is equal to the representation on any of the vertices.

At half-filling the first option includes the Dirac sea state for the fermions—the right site x = 1 is fully occupied, and the left site (x = 0) is empty and the Gauss's law implies the gauge field to be $\left\vert 000\right\rangle$. The entire state becomes

$$\begin{equation}\vert D\rangle ={\psi }_{1}^{{\dagger}}\left(1\right){\psi }_{2}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \left\vert 000\right\rangle =\frac{1}{2}{{\epsilon}}_{mn}{\psi }_{m}^{{\dagger}}\left(1\right){\psi }_{n}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \left\vert 000\right\rangle .\end{equation} \tag{ 109 }$$

The second possibility to distribute the fermions is one fermion per site. In principle there are four configurations, however, gauge invariance forces to create this state from $\left\vert 0\right\rangle$ by using a gauge invariant operator. One possibility is to use ${\hat{\psi }}_{m}^{{\dagger}}\left(0\right){U}_{mn}{\psi }_{n}\left(1\right)$, which produces only one particular superposition of the four states, which will be gauge invariant. Normalizing the state results in

$$\begin{equation}\vert p\bar{p}\rangle =\frac{1}{\sqrt{2}}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){U}_{mn}{\psi }_{n}\left(1\right)\vert D\rangle =\frac{1}{2}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){{\epsilon}}_{nk}{\psi }_{k}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert \frac{1}{2}mn\rangle .\end{equation} \tag{ 110 }$$

The third option consists of having two fermions on the left site, where both sites are in the j = 0 representation. No particles or anti-particles are present and the link is not excited. The state is given by

$$\begin{equation}\vert \bar{D}\rangle ={\hat{\psi }}_{1}^{{\dagger}}\left(0\right){\hat{\psi }}_{2}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \left\vert 000\right\rangle =\frac{1}{2}{{\epsilon}}_{mn}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{\psi }}_{n}^{{\dagger}}\left(0\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \left\vert 000\right\rangle .\end{equation} \tag{ 111 }$$

The three states span the Hilbert space sector, because of the gauge symmetry. If we order the basis of our sector as $\vert D\rangle ,\vert p\bar{p}\rangle ,\vert \bar{D}\rangle$, the matrix representation of the Hamiltonian restricted to this sector is given by

$$\begin{equation}H{\vert }_{0}=\left(\begin{pmatrix}\hfill -2M\hfill & \hfill \sqrt{2}{\epsilon}\hfill & \hfill 0\hfill \\ \hfill \sqrt{2}{\epsilon}\hfill & \hfill \frac{3{g}^{2}}{8}\hfill & \hfill \sqrt{2}{\epsilon}\hfill \\ \hfill 0\hfill & \hfill \sqrt{2}{\epsilon}\hfill & \hfill 2M\hfill \end{pmatrix}\right).\end{equation} \tag{ 112 }$$

The spectrum and the dynamics generated by this Hamiltonian is illustrated in figure 5.

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A possible experimental realization of the Hamiltonian (112) can be achieved by using alkali atoms in the F = 1 manifold. In particular, one has to engineer an effective magnetic field B_eff ∝ (2, 0, −2M) generated by Raman or IR coupling, and the term 3g²/8 is described by the quadratic Zeeman effect [83, 84].

In the case of zero coupling ( = 0) between fermions and gauge fields, the basis states $\vert D\rangle ,\vert p\bar{p}\rangle ,\vert \bar{D}\rangle$ are the exact eigenstates, and are visualized in figure 5. |D⟩, the ground state in the absence of interactions, involves the Dirac sea (no particles and anti-particles) in a product with the gauge field singlet state. The state $\vert p\bar{p}\rangle$ is a meson, an excitation which includes a particle and an anti-particle connected by an electric flux tube in a gauge invariant fashion. Finally, $\vert \bar{D}\rangle$ includes two particles on the left site and two anti-particles (the absence of fermions) on the right one, and may be interpreted as a pair of a baryon and an anti-baryon in our toy model. A baryon is an antisymmetrized (and therefore colorless) combination of particles of all possible colors, (the particles on the left vertex), and the right site realizes the anti-baryon. Since baryons are color neutral, they are singlets and need not be connected by any flux tubes, manifested by the absence of gauge field excitations in $\vert \bar{D}\rangle$. With the increase of the interaction strength , the vacuum becomes more and more dressed by the meson state $\left\vert p\bar{p}\right\rangle$. At $M\approx \frac{3{g}^{2}}{16}$ the meson state $\left\vert p\bar{p}\right\rangle$ becomes resonant with the baryon-antibaryon state $\vert \bar{D}\rangle$. The resulting resonant dynamics are sketched in figure 5(B).

In the next step we introduce static charges into the single link model. Suppose we place on the left site some static charge with j_q, m_q and j_q > 0, and another static charge j'_q, n_q on the right site, which also determine the subsector. As before, the number of fermionic possibilities is highly restricted. We begin with the configuration that includes the Dirac sea—both fermions are on the right site denoted by |0(j_q, m_q, j'_q, n_q)⟩ a state which satisfies that. The fermion configuration determines the eigenvalues of the charge operators

$$\begin{equation}{\hat{Q}}_{a}\left(0\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle ={\hat{Q}}_{a}\left(1\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle =0\end{equation} \tag{ 113 }$$

and the Gauss's law implies that

$$\begin{equation}{\hat{L}}_{a}\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle ={\hat{G}}_{a}\left(0\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle ,\end{equation} \tag{ 114 }$$

$$\begin{equation}-{\hat{R}}_{a}\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle ={\hat{G}}_{a}\left(1\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle .\end{equation} \tag{ 115 }$$

Due to ${\hat{\mathbf{L}}}^{2}\left(0\right)={\hat{\mathbf{R}}}^{2}\left(0\right)$ we conclude that

$$\begin{equation}{\hat{\mathbf{G}}}^{2}\left(0\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle ={\hat{\mathbf{G}}}^{2}\left(1\right)\vert 0\left({j}_{q},{m}_{q},{j}_{q}^{\prime },{n}_{q}\right)\rangle \end{equation} \tag{ 116 }$$

and since the eigenvalues of all ${\hat{\mathbf{G}}}^{2}$ are fixed within the sector, a similar condition applies to all the states of the sector and we conclude that j_q = j'_q in the one link case. The Gauss's law also dictates exactly the state of the gauge field for $\left\vert 0\left({j}_{q},{m}_{q},{n}_{q}\right)\right\rangle$, and we conclude that there is only one such state per subsector

$$\begin{equation}\vert D\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle =\frac{1}{2}{{\epsilon}}_{mn}{\hat{\psi }}_{m}^{{\dagger}}\left(1\right){\hat{\psi }}_{n}^{{\dagger}}\left(1\right){\left\vert 0\right\rangle }_{\text{F}}\otimes \vert {j}_{q},{m}_{q},-{n}_{q}\rangle \end{equation} \tag{ 117 }$$

—a state with Dirac sea and an electric field originating from the static charges on top of it.

Employing again the Gauss's law we can define a state for the case where both fermions are on the right site

$$\begin{equation}\vert \bar{D}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle =\frac{1}{2}{{\epsilon}}_{mn}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{\psi }}_{n}^{{\dagger}}\left(0\right){\left\vert 0\right\rangle }_{F}\otimes \vert {j}_{q},{m}_{q},-{n}_{q}\rangle \end{equation} \tag{ 118 }$$

—representing a baryon-antibaryon pair, again, but this time with the electric field originating from the static charges.

Finally, we have to consider the action of the mesonic operator on any of these two states, to obtain the other basis elements of the subsector. Acting on $\vert 0\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle$, we obtain

$$\begin{equation}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{U}}_{mn}\left(0\right){\hat{\psi }}_{n}\left(1\right)\left\vert 0\left({j}_{q}{m}_{q}{n}_{q}\right)\right\rangle ={\hat{\psi }}_{m}^{{\dagger}}\left(0\right){{\epsilon}}_{nk}{\hat{\psi }}_{k}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes {\hat{U}}_{mn}\vert {j}_{q},{m}_{q},-{n}_{q}\rangle .\end{equation} \tag{ 119 }$$

This creates a superposition of two states with two different gauge field representations for ${j}_{q}{\pm}\frac{1}{2}$ resulting in

$$\begin{equation}{\hat{U}}_{mn}^{1/2}\left(0\right)\vert {j}_{q},{m}_{q},-{n}_{q}\rangle ={C}_{mn}^{{j}_{q}+\frac{1}{2}}\vert {j}_{q}+\frac{1}{2},{m}_{q}+m,-{n}_{q}+n\rangle +{C}_{mn}^{{j}_{q}-\frac{1}{2}}\vert {j}_{q}-\frac{1}{2},{m}_{q}+m,-{n}_{q}+n\rangle \end{equation} \tag{ 120 }$$

where

$$\begin{equation}{C}_{mn}^{{j}_{q}+\frac{1}{2}}=\sqrt{\frac{2{j}_{q}+1}{2{j}_{q}+2}}\langle \enspace {j}_{q}{m}_{q};\frac{1}{2}m\vert {j}_{q}+\frac{1}{2},{m}_{q}+m\rangle \langle \enspace {j}_{q}+\frac{1}{2},-{n}_{q}+n\vert {j}_{q},-{n}_{q};\frac{1}{2}n\rangle ,\end{equation} \tag{ 121a }$$

$$\begin{equation}{C}_{mn}^{{j}_{q}-\frac{1}{2}}=\sqrt{\frac{2{j}_{q}+1}{2{j}_{q}}}\langle \enspace {j}_{q}{m}_{q};\frac{1}{2}m\vert {j}_{q}-\frac{1}{2},{m}_{q}+m\rangle \langle \enspace {j}_{q}-\frac{1}{2},-{n}_{q}+n\vert {j}_{q},-{n}_{q};\frac{1}{2}n\rangle .\end{equation} \tag{ 121b }$$

With the proper normalizations and Clebsch–Gordan calculus we obtain that

$$\begin{equation}{\hat{\psi }}_{m}^{{\dagger}}\left(0\right){\hat{U}}_{mn}\left(0\right){\hat{\psi }}_{n}\left(1\right)\left\vert 0\left({j}_{q}{m}_{q}{n}_{q}\right)\right\rangle =\sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}\vert p{\bar{p}}^{\left(+\right)}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle +\sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}\vert p{\bar{p}}^{\left(-\right)}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle ,\end{equation} \tag{ 122 }$$

where the two remaining basis states are

$$\begin{equation}\vert p{\bar{p}}^{\left({\pm}\right)}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle ={\hat{\psi }}_{m}^{{\dagger}}\left(0\right){C}_{mn}^{{j}_{q}+\frac{1}{2}}{{\epsilon}}_{nk}{\hat{\psi }}_{k}^{{\dagger}}\left(1\right)\vert {\mathrm{v}\mathrm{a}\mathrm{c}\rangle }_{\text{F}}\otimes \vert {j}_{q}{\pm}\frac{1}{2},{m}_{q}+m,-{n}_{q}+n\rangle .\end{equation} \tag{ 123 }$$

The restriction of the Hamiltonian to any other subsector of the j_q > 0 sector, leads to

$$\begin{equation}\begin{aligned}\hfill & H{\vert }_{{j}_{q}{m}_{q}{n}_{q}}=\hfill \\ \hfill & \left(\begin{pmatrix}\hfill -2M+\frac{{g}^{2}}{2}{j}_{q}\left({j}_{q}+1\right)\hfill & \hfill \sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill \sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill 0\hfill \\ \hfill \sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill \frac{{g}^{2}}{2}\left({j}_{q}+\frac{1}{2}\right)\left({j}_{q}+\frac{3}{2}\right)\hfill & \hfill 0\hfill & \hfill \sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}{\epsilon}\hfill \\ \hfill \sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill 0\hfill & \hfill \frac{{g}^{2}}{2}\left({j}_{q}-\frac{1}{2}\right)\left({j}_{q}+\frac{1}{2}\right)\hfill & \hfill \sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}{\epsilon}\hfill \\ \hfill 0\hfill & \hfill \sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill \sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}{\epsilon}\hfill & \hfill 2M+\frac{{g}^{2}}{2}{j}_{q}\left({j}_{q}+1\right)\hfill \end{pmatrix}\right)\hfill \end{aligned},\end{equation} \tag{ 124 }$$

where the basis states are ordered as $\vert D\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle$, $\vert p{\bar{p}}^{\left(+\right)}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle$, $\vert p{\bar{p}}^{\left(-\right)}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle$, $\vert \bar{D}\left({j}_{q}{m}_{q}{n}_{q}\right)\rangle$. While this matrix representation holds for j_q > 0, one observes a decoupling of the third row and third column for j_q = 0, and the the problem becomes effectively three dimensional i.e. H|₀. Altogether, the Hamiltonian of the single link case is given by the direct sum of blocks:

$$\begin{equation}H=\underset{{j}_{q},{m}_{q},{n}_{q}}{\bigoplus}H{\vert }_{{j}_{q}{m}_{q}{n}_{q}},\end{equation} \tag{ 125 }$$

where $H{\vert }_{{j}_{q}{m}_{q}{n}_{q}}=H{\vert }_{{j}_{q}{m}_{q}^{\prime }{n}_{q}^{\prime }}$ is the same for all the subsectors belonging to the same sector j_q and the spectrum and dynamics of a single block is illustrated in figure 5. The constituting 4 × 4 matrix-Hamiltonians $H{\vert }_{{j}_{q}{m}_{q}{n}_{q}}$ given by equation (124) describe systems of two interacting qubits. Ignoring a common energy offset, we can explicitly write out the Hamiltonian (124) in terms of two interacting qubits as

$$\begin{align}\hfill {\hat{H}}_{j}& =\left(2M-\frac{{g}^{2}}{4}\left(2j+1\right)\right){\hat{s}}_{z,1}+\left(2M+\frac{{g}^{2}}{4}\left(2j+1\right)\right){\hat{s}}_{z,2}+\sqrt{\frac{2{j}_{q}+2}{2{j}_{q}+1}}{\epsilon}{\hat{s}}_{x,1},\hfill \\ \hfill & \quad +\sqrt{\frac{2{j}_{q}}{2{j}_{q}+1}}{\epsilon}{\hat{s}}_{x,2}-\left(\frac{{g}^{2}}{2}j\left(j+1\right)+\frac{{g}^{2}}{4}\right){\hat{s}}_{z,1}{\hat{s}}_{z,2},\hfill \end{align} \tag{ 126 }$$

where we directly see that J < 0 for all j, hence favoring a ferromagnetic configuration. Such a system of two interacting qubits can be implemented in quantum computing and simulation architectures like Rydberg systems, trapped ions or superconducting qubits with local control of detuning and coupling. To be explicit we briefly discuss the experimental realization of the Hamiltonian equation (126) with two Rydberg atoms trapped separately in an optical tweezer. Each Rydberg atom realizes a two-level system which corresponds to a spin-1/2 system and the detuning of the dressing laser from resonance allows to tune the local ${\hat{s}}_{z,i}$ terms. The locally adjustable light-dipole Rabi coupling is described by the spin-flip operator ${\hat{s}}_{x,i}$. Moreover, the dipole–dipole interaction of two Rydberg atoms leads to an interaction of the form $\enspace {\hat{s}}_{z,1}{\hat{s}}_{z,2}$ which can be tuned by the distance of the Rydberg atoms [85]. For comparison, the abelian U(1) link provides the 2 × 2 q-dependent matrix Hamiltonian (41) describing a single qubit or equivalently a single spin. Thus the non-abelian link model yields an additional complexity to quantum simulations.

In a similar fashion as for the U(1) case in section 3.1, we can observe that the energies of |D, j⟩ and $\vert p{\bar{p}}^{\left(-\right)},j\rangle$ invert for sufficiently strong static charges:

$$\begin{equation}{j}_{\text{s}}=\frac{8M}{3{g}^{2}}-\frac{1}{2}.\end{equation} \tag{ 127 }$$

This equation is the non-abelian analog of equation (42), at which the ground state changes. We visualized this behavior in figure 5(C), where the two states become energetically degenerate. In the next subsection we extend the discussion of the single link to a one-dimensional system of non-abelian links (figure 6).

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We introduce once again a one dimensional lattice with open boundaries, with an even number of sites, $\mathcal{N}=2\mathcal{M}+2$, labeled from 0 to $\mathcal{N}-1=2\mathcal{M}+1$, and $\mathcal{N}-1=2\mathcal{M}+1$ intermediate links. Each vertex may host at most two fermionic species, created by the operators ${\left\{{\hat{\psi }}_{m}^{{\dagger}}\left(x\right)\right\}}_{m=1,2}$, which form a fundamental SU(2) spinor. Each link hosts a rigid rotor Hilbert space. The Hamiltonian, which can be seen as a direct generalization of the U(1) case, see equation (43), takes the form

$$\begin{equation}\hat{H}=\frac{{g}^{2}}{2}\sum _{x}{\hat{\mathbf{J}}}^{2}\left(x\right)+M\sum _{x}{\left(-1\right)}^{x}{\hat{\psi }}_{m}^{{\dagger}}\left(x\right){\hat{\psi }}_{m}\left(x\right)+{\epsilon}\sum _{x}\left[{\hat{\psi }}_{m}^{{\dagger}}{\hat{U}}_{mn}\left(x\right){\hat{\psi }}_{n}\left(x+1\right)+\text{h.c.}\right],\end{equation} \tag{ 128 }$$

where we also sum over gauge-indices. In the second term the summation over x covers all the lattice sites (all the vertices): $x=0,\dots ,\mathcal{N}-1$. On the other hand, the summation is over all the links ($x=0,\dots ,\mathcal{N}-2$) in the first and the third terms. In the single link case ($\mathcal{N}=2$) equation (128) reduces to the Hamiltonian given by equation (96) subject to replacement ${\hat{\mathbf{J}}}^{2}\left(0\right)\to {\hat{\mathbf{J}}}^{2}$ and ${\hat{U}}_{mn}\left(0\right)\to {\hat{U}}_{mn}$.

The first term in the Hamiltonian (128) corresponds to the electric energy. The second is the (staggered) mass term as before: for SU(2), staggering is simpler and is manifested only in the mass term, while the charges are defined exactly the same at each vertex with operators $\hat{Q}\left(x\right)={\psi }_{m}^{{\dagger}}{\left({T}_{a}\right)}_{mn}{\psi }_{n}\left(x\right)$ as defined above. The last term is the interaction between the matter and the gauge field. As in the previous section we suppress representation index j when referring to the fundamental representation i.e. for SU(2) one has ${\hat{\psi }}_{m}^{{\dagger}}\equiv {\hat{\psi }}_{m}^{1/2{\dagger}}$, ${\hat{U}}_{mn}\equiv {\hat{U}}_{mn}^{1/2}$ and ${D}_{mn}\equiv {D}_{mn}^{1/2}$.

A gauge transformation with respect to a set of locally chosen group element $g\left(x\right)\in \text{SU}\left(2\right)$, parametrized by Euler angles $\left\{\alpha \left(x\right),\beta \left(x\right),\gamma \left(x\right)\right\}$ acts on the various degrees of freedom as follows:

$$\begin{equation}{\hat{\psi }}_{m}\left(x\right)\to {D}_{m{m}^{\prime }}\left(g\left(x\right)\right){\hat{\psi }}_{{m}^{\prime }}\left(x\right),\end{equation} \tag{ 129a }$$

$$\begin{equation}{\hat{U}}_{mn}\left(x\right)\to {D}_{m{m}^{\prime }}\left(g\left(x\right)\right){\hat{U}}_{{m}^{\prime }{n}^{\prime }}\left(x\right),\end{equation} \tag{ 129b }$$

$$\begin{equation}{\hat{U}}_{mn}\left(x-1\right)\to {\hat{U}}_{{m}^{\prime }{n}^{\prime }}\left(x\right){D}_{{n}^{\prime }n}^{{\dagger}}\left(g\left(x\right)\right).\end{equation} \tag{ 129c }$$

Due to the unitarity of the Wigner matrices this local transformation leaves the Hamiltonian $\hat{H}$ invariant. The gauge transformation is generated by the operators

$$\begin{equation}{\hat{G}}_{a}\left(x\right)={\hat{L}}_{a}\left(x\right)-{\hat{R}}_{a}\left(x-1\right)-{\hat{Q}}_{a}\left(x\right)\end{equation} \tag{ 130 }$$

for $0{< }x{< }\mathcal{N}-1$, and

$$\begin{equation}{\hat{G}}_{a}\left(0\right)={\hat{L}}_{a}\left(0\right)-{\hat{Q}}_{a}\left(0\right),\end{equation} \tag{ 131 }$$

$$\begin{equation}{\hat{G}}_{a}\left(\mathcal{N}-1\right)=-{\hat{R}}_{a}\left(\mathcal{N}-2\right)-{\hat{Q}}_{a}\left(\mathcal{N}-1\right)\end{equation} \tag{ 132 }$$

- at each site x we have a set of non-commuting generators—three in our SU(2) case, satisfying the group's left algebra commute with the Hamiltonian.

$$\begin{equation}\left[{\hat{G}}_{a}\left(x\right),\hat{H}\right]=0\quad \forall \enspace x,a.\end{equation} \tag{ 133 }$$

After analyzing the Hilbert space of the single link case we determine an upper bound on the dimension of the SU(2) one dimensional open system for a given static charge subsector $\left\{{j}_{q}\left(x\right),{n}_{q}\left(x\right)\right\}$. The system has a $\mathcal{N}=2\mathcal{M}+2$ vertices labeled from 0 to $2\mathcal{M}+1$, and the central link is located between the vertices $\mathcal{M}$ and $\mathcal{M}+1$.

The Gauss's law on the first vertex implies that the possible representations are given by ${j}_{q}\left(0\right),{j}_{q}\left(0\right){\pm}\frac{1}{2}$-adding the representation of the static charge on the first site with the possible representations of the dynamical matter there-$0,\frac{1}{2}$. At the next site, we have a static charge of ${j}_{q}\left(1\right)$, implying that the next link can host all the representations from $\vert {j}_{q}\left(1\right)-{j}_{q}\left(0\right)\vert -\frac{1}{2}$ to ${j}_{q}\left(1\right)+{j}_{q}\left(0\right)+\frac{1}{2}$ with half integer increments. We can move on and also perform a similar procedure from the end to the right, until we obtain that the middle link, $\mathcal{M}$, can host the maximal number of representations.

In the following we focus on states without static charges i.e. j_q(x) = 0 everywhere. In the case of no charges the Gauss's law at vertex 0 implies that the first link may host the representations 0 and $\frac{1}{2}$. The same argument can be applied starting from the Gauss's law at vertex $\mathcal{N}$. Using this recursive argument the middle link $\mathcal{M}$ can host the maximal number of representations. Each link n, there will be a finite number of allowed representations, ranging from 0 to ${J}_{\text{max}}\left(n\right)$, with increments of $\frac{1}{2}$. Using both directions, we obtain that

$$\begin{equation}{J}_{\text{max}}\left(n\right)={J}_{\text{max}}\left(2\mathcal{M}-n\right)\end{equation} \tag{ 134 }$$

and for $0{\leqslant}n{\leqslant}\mathcal{M}$,

$$\begin{equation}{J}_{\text{max}}\left(n\right)=\frac{n+1}{2}.\end{equation} \tag{ 135 }$$

Thus, we obtain for $0{\leqslant}n{\leqslant}\mathcal{M}$ that

$$\begin{equation}\mathrm{dim}\left(n\right)=\underset{j=0}{\sum ^{{J}_{\text{max}}\left(n\right)}}{\left(2j+1\right)}^{2}=\frac{1}{6}\left(n+2\right)\left(n+3\right)\left(2n+5\right),\end{equation} \tag{ 136 }$$

where the summation is over integers and half-integers. The maximal Hilbert space dimension of all the links is

$$\begin{equation}\begin{aligned}\hfill {\mathcal{D}}_{\text{Gauge}}\left(\mathcal{M}\right)& \equiv {\left(\prod _{n=0}^{\mathcal{M}-1}\left(\frac{1}{6}\left(n+2\right)\left(n+3\right)\left(2n+5\right)\right)\right)}^{2}\left(\frac{1}{6}\left(\mathcal{M}+2\right)\left(\mathcal{M}+3\right)\left(2\mathcal{M}+5\right)\right)\hfill \\ \hfill & =\frac{\left(\mathcal{M}+2\right)\left(2\mathcal{M}+5\right)\left(\mathcal{M}+2\right)!\left(\mathcal{M}+3\right)!\left(2\mathcal{M}+3\right){!}^{2}}{{2}^{4\mathcal{M}+5}{3}^{2\mathcal{M}+3}},\hfill \end{aligned}\end{equation} \tag{ 137 }$$

which is finite for a finite system size. The fermionic Fock space dimension reduces in the half filling case to $\left(\genfrac{}{}{0.0pt}{}{4\mathcal{M}+4}{2\mathcal{M}+2}\right)$-number of options to distribute $2\mathcal{M}+2$ fermions in $4\mathcal{M}+4$ positions. Therefore, we conclude that the following upper bound for the dimension of the Hilbert space of a (1 + 1) dimensional SU(2) lattice gauge theory

$$\begin{equation}\begin{aligned}\hfill \mathcal{D}\left(\mathcal{M}\right)& {< }\frac{\left(\mathcal{M}+2\right)\left(\mathcal{M}+2\right)!\left(\mathcal{M}+3\right)!{\left(2\mathcal{M}+3\right)}^{2}\left(2\mathcal{M}+5\right)\left(4\mathcal{M}+4\right)!}{{3}^{2\mathcal{M}+3}{4}^{4\mathcal{M}+5}}\hfill \\ \hfill & =\frac{\left(\mathcal{N}/2+1\right)\left(\mathcal{N}/2+1\right)!\left(\mathcal{N}/2+2\right)!{\left(\mathcal{N}+1\right)}^{2}\left(\mathcal{N}+3\right)\left(2\mathcal{N}\right)!}{{3}^{\mathcal{N}+1}{4}^{2\mathcal{N}+1}}.\hfill \end{aligned}\end{equation} \tag{ 138 }$$

This result is not independent of the charge sector, and is valid only for the sector without static charges. As in the U(1) case, it is an inequality and not an equation, since gauge invariance has not been fully exploited yet, and doing it would introduce nonlocality. Inserting $\mathcal{N}=2$ or $\mathcal{M}=0$ into the above formula for the single link case, we obtain 5, while we know that the dimension of the j_q = 0 sector there is 3, from taking full gauge invariance into account. An analysis taking into account full gauge invariance is discussed in [86] which is based on ideas from [87].