Spectra of Eigenstates in Fermionic Tensor Quantum Mechanics

We study the $O(N_1)\times O(N_2)\times O(N_3)$ symmetric quantum mechanics of 3-index Majorana fermions. When the ranks $N_i$ are all equal, this model has a large $N$ limit which is dominated by the melonic Feynman diagrams. We derive an integral formula which computes the number of $SO(N_1)\times SO(N_2)\times SO(N_3)$ invariant states for any set of $N_i$. For equal ranks the number of singlets is non-vanishing only when $N$ is even, and it exhibits rapid growth: it jumps from $36$ in the $O(4)^3$ model to $595354780$ in the $O(6)^3$ model. We derive bounds on the values of energy, which show that they scale at most as $N^3$ in the large $N$ limit, in agreement with expectations. We also show that the splitting between the lowest singlet and non-singlet states is of order $1/N$. For $N_3=1$ the tensor model reduces to $O(N_1)\times O(N_2)$ fermionic matrix quantum mechanics, and we find a simple expression for the Hamiltonian in terms of the quadratic Casimir operators of the symmetry group. A similar expression is derived for the complex matrix model with $SU(N_1)\times SU(N_2)\times U(1)$ symmetry. Finally, we study the $N_3=2$ case of the tensor model, which gives a more intricate complex matrix model whose symmetry is only $O(N_1)\times O(N_2)\times U(1)$. All energies are again integers in appropriate units, and we derive a concise formula for the spectrum. The fermionic matrix models we studied possess standard 't Hooft large $N$ limits where the ground state energies are of order $N^2$, while the energy gaps are of order $1$.


Introduction and Summary
In recent literature there has been considerable interest in the quantum mechanical models where the degrees of freedom are fermionic tensors of rank 3 or higher [1,2].These models have solvable large N limits dominated by the so-called melonic diagrams.Such novel large N limits were discovered and developed in [3][4][5][6][7][8][9][10][11], mostly in the context of zero-dimensional tensor models with multiple U (N ) or O(N ) symmetries (for reviews, see [12][13][14]).The quantum mechanical tensor models are richer: they have interesting spectra of energy eigenstates and may have connections with physical systems like the quantum dots.More amibitiously, large N tensor quantum mechanics may provide a dual description of two-dimensional black holes [15][16][17][18], in the sense of the gauge/gravity duality [19][20][21].The original motivation [1] for introducing the tensor quantum mechanics is that they have a large N limit similar to the one in the Sachdev-Ye-Kitaev (SYK) model [22][23][24][25], but without the necessity of the disorder.Indeed, as shown explicitly in [2], the 2-and 4-point functions in the large N tensor models are governed by the same Schwinger-Dyson equations as were derived earlier for the SYK-like models [25][26][27][28][29].
At the same time, there are significant differences between the tensor and SYK-like models.An early hint was the different scaling of the corrections to the large N limit [1] (see also the further work in [30][31][32][33]); more recently, additional evidence for the differences is emerging in the operator spectra and Hagedorn transition [34][35][36].The formal structure of the two types of models is indeed quite different: the SYK-like models containing a large number of fermions, N SYK , have no continuous symmetries (although an O(N SYK ) symmetry appears in the replica formalism), while in the tensor models one typically encounters multiple symmetry groups.For example, in the Gurau-Witten (GW) model [1] containing 4 Majorana rank-3 tensors, the symmetry is O(N ) 6 ; there is evidence [30,34] that this model is the tensor counterpart of a 4-flavor generalization of the SYK model introduced in [29].
A simpler tensor quantum mechanics with a single rank-3 Majorana tensor has O(N ) 3 symmetry [2] and is the tensor counterpart of the basic SYK model with real fermions.The quantum mechanics of complex rank-3 fermionic tensor, which has SU (N ) 2 × O(N ) × U (1) symmetry [2], is the tensor counterpart of the variant of SYK model where real fermions are replaced by complex ones [37].
The absence of disorder and the presence of the continuous symmetry groups in the tensor models endows them with a number of theoretical advantages, but also makes them quite difficult to study.In the tensor models any invariant operator should be meaningful and be assigned a definite scaling dimension in the large N limit.While the simplest scaling dimensions coincide with those in the corresponding SYK-like models, the operator spectrum in tensor models is much richer: the number of 2k-particle operators grows as 2 k k! [34][35][36].
Beyond the operator spectrum, it is interesting to investigate the spectrum of eigenstates of the Hamiltonian.While this spectrum is discrete and bounded for finite N , the low-lying states become dense for large N leading to the (nearly) conformal behavior where it makes sense to calculate the operator scaling dimensions.In the SYK model, the number of states is 2 N SYK /2 , and numerical calculations of spectra have been carried out for rather large values of N SYK [38,39].They reveal a smooth distribution of energy eigenvalues, which is almost symmetric under E → −E; it exhibits little sensitivity to the randomly chosen coupling constants J ijkl .Such numerical studies of the SYK model have revealed various interesting physical phenomena, including the quantum chaos.
The corresponding studies of spectra in the GW model [1] and the O(N ) 3 model [2] have been carried out in [40][41][42][43][44][45][46], but in these cases the numerical limitations have been more severe -the number of states grows as 2 N 3 /2 in the O(N ) 3 model and as 2 2N 3 in the GW model.This is why only the N = 2 GW model and N = 2, 3 O(N ) 3 models have been studied explicitly so far. 1 Furthermore, in the tensor models the states need to be decomposed into various representations of the symmetry groups.As a result, the details of the energy spectrum in the O(N ) 3 tensor model are quite different from those in the corresponding SYK model with N SYK = N 3 fermion species.
The goal of this paper is to improve our understanding of energy spectra in the tensor models.We will mostly focus on the simplest tensor model with O(N ) 3 symmetry [2] and its generalization to O(N 1 ) × O(N 2 ) × O(N 3 ), where the Majorana tensor degrees of freedom are ψ abc with a = 1, . . ., N 1 ; b = 1, . . ., N 2 ; c = 1, . . ., N 3 , and anti-commutation relations {ψ abc , ψ a b c } = δ aa δ bb δ cc . ( The Hamiltonian is taken to be of the "tetrahedral" form [2,10] H = g 4 ψ abc ψ ab c ψ a bc ψ a b c − g 16 and we have added a shift to make the spectrum traceless.In section 2 we discuss some essential features of this model, including its discrete symmetries.In section 3 we will derive lower bounds on the energy in each representation of SO(N 1 ) × SO(N 2 ) × SO(N 3 ).We will show that, in the melonic large N limit where gN 3/2 = J is kept constant, the most stringent bounds (3.25) scale as JN 3 , in agreement with expectations for a system with N 3 degrees of freedom.On the other hand, the splitting between lowest states in different representations is found to be of order J/N .Another derivation of this fact, based on effective action considerations, is presented in section 4. While this gap vanishes in the large N limit, we expect the splitting between states in the same representation to vanish much faster, i.e. as c −N 3 , where c is a positive constant.Such small singlet sector gaps are needed to account for the large low-temperature entropy, which is given by the sum over melonic graphs and, therefore, has to be of order N 3 .
If the global symmetry of the quantum mechanical model is gauged, this simply truncates the spectrum to the SO(N 1 ) × SO(N 2 ) × SO(N 3 ) invariant states.In section 5 we derive integral formulae for the number of singlets as functions of the ranks N i .They lead to the conclusion that the singlets are present only when all N i are even.The absence of singlets when some of N i are odd can often be related to anomalies, which we discuss in section 5.2.
For the O(N ) 3 model, the number of singlet states is shown in Table 1; it exhibits rapid growth from 2 for N = 2, to 36 for N = 4, to 595354780 for N = 6.Thus, even though the O(4) 3 model is out of reach of complete numerical diagonalization because it has 64 Majorana fermions, in contrast to the SYK model with N SYK = 64, it is far from the nearly conformal large N limit.Indeed, since the spectrum is symmetric under E → −E [34], the number of distinct singlet eigenvalues with E < 0 cannot exceed 18.Therefore, there are significant gaps in the singlet spectrum of the O(4) 3 model.On the other hand, the presence of the vast number of singlet states for the O(6) 3 model suggests that the low-lying singlet spectrum should be dense for N = 6 and higher.For large N the number of singlets grows as exp (N 3 log 2/2 − 3N 2 log N/2).Since all of these states must fit in an energy interval of order N 3 , it is plausible that the gaps between low-lying singlet states vanish as c −N 3 . The 2) may be viewed as N 3 coupled Majorana [47,48].As discussed in section 6.1, for N 3 = 1 we find a one-matrix model with O(N 1 ) × O(N 2 ) symmetry, which is exactly solvable because the Hamiltonian may be written in terms of a quadratic Casimir.When we set N 3 = 2 we find a complex N 1 × N 2 matrix model with O(N 1 )×O(N 2 )×U (1) symmetry.It may be studied numerically for values of N 1 and N 2 as large as 4 and reveals a spectrum which is integer in units of g/4.In section 6.3 we explain why this fermionic matrix model is again exactly solvable and derive a concise expression (6.23) for its spectrum.When both N 1 and N 2 are even, so that the spectrum contains singlet states, we show that the ground state is a singlet.In section 6.2 we apply similar methods to another complex matrix model, which was introduced in [49] and has symmetry [2].We show that the Hamiltonian of this model may be expressed in terms of the symmetry charges.The solvable matrix models presented in section 6 have standard 't Hooft limits when N 1 = N 2 = N is sent to infinity while λ = gN is held fixed.Then the low-lying states have energies ∼ λN2 , while the splittings are of order λ.So, in contrast to the melonic large N limit, the energy levels don't become dense in the 't Hooft limit of the matrix models.Nevertheless, these fermionic matrix models are nice examples of exactly solvable 't Hooft limits.
2 The rank-3 tensor model and its symmetries The O(N 1 ) × O(N 2 ) × O(N 3 ) tensor model is specified by the action where H is given in (1.2).Sometimes it will be convenient to use capital letters A, B, . . . to denote the multi-index, i.e.A = (a, b, c).It is easy to see that the Hamiltonian (1.2) has a traceless spectrum: where d i is the degeneracy of eigenvalue E i .
We can make some general restrictions on the possible values of the energies.Operators ψ obeying the anti-commutation relation (1.1) may be represented as the Majorana γ-matrices in N 1 N 2 N 3 -dimensional Euclidean space.They have entries which, in our conventions, are integers divided by √ 2. As a result, the Hamiltonian is an integer matrix times g/16.
It is a well-known mathematical fact that such matrices cannot have rational eigenvalues.
Therefore, in units of g/16, the energy eigenvalues have to be either integer or irrational numbers.The explicit results we will find are in agreement with this.
The discrete symmetries of the theory depend on whether some of the ranks are equal.
In a O(N 1 ) × O(N ) 2 theory, N 1 = N , we may study interchange of the two O(N ) groups, which acts as ψ abc → ψ acb .The invariant operators can be divided into even or odd under the interchange.The Hamiltonian (1.2) is odd [34], which implies that the energy spectrum is symmetric under E → −E.
Let us construct the operator which implements the interchange ψ abc → ψ acb : This operator has the following properties Due to the last relation one can check where we have renamed the repeated indices in the second line and used the anti-commutation relations (1.1) in the third line.Let us consider any state that is an eigenvector of the P 23 , it exists because P 23 is either hermitian or antihermitian The energy of such state is equal to zero.Indeed, Let us now discuss the case when all three ranks are equal and we have O(N ) 3 symmetry.
Then the invariant operators form irreducible representations of the group S 3 which interchanges the 3 O(N ) groups.The Hamiltonian is in the sign representation of degree 1: it is invariant under the even permutations and changes sign under the odd ones.Therefore, the symmetry of the Hamiltonian is the alternating group A 3 , which is isomorphic to Z 3 .
The SO(N i ) symmetry charges are In addition, each O(N i ) group contains Z 2 parity symmetries which are axis reflections.
Inside O(N 1 ) there are parity symmetries P a : for a given a , P a sends ψ a bc → −ψ a bc for all b, c and leaves all ψ abc , a = a invariant.It is not hard to see that the corresponding charges are One can indeed check that The anti-unitary time reversal symmetry T is a general feature of systems of Majorana fermions; it commutes with them and, therefore, with the Hamiltonian (1.2): The action of T on the eigenstates depends on the total number of the Majorana fermions N 1 N 2 N 3 and is well-known in the theory of topological insulators and superconductors [50].
If the total number of fermions is divisible by 8, the operator T acts trivially, so the ground state may be non-degenerate.Otherwise T acts non-trivially and one finds that the ground state must be degenerate.
3 Energy bounds for the Since the Hilbert space of the model is finite dimensional, it is interesting to put an upper bound on the absolute value of the energy eigenvalues in each representation of the symmetry group.In this section we address this question in two different ways.We first derive a basic linear relation between the Hamiltonian, a quadratic Casimir operator, and a square of a Hermitian operator which is positive definite.This gives bounds which are useful for the representations where the quadratic Casimir of one of the orthogonal groups is near its maximum allowed value.We also find that the bounds are exactly saturated for N 3 = 2, but are not stringent when equal ranks become large.Then in section 3.2 we derive more refined bounds which are more stringent in the large N limit and give the expected scaling of the ground state energy.Furthermore, we derive a finite multiplicative factor which corrects the refined bound and allows us to deduce the ground state energy in the large N limit.

Basic bounds
To derive an energy bound we introduce the hermitian tensor If we think of bc as a combined index which takes N 2 N 3 values, then A bc,b c are the generators and the quadratic Casimir of the O(N 1 ) symmetry, Therefore, for the states which appear in the model, we find the upper bound: = 0 so that the state is invariant under SO(N 2 N 3 ).
The Hamiltonian may be written as Now we note the inequality Combining this with (3.4) we get In an analogous fashion we can also derive the bounds in terms of C 2 : and similarly in terms of C O(N 3 ) 2 .
An interesting special case, which we will consider in section 6, is N 3 = 2 where we find a complex N 1 × N 2 matrix model.For the singlet states where = 0 the most stringent bound we get from (3.8) and (3.9) is In section 6 we will show that these bounds are saturated by the exact solution for even For N 1 = N 2 = N we have a N × N matrix quantum mechanics which possesses a 't Hooft large N limit where gN = λ is held fixed.In this limit, the ground state energy is , which has the expected scaling with N for a matrix model.More generally, if at least one of the ranks is even (we will call it N 3 ), we may introduce the operators [44] where where The Hamiltonian is invariant under the charge conjugation symmetry which interchanges c abk with cabk .
For any even N 3 , using the basis (3.11) we define the oscillator vacuum state by the condition c abk |0 = 0. Since this condition is invariant under O(N 1 N 2 ), so is |0 .Furthermore, all the states that are created by operators that are O(N 1 N 2 ) invariant are also O(N 1 N 2 ) invariant and have energy g ).The number of such states is estimated to be the dimension of the maximal representation for the (see apendix A for details).The relation (3.4) also simplifies the search for the singlets.
For example, we can first forget about the group nature of the third index in the approach of [44] and impose the vanishing of the Casimir of the third group afterwards.By studying the charges under U (1) ∈ U (N 3 /2) we find that the singlet states must have N 1 N 2 N 3 /4 creation operators acting on |0 .
Specifying the bound (3.8) to the equal ranks When the bound (3.5) is saturated, the corresponding state must have zero energy.This shows that all the states invariant under For the singlet states (3.13) gives For N = 2, exact diagonalization gives that the ground states is a singlet with energy E 0 = −2g; this saturates the bound (3.14).For N = 3, exact diagonalization gives a ground state with energy Since for the 2 of SO(3), C 1 = 3/4, the bound (3.13) is E 0 ≥ − 33 2 g.This is satisfied and is far from being saturated.
In the large N limit, J = gN 3/2 is held fixed.Thus, we obtain a bound on the lowest singlet energy E 0 , which is E 0 ≥ −cJN 7/2 , where c is a positive constant.Since we expect the ground state energy to be of order N 3 , this bound is not very informative at large N .A better bound at large N will be derived in the next section.

Refined bounds
In this section we present another approach to deriving energy bounds for the O(N 1 ) × O(N 2 ) × O(N 3 ) invariant states, which gives a more stringent bound in the large N limit than the ones in the previous section.
It is easy to see, that this density matrix is invariant under rotations ).We can calculate the mean value of the energy for this density matrix as For a fixed a, b, c we can act by the rotation matrices (that act trivially on the singlet density matrix ρ s ) and make the interchange a ↔ 1, b ↔ 1, c ↔ 1.This argument gives us that where we sum over the repeated indexes.From now on we consider the density matrix to be of the form (3.15).Now we can estimate the trace in the formula (3.17).With the use of Cauchy -Schwarz inequality, we have Because the density matrix ρ R is a singlet we can rotate indices back to get We can express it is the following way The square of the operator q abc can be expressed as a sum of Casimir operators due to the virtue of the anticommutation relations.That gives us the bound on the energies of states in representation R: where C R i is the value of Casimir operator in the representation R. For the singlet states this gives Since C i ≥ 0 this bound applies to all energies.Let us note that for N 3 = 2 the square root may be taken explicitly: which is identical to the earlier result (3.10).In section 6 we will show that this is saturated when N 1 , N 2 are even and the ground state is a singlet.
For the case when In the large N limit, E bound → JN 3 /16, which is the expected behavior of the ground state energy; in the melonic limit it scales as N 3 .We may expand (3.21) for large N to find The discussion of the splittings between non-singlet and singlet states in section 4 will be in agreeement with the scaling of the second term.
We can try to estimate how close the singlet ground state |vac comes to the bound (3.24) by using the exact propagator G(t) = T ψ abc (t)ψ def (0) in the large N limit.To do it let us consider the two states where we have introduced ∂ t ψ abc = i[H, ψ abc ].We can introduce the angle θ between these states where we have rotated back the indexes back by using the fact that the |vac is a singlet state.One can notice H = iψ abc ∂ t ψ abc , while vac| ∂ t ψ abc 2 |vac is just equal to the bound where E bound is the bound on the singlet ground state energy (3.24).The other way to estimate this angle θ can be done in the following way.We shift the Hamiltonian, such that the ground state has the zero energy (H − E 0 ) |vac = 0 and calculate the expectation value for the energy for the state |1 : at the same time the second moment of the energy is Where we have used the fact that ∂ t ψ abc = i[H, ψ abc ].After that we can notice that (3.27) can be rewritten as But as we know the solution for the propagator in the large N limit is a conformal propagator.
From this we deduce that the bound can not be saturated.Nevertheless we can estimate the angle cos 2 θ.Indeed, in the large N limit the propagator can be calculated numerically or approximated by a conformal one.From this we can calculate the E 1 and E 2 1 .We assume t > t = 0, a = a , b = b , c = c and insert the full basis |E n of eigenstates of the Hamiltonian in the propagator to get where The function ρ(E) is known as a structure factor.From this we can calculate One can use conformal propagator to estimate this angle, which gives cos θ ≈ 0.745, while the numerical calculation [22] gives cos θ = 0.6608.From this and the formula (3.28) we get the ground state energy in the large N limit: This answer is close to the numerical result for the ground state energy in the SYK model [39]: One can make analogous calculations for the other representations.It gives us in the large N limit the following formula for the gap to the lowest state in a representation R: 4 Sigma model and energy gaps In the large N limit the model 2.1 is dominated by melonic diagrams.This allows one to write down a closed system of Schwinger-Dyson equations for the Green function For simplicity we shall introduce the multi-index A = (a, b, c).We can look for a solution in the diagonal form G AB = G(t)δ AB and Σ AB = Σ(t)δ AB .Then we have the following set of equations: These equations exactly coincide with the Schwinger-Dyson equations of the SYK model and have a conformal solution.
It was argued in [35] that the system of equations (4.1) can be obtained from the effective action 3 : This action was recently derived from two-particle irreducible diagrams in [33].
In the strong coupling limit J → ∞ one can neglect the bare Green function.Then, as first noticed in [35], the global symmetry O(N ) 3 is promoted to the gauged symmetry O(N ) 3 .Indeed, if we neglect G AB 0 (ω) in (4.1) then it is easy to see that we can generate a series of solutions by doing a gauge transformation: where we introduce matrix O in O(N ) 3 which equals to γγ .The effective action (5.18) is also invariant under these transformations if one omits the term ∂ t in the Pfaffian.For finite J, the action ceases to be invariant.If we plug the gauge transformation (4.4) into the effective action (5.18), the potential does not change, while we will get a kinetic term for matrices O i of order 1/J.Indeed, for the conformal solution we have Σ AB = − (1/G) AB and we can rewrite the kinetic part of the action as The second term log Pf (Σ AB ) is invariant under gauge transformations.Then expanding 3 For clarity, we have omitted the indices in the G 4 term.Explicitly, this term reads as the Pfaffian in the leading order in derivatives we get Making the gauge transformation (4.4) of the conformal solution G AB = δ AB G and plugging into (4.6)we get: Factors N 2 come from Tr(O T 1 O 1 ) = N .Now one has to regularize the limit t 2 → t 1 but this does not going to affect N 2 factors.The details are worked out in [33,51].The upshot is that G(t − t )O i (t ) becomes ∂ t O i (t)/J up to a normalization constant.This leads to the sigma model action The spectrum of such a quantum mechanical sigma model is well-known: the Hamiltonian is proportional to the quadratic Casimir and the eigenstates are simply representations of O(N ) 3 .In our case: We note that this has the same structure as the Casimir correction to the energy bound (3.25).Since for the lowest non-singlet representations C 2 ∼ N , we find the energy gap between singlets and non-singlets to be of the order ∼ J/N .

Counting singlet states
Suppose we have a free fermionic system of N Majorana fermions ψ I , I = 1, . . ., M transforming under some representation R of the gauge group G.We want to compute the number of singlet states in such a system.In order to do it, we use the following method.
The Lagrangian in the Euclidean space reads as: where A IJ is a real gauge field in the representation R. Since Majorana fermions anticommute with each other, A IJ must be anti-symmetric A IJ = −A JI .The partition function of the gauged system at the temperature β is where we have put the fermionic system on a circle with the circumference β with antiperiodic boundary conditions ψ(t) = −ψ(t + β).The normalization factor N can be easily recovered if we study the ungauged model.The integration over DA gives the volume of the gauge group and the integral over the fermion variables will yield just the dimension of the Hilbert space because the Hamiltonian of the ungauged theory is equal to zero H ungauge = 0.In this case the total number of states is simply 2 M/2 : (5.3) is Gaussian; therefore, the problem boils down to computing the Pfaffian: As discussed above, we fix A to be a constant matrix in the Cartan subalgebra.The Faddeev-Popov determinant then yields the normalized Haar measure dλ N G on the gauge group G [52]: Also, since A is anti-symmetric, the eigenvalues of A are pairs of pure-imaginary numbers iλ a , −iλ a , a = 1, . . ., N/2 .The ratio of the Pfaffians is cos(λ a /2) . (5.8) There are different ways to derive this formula.One is to compute the ratio of determinants: After that we note that if we go to the Fourier space, both ∂ t and A are real anti-symmetric matrices, so the ratio of Pfaffians must be a real smooth function of λ a .Therefore, taking the square root of eq. ( 5.9) we get eq.(5.8).Alternatively, we can calculate the Pfaffian of ∂ t + A in Fourier space.The result is the following formula: where we have got the normalization by studying the ungauged theory (5.3).
Let us apply this approach to the case when Majorana fermions live in the fundamental representation of several orthogonal groups.It is important to distinguish between SO(2n) and SO(2n+1).The Cartan subalgebra in the SO(2n) algebra consists of the block diagonal matrices with 2 × 2 blocks 0 where x i is a rotation phase ranging from 0 to 2π.Geometrically it means that for a fixed SO(2n) transformation, there is a basis in which this transformation looks like a set of rotations in independent two-planes.In the SO(2n + 1) case the last column/row is zero.
It corresponds to a fixed one-dimensional subspace.Non-normalized Haar measure in these two cases reads as: ) (5.13) Now we discuss the case where the gauge group is the product of three orthogonal groups SO(N 1 ) × SO(N 2 ) × SO(N 3 ), so that the gauge field decomposes as (5.14) For even N i in eq.(5.10) eigenvalues λ a are given by x i , y j , z k are rotation phases for SO(N 1 ), SO(N 2 ) and SO(N 3 ) respectively.In the case when one of the N i is odd we have to add a zero eigenvalue to this list.With the use of the equation For the O(N ) 3 model the number of singlets for various even N is given in Table 1.For odd N it is not hard to see that the integral which gives the number of singlets vanishes; this is related to the fact that each group exhibits an individual anomaly, which we discuss in the next section. 4In the next section 5.1 we will show that the number of singlets grows as exp (N 3 log 2/2 − 3N 2 log N/2) for large even N .
Using similar methods, the number of singlets can be calculated in the O(N ) 6 GW model for low values of N , and the results are presented in Table 2.The fact that there are 140 states for N = 2 is in agreement with the direct construction of singlet states in [46].
We may similarly calculate the number of singlets for the O(N 1 )×O(N 2 )×O(N 3 ) models.  Gurau-Witten model When N 2 = N 3 = 2, while N 1 is even, there are 2 singlets.For the cases where N 3 = 2, while N 1 and N 2 are even, some answers are listed in Table 3.We note that the growth of the

Number of singlets for large N
In this section we will estimate the number of singlets in the SO(N ) 3 model in the large N limit, assuming N to be odd N = 2M .For general N , the number of singlets is given by the following integral: 16 cos Where V is the volume of SO(N ).When N is large, cosine functions oscillate very rapidly, so the integral localizes near x i = y j = z k = 0. Near this point the integrand is positive, so we can exponentiate it: (5.17) Notice that we have introduced a "regulator" t which we have to send to one: t → 1.Similar integrals count operators in theories with tri-fundamental fields [36].In such cases t = e −1/T , where T is the temperature.So we are interested in the infinite temperature limit.This case has been studied in detail in [36].Here we perform a similar analysis.As usual, we will encode the saddle-point configuration of the angles x, y, z using the density function ρ(x) (obviously it is the same function for the three SO(N ) groups).Moreover this function is symmetric ρ(x) = ρ(−x).It would be convenient to work with the normalized density π −π dxρ(x) = 1.The effective action now reads as: (5.18) In the infinite temperature limit the saddle-point density is non-zero only on a small interval [−x 0 , x 0 ] where x 0 ∼ 2 N .The leading contribution is coming from the first term and it equals to 1  2 N 3 log 2. But this yields simply the dimensions of the Hilbert space, which is 2 1 2 N 3 .The subleading term is coming from the second term in (5.18).Fortunately, we will not need the exact value of x 0 because of the logarithmic behaviour: Therefore the subleading term is − 3 4 N 2 log N .So, in total we have

Anomalies
Since we are studying fermions on a compact space S 1 there is a potential global anomaly associated with π 1 (G).And indeed it is well-known that π 1 (SO(N )) = Z 2 .Corresponding "large" gauge transformation has a simple description: the gauge transformation matrix is the identity matrix, apart from one 2 × 2 block cos(2πt) − sin(2πt) sin(2πt) cos(2πt) . (5.21) It is easy to see that after such transformation one chosen rotation phase x i will be shifted by 2π: x i → x i + 2π.It does not matter which x i to pick up, since an even number of 2π-rotation blocks gives, in fact, a trivial element in π 1 (SO(N )).It has been known for some time [53] that a theory of a single Majorana fermion in the fundamental representation of SO(N ) is suffering from this Z 2 anomaly.It is instructive to see it using our machinery.
The Pfaffian in this case reads as: Under the shift x j → x j + 2π it changes sign.Therefore the theory is not invariant under large gauge transformations.In our case of O(N 1 ) × O(N 2 ) × O(N 3 ) group it means that at least two out of three N i should be even, otherwise we will have an odd number of anomalous multiplets.Since this anomaly is associated with only one group we will refer to it as "individual anomaly".It is easy to see that this anomaly is always Z 2 (in other words, it squares to one), even if we add more gauge groups.
If the gauge group is a product SO(2n 1 ) × SO(2n 2 ) there is a new anomaly mixing these two groups.For each group in the product, the large gauge transformation consists of identical 2 × 2 blocks: cos(πt) − sin(πt) sin(πt) cos(πt) . (5.23) Since there are two gauge groups, at t = 1 overall −1 will cancel.Now all phases x i and y j are shifted by π: x i → x i + π, y j → y j + π.The Pfaffian reads as: (5.24) Under the large gauge transformation the Pfaffian acquires (−1) n 1 n 2 .This anomaly means that for G = SO(2n 1 ) × SO(2n 2 ) × SO(N 3 ), N 3 can be odd only if the product N 1 N 2 is even.We will call this anomaly "mixed anomaly".This anomaly is not always Z 2 as we will see shortly.
We do not find any more anomalies: using the long exact sequence in homotopy groups one can show that the fundamental group of SO(2n 1 )×SO(2n 2 )/Z 2 5 is equal to Z 2 ×Z 2 ×Z 2 or Z 4 × Z 2 depending on n 1 and n 2 .Using the above explicit descriptions of the individual anomalies and the mixed anomaly we see that: • If n 1 and n 2 are both even, then the square of the mixed anomaly gives a trivial gauge transformation.Indeed, for each gauge group the number N i of 2π-rotation blocks (5.21) is even.Therefore, this is the case of • If only one of n i , say n 1 , is odd, then the mixed anomaly squares to the individual anomaly of SO(2n 1 ), since this group will have an odd number of 2π rotation blocks.
• Finally, when both n 1 and n 2 are odd, then the mixed anomaly squares to the sum of the individual anomalies.This is again Z 4 × Z 2 .
5 One has to divide by Z 2 because g 1 × g 2 acts on ψ in the same way as (−g 1 ) × (−g 2 ) 6 Solution of some fermionic matrix models and becomes a fermionic N 1 × N 2 matrix model.In this section we discuss the solution of these models.For the O(N 1 ) × O(N 2 ) real matrix model the Hamiltonian may be expressed in terms of the quadratic Casimir operators, which shows that all the states within the same group representation have the same energy.This also applies to the SU (N 1 )×SU (N 2 )×U (1) symmetric complex fermionic matrix model, which was considered in [49], [54] (see also [55]), and will be further discussed in section 6.2.However, the O(N 1 ) × O(N 2 ) × U (1) complex fermionic matrix model is more complicated in that there are energy splittings within the same representation of the symmetry group.Nevertheless, as we show in section 6.3 this model is solvable.

The
2) we find a real matrix model with O(N 1 ) × O(N 2 ) symmetry: Using the SO(N 1 ) and SO(N 2 ) charges the Hamiltonian may be expressed in terms of the quadratic Casimirs: This shows that, under the interchange of N 1 and N 2 , H → −H; therefore, for N 1 = N 2 the spectrum is symmetric around zero.The sum of this Casimir operators is fixed: This shows that there are no states which are singlets under both SO(N 1 ) and SO(N 2 ).The irreducible representations (r 1 , r 2 ) which appear in the spectrum must satisfy the condition (6.4).In appendix B.1 we list these representations for a few low values of N 1 and N 2 .The complete lists of the energies and degeneracies are shown in Table 4.
For O(N ) × O(N ) with even N , we find that the ground state is a singlet under O(N ) 1 and transforms in the SO(N ) 2 representation whose Young diagram is a N 2 × N 2 square.The ground state has energy E 0 = −gN 2 (N − 1)/16, while the first excited state is in the fundamental of O(N ) 1 which has quadratic Casimir N − 1.Therefore, the energy gap In the 't Hooft large N limit, g ∼ 1/N and the gap stays finite.Therefore, unlike the SYK and tensor models, the matrix model cannot exhibit quasi-conformal behavior. (

6.2
The SU (N 1 ) × SU (N 2 ) × U (1) model In [2] a class of complex tensor quantum mechanical models with SU (N 1 )×SU (N 2 )×O(N 3 )× U (1) symmetry was introduced.We will use the Hamiltonian where ψ abc with a = 1, . . ., N 1 , b = 1, . . ., N 2 and c = 1, . . ., N 3 are complex fermions with anti-commutation relations { ψabc , ψ a b c } = δ aa δ bb δ cc .The second and third terms were added to the Hamiltonian to make it traceless and invariant under the charge conjugation symmetry, which interchanges ψ abc and ψabc .This means it is invariant under where Q is the U (1) charge: If we set N 3 = 1 we obtain a complex matrix model with SU (N 1 ) × SU (N 2 ) × U (1) which is the subject of this section.Note that the index contraction in the first term is different from those in (6.20); the SU (N 1 ) × SU (N 2 ) × U (1) symmetry fixes it uniquely.
This matrix model has some features in common with the O(N 1 ) × O(N 2 ) from the previous section.In both of them the energy is completely fixed by the quadratic Casimir operators of the symmetry group factors.Also, neither model contains states invariant under the entire symmetry group.
The SU (N i ) charges with i = 1, 2 are where we used the Hermitian SU (N i ) generators Using the completeness relation (no sum over i): we find that the quadratic Casimirs of SU (N 2 ) and SU (N 2 ): Adding them, we obtain the constraint To have the singlets of SU (N 1 ) and SU (N 2 ), we need the RHS to vanish.This means that there are only two SU (N 1 ) × SU (N 2 ) singlet states: the ones with Q = ± N 1 N 2 2 .These are the oscillator vacuum |0 , which is annihilated by all ψ ab , and the state |0 = a,b ψab |0 , which is annihilated by all ψab .
The absence of singlets for other values of Q may be seen explicitly as follows.The states but there is no way to contract the indices of SU (N 1 ) and of SU (N 2 ); in contrast to the O(N ) case, the tensor δ a 1 a 2 is not available.If N 1 = N 2 = N there seems to be a state at level N obtained by contracting (6.14) with a 1 ...a N b 1 ...b N , but this state vanishes due to the Fermi statistics.
Using (6.12) we can express the Hamiltonian (6.8) in terms of the Casimirs: Therefore, all the states in the same representation of SU (N 1 ) × SU (N 2 ) × U (1) are degenerate, which makes this matrix model very simple.In table 5 we list the spectra of the the Hamiltonian (6.8) for a few different values of N 1 and N 2 .

The
where a = 1, 2, . . .N 1 and b = 1, 2 . . .N 2 .In this basis, the O(2) charge is while the SO(N 1 ) and SO(N 2 ) charges are Squaring these charges, we find the following expressions for quadratic Casimirs: Setting k = 1 in (3.12), we find that the traceless form of the Hamiltonian is This Hamiltonian exhibits the charge conjugation symmetry which acts as ψab ↔ ψ ab .This means that states with opposite eigenvalues of Q have the same energy.
There is a "Clifford vacuum" state, which satisfies There is also the conjugate vacuum |0 = ab ψab |0 which satisfies In fact, the bound obtained from (3.8) completely fixes the energy to be g  Now we take the difference between equations (6.19), and also use the difference of equations (6.12), to find the following nice expression for the Hamiltonian: where we used (6.13) to obtain the second line from the first.Due to the C terms, the spectrum is not symmetric under SU (N 1 ) × SU (N 2 ).
Using (6.23) To calculate the energies of all states, we need to first decompose the Hilbert space into SU (N 1 ) L × SU (N 2 ) R representations and then, in turn, decompose these representations into SO(N 1 ) L × SO(N 2 ) R representations.To find which SU (N 1 ) L × SU (N 2 ) R representations (L, R) we have in the Hilbert space, we need to compute the following integral over We can always put U 1 and U 2 in a diagonal form: U 1 = diag(w 1 , . . ., w N 1 ), U 2 = diag(q 1 , . . ., q N 2 ).
w i and q i are corresponding SU holonomies, i.e.
Actually, it is not neccessary to compute the above integral for various representations.It is very well-known that characters of SU (N 1 ) representations are Schur polynomials Tr L U 1 = χ L (w) which form a basis in the space of symmetric functions of N 1 variables.This space also contains the so-called power series polynomials TrU n A conversion from power series p n to χ L can be easily done on a computer.For example, .25)This suggests the following simple procedure yielding the list of all representations directly.
As a further check, in appendix B.2 we calculate the spectrum of the O model using this method.The results for the energies and their degeneracies agree with the direct diagonalization of the Hamiltonian, whose results are assembled in Table 6.We also note that, due to the charge conjugation symmetry, the energies and representations at oscillator level n are the same as at level N 1 N 2 − n.
1620059.The work of GT was supported in part by the MURI grant W911NF-14-1-0003 from ARO and by DOE grant de-sc0007870.
A The eigenvalues of the quadratic Casimir operator In this appendix we describe the value of quadratic Casimir operator for the representations of O(N ) and SU (N ) groups in terms of Young diagrams.To extract the irreducible representation corresponding to a Young diagram from a generic tensor, we first fill in the boxes with this tensor indices, then we symmetrize over the indexes in the rows and after that antisymmetrize the indexes in the columns.In the case of the orthogonal group we additionally subtract all possible traces from the tensor.
For the representation of the group O(N ) that is described by the Young diagram Y with row lengths λ i , the quadratic Casimir operator is equal to The dimension of this representation reads as: where h λ is the product of all hook lengths.For each box the hook length is defined as: (hook length) = (number of boxes to the right) + (number of boxes below) + 1 (A. 3) The following lemma will be useful for studying the matrix models.
After that it is easy to see Consider an arbitrary singlet density matrix ρ; this means a density matrix invariant under the O(N 1 ) × O(N 2 ) × O(N 3 ) rotations.For example, it can be ρ s = |s s|, where |s is an singlet state, or if we have some representation R of the O(N 1 ) × O(N 2 ) × O(N 3 ) with an orthonormal basis |e i , i = 1.. dim R we can define the projector on this subspace of the Hilbert space and N > 2 the bound (3.22) is more stringent than the earlier bound (3.14): ψ a b c (t 2 ) and self-energy Σ abc a b c and the bare Green function G abc a b c ,0

. 3 )
From now on, we will put β = 1.If we fix Lorentz gauge ∂ t A = 0 with A in the Cartan subalgebra, the Faddeev-Popov determinant gives the Haar measure, while the path integral over Majoranna fermions gives the partition function of the system with Hamiltonian H = −ψ I A IJ ψ J .Therefore the (5.2) can be rewritten asZ gauged = DA Tr exp −ψ I A IJ ψ J ,(5.4)The expression under the trace is an operator of rotations and can be interpreted as a char-acterof the group acting in the Hilbert space of fermions.By the virtue of the representation theory we know that the integral of the character over a group is equal to the number of the trivial representations, i.e. the number of the singlet states.Therefore, Z gauged equals the number of singlet states.If we insert in (5.2) a Wilson line in some representation R , it gives the character of this representation: Tr R exp A dt = #states in the representation R .(5.5)One can compute the partition function because the integral over ψ in both (5.2) and

( 5 .
10) we can write expicit form of the character of the representation and decompose it in terms of characters of the irreducible representations.For example, for the O(2) 3 model the number of singlets is given by the symmetric tensor model (1.2) we find a complex matrix model with O(N 1 ) × O(N 2 ) × U (1) symmetry.This model has some features in common with the SU (N 1 ) × SU (N 2 ) × U (1) model discussed in the previous section; they possess the same 2 N 1 N 2 dimensional Hilbert space.However, in the present model the symmetry is broken to O(N 1 ) × O(N 2 ) × U (1) by the Hamiltonian.Although the model is still exactly solvable, it is quite interesting in that the energy is not completely fixed by the quadratic Casimir operators of O(N 1 ) × O(N 2 ) × U (1).Also, as we have seen in section 5, for even N 1 and N 2 the model contains singlet states.

8 N 1 N 2 (N 2 −
N 1 ) because the states are O(N 1 N 2 ) invariant and C O(N 1 N 2 ) 2 = 0.The states with vanishing O(2) charge Q are obtained by acting on |0 with N 1 N 2 2 creation operators ψab .Then, to insure that the state is also a singlet under SO(N 1 ) × SO(N 2 ), we have to contract the indices using the invariant tensors a 1 ,...a N 1 , δ a 1 a 2 and b 1 ,...b N 2 , δ b 1 b 2 .Some states invariant under SO(N 1 ) × SO(N 2 ) × O(2) are listed in Appendix B.3.For low values of N 1 and N 2 it is possible to construct the complete spectrum via direct numerical diagonalization.If N 1 = N 2 or if one or both N i are equal to 2, the spectrum is symmetric under E → −E due to the fact that the interchange of two O(N ) groups send H → −H.For all other values of N i the spectrum is not symmetric under E → −E.The results for some low values of N 1 , N 2 are shown in table 6.For the O(4) 2 × O(2) model the spectrum is plotted in figure 1.

Figure 1 :
Figure 1: Spectrum of the O(4) 2 × O(2) model.There are four singlet states, and the stars mark their energies.

. 26 )
in power series in x.Then at each level x k we have a polynomial in p l (w) and p m (q).It can be re-expressed in terms of Schur polynomials.This gives the list of representations underSU L (N 1 ) × SU R (N 2 ) at level k, i.e. for states where there are k raising operators ψ acting on the vacuum.After finding the representations under SU (N 1 ) L × SU (N 2 ) R , we need to decompose then in terms of SO(N 1 ) L × SO(N 2 ) R representations.Recall that both SU and SO representations are classified by Young diagrams.The only difference is that for SO representations one has to subtract all the traces in each row, where indices are symmetric.It means that if we want to extract SO representations from a given SU representation λ, we need to consecutivly remove all possible pairs of boxes in each row.The resulting sequence of Young diagrams give SO representations.

C Ym 2 = n 2 m
Let us consider two groups O(2n) and O(2m) and Young diagram Y n for group O(2n) such that the length of the rows is less then m.There is a maximal Young diagram -a rectangular n × m, that we shall denote as Y n×m .We would like to consider a specific Young diagram Y m = (Y n×m /Y n ) T for a group O(2m), where T stands for transposition.Then C Yn 2 + + nm 2 − nm .(A.4)The proof goes as following.Let λ i be the length of rows of the diagram Y n , we introduceλ 0 = m, λ n+1 = 0. Then C Yn 2 = n i=1 λ i (λ i + 2(n − i)) (A.5)The value of Casimir operator of C Ym 2 can be expressed as the following.The difference λ i − λ i+1 is just equal to the number of the rows that has length n − i.

Table 1 :
they transform in the spinorial (2, 2, 2) representation.Substuting the value C i = 3/4 into the bound (3.21) for the energy gives −11.53g, which is quite close to the actual value.Number of singlet states in the O(N ) 3 model √41g ≈ −8.00391g;

Table 2 :
Number of singlet states in the O(N )

Table 3 :
Number of singlet states in the O(N 1 ) × O(N 2 ) × O(2) model number of singlets for the O(N ) 2 × O(2) model is much slower than for the O(N ) 3 model.For low values of N it is not hard to write down explicit expressions for all the singlet states in the oscillator basis; see appendix B.3.For example, for the O(4) 2 × O(2) model we find that the 4 singlet energies are ±16g and ±4g.
.22) Both of these states are invariant not only under O(N 1 ) × O(N 2 ), but under the enhanced symmetry O(N 1 N 2 ).It is interesting to note that the states |0 and |0 saturate the energy bound (3.21).Indeed, substituting N 3 = 2, C we can show that the lowest singlet saturates the energy bound (3.23), i.e. +N 2 )N 1 N 2 , i.e. it saturates the bound(3.23).This value of Casimir corresponds to the rectangular Young diagram N 1 /2 × N 2 for SU (N 1 ).Similarly, the singlet state with the highest possible energy, E = g 8 (2)is a ground state.For a singlet, Q and the quadratic Casimir operators of SO(N 1 ) and SO(N 2 ) vanish.To minimize the energy we should take a state which has the greatest possible value of C i.e. it is invariant under SO(N 1 ) × SU (N 2 ) × O(2).Substituting this into (6.23)we see that this state has E = − g 8 (N 1

Table 7 :
Energy spectrum of the O(2) × O(3) × O(2) model.Due to the charge conjugation symmetry for the last O(2) charge, the energies and representations are invariant under transformation level → 6 − level.