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I'm excited to better understand the holography within the SYK model. I enjoy a rigorous understanding, which I'm trying to achieve via the 'review' papers from Rosenhaus, Maldacena, Trunin and Sarosi.

Question: I am quite stuck and it boils down to understanding how the different definitions for the SYK Hamiltonians agree:

$$H^{(1)}_\text{SYK}=\sum_{i,j,k,l=1}^NJ_{ijkl}\psi_i\psi_j\psi_k\psi_l,\tag1$$ $$ H^{(2)}_\text{SYK}=\frac{1}{4!}\sum_{i,j,k,l=1}^NJ_{ijkl}\psi_i\psi_j\psi_k\psi_l,\tag2$$ $$ H_\text{SYK}^{(3)}=\sum_{1\leq i<j<k<l\leq N}J_{ijkl}\psi_i\psi_j\psi_k\psi_l.\tag3$$ (1) is used in Sarosi (93); (2) in Rosenhaus (2.8) & Trunin (3.1); (3) in Maldacena (2.2). Here $$\{\psi_n,\psi_m\}=\delta_{nm}.$$

I tried: It is claimed the coupling $J_{ijkl}$ is totally anti-symmetric (given the anti-commutation relation/Clifford algebra). Using this, I can show definitions (2) and (3) are equivalent except for terms in (2) that have terms like $\psi_i\psi_j\psi_k\psi_k=\psi_i\psi_j/2$ which is formally allowed the way the notation is denoted (here I used that $\psi_n^2=1/2$ via the algebra). Do I need to understand the summation in (2) to have $i\neq j\neq k\neq l$? If so, this seems to conflict with equations in Trunin (such as on p12 and p13). I don't see how (1) and (2) are the same: they don't even generate the same Euler-Lagrange equations.

Additionally, I cannot convince myself $J_{ijkl}$ is anti-symmetric. What I did was $$J_{2134}\psi_2\psi_1\psi_3\psi_4=-J_{2134}\psi_1\psi_2\psi_3\psi_4:=J_{1234}\psi_1\psi_2\psi_3\psi_4,$$ where the last step is done on physical grounds 'the number in front of $\psi_1\psi_2\psi_3\psi_4$ should be the same in all cases since the interaction strength is the same between the fields'. But this doesn't make any sense: I could equally well just add them up $$J_{1234}/N-J_{2134}/N+...:=J^\text{eff}_{1234}.$$ Doing this procedure for all fermions gets me from definition (2) to (3) and $J^\text{eff}_{ijkl}$ is anti-symmetric, but $J_{ijkl}$ is not.

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jrobins
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1 Answers1

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  1. In the operator formalism there is an implicitly written normal ordering of the 4-fermion term. The fermions anticommute under the normal order symbol.

  2. In the path integral formulation, the fermions also anticommute in the action.

  3. The symmetric group $S_4$ acts on the 4 indices of the coupling constants $J_{ijkl}$, thereby establishing a group representation. We can decompose $J_{ijkl}$ according to irreducible representations (irreps) of $S_4$. The anticommuting fermions annihilate all but the sign representation, which has Young tableau $$\begin{array}{c} [~~]\cr [~~] \cr [~~]\cr [~~]\end{array}.$$ (All other Young tableaux for irreps of $S_4$ have a row with at least 2 elements, which correspond to symmetric indices.)

  4. Equivalently, we may assume from the onset that the coupling constants $J_{ijkl}$ are totally antisymmetric.

  5. With assumption 4, the normal order symbol is superfluous.

  6. It follows that $\frac{1}{4!}H^{(1)}_\text{SYK}=H^{(2)}_\text{SYK}=H^{(3)}_\text{SYK}.$

Qmechanic
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