Action of Sachdev-Ye-Kitaev (SYK) model Consider the following hamiltonian for non-interacting phonons and fermions
\begin{equation}
H = -\sum_{i} \mu c^\dagger_{i} c_i + \frac{1}{2} \sum_k (\pi^2_k + \omega_0^2 \phi_k^2) 
\end{equation}
where $\pi_k$ is the canonical momentum conjugated to $\phi_k$ ( $[\phi_k,\pi_{k'}]= i \delta_{kk'} $).

How do I derive the Lagrangian?

Since there is no variable canonically conjugated to the $c_i$ fields in the above Hamiltonian, how do we write down the Legendre transformation and the term $\propto \partial_\tau c$?
These questions arise from the analysis of this SYK paper. I do not quite get how they write down the free action Eq.(A3) from the model Eq.(5).
 A: Revised answer: The first one was wrong even if three people liked it!
The "Lagrangian" derived in the cited paper is  first order in the time derivative of $c_i$ and so the "Lagrangian" is really that of the
Hamiltonian action principle. This  starts from the action functional
$$
S[p,q] = \int \{\sum_i p_i\dot q_i -H(p,q)\}dt
$$
whose variation wrt $p_i$ gives us
$$
\dot q_i = \frac{\partial H}{\partial p_i}
$$
and the  variation of $q_i$ (and an integration by parts) gives
$$
\dot p_i= - \frac{\partial H}{\partial q_i}
$$
The fermionic action
$$
i \sum_i c^\dagger_i \partial_t c_i - H(c,c^\dagger)
$$
works in the same way so  the "momentum" conjugate to $c_i$ is $\pi_i=ic^\dagger_i$. The Bose commutator
$$
[q,p]=i
$$
is replaced by the fermionic
anticommutator
$$
\{c_i,\pi_j\}=\{c_i,  ic^\dagger_j\}=i\hbar \delta_{ij},
$$
or
$$
\{c_i,  c^\dagger_j\}=\delta_{ij}.
$$
and we read off that the Hamitonian is
$$
H= \mu c_i^\dagger c_i
$$
which (up to sign) is the expression in the question.
