Canonical momentum combinations in covariant Hamiltonian formalism I am interested in a statement of the following paper (arxiv:hep-th/9802115), but I will describe the simplest case.
I am interested in a free scalar Lagrangian with mostly plus signature (the above paper describes the case of higher-derivatives, but the main features of my question appear in this simpleste case)
$$
L = \frac{1}{2}\, \phi \Delta \phi, \qquad
\Delta = \partial_\mu \partial^\mu.
$$
Written in this way the Lagrangian depends on higher-order derivatives so one will have higher-order momenta. Following the covariant Hamiltonian formalism à la De Donder–Weyl one defines a conjugate momentum for every field
$$
p^{\mu} = \frac{\partial L}{\partial(\partial_\mu \phi)} = 0, \qquad
p^{\mu\nu} = \frac{\partial L}{\partial(\partial_\mu \partial_\nu \phi)}
= \frac{\phi}{2}\, \eta^{\mu\nu}.
$$
The covariant Hamiltonian follows as
$$
H = p^\mu \partial_\mu \phi + p^{\mu\nu} \partial_\mu \partial_\nu \phi - L.
$$
In the paper they say that one can use a combination of the various momenta (since the Lagrangian depends only on the Laplacian) in order to write a single momentum
$$
\pi = \frac{\partial L}{\partial(\Delta \phi)}
= \frac{\phi}{2}.
$$
Note that this moment and the other ones are correctly linked by the chain rule
$$
p^{\mu\nu} = \frac{\partial L}{\partial(\Delta \phi)}\, \frac{\partial(\Delta \phi)}{\partial(\partial_\mu \partial_\nu \phi)}
= \pi \eta^{\mu\nu}.
$$
Then let me define $p = \eta_\mu\nu p^{\mu\nu}$ then the Hamiltonian is
$$
H = 2\, p^{\mu\nu} \partial_\mu \partial_\nu p - p \Delta p
= - 8\, \pi \Delta \pi.
$$
I am a bit disturbed by the fact that $\pi \sim \phi$, which reminds me the case of the Dirac fermion. So in the above Hamiltonian I have used $\pi$ this looks a bit strange (and I could have written $H \sim \phi \Delta \pi$).
This paper discusses a bit how to rewrite the Euler–Lagrange equations in terms of the Laplacian, but it is not really helpful for the Hamiltonian formalism.
I did not find any other reference to this procedure elsewhere, so I was wondering 1) if the above computations are correct and 2) if you knew something or if you had any intuition on the use of such parametrization, for example in view of canonical quantization, Poisson bracket, canonical transformations, etc.
For comparison, note that in terms of the Lagrangian
$$
L_2 = \frac{1}{2}\, (\partial\phi)^2
$$
the conjugate momentum and covariant Hamiltonian are
$$
p_2^\mu = \frac{\partial L}{\partial(\partial_\mu \phi)} = - \partial^\mu \phi, \qquad
H_2 = p_2^\mu \partial_\mu - L_2 = - \frac{p_2^2}{2}.
$$
 A: Take a closer look at Ostrogadski's method in the paper you referenced at either (2.16) or the non-field theory case (2.3). Only the momentum conjugate to the second highest derivative (in this case $\partial \phi$) has a form like your second line
$$p^{\mu\nu} = \frac{\partial L}{\partial(\partial_\mu \partial_\nu \phi)}
= \frac{\phi}{2}\, \eta^{\mu\nu}.$$
The momentum conjugate to lower derivatives (e.g. $\phi$ itself) is defined as
$$p^{\mu} = \frac{\partial L}{\partial(\partial_\mu \phi)}-\partial_\nu p^{\mu\nu} = -\frac{1}{2}\partial^\mu\phi$$
So as you wrote, the Hamiltonian is
$$H = p^\mu \partial_\mu \phi + p^{\mu\nu} \partial_\mu \partial_\nu \phi - L$$
we are unable to solve for $\partial^2\phi$ in terms of the phase space variables, but in any case $L$ cancels with the second term in H and we have $H=-\frac{1}{2}(\partial\phi)^2$, as it should be.
Note that our inability to solve for $\partial^2\phi$ goes hand in hand with constraints on the phase space $(\phi,\dot\phi,p^0,p^{00})$ but if we reduce it to just $(\phi,p^0)$ we have no trouble. This is something you see often even in Lagrangians with only first derivatives. You might be interested in searching about Dirac brackets and related topics (though it wasn't needed here).
