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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 $$ p_2^\mu = \frac{\partial L}{\partial(\partial_\mu \phi)} = - \partial^\mu \phi, \qquad H_2 = p_2^\mu \partial_\mu - L_2 = - \frac{p_2^2}{2}. $$

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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).

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  • $\begingroup$ Indeed, I made a mistake while reading the formula, I will edit my post above. $\endgroup$ – Harold Jan 8 '17 at 17:50
  • $\begingroup$ I learned about Dirac brackets some time ago, but I am not willing (for now) to engage in the full derivation since the above problem is not of very high importance (so it is why I was looking for a reference). If I understand your last paragraph, for a more general Lagrangian some $\partial^2 \phi$ would remain and we would not be able to solve? So what does it mean if these $\partial^2 \phi$ remains in $H$? $\endgroup$ – Harold Jan 8 '17 at 17:56
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    $\begingroup$ No for a non-degenerate case $\partial^2 \phi$ would appear in your formula for $p^{\mu\nu}$ and you could implicitly solve for it and replace by some combination of phase space variables. Just like how $\dot{q}$ is usually a function of $p$. $\endgroup$ – octonion Jan 8 '17 at 18:57
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    $\begingroup$ But in your actual example you get that $p^{\mu\nu}$ is proportional to $\phi$. This is a constraint on the phase space. You know you ought to get a constraint on the phase space, because this is 'really' an ordinary system that can be written in terms of first derivatives so the expanded phase space in Ostrogadsky's method is unnecessary. Here it is obvious which phase space variables are extraneous, but in more complicated cases of constraints you could use Dirac brackets. Again, in a nontrivial example you might not have any constraints at all. $\endgroup$ – octonion Jan 8 '17 at 19:04
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    $\begingroup$ If it's non-degenerate, you might have a term like $(\Delta\phi)^2$. If you put it in the form of (2.20), they would consider that an "N=2" theory, and you can follow their discussion (I don't believe their new method works for N=1). Note that their 'Hamilton's equations' are second order in time, and that's why they have less variables than the Ostrogadski method. For this reason they aren't constructing an ordinary phase space, and so I wouldn't use their new method directly as a basis for quantization. $\endgroup$ – octonion Jan 12 '17 at 12:07

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