I am reading a book about non relativistic quantization of E.M field. But first we do classical field theory.

We directly wrote the Hamiltonian of our study, and a part of our Hamiltonian is the following (we are in Coulomb gauge).

$$ H_{trans} = \frac{\epsilon_0}{2} \int d^3 k \sum_{\epsilon} \dot A_{\bot, \epsilon}^* (\vec{k},t) \dot A_{\bot, \epsilon} (\vec{k},t) + \omega^2 A_{\bot, \epsilon}^*(\vec{k},t) A_{\bot, \epsilon}(\vec{k},t) $$

Where $\epsilon$ denote a given polarization. Thus, we have 4 generalized coordinates : $A_{\bot, \epsilon_1},A_{\bot, \epsilon_2},A_{\bot, \epsilon_1}^*,A_{\bot, \epsilon_2}^*$ (one for each value of $\epsilon$) and four velocities (I don't know the exact denomination in english) : $\dot A_{\bot, \epsilon_1},\dot A_{\bot, \epsilon_2} ,\dot A_{\bot, \epsilon_1}^*, \dot A_{\bot, \epsilon_2}^*$.

In the book, they say that this Hamiltonian looks like a sum of harmonic oscillators hamiltonian. Even though I agree, I would like to see it properly by finding what the momentum is, to end up with something like $$\frac{p^2}{2m}+\frac{1}{2} m \omega^2 q^2$$ i.e. the field version of it.

Given a Lagrangian, I know how to find the momentum, but how is it done when given the Hamiltonian expressed in position/velocity?

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    $\begingroup$ Which book is this? $\endgroup$ – Emilio Pisanty Apr 2 at 20:09
  • $\begingroup$ @EmilioPisanty Mécanique quantique - Tome III - Claude Cohen Tanoudji - page 399 (it is in French) $\endgroup$ – StarBucK Apr 2 at 20:12
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    $\begingroup$ That is in fact not the Hamiltonian, rather the energy associated to a Lagrangian (which is a function of the generalised velocities and positions): some books still call this quantity the Hamiltonian, with abuse of terminology. The true Hamiltonian is the Legendre transform of the Lagrangian as function on the cotangent bundle of the configuration space. $\endgroup$ – gented Apr 2 at 20:43
  • $\begingroup$ @gented thus it is not possible in practice, given this quantity to guess what the momentum will be ? The only way to do it is to have started from the Lagrangian ? $\endgroup$ – StarBucK Apr 6 at 17:48
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    $\begingroup$ Correct, the momentum is defined only given a Lagrangian. $\endgroup$ – gented Apr 6 at 18:34

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