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My question is essentially how to extract the canonical momentum out of an on-shell action.

The Hamilton-Jacobi formalism tells us that Hamilton's principal function is the on-shell action, which depends only on the coordinates $q$. Therefore, if $S$ refers to the off-shell action with Lagrangian $L$, a result for the canonical momentum is

$$ \frac{\partial L}{\partial \dot q} = p = \frac{\partial S^\text{on-shell}}{\partial q} $$

In theory, when evaluating the on-shell action, one plugs in the EOM into $S$ and all that's left are boundary terms. If we illustrate this with the harmonic oscillator in 1D, one gets

\begin{align} S^\text{on-shell} &= \int_{t_i}^{t_f} L dt \\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \dot q^2 - \omega^2 q^2 \right) dt\\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \frac{d}{dt} (q \dot q) - q \ddot q - \omega^2 q^2 \right) \\ &= \frac{m}{2} \left[ q(t_f) \dot q(t_f) - q(t_i) \dot q(t_i) \right] \end{align}

Now, this depends on $\dot q$, whereas it should depend only on generalized coordinates. Also, the on-shell action depends on the boundary values of our coordinate $q$, so we can't differentiate w.r.t. $q(t)$ to get a conjugate momentum $p(t)$ for all times, like it is usually done in the Lagrangian formalism.

What am I not seeing?

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Comment to the question (v1): OP's sought-for relation is eq. (11) in this Phys.SE answer, where a proof is also given. –  Qmechanic Dec 2 '13 at 23:10
    
Thanks for the link. I guess that what I don't understand now is the extra factor of $1/2$ appearing in the on-shell action. –  physguy Dec 2 '13 at 23:28

1 Answer 1

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$:

  1. The off-shell action $S[q;t_i,t_f]$,

  2. The (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$, and

  3. Hamilton's principal function $S(q,\alpha, t).$

For their definitions and how they are related, see e.g. this Phys.SE answer.

II) OP's sought-for relations $$\tag{A} p_f~=~\frac{\partial S}{\partial q_f}, \qquad p_i~=~-\frac{\partial S}{\partial q_i}, $$ for the on-shell action are eq. (11) in this Phys.SE answer, where a proof is also given.

III) For the Harmonic oscillator it is indeed true that

$$\tag{B} S(q_f,t_f;q_i,t_i)~=~\frac{ q_f p_f - q_i p_i }{2}, $$

with a half on the RHS, as OP mentions.

IV) Eqs.(A) and (B) are not inconsistent with each other even in the presence of the factor half, because the initial and final momentum $p_i=p_i(q_f,t_f;q_i,t_i)$ and $p_f=p_f(q_f,t_f;q_i,t_i)$ in eq. (B) depend on the value of the Dirichlet boundary conditions $q_i$ and $q_f$.

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