How to obtain commutation relations from symplectic potential?

Question

I am studying the notes on susy qm of David Skinner (http://www.damtp.cam.ac.uk/user/dbs26/SUSY.html) (which itself follows the mirror symmetry book by Vafa and Hori (relevant pp. 206 - 210)) and had a trouble with the footnote on p.45 in chapter 3 which suggests that one can obtain the canonical commutation relation from the `symplectic potential' that is obtained by the boundary term. How to see this ? What point am I missing? I would appreciate any direct help on this.

Answer 1

The central idea is that the Dirac quantization can be implemented by replacing the Poisson brackets of functions $\{f,g\}$ with a commutator of operators $[\hat{f}, \hat{g}]$. So you need to know how to compute the Poisson bracket of two functions in the symplectic formulation of Hamiltonian mechanics.

The central object in this formulation (see, for instance, Sec 11.4.2 of Mathematics for Physics by Stone and Goldbart.) is the symplectic form $\omega = d\eta$, where $\eta$ is the symplectic potential. Given a function $f$, one first needs to compute the corresponding velocity vector fields by the Cartan equation $d f = - \omega(v_f, \cdot)$, which can be solved for $v_f$. The Poisson bracket is then defined in terms of these velocity vectors as $\{f,g\} = \omega(v_f, v_g)$.

This is equivalent to the conventional formulation of classical mechanics for the symplectic form $\omega = dp^i \wedge dx^i$. Explicitly, given $f$, setting $v_f \equiv a_f^i \frac\partial{\partial x^i} + b_f^i \frac\partial{\partial p^i}$, the Cartan equation becomes $$ \frac{\partial f}{\partial x^i} dx^i + \frac{\partial f}{\partial p^i} dp^i = - \left( b_f^i dx^i - a_f^i dp^i \right), $$ which can be solved as $$ v_f = \frac{\partial f}{\partial p^i} \frac\partial{\partial x^i} - \frac{\partial f}{\partial x^i} \frac\partial{\partial p^i}. $$ After a similar computation for $g$, we can compute the Poisson bracket of $f$ and $g$ as $$ \{f,g\} = dp^i \wedge dx^i (v_f, v_g) = \frac{\partial f}{\partial x^i}\frac{\partial g}{\partial p^i} - \frac{\partial g}{\partial x^i}\frac{\partial f}{\partial p^i}, $$ which is the conventional definition of the Poisson bracket.

Answer 2

1. The infinitesimal variation of a general action $S=\int_{t_i}^{t_f}\! dt~L $ is of the form $$ \delta S~=~\int_{t_i}^{t_f}\!dt \left(\frac{\delta S}{\delta q} \delta q +\frac{d}{dt}\left(\underbrace{p_j \delta q^j}_{=\Theta} \right)\right), \qquad p_j~=~\frac{\partial L}{\partial q^j}.$$ Here the boundary term $$ \Theta~=~p_j \delta q^j$$ is the pre-symplectic 1-form potential. The pre-symplectic 2-form is $$ \Omega~=~\delta\Theta.$$

2. If $\Omega$ is non-degenerate/invertible, $\Omega$ is a symplectic 2-form, and the inverse structure is the Poisson structure, which determines the CCR.

3. For more information, see e.g. this & this related Phys.SE posts.

References:

1. D. skinner, SUSY lecture notes; footnote on p. 45 in chap. 3.