Generating functions

This post has been migrated from my old blog, the math-physics learning seminar.

Method of Generating Functions

Let (X) and (Y) be two smooth manifolds, and let (M = T^\ast X, N = T^\ast Y) with corresponding symplectic forms (\omega_M) and (\omega_N).

Question: How can we produce symplectomorphisms (\phi: M \to N)?

The most important construction from classical mechanics is the method of generating functions. I will outline this method, shameless stolen from Ana Cannas da Silva’s lecture notes.

Suppose we have a smooth function (f \in C^\infty(X \times Y)). Then its graph (\Gamma) is a submanifold of (M \times N): ( \Gamma = { (x,y, df_{x,y}) \in M \times N }). Since (M \times N) is a product, we have projections (\pi_M, \pi_N), and this allows us to write the graph as

[ \Gamma = { (x, y, df_x, df_y) }]

Now there is a not-so-obvious trick: we consider the twisted graph (\Gamma^\sigma) given by

[ \Gamma^\sigma = {(x,y, df_x, -df_y) } ]

Note the minus sign.

Proposition If (\Gamma^\sigma) is the graph of a diffeomorphism (\phi: M \to N), then (\phi) is a symplectomorphism.

Proof By construction, (\Gamma^\sigma) is a Lagrangian submanifold of (M \times N) with respect to the twisted symplectic form (\pi_M^\ast \omega_M - \pi_N^\ast \omega_N). It is a standard fact that a diffeomorphism is a symplectomorphism iff its graph is Lagrangian with respect to the twisted symplectic form, so we’re done.

Now we have:

Modified question: Given (f \in C^\infty(M \times N)), when is its graph the graph of a diffeomorphism (\phi: M \to N)?

Pick coordinates (x) on (X) and (y) on (Y), with corresponding momenta (\xi) and (\eta). Then if (\phi(x,\xi) = (y,\eta)), we obtain

[ \xi = d_x f, \ \eta = -d_y f ]

Note the simlarity to Hamilton’s equations. By the implicit function theorem, we can construct a (local) diffeomorphism (\phi) as long as (f) is sufficiently non-degenerate.

Different Types of Generating Functions

We now concentrate on the special case of (M = T^\ast \mathbb{R} = \mathbb{R} \times \mathbb{R}^\ast). Note that this is a cotangent bundle in two ways: (T^\ast \mathbb{R} \cong T^\ast \mathbb{R}^\ast). Hence we can construct local diffeomorphisms (T^\ast \mathbb{R} \to T^\ast \mathbb{R}) in four ways, by taking functions of the forms

[ f(x_1, x_2), \ f(x_1, p_2), \ f(p_1, x_2), \ f(p_1, p_2) ] Origins from the Action Principle, and Hamilton-Jacobi

Suppose that we have two actions

[ S_1 = \int p_1 \dot{q}_1 - H_1 dt, \ S_2 = \int p_2 \dot{q}_2 - H_2 dt ]

which give rise to the same dynamics. Then the Lagrangians must differ by a total derivative, i.e.

[ p_1 \dot{q}_1 - H_1 = p_2 \dot{q}_2 - H_2 + \frac{d f}{dt} ]

Suppose that (f = -q_2 p_2 + g(q_1, p_2, t)). Then we have

[ p_1 \dot{q}_1 - H_1 = -q_2 \dot{p}_2 - H_2 + \frac{\partial g}{\partial t} + \frac{\partial g}{\partial q_1}\dot{q}_1 + \frac{\partial g}{\partial p_2} \dot{p_2} ]

Comparing coefficients, we find

[ p_1 = \frac{\partial g}{\partial q_1}, \ q_2 = \frac{\partial g}{\partial p_2}, \ H_2 = H_1 + \frac{\partial g}{\partial t} ]

Now suppose that the coordinates ((q_2, p_2)) are chosen so that Hamilton’s equations become

[ \dot{q_2} = 0, \ \dot{p}_2 = 0 ]

Then we must have (H_2 = 0), i.e.

[ H_1 + \frac{\partial g}{\partial t} = 0 ]

Now we also have (\partial H_2 / \partial p_2 = 0), so this tells us that (g) is independent of (p_2), i.e. (g = g(q_1, t)). Since (p_1 = \partial g / \partial q_1), we obtain

[ \frac{\partial g}{\partial t} + H_1(q_1, \frac{\partial g}{\partial q_1}) = 0 ]

This is the Hamilton-Jacobi equation, usually written as

[ \frac{\partial S}{\partial t} + H(x, \frac{\partial S}{\partial x}) = 0 ]

Note the similarity to the Schrodinger equation! In fact, one can derive the Hamilton-Jacobi equation from the Schrodinger equation by taking a wavefunction of the form

[ \psi(x,t) = A(x,t) \exp({\frac{i}{\hbar} S(x,t)}) ]

and expanding in powers of (\hbar). This also helps to motivate the path integral formulation of quantum theory.