Generating Functions

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.