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.

Classical Mechanics 6: Poisson brackets and the Heisenberg picture

Last time we saw that a classical mechanical system which has a Lagrangian formulation can (under some mild assumptions) be repackaged as a symplectic manifold \((M, \omega)\) together with a smooth function \(H\) called the Hamiltonian. The equations of motion then become
\[ \dot{x} = X_H \]
where \(X_H\) is the Hamiltonian vector field associated with \(H\) (sometimes called the symplectic gradient of \(H\)). This identifies solutions to the equations of motion with certain curves in the manifold \(M\), which together form a 1-parameter group of diffeomorphisms (in fact, symplectomorphisms) of \(M\).

Today, I'd like to discuss a dual formulation. Instead of thinking of the equations of motion as describing evolution of the points of \(M\), we will instead think of the equations of motion as describing evolution of the functions on \(M\). We will see later that this is the classical analog of the Heisenberg picture in quantum mechanics.

Definition An observable on \(M\) is a smooth real-valued function on \(M\).

Suppose we have a classical mechanical system \(M, \omega, H\). By integrating the equations of motion, we obtain a 1-parameter family of symplectomorphisms \(\phi_t\). For any point \(x \in M\), the curve \(x(t)\) defined by
\[ x(t) = \phi_t(x) \]
solved Hamilton's equations.

Now suppose we have some observable \(f\). Its value along any solution to the equations of motion is
\[ f(x(t) = f(\phi_t(x)) = f_t(x) \]
where \(f_t := f \circ \phi_t\). So, if we only care about the values of observables along any solution to the equations of motion, the transformation
\[ x \mapsto x(t) = \phi_t(x) \]
\[ f \mapsto f \]
which is the Schrodinger picture, is indistinguishable from the transformation
\[ x \mapsto x \]
\[ f \mapsto f_t = f \circ \phi_t \]
which we will call the Heisenberg picture. What is the analog of Hamilton's equations for the Heisenberg picture? Let us compute:
\[ \frac{d}{dt} f_t = \frac{d}{dt}f(\phi_t) = df \circ \frac{d}{dt}\phi_t \]
Since \(\phi_t\) is generated by the vector field \(X_H\), we obtain
\[ \frac{d}{dt} f_t = X_H(f) = df(X_H) = \omega^{-1}(dH, df) \]
For two observables \(f\) and \(g\), let us define the Poisson bracket of \(f\) and \(g\) as
\[ \{f, g\} := \omega^{-1}(df, dg). \]
Then we have
\[ \frac{d}{dt} f_t = \{H, f_t \} \]
which is called the Heisenberg equation of motion.

Theorem The Poisson bracket \(\{\cdot, \cdot\}\) satisfies the following properties:
1. It is \(\mathbb{R}\)-linear and skew-symmetric.
2. It satisfies the Jacobi identity.
3. It satisfies the Leibniz rule \(\{fg,h\}) = f\{g,h\} + \{f,h\}g\).

Proof Too lazy for now! But it's really easy.

Now let \(\mathscr{A} = C^\infty(M, \mathbb{R})\) be the commutative algebra of observables. This has an additional structure: the Poisson bracket \(\{,\}\), so we will call \(\mathscr{A}\) a Poisson algebra.

Now let us consider something completely crazy. Consider the following generalization of mechanical system.

Tentative Definition A mechanical system is an algebra \(\mathscr{A}\) together with a Poisson bracket \(\{\}\) on \(\mathscr{A}\) and an element \(H \in \mathscr{A}\) called the Hamiltonian. The Heisenberg equations of motion are
\[ \frac{d}{dt} f_t = \{H, f_t\} \]
for any \(f \in \mathscr{A}\).

This definition is a little too vague at the moment, since without specifying a topology on \(\mathscr{A}\) we have no way of making sense of the Heisenberg equation. However, up to this caveat, this definition of mechanical system captures the essence of all of classical mechanics, classical field theory, quantum mechanics, and quantum field theory!

Classical Mechanics 5: Symplectic structures

As we saw in the previous post, the equations of motion for a mechanical system can be cast into a 1st order form called Hamilton's equations, which are naturally interpreted as describing a path in the phase space \(T^\ast M\) associated to the configuration space \(M\). Let us investigate the geometry of \(T^\ast M\) see why Hamilton's equations are so nice.

Definition The canonical (or sometimes tautological) 1-form on the cotangent bundle \(T^\ast M\) is the 1-form \(\theta\) defined by
\[ \theta_{(q,p)}(X) = p(\pi_\ast X), \]
where \(\pi_\ast\) is the pushforward induced by the natural projection \(\pi: T^\ast M \to TM\). In other words, the form is defined by
\[ \theta_{(q,p)} = \pi^\ast p. \]

Definition The canonical symplectic form on the cotangent bundle \(T^\ast M\) is the 2-form \(\omega\) defined by
\[ \omega = -d\theta. \]

Let \(\omega_\flat: T M \to T^\ast M\) be the map given by \(X \mapsto \iota(X)\omega\).

Proposition The canonical symplectic form satisfies the following two conditions:
1. It is closed, i.e. \(d\omega = 0\).
2. It is nondegenerate, i.e. the map \(\omega_\flat\) is invertible with inverse \(\omega^\sharp: T^\ast M \to TM\).

Proof The first property follows from \(d^2 = 0\). To prove the second, suppose we have local coordinates \(q^i\) on \(M\) with cotangent coordinates \(p^i\). Then it is easily seen that
\[ \theta = p^i dq^i, \]
so that
\[ \omega = dq^i \wedge dp^i, \]
from which nondegeneracy is obvious.

Definition Any 2-form on a manifold \(N\) (not necessarily a cotangent bundle) which satisfies the above two properties will be called symplectic. A pair \((N, \omega)\) will be called symplectic if \(\omega\) is a symplectic 2-form on \(N\).

Definition Given a function \(H\) on a symplectic manifold \((N, \omega)\), the Hamiltonian vector field associated to \(H\) is the vector field \(X_H\) uniquely defined by
\[ dH = \omega_\flat X_H. \]

Proposition For \(N = T^\ast M\) a cotangent bundle with the canonical symplectic form, Hamilton's equations with respect to a Hamiltonian function \(H\) describe the flow of the vector field \(X_H\).

Proof Again pick local coordinates \(q\) and \(p\). Then the inverse map \(\omega^\sharp\) is given by
\[ dq \mapsto -\frac{\partial}{\partial p} \]
\[ dp \mapsto \frac{\partial}{\partial q} \]
Since
\[ dH = \frac{\partial H}{\partial q} dq + \frac{\partial H}{\partial p} dp, \]
we see that
\[ X_H = \frac{\partial H}{\partial p} \frac{\partial}{\partial q} - \frac{\partial H}{\partial q} \frac{\partial}{\partial p} \]
But then the equation describing the flow of \(X_H\) is (in components)
\[ \dot{q} = \frac{\partial H}{\partial p} \]
\[ \dot{p} = -\frac{\partial H}{\partial q} \]
which are exactly Hamilton's equations.

Classical Mechanics 4: Hamilton's Equations

Recall from last time that a classical mechanical system consists of a manifold \(M\) (the configuration space) and a function \(L\) on the tangent bundle \(TM\). The equations of motion for a path \(x(t)\) in \(M\) are the 2nd order Euler-Lagrange equations:
\[ \frac{d}{dt}\left( \frac{\partial L}{\partial v}(x, \dot{x}) \right) = \frac{\partial L}{\partial x}(x, \dot{x}) \]

Hamilton discovered a way of recasting these equations as a first order system for a path in a related manifold, the cotangent bundle \(T^\ast M\). The benefits are twofold: in addition to reducing the equations to a first order system (at the cost of introducing new variables), it turns out that this framework makes it much easier to find conserved quantities and prove theorems about mechanical systems. So let's see what he did.

Theorem Under mild assumptions on \(L\), there is a function \(H\) on \(T^\ast M\) constructed canonically out of \(L\) such that solutions of the Euler-Lagrange equations are in 1-1 correspondence with solutions \(q(t), p(t)\) on \(T^\ast M\) of Hamilton's equations
\[ \dot{q} = \frac{\partial H}{\partial p}(q, p) \]
\[ \dot{p} = -\frac{\partial H}{\partial q}(q,p) \]
Furthermore, if the original Lagrangian function \(L\) is not explicitly time-dependent, then the function \(H\) is constant for any solution of the equations of motion.

To start with, introduce coordinates \(x, v\) on \(TM\) as before. We will define new coordinates \(q,p\) as follows:
\[ q(x,v) = x \]
\[ p(x,v) = \frac{\partial L}{\partial x}(x,v) \]
Our assumption on \(L\) will be the following: the above formulas can be inverted to obtain \(x\) and \(v\) as functions of \(q\) and \(p\). It is easily seen from the definition of \(p\) that under a coordinate transformation (of \(x\)), it behaves as a 1-form, so the coordinates \(q\) and \(p\) can be interpreted as local coordinates on the cotangent bundle \(T^\ast M\). We construct the Hamiltonian as
\[ H(q,p) = p\cdot v(q,p) - L(x(q,p), v(q,p)) \]
(this is the Legendre transform--more later). Of course, the above formula is not well-defined if we cannot solve for \(x\) and \(v\) in terms of \(p\) and \(q\)--hence the assumption. Now we check:
\[ \frac{\partial H}{\partial p} = v + p \frac{\partial v}{\partial p} - \frac{\partial L}{\partial x} \frac{\partial x}{\partial p} -\frac{\partial L}{\partial v}\frac{\partial v}{\partial p} = v + p\frac{\partial v}{\partial p} - p\frac{\partial v}{\partial p} = v\]
Since \(\dot{x} = v\), this is the first of Hamilton's equations.

For the second, we perform a similar computation:
\[ \frac{\partial H}{\partial q} = -\frac{\partial L}{\partial x} \frac{\partial x}{\partial q} - \frac{\partial L}{\partial v} \frac{\partial v}{\partial q} = -\frac{\partial L}{\partial x} \]
But the Euler-Lagrange equations say that
\[ \dot{p} = \frac{d}{dt}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial x} = -\frac{\partial H}{\partial x}, \]
so we've obtained the second of Hamilton's equations.

For the last part, suppose that the Lagrangian does not depend explicitly on time, i.e.
\[ \frac{\partial L}{\partial t} = 0. \]
Then we compute:
\[ \frac{d}{dt}H = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} = \frac{\partial H}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial H}{\partial p} \frac{\partial H}{\partial q} = 0. \]
Hence \(H\) is automatically conserved. For this reason, \(H\) is often called the energy of the system.

Later, we will see that conserved quantities correspond to symmetries, and conservation of energy is a statement about the symmetry corresponding to time translation.

Classical Mechanics 3: Hamilton's action principle.

We saw before that Newton's 2nd law can be written in a more general form as
\[ \frac{d}{dt} \frac{\partial L}{\partial v}(x, \dot{x}) = \frac{\partial L}{\partial x}(x, \dot{x}), \]
known as the Euler-Lagrange equations. Hamilton discovered a principle that explains the origin of these equations. Consider some path of the system given by a curve \(\gamma\), i.e.
\[ x(t) = \gamma(t) \]
\[ \dot{x}(t) = \frac{d}{dt}\gamma(t) \]
Then we may define a quantity associated with the path \(\gamma\):
\[ S = \int L(\gamma, \dot{\gamma})dt \]
called the action. Hamilton discovered the following.

Theorem The path taken by a mechanical system is one which extremizes the action.

To prove this, suppose we perturb the path a small amount, while leaving the endpoints fixed, i.e. \(\gamma \mapsto \gamma + \epsilon (\delta\gamma)\) with \(\epsilon > 0\) small and \(\delta\gamma\) a path that is \(0\) at its endpoints. Then
\[ L(\gamma + \epsilon\delta\gamma, \dot\gamma + \epsilon\delta\dot{\gamma}) = L(\gamma, \dot{\gamma}) + \epsilon \frac{\partial L}{\partial x}\delta\gamma + \epsilon \frac{\partial L}{\partial v} \delta\dot\gamma + o(\epsilon^2) \]
Thus
\[ S[\gamma + \epsilon\delta\gamma] = S[\gamma] + \epsilon \int \frac{\partial L}{\partial x} \delta \gamma dt + \epsilon \int \frac{\partial L}{\partial v} \delta \dot\gamma dt + o(\epsilon^2) \]
Integrating by parts, and using the fact that \(\delta\gamma\) is \(0\) on the endpoints, we have
\[ \int \frac{\partial L}{\partial v}\delta\dot\gamma dt = -\int \frac{d}{dt} \frac{\partial L}{\partial v} \delta\gamma dt \]

Combining the above, we have
\[ \frac{\delta S}{\delta \gamma}(\delta \gamma) = \int \left(\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial v} \right) \delta\gamma dt \]
Thus the variational derivative of \(S\) is
\[ \frac{\delta S}{\delta \gamma} = \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial v} \]
So a path \(\gamma\) is a critical point of \(S\) (i.e. it extremizes \(S\)) if and only if the Euler-Lagrange equations are satisfied.

Classical Mechanics 2: The Euler-Lagrange equations from Newton's 2nd law.

After the previous post, we are now familiar with Newton's 2nd law

\[ \mathbf{F} = m\mathbf{a}, \]

which (suitably interpreted) holds for any system of \(N\) particles. However, this equation requires the use of cartesian coordinates, which for many systems may not be the most convenient choice. Suppose we have some other coordinates \(q^i = q^i(x^j)\). What is the correct analogue of Newton's 2nd law for the \(q\)-coordinates?

To make life easier, we will assume for now that the force \(\mathbf{F}\) is conservative; i.e. 

\[ \mathbf{F} = -\nabla V(x) \]

for some potential function \(V(x)\). Under this assumption, Newton's 2nd law is

\[ m\mathbf{a} + \nabla V(x) = 0. \]

Let us define the function \(T\) by

\[ T(x,v) = \frac{1}{2}m |v|^2, \]

and define the function \(L\) as

\[ L(x,v) = T(x,v) - V(x). \]

Then we have immediately that Newton's 2nd law is equivalent to

\[ \frac{d}{dt}\frac{\partial L}{\partial v}(x, \dot{x}) - \frac{\partial L}{\partial x}(x,\dot{x})  = 0. \]

Why go through the trouble of introducing these auxiliary functions and rewriting Newton's 2nd law in this way? The answer lies in the following theorem.

Theorem For any choice of coordinates \(y = y(x)\), Newton's 2nd law is equivalent to the equations

\[ \frac{d}{dt}\frac{\partial \tilde{L}}{\partial w}(y, \dot{y}) - \frac{\partial \tilde{L}}{\partial y}(y,\dot{y})  = 0, \]

where \(w = dY_x(v)\) and \(\tilde{L}(y,w) = L(x(y,w), v(y,w)\). These equations are called the Euler-Lagrange equations.

The proof of this theorem is a straightforward calculation using the chain rule. Let \(M\) denote the manifold \(\mathbb{R}^{3N}\) (or some open subset thereof). The coordinate change \(y = y(x) \) can be thought of as a diffeomorphism \(Y: M \to M\) given by \(x \mapsto y(x) \). The differential \( dY: TM \to TM\) is also a diffeomorphism. In coordinates, we have

\[ y = y(x) \]

\[ w = dY_x(v)  = Jv \]

where \(y,w\) are coordinates on the target \(TM\). 

Now we need to compute the derivatives of \(\tilde{L}\). 

\[ \frac{\partial L}{\partial x} = \frac{\partial \tilde{L}}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial \tilde{L}}{\partial w}\frac{\partial w}{\partial x} = \tilde{L}_y J + \tilde{L}_w H v \]

where \(J\) is the matrix of mixed partials and \(H\) is the Hessian matrix.

\[ \frac{\partial L}{\partial v} = \frac{\partial \tilde{L}}{\partial y} \frac{\partial y}{\partial v} + \frac{\partial \tilde{L}}{\partial w} \frac{\partial w}{\partial v} = \tilde{L}_w J \]

Then we have

\[ \frac{d}{dt}\left( \tilde{L}_w J \right) = \frac{d}{dt}\tilde{L}_w J + \tilde{L}_w H \dot{x}  \]

Subtracting \(\frac{\partial L}{\partial x}\) from this and using the calculation above, we obtain

\[ \left( \frac{d}{dt} \tilde{L}_w - \tilde{L}_y \right) J \]

and since \(J\) is invertible, this is \(0\) if and only if \(\frac{d}{dt} \tilde{L}_w - \tilde{L}_y\) is.

But this is exactly what we wanted to prove!

Classical Mechanics 1: Newton's Laws.

This is a surpisingly hard topic, and someday I would like to make an honest attempt at it. But Feynman did it better than I ever could! See "What is a Force?" from the Feynman Lectures on Physics, available here: here (pdf).

I promise to update eventually.