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Hamiltonian constraints

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The topic of Hamiltonian constraints is a particularly thorny one as it is both rarely brought up (it really only starts being brought up when BRST quantization for gauge system is done), and when it is, it's usually only done in the abstract, with very little examples given. The situation isn't helped by the fact that practical examples are usually fairly complex already, such as electrodynamics, with fields involved, or point particle systems, where the gauge involves reparametrization invariance. To remedy this situation a bit, I'll try to bring up some known toy models on the topic.

1. Reminder on Hamiltonian mechanics

Before digging in, here's a (fairly basic) introduction to Hamiltonian systems.

We have a linear functional $S$ called the action, depending on some functions of time $q^i(t)$ and its derivatives. This action stems from a function $L$ called the Lagrangian :

\begin{equation} S[q^i(t), \dot{q}^i(t), t] = \int_{t_a}^{t_b} dt\ L(q^i(t), \dot{q}^i(t), t) \end{equation}

We want each $q^i$ to be an extremal function for the action functional, that is,

\begin{equation} \frac{\delta S}{\delta q^i(t)} = 0 \end{equation} \begin{equation} \frac{\partial}{\partial t} (\frac{\partial L}{\partial \dot{x}_i}) - \frac{\partial L}{\partial x_i} = 0 \end{equation

An important feature of Lagrangian mechanics is the concept of Lagrange multipliers. If we have a variable $\lambda(t)$ that only appears linearly, as in

\begin{equation} S[q^i, \lambda] = \int_{t_a}^{t_b}dt\ \left[ L(q^i(t), t) + \lambda(t) g(q^i(t), t) \right] \end{equation}

Then, from the Euler-Lagrange equation, we get

\begin{equation} g(x^i(t), t) = 0 \end{equation}

This will be useful to defining constraints later on.

The Hamiltonian is defined at its core by considering the Legendre transform of our Lagrangian

\begin{equation} H = \sum_i \dot{q}_i p_i - L \end{equation}

What we want from the Hamiltonian is to remove the dependence on derivatives, so that we can consider our system of equations of motion as depending only on first time derivatives. To do this, we introduce two new variables, $p^i(t)$ and $v^i(t)$. $v$ will correspond to the velocity, so that $v^i = \dot{q}^i$, and $p$ will be a Lagrange multiplier. What we want is therefore

\begin{equation} S[q^i, p^i, v^i] = \int_{t_a}^{t_b} dt\ \left[ L(q^i(t), v^i(t), t) + p_i(t)(\dot{q}^i(t) - v^i(t)) \right] \end{equation}

This leads immediately to the equations of motion

\begin{eqnarray} \dot{x}^i(t) &=& v^i(t) p_i(t) &=& \frac{\partial L}{\partial v^i}\\ \frac{\partial L}{\partial q^i} &=& \dot{p}_i(t) \end{eqnarray}

Performing all the proper substitutions, it can be seen that these equations of motion are equivalent. The Hamiltonian is defined then by

\begin{equation} H() = p_i(t) v^i(t) - L \end{equation}

In which case we get the simpler form

\begin{equation} S[q^i, \dot{q}, p^i, v^i] = \int_{t_a}^{t_b}dt\ \left[p_i(t) \dot{q}^i(t) - H(p, v, q) \right] \end{equation}

with the following conditions for the variation (assuming proper boundary conditions) :

\begin{eqnarray} \vec{v}(t) &=& \dot{\vec{x}}(t)\\ x \end{eqnarray}

Things are not particularly more simple now, but if we have the good luck of having the functions $v^i$ expressible as functions of $p_i$, that is,

\begin{equation} v^i(t) = v^i(p_j(t)) \end{equation}

Then we can simply drop $v$ as a variable and just express the action as

\begin{equation} S[q^i, \dot{q}, p^i] = \int_{t_a}^{t_b}dt\ \left[p_i(t) \dot{q}^i(t) - H(p, q) \right] \end{equation}

with

\begin{equation} H() = p_i(t) v^i(p(t)) - L \end{equation}

The condition for this is that

Constrained Hamiltonians

The situation we have presented breaks down somewhat when the system is under-determined. To get a feeling of how such a thing may happen, consider the Euler-Lagrange equations, using the chain rule on the time derivative :

\begin{eqnarray} 0 &=& \frac{\partial L}{\partial q^i}(q(t), \dot{q}(t)) - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}^i}(q(t), \dot{q}(t)))\\ &=& \frac{\partial L}{\partial q^i} - \dot{q}^j \frac{\partial^2 L}{\partial q^j \partial \dot{q}^i} - \ddot{q} \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} \end{eqnarray}

Rearranging things a bit, we obtain

\begin{equation} \ddot{q}^j \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} = \frac{\partial L}{\partial q^i} - \dot{q}^j \frac{\partial^2 L}{\partial q^j \partial \dot{q}^i} \end{equation}

For brevity let's call the Hessian of $L$ with respect to $\dot{q}$ $W_{ij}$. From there, it turns out that we can express the second derivatives of our degrees of freedom entirely from their values and first derivatives if $W$ is invertible, in which case,

\begin{equation} \ddot{q}^j = W_{ij}^{-1} \left( \frac{\partial L}{\partial q^i} - \dot{q}^j \frac{\partial^2 L}{\partial q^j \partial \dot{q}^i} \right) \end{equation}

If $W$ is not invertible (that is, $\det(W) = 0$), solutions may fail to be unique. There may be variables which can be expressed in terms of one another. In the configuration space of $(q^i, \dot{q}^i)$, this corresponds to some constraints of the form

\begin{equation} \phi_m(q^i, \dot{q}^i) = 0 \end{equation}

This will define a submanifold in the configuration space. We'll ask that, in this submanifold,

A very simple example (as found in [2]) of a constrained system is the system with two degrees of freedom $(x(t), y(t))$ described by the Lagrangian

\begin{eqnarray} L &=& \frac{1}{2} (\dot{x} - y)^2\\ &=& \frac{1}{2} (\dot{x}^2 - 2y\dot{x} + y^2) \end{eqnarray}

The Euler-Lagrange equations are fairly simple then :

\begin{eqnarray} \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} &=& \ddot{x} - \dot{y} = 0\\ \frac{\partial L}{\partial y} - \frac{d}{dt} \frac{\partial L}{\partial \dot{y}} &=& \dot{x} - y = 0 \end{eqnarray}

It is fairly obvious here that the system here is under-determined : for two variables to solve, we only have a single equation of motion (the first equation is simply the time derivative of the first).

The momenta of the system are simply

\begin{eqnarray} p_x &=& \dot{x} - y\\ p_y &=& 0 \end{eqnarray}

Electromagnetism

The simplest case of a realistic Lagrangian experiencing gauge freedom is the case of electromagnetism. A simple way to do this is by considering a charged point particle, with Lagrangian

\begin{equation} L = \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} - q \phi(\vec{x}) \end{equation}

Bibliography

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Last updated : 2019-07-26 19:38:05