path: root/source/know/concept/hamiltonian-mechanics/index.md

---
title: "Hamiltonian mechanics"
sort_title: "Hamiltonian mechanics"
date: 2021-07-03
categories:
- Physics
- Classical mechanics
layout: "concept"
---

**Hamiltonian mechanics** is an alternative formulation of classical mechanics,
which equivalent to Newton's laws,
but often mathematically advantageous.
It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-mechanics/),
which is in turn built on [variational calculus](/know/concept/calculus-of-variations/).



## Definitions

In Lagrangian mechanics, use a Lagrangian $$L$$,
which depends on position $$q(t)$$ and velocity $$\dot{q}(t)$$,
to define the momentum $$p(t)$$ as a derived quantity.
Hamiltonian mechanics switches the roles of $$\dot{q}$$ and $$p$$:
the **Hamiltonian** $$H$$ is a function of $$q$$ and $$p$$,
and the velocity $$\dot{q}$$ is derived from it:

$$\begin{aligned}
    \pdv{L(q, \dot{q})}{\dot{q}} = p
    \qquad \quad
    \pdv{H(q, p)}{p} \equiv \dot{q}
\end{aligned}$$

Conveniently, this switch turns out to be
[Legendre transformation](/know/concept/legendre-transform/):
$$H$$ is the Legendre transform of $$L$$,
with $$p = \partial L / \partial \dot{q}$$ taken as
the coordinate to replace $$\dot{q}$$.
Therefore:

$$\begin{aligned}
    \boxed{
        H(q, p) \equiv \dot{q} \: p - L(q, \dot{q})
    }
\end{aligned}$$

This almost always works,
because $$L$$ is usually a second-order polynomial of $$\dot{q}$$,
and thus convex as required for Legendre transformation.
In the above expression,
$$\dot{q}$$ must be rewritten in terms of $$p$$ and $$q$$,
which is trivial, since $$p$$ is proportional to $$\dot{q}$$ by definition.

The Hamiltonian $$H$$ also has a direct physical meaning:
for a mass $$m$$, and for $$L = T - V$$,
it is straightforward to show that $$H$$ represents the total energy $$T + V$$:

$$\begin{aligned}
    H
    = \dot{q} \: p - L
    = m \dot{q}^2 - L
    = 2 T - (T - V)
    = T + V
\end{aligned}$$

Just as Lagrangian mechanics,
Hamiltonian mechanics scales well for large systems.
Its definition is generalized as follows to $$N$$ objects,
where $$p$$ is shorthand for $$p_1, ..., p_N$$:

$$\begin{aligned}
    \boxed{
        H(q, p)
        \equiv \bigg( \sum_{n = 1}^N \dot{q}_n \: p_n \bigg) - L(q, \dot{q})
    }
\end{aligned}$$

The positions and momenta $$(q, p)$$ form a phase space,
i.e. they fully describe the state.

An extremely useful concept in Hamiltonian mechanics
is the **Poisson bracket** (PB),
which is a binary operation on two quantities $$A(q, p)$$ and $$B(q, p)$$,
denoted by $$\{A, B\}$$:

$$\begin{aligned}
    \boxed{
        \{ A, B \}
        \equiv \sum_{n = 1}^N \Big( \pdv{A}{q_n} \pdv{B}{p_n} - \pdv{A}{p_n} \pdv{B}{q_n} \Big)
    }
\end{aligned}$$



## Canonical equations

Lagrangian mechanics has a single Euler-Lagrange equation per object,
yielding $$N$$ second-order equations of motion in total.
In contrast, Hamiltonian mechanics has $$2 N$$ first-order equations of motion,
known as **Hamilton's canonical equations**:

$$\begin{aligned}
    \boxed{
        - \pdv{H}{q_n} = \dot{p}_n
        \qquad
        \pdv{H}{p_n} = \dot{q}_n
    }
\end{aligned}$$


{% include proof/start.html id="proof-canonical" -%}
For the first equation,
we differentiate $$H$$ with respect to $$q_n$$,
and use the chain rule:

$$\begin{aligned}
    \pdv{H}{q_n}
    &= \pdv{}{q_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
    \\
    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{q_n} + p_j \pdv{\dot{q}_j}{q_n} \Big)
    - \Big( \pdv{L}{q_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{q_n} \Big) \bigg)
    \\
    &= \sum_{j} \Big( p_j \pdv{\dot{q}_j}{q_n} - \pdv{L}{q_n} - p_j \pdv{\dot{q}_j}{q_n} \Big)
    = - \pdv{L}{q_n}
\end{aligned}$$

We use the Euler-Lagrange equation here,
leading to the desired equation:

$$\begin{aligned}
    - \pdv{L}{q_n} = - \dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = - \dv{p_n}{t} = - \dot{p}_n
\end{aligned}$$

The second equation is somewhat trivial,
since $$H$$ is defined to satisfy it in the first place.
Nevertheless, we can prove it by brute force,
using the same approach as above:

$$\begin{aligned}
    \pdv{H}{p_n}
    &= \pdv{}{p_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
    \\
    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big)
    - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg)
    \\
    &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n}
    - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
    = \dot{q}_n
\end{aligned}$$
{% include proof/end.html id="proof-canonical" %}


Just like in Lagrangian mechanics, if $$H$$ does not explicitly contain $$q_n$$,
then $$q_n$$ is called a **cyclic coordinate**, and leads to the conservation of $$p_n$$:

$$\begin{aligned}
    \dot{p}_n = - \pdv{H}{q_n} = 0
    \quad \implies \quad
    p_n = \mathrm{conserved}
\end{aligned}$$

Of course, there may be other conserved quantities.
Generally speaking, the $$t$$-derivative of an arbitrary quantity $$A(q, p, t)$$ is as follows,
where $$\ipdv{}{t}$$ is a "soft" derivative
(only affects explicit occurrences of $$t$$),
and $$\idv{}{t}$$ is a "hard" derivative
(also affects implicit $$t$$ inside $$q$$ and $$p$$):

$$\begin{aligned}
    \boxed{
        \dv{A}{t}
        = \{ A, H \} + \pdv{A}{t}
    }
\end{aligned}$$


{% include proof/start.html id="proof-dv-t" -%}
We differentiate via the multivariate chain rule,
insert the canonical equations,
and eventually recognize the PB definition:

$$\begin{aligned}
    \dv{A}{t}
    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t}
    \\
    &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t}
    \\
    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
\end{aligned}$$
{% include proof/end.html id="proof-dv-t" %}


Assuming that $$H$$ does not explicitly depend on $$t$$,
the above property naturally leads us to an alternative
way of writing Hamilton's canonical equations:

$$\begin{aligned}
    \dot{q}_n = \{ q_n, H \}
    \qquad \quad
    \dot{p}_n = \{ p_n, H \}
\end{aligned}$$



## Canonical coordinates

So far, we have assumed that the phase space coordinates $$(q, p)$$
are the *positions* and *canonical momenta*, respectively,
and that led us to Hamilton's canonical equations.

In theory, we could make a transformation of the following general form:

$$\begin{aligned}
    q \to Q(q, p)
    \qquad \quad
    p \to P(q, p)
\end{aligned}$$

However, most choices of $$(Q, P)$$ would not preserve Hamilton's equations.
Any $$(Q, P)$$ that do keep this form
are known as **canonical coordinates**,
and the corresponding transformation is a **canonical transformation**.
That is, any $$(Q, P)$$ that satisfy:

$$\begin{aligned}
    - \pdv{H}{Q_n} = \dot{P}_n
    \qquad \quad
    \pdv{H}{P_n} = \dot{Q}_n
\end{aligned}$$

Then we might as well write $$H(q, p)$$ as $$H(Q, P)$$.
So, which $$(Q, P)$$ fulfill this?
It turns out that the following must be satisfied for all $$n, j$$,
where $$\delta_{nj}$$ is the Kronecker delta:

$$\begin{aligned}
    \boxed{
        \{ Q_n, Q_j \} = \{ P_n, P_j \} = 0
        \qquad
        \{ Q_n, P_j \} = \delta_{nj}
    }
\end{aligned}$$


{% include proof/start.html id="proof-transformation" -%}
Assuming that $$Q_n$$, $$P_n$$ and $$H$$ do not explicitly depend on $$t$$,
we use our expression for the $$t$$-derivative of an arbitrary quantity,
and apply the multivariate chain rule to it:

$$\begin{aligned}
    \dot{Q}_n
    &= \{Q_n, H\}
    = \sum_{n} \bigg( \pdv{Q_n}{q_n} \pdv{H}{p_n} - \pdv{Q_n}{p_n} \pdv{H}{q_n} \bigg)
    \\
    &= \sum_{n, j} \bigg( \pdv{Q_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
    - \pdv{Q_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
    \\
    &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{Q_n}{q_n} \pdv{Q_j}{p_n} - \pdv{Q_n}{p_n} \pdv{Q_j}{q_n} \Big)
    + \pdv{H}{P_j} \Big( \pdv{Q_n}{q_n} \pdv{P_j}{p_n} - \pdv{Q_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
    \\
    &= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg)
\end{aligned}$$

This is equivalent to Hamilton's equation $$\dot{Q}_n = \ipdv{H}{P_n}$$
if and only if $$\{Q_n, Q_j\} = 0$$ for all $$n$$ and $$j$$,
and if $$\{Q_n, P_j\} = \delta_{nj}$$.

Next, we do the exact same thing with $$P_n$$ instead of $$Q_n$$,
giving an analogous result:

$$\begin{aligned}
    \dot{P}_n
    &= \{P_n, H\}
    = \sum_{n} \bigg( \pdv{P_n}{q_n} \pdv{H}{p_n} - \pdv{P_n}{p_n} \pdv{H}{q_n} \bigg)
    \\
    &= \sum_{n, j} \bigg( \pdv{P_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
    - \pdv{P_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
    \\
    &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{P_n}{q_n} \pdv{Q_j}{p_n} - \pdv{P_n}{p_n} \pdv{Q_j}{q_n} \Big)
    + \pdv{H}{P_j} \Big( \pdv{P_n}{q_n} \pdv{P_j}{p_n} - \pdv{P_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
    \\
    &= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg)
\end{aligned}$$

Which is equivalent to Hamilton's equation $$\dot{P}_n = -\ipdv{H}{Q_n}$$
if and only if $$\{P_n, P_j\} = 0$$,
and $$\{Q_n, P_j\} = - \delta_{nj}$$.
The PB is anticommutative,
i.e. $$\{A, B\} = - \{B, A\}$$.
{% include proof/end.html id="proof-transformation" %}


If you have experience with quantum mechanics,
the latter equation should look suspiciously similar
to the *canonical commutation relation* $$[\hat{Q}, \hat{P}] = i \hbar$$.



## References
1.  R. Shankar,
    *Principles of quantum mechanics*, 2nd edition,
    Springer.