path: root/content/know/concept/hamiltonian-mechanics/index.pdc

---
title: "Hamiltonian mechanics"
firstLetter: "H"
publishDate: 2021-07-03
categories:
- Physics
- Classical mechanics

date: 2021-07-03T14:39:14+02:00
draft: false
markup: pandoc
---

# Hamiltonian mechanics

**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**:

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

<div class="accordion">
<input type="checkbox" id="proof-canoneq"/>
<label for="proof-canoneq">Proof</label>
<div class="hidden">
<label for="proof-canoneq">Proof.</label>
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:
$$\begin{aligned}
    \pdv{H}{p_n}
    &= \pdv{p_n} \Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
    \\
    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big)
    - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg)
    \\
    &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n}
    - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
    = \dot{q}_n
\end{aligned}$$
</div>
</div>

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 $\pdv*{t}$ is a "soft" derivative
(only affects explicit occurrences of $t$),
and $\dv*{t}$ is a "hard" derivative
(also affects implicit $t$ inside $q$ and $p$):

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

<div class="accordion">
<input type="checkbox" id="proof-diff-t"/>
<label for="proof-diff-t">Proof</label>
<div class="hidden">
<label for="proof-diff-t">Proof.</label>
We differentiate via the multivariate chain rule,
insert the canonical equations,
and eventually recognize the PB definition:

$$\begin{aligned}
    \dv{A}{t}
    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t}
    \\
    &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t}
    \\
    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
\end{aligned}$$
</div>
</div>

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:

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

<div class="accordion">
<input type="checkbox" id="proof-cantrans"/>
<label for="proof-cantrans">Proof</label>
<div class="hidden">
<label for="proof-cantrans">Proof.</label>
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 $\dot{Q}_n = \pdv*{H}{P_n}$
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 $\dot{P}_n = -\pdv*{H}{Q_n}$
if and only if $\{P_n, P_j\} = 0$,
and $\{Q_n, P_j\} = - \delta_{nj}$.
The PB is anticommutative,
i.e. $\{A, B\} = - \{B, A\}$.
</div>
</div>

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.