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+---
+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.