04 May 2026

Least Action and Hamilton Equations

Principle of least action, Legendre transform to the Hamiltonian, Hamilton equations, and examples.

msc semester-i classical-mechanics hamiltonian-mechanics least-action

Lagrangian mechanics describes motion using coordinates and velocities. Hamiltonian mechanics rewrites the same dynamics using coordinates and momenta. This change is important because the state of a system is then represented by a point in phase space $(q,p)$.

Hamilton’s equations are first-order equations in phase space. They are especially useful in canonical transformations, Poisson brackets, Hamilton-Jacobi theory, and statistical mechanics.

Principle of least action

The action is

\[S=\int_{t_1}^{t_2}L(q_i,\dot q_i,t)\,dt.\]

The physical path satisfies

\[\delta S=0.\]

Although the phrase “least action” is common, the more accurate statement is stationary action.

The action is a number assigned to an entire path. The physical path is selected by the condition that the first-order change in this number vanishes under small endpoint-fixed variations.

Conjugate momentum

For each generalized coordinate $q_i$, define the conjugate momentum

\[\boxed{ p_i=\frac{\partial L}{\partial\dot q_i}. }\]

The Hamiltonian is obtained by a Legendre transform:

\[\boxed{ H(q_i,p_i,t)=\sum_i p_i\dot q_i-L(q_i,\dot q_i,t). }\]

After forming $H$, the velocities must be expressed in terms of $(q_i,p_i,t)$.

This last step is essential. A Hamiltonian is not complete if it still contains $\dot q_i$ as an independent variable.

Hamilton equations

The differential of $H$ gives

\[\boxed{ \dot q_i=\frac{\partial H}{\partial p_i}, \qquad \dot p_i=-\frac{\partial H}{\partial q_i}. }\]

These are Hamilton’s equations of motion. They replace $n$ second-order Lagrange equations by $2n$ first-order equations.

Energy conservation

If $H$ has no explicit time dependence,

\[\frac{\partial H}{\partial t}=0,\]

then

\[\frac{dH}{dt}=0.\]

Thus the Hamiltonian is conserved.

Example: one-dimensional harmonic oscillator

For

\[L=\frac12m\dot x^2-\frac12kx^2,\]

the conjugate momentum is

\[p=m\dot x.\]

The Hamiltonian is

\[H=\frac{p^2}{2m}+\frac12kx^2.\]

Hamilton’s equations are

\[\dot x=\frac{p}{m}, \qquad \dot p=-kx.\]

Combining them,

\[\ddot x+\frac{k}{m}x=0.\]

Example: free particle

For

\[H=\frac{p^2}{2m},\]

Hamilton’s equations give

\[\dot x=\frac{p}{m}, \qquad \dot p=0.\]

So $p$ is constant and the motion is uniform.

From the differential of $H$

Starting with

\[H=\sum_i p_i\dot q_i-L,\]

take the differential:

\[dH=\sum_i \dot q_i\,dp_i+\sum_i p_i\,d\dot q_i-dL.\]

Since

\[dL=\sum_i\frac{\partial L}{\partial q_i}dq_i +\sum_i\frac{\partial L}{\partial\dot q_i}d\dot q_i +\frac{\partial L}{\partial t}dt,\]

and $p_i=\partial L/\partial\dot q_i$, the $d\dot q_i$ terms cancel. Using Lagrange’s equation,

\[\frac{\partial L}{\partial q_i}=\dot p_i,\]

we get

\[dH=\sum_i\dot q_i\,dp_i-\sum_i\dot p_i\,dq_i -\frac{\partial L}{\partial t}dt.\]

Comparing this with

\[dH=\sum_i\frac{\partial H}{\partial q_i}dq_i +\sum_i\frac{\partial H}{\partial p_i}dp_i +\frac{\partial H}{\partial t}dt\]

gives Hamilton’s equations.

Standard form for one-dimensional motion

For

\[H=\frac{p^2}{2m}+V(x),\]

Hamilton’s equations are

\[\dot x=\frac{p}{m}, \qquad \dot p=-\frac{dV}{dx}.\]

Combining them gives

\[m\ddot x=-\frac{dV}{dx},\]

which is Newton’s equation in potential form.

Main points

• The action principle is stationary action.
• Momentum is $p_i=\partial L/\partial\dot q_i$.
• The Hamiltonian is the Legendre transform of $L$.
• Hamilton’s equations are first-order equations in phase space.
• Time-independent $H$ is conserved.

Practice questions

1. Define conjugate momentum.
2. Obtain $H$ from $L$ using a Legendre transform.
3. Derive Hamilton’s equations from $dH$.
4. Write Hamilton’s equations for the harmonic oscillator.
5. Explain why a time-independent Hamiltonian is conserved.