Action Principles

An action principle says that the physical trajectories of a system are stationary points of an action functional under a specified class of variations. For ordinary Lagrangian mechanics, the action is

$$S[q] = \int_{t_i}^{t_f} L(q,\dot q,t)\,dt,$$

and Hamilton’s principle is

$$\delta S=0$$

for all allowed variations with the chosen boundary data. This page collects the conceptual rules: what is varied, which endpoints are fixed, why boundary terms matter, how the phase-space action works, and why stationary action becomes stationary phase in semiclassical quantum mechanics.

Helpful background. Calculus of Variations explains endpoint-fixed variations; Hamiltonian Mechanics Review introduces the phase-space variables.

Stationary action and quantum phases

To apply stationary action, specify which paths may vary and what stays fixed at their endpoints. This is the step that determines whether an integration-by-parts boundary term vanishes. In a real-time path integral, stationary action becomes a stationary-phase condition, not a maximum of a probability.

Stationary, Not Always Least

The traditional phrase “principle of least action” is often misleading. The classical path makes the action stationary:

$$\left. \frac{d}{d\epsilon} S[q+\epsilon\eta] \right\rvert_{\epsilon=0} = 0$$

for every allowed variation $\eta$. A stationary point can be a minimum, maximum, or saddle.

Real-time actions commonly have saddle behavior. This is especially important in path integrals, where the classical path is a stationary phase point of an oscillatory amplitude, not the most probable path of a positive probability measure.

Ingredients of an Action Principle

Every action principle needs four pieces:

• the space of histories, such as paths $q(t)$ or fields $\phi(x)$;
• the action functional $S$;
• the allowed variations, such as fixed-endpoint variations;
• the boundary conditions or endpoint terms.

For a particle path, an allowed fixed-endpoint variation is

$$q^i(t) \longmapsto q^i(t)+\epsilon\eta^i(t),$$

with

$$\eta^i(t_i)=\eta^i(t_f)=0.$$

Changing the allowed variations changes the variational problem. If endpoints are free, boundary terms that vanish in the fixed-endpoint problem can become physical conditions.

Configuration-Space Action

For a Lagrangian

$$L(q,\dot q,t),$$

the configuration-space action is

$$S[q] = \int_{t_i}^{t_f} L(q,\dot q,t)\,dt.$$

Its first variation has the form

$$\delta S = \int_{t_i}^{t_f} \sum_i \left[ \frac{\partial L}{\partial q^i} - \frac{d}{dt} \left( \frac{\partial L}{\partial\dot q^i} \right) \right] \eta^i(t)\,dt + \left[ \sum_i p_i\eta^i \right]_{t_i}^{t_f},$$

where

$$p_i = \frac{\partial L}{\partial\dot q^i}.$$

For fixed endpoints, the boundary term vanishes. Since the remaining variations are arbitrary in the interior, stationary action gives the Euler–Lagrange equations:

$$\frac{\partial L}{\partial q^i} - \frac{d}{dt} \left( \frac{\partial L}{\partial\dot q^i} \right) = 0.$$

The full mathematical derivation belongs to Calculus of Variations. The mechanics review is Lagrangian Mechanics Review.

Boundary Terms

Boundary terms tell us what is fixed at an endpoint. If an endpoint is allowed to move, its variation must also be included in the stationary-action condition.

For fixed endpoints, the variation satisfies $\eta(t_i)=\eta(t_f)=0$, so a term such as

$$\left[ p_i\eta^i \right]_{t_i}^{t_f}$$

vanishes. If an endpoint is free, the same term may imply a natural boundary condition.

Adding a total time derivative to the Lagrangian,

$$L \longmapsto L+\frac{dF(q,t)}{dt},$$

changes the action by an endpoint term:

$$S \longmapsto S+F(q_f,t_f)-F(q_i,t_i).$$

The interior equations of motion are unchanged for fixed endpoints, but the endpoint phase in a path integral and the generating function for a canonical transformation can change.

Phase-Space Action

Hamiltonian mechanics has its own action principle. Treat $q^i(t)$ and $p_i(t)$ as independent paths and define

$$S[q,p] = \int_{t_i}^{t_f} \left[ \sum_i p_i\dot q^i - H(q,p,t) \right]dt.$$

Varying with respect to $p_i$ gives

$$\dot q^i = \frac{\partial H}{\partial p_i}.$$

Varying with respect to $q^i$ and using fixed endpoint variations for $q^i$ gives

$$\dot p_i = - \frac{\partial H}{\partial q^i}.$$

Thus the phase-space action produces Hamilton’s equations directly. Only the endpoint positions are fixed here; the endpoint momenta need not be fixed, because no integration by parts produces a $\delta p$ boundary term.

The term

$$\sum_i p_i\,dq^i$$

is the canonical one-form $\theta$. With the convention used throughout this chapter, the symplectic form is $\omega=-d\theta=\sum_i dq^i\wedge dp_i$. The minus sign matters: $d(p_i\,dq^i)=dp_i\wedge dq^i$.

Classical Paths in Path Integrals

The real-time path-integral weight is

$$\exp\left( \frac{i}{\hbar}S[q] \right).$$

When the action scale is large compared with $\hbar$, phases vary rapidly as the path is varied. Contributions away from stationary paths cancel by destructive interference, while neighborhoods of stationary paths contribute coherently.

The leading semiclassical approximation is therefore organized around solutions of

$$\delta S=0.$$

Those are classical paths, but the approximation remains quantum. The amplitude also contains fluctuation determinants, Maslov phases, tunneling exponents, and interference between multiple stationary paths.

Relation to Hamilton–Jacobi Theory

If the action is evaluated on a classical path connecting endpoints, it becomes Hamilton’s principal function:

$$S(q_f,t_f;q_i,t_i) = \int_{t_i}^{t_f} L(q_{\mathrm{cl}},\dot q_{\mathrm{cl}},t)\,dt.$$

Endpoint derivatives of this function give endpoint momenta, and the function satisfies the Hamilton–Jacobi equation. Thus action principles define the classical path, while Hamilton–Jacobi theory studies the endpoint-dependent action function built from that path.

Quantum Variational Principles

Not every action principle in quantum mechanics is a path integral. Variational methods also extremize functionals over trial states. For example, the stationary condition of

$$\langle\psi\rvert H\lvert\psi\rangle - E \left( \langle\psi\rvert\psi\rangle-1 \right)$$

with respect to $\langle\psi\rvert$ gives

$$H\lvert\psi\rangle=E\lvert\psi\rangle.$$

Time-dependent variational principles restrict Schrödinger evolution to a trial manifold. These methods are different from Hamilton’s classical action principle, but they share the same variational discipline: specify the functional, constraints, allowed variations, and error interpretation.

Common Mistakes

• Saying “least action” when the result is only stationary.
• Forgetting to specify endpoint conditions.
• Dropping boundary terms automatically.
• Varying $q$ and $\dot q$ as if they were independent in the configuration-space action.
• Forgetting that $q$ and $p$ are independent variables in the phase-space action.
• Treating stationary phase in a path integral as a positive-probability maximum.
• Assuming adding a total derivative changes nothing in every context.
• Confusing a classical action principle with a quantum variational approximation over states.

Exercises

1. Why does fixed-endpoint variation make the boundary term $\left[p_i\eta^i\right]_{t_i}^{t_f}$ vanish?

Solution

Fixed endpoints mean the varied path agrees with the original path at $t_i$ and $t_f$. Therefore the variation satisfies

$$\eta^i(t_i)=\eta^i(t_f)=0.$$

Substituting into the boundary term gives zero at both endpoints:

$$\left[ p_i\eta^i \right]_{t_i}^{t_f} = p_i(t_f)\eta^i(t_f) - p_i(t_i)\eta^i(t_i) = 0.$$

2. Derive Hamilton’s equations from the phase-space action

$$S[q,p] = \int \left( p\dot q-H(q,p,t) \right)dt$$

for one degree of freedom.

Solution

Varying $p$ gives

$$\delta_p S = \int \left( \dot q-\frac{\partial H}{\partial p} \right)\delta p\,dt.$$

Since $\delta p$ is arbitrary,

$$\dot q=\frac{\partial H}{\partial p}.$$

Varying $q$ gives

$$\delta_q S = \int \left( p\,\delta\dot q - \frac{\partial H}{\partial q}\delta q \right)dt.$$

Integrating $p\,\delta\dot q$ by parts and using fixed endpoint variations,

$$\delta_q S = \int \left( - \dot p - \frac{\partial H}{\partial q} \right)\delta q\,dt.$$

Thus

$$\dot p=-\frac{\partial H}{\partial q}.$$

3. Explain why adding $dF/dt$ to a Lagrangian does not change the Euler–Lagrange equations for fixed endpoints.

Solution

The action changes by

$$\int_{t_i}^{t_f}\frac{dF}{dt}\,dt = F(q_f,t_f)-F(q_i,t_i).$$

For fixed endpoints, this endpoint contribution is fixed under allowed variations, so its first variation vanishes. Therefore the interior stationary-action equations are unchanged.

4. Why does stationary action become stationary phase in a real-time path integral?

Solution

The path-integral weight is

$$\exp\left( \frac{i}{\hbar}S[q] \right).$$

When action differences across the varied paths are large compared with $\hbar$, nonstationary contributions tend to cancel through rapid phase variation. Adding the same constant to every action changes only a common phase. Near a stationary path, $\delta S=0$, so the phase is stable to first order and nearby paths add coherently. This is stationary phase, not probability maximization.

References

• I. M. Gelfand and S. V. Fomin, Calculus of Variations, Dover, 2000.
• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.