Liouville's Theorem and the Flow in Phase Space
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You should be comfortable with Hamilton's equations of motion and the definition of the Poisson bracket from the previous lecture. Familiarity with the divergence of a vector field in multi-dimensional space is assumed.
Why this matters
Liouville's theorem reveals that phase space is incompressible, a geometric constraint that underpins statistical mechanics and quantum theory.
In Newtonian mechanics, we often think of particles moving through physical space. In Hamiltonian mechanics, the state of a system is a point in a higher-dimensional abstract space called phase space. Liouville's theorem states that the "volume" occupied by a cloud of points in this phase space remains constant as the system evolves. This is not merely a mathematical curiosity; it is the foundation of the microcanonical ensemble in statistical mechanics and ensures that probability is conserved in classical dynamics. It tells us that information about the initial state is never lost, only scrambled, which is a precursor to the unitarity of time evolution in quantum mechanics.
The Idea in Plain Language
Imagine a drop of ink in a perfectly incompressible fluid, like water. As the water flows, stretches, and swirls, the drop changes shape. It might become a long, thin filament or a complex blob. However, if the fluid is incompressible, the total volume of the ink drop remains exactly the same. It cannot shrink to a point, nor can it expand to fill the entire container. The fluid elements move around, but they do not create or destroy space. Liouville's theorem asserts that phase space behaves exactly like this incompressible fluid. The "flow" of the system's state is governed by Hamilton's equations, and this flow has zero divergence. Consequently, any region of phase space, no matter how distorted it becomes over time, preserves its initial volume. This is a profound statement about the conservation of information in classical mechanics.
Phase Space and the Flow Velocity
To formalize this, we consider a system with \( N \) degrees of freedom. The phase space is a \( 2N \)-dimensional manifold with coordinates \( (q_1, \dots, q_N, p_1, \dots, p_N) \). The state of the system at any time \( t \) is a point \( \mathbf{z}(t) \) in this space. The time evolution of this point is determined by Hamilton's equations:
We can define a velocity vector field \( \mathbf{u} \) in phase space, where the components are the time derivatives of the coordinates:
This vector field describes the direction and speed at which every point in phase space moves. The key question is: does this flow compress or expand the volume of a region? In fluid dynamics, this is determined by the divergence of the velocity field. If the divergence is zero, the flow is incompressible.
Calculating the Divergence
The divergence of the phase-space velocity field \( \mathbf{u} \) is defined as the sum of the partial derivatives of the velocity components with respect to their corresponding coordinates:
Substituting Hamilton's equations into this expression, we get:
Assuming the Hamiltonian \( H \) is sufficiently smooth (specifically, that its second partial derivatives are continuous and the mixed partials are equal), we can interchange the order of differentiation:
Since \( \frac{\partial^2 H}{\partial q_i \partial p_i} = \frac{\partial^2 H}{\partial p_i \partial q_i} \), the terms inside the summation cancel out exactly:
Thus, the divergence of the Hamiltonian flow is identically zero. This is the core mathematical statement of Liouville's theorem.
Volume Conservation and Density
The condition \( \nabla \cdot \mathbf{u} = 0 \) implies that the flow is incompressible. Consider a small region \( V \) in phase space with a boundary surface \( S \). The rate of change of the volume of this region is given by the flux of the velocity field through its boundary:
By the divergence theorem, this surface integral is equal to the volume integral of the divergence over the region \( V \):
Since \( \nabla \cdot \mathbf{u} = 0 \) everywhere, the integral vanishes, and we conclude that \( \frac{dV}{dt} = 0 \). The volume of any co-moving region in phase space is constant in time.
This result has a direct implication for the phase-space density \( \rho(\mathbf{z}, t) \), which represents the probability density of finding the system in a particular state. The continuity equation for this density is:
Expanding the divergence term and using the product rule:
Since \( \nabla \cdot \mathbf{u} = 0 \), the last term vanishes. The remaining terms form the total time derivative of \( \rho \) along the trajectory (the material derivative):
Thus, the phase-space density is conserved along the trajectories of the system. This is the "strong" form of Liouville's theorem.
The derivation
Liouville's Theorem on Phase-Space Volume
For an autonomous Hamiltonian system with Hamiltonian H(q, p) on the 2N-dimensional phase space with canonical coordinates (q1,…,qN, p1,…,pN), the phase-flow velocity field is divergence-free, so the flow is incompressible: any co-moving region of phase space retains its 2N-volume for all time, dV/dt = 0, and equivalently the phase-space density is conserved along every trajectory, dρ/dt = 0.
A common misconception is that Liouville's theorem implies that the system's state cannot change or that the distribution function \( \rho \) is static in space. It is not. The density \( \rho \) at a fixed point in phase space can change over time if the flow carries high-density regions into or out of that point. What is conserved is the density *following* the flow. The shape of the distribution can become extremely complex and filamentary, but its total "amount" (integral over phase space) and local volume elements remain invariant. This is distinct from dissipative systems where phase-space volume contracts, which is not possible for standard Hamiltonian systems. C
Worked example
Problem. Consider a one-dimensional harmonic oscillator with Hamiltonian \( H = \frac{p^2}{2m} + \frac{1}{2}kq^2 \). Verify that the phase-space flow is incompressible by calculating the divergence of the velocity field. Given \( m = 2.0\ \mathrm{kg} \) and \( k = 8.0\ \mathrm{N/m} \), determine the period of oscillation and confirm that the area enclosed by the trajectory in phase space is conserved.
- Write down Hamilton's equations for \( q \) and \( p \): \[ \dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m}, \quad \dot{p} = -\frac{\partial H}{\partial q} = -kq \]
- Calculate the divergence of the velocity field \( \mathbf{u} = (\dot{q}, \dot{p}) \) in the 2D phase space: \[ \nabla \cdot \mathbf{u} = \frac{\partial \dot{q}}{\partial q} + \frac{\partial \dot{p}}{\partial p} \]
- Compute the partial derivatives: \[ \frac{\partial \dot{q}}{\partial q} = \frac{\partial}{\partial q} \left( \frac{p}{m} \right) = 0 \] \[ \frac{\partial \dot{p}}{\partial p} = \frac{\partial}{\partial p} (-kq) = 0 \]
- Sum the derivatives to find the divergence: \[ \nabla \cdot \mathbf{u} = 0 + 0 = 0 \] This confirms the flow is incompressible.
- Calculate the angular frequency \( \omega \) and period \( T \): \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{8.0}{2.0}} = \sqrt{4.0} = 2.0\ \mathrm{rad/s} \] \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2.0} = \pi\ \mathrm{s} \approx 3.14\ \mathrm{s} \]
Area check. The orbit is the ellipse \( p^2/(2m) + kq^2/2 = E \), with semi-axes \( \sqrt{2E/k} \) in \( q \) and \( \sqrt{2mE} \) in \( p \), so the enclosed area is \( \pi\sqrt{2E/k}\,\sqrt{2mE} = 2\pi E/\omega \). It cannot change, because \( E \) is conserved. Units check. Each term of the divergence, \( \partial\dot q/\partial q \) and \( \partial\dot p/\partial p \), has units \( \mathrm{s^{-1}} \), so the divergence is a rate (here zero); \( \sqrt{k/m} \) is in \( \mathrm{s^{-1}} \), so \( T \) is in seconds, and \( E/\omega \) is in \( \mathrm{J\,s} \), the units of an area in the \( (q, p) \) plane.
Three quick tests
- What is the physical interpretation of \( \nabla \cdot \mathbf{u} = 0 \) in phase space?
Answer
The phase-space flow is incompressible; the volume of any region in phase space remains constant over time. - If the Hamiltonian depends explicitly on time, does Liouville's theorem still hold for the phase-space volume?
Answer
Yes, the divergence is still zero because the partial derivatives of \( \dot{q} \) w.r.t \( q \) and \( \dot{p} \) w.r.t \( p \) still cancel, provided the Hamiltonian is smooth. The volume conservation holds for non-autonomous systems as well. - How does Liouville's theorem relate to the conservation of probability in quantum mechanics?
Answer
It is the classical precursor to the unitarity of the time-evolution operator in quantum mechanics, ensuring that the total probability (norm of the wavefunction) is conserved.
Problem set for PU-201 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.