In mechanics, a constant of motion is a physical quantity conserved throughout the motion, imposing in effect a constraint on the motion. However, it is a mathematical constraint, the natural consequence of the equations of motion, rather than a physical constraint (which would require extra constraint forces). Common examples include energy, linear momentum, angular momentum and the Laplace–Runge–Lenz vector (for inverse-square force laws).

Applications

Constants of motion are useful because they allow properties of the motion to be derived without solving the equations of motion. In fortunate cases, even the trajectory of the motion can be derived as the intersection of isosurfaces corresponding to the constants of motion. For example, Poinsot's construction shows that the torque-free rotation of a rigid body is the intersection of a sphere (conservation of total angular momentum) and an ellipsoid (conservation of energy), a trajectory that might be otherwise hard to derive and visualize. Therefore, the identification of constants of motion is an important objective in mechanics.

Methods for identifying constants of motion

There are several methods for identifying constants of motion.

• The simplest but least systematic approach is the intuitive ("psychic") derivation, in which a quantity is hypothesized to be constant (perhaps because of experimental data) and later shown mathematically to be conserved throughout the motion.
• The Hamilton–Jacobi equations provide a commonly used and straightforward method for identifying constants of motion, particularly when the Hamiltonian adopts recognizable functional forms in orthogonal coordinates.
• Another approach is to recognize that a conserved quantity corresponds to a symmetry of the Lagrangian. Noether's theorem provides a systematic way of deriving such quantities from the symmetry. For example, conservation of energy results from the invariance of the Lagrangian under shifts in the origin of time, conservation of linear momentum results from the invariance of the Lagrangian under shifts in the origin of space (translational symmetry) and conservation of angular momentum results from the invariance of the Lagrangian under rotations. The converse is also true; every symmetry of the Lagrangian corresponds to a constant of motion, often called a conserved charge or current.
• A quantity ${\displaystyle A}$ is a constant of the motion if its total time derivative is zero
  $${\displaystyle 0={\frac {dA}{dt}}={\frac {\partial A}{\partial t}}+\{A,H\},}$$

  
  which occurs when ${\displaystyle A}$'s Poisson bracket with the Hamiltonian equals minus its partial derivative with respect to time^[1]
  $${\displaystyle {\frac {\partial A}{\partial t}}=-\{A,H\}.}$$

  

Another useful result is Poisson's theorem, which states that if two quantities ${\displaystyle A}$ and ${\displaystyle B}$ are constants of motion, so is their Poisson bracket ${\displaystyle \{A,B\}}$.

A system with n degrees of freedom, and n constants of motion, such that the Poisson bracket of any pair of constants of motion vanishes, is known as a completely integrable system. Such a collection of constants of motion are said to be in involution with each other. For a closed system (Lagrangian not explicitly dependent on time), the energy of the system is a constant of motion (a conserved quantity).

In quantum mechanics

An observable quantity Q will be a constant of motion if it commutes with the Hamiltonian, H, and it does not itself depend explicitly on time. This is because

$${\displaystyle {\frac {d}{dt}}\langle \psi |Q|\psi \rangle =-{\frac {1}{i\hbar }}\left\langle \psi \right|\left\left|\psi \right\rangle +\left\langle \psi \right|{\frac {dQ}{dt}}\left|\psi \right\rangle \,}$$

$${\displaystyle =HQ-QH\,}$$

Derivation

Say there is some observable quantity Q which depends on position, momentum and time,

$${\displaystyle Q=Q(x,p,t)}$$

And also, that there is a wave function which obeys Schrödinger's equation

$${\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=H\psi .}$$

Taking the time derivative of the expectation value of Q requires use of the product rule, and results in

$${\displaystyle {\begin{aligned}{\frac {d}{dt}}\left\langle Q\right\rangle &={\frac {d}{dt}}\left\langle \psi \right|Q\left|\psi \right\rangle \\&=\left({\frac {d}{dt}}\left\langle \psi \right|\right)Q\left|\psi \right\rangle +\left\langle \psi \right|{\frac {dQ}{dt}}\left|\psi \right\rangle +\left\langle \psi \right|Q\left({\frac {d}{dt}}\left|\psi \right\rangle \right)\\&=-{\frac {1}{i\hbar }}\left\langle H\psi \right|Q\left|\psi \right\rangle +\left\langle \psi \right|{\frac {dQ}{dt}}\left|\psi \right\rangle +{\frac {1}{i\hbar }}\left\langle \psi \right|Q\left|H\psi \right\rangle \\&=-{\frac {1}{i\hbar }}\left\langle \psi \right|HQ\left|\psi \right\rangle +\left\langle \psi \right|{\frac {dQ}{dt}}\left|\psi \right\rangle +{\frac {1}{i\hbar }}\left\langle \psi \right|QH\left|\psi \right\rangle \\&=-{\frac {1}{i\hbar }}\left\langle \psi \right|\left\left|\psi \right\rangle +\left\langle \psi \right|{\frac {dQ}{dt}}\left|\psi \right\rangle \end{aligned}}}$$

So finally,

$${\displaystyle {\frac {d}{dt}}\langle \psi |Q|\psi \rangle ={\frac {-1}{i\hbar }}\langle \psi |\leftH,Q\right|\psi \rangle +\langle \psi |{\frac {dQ}{dt}}|\psi \rangle \,}$$

Commentedit

For an arbitrary state of a Quantum Mechanical system, if H and Q commute, i.e. if