Mathematically I can understand Canonical Transformations but I don't have an intuitive understanding of them.

  • Why do we need to a canonical transformation? Is it to simplify the form of Hamilton's equation?
  • We can view the Hamiltonian as the generator for infinitesimal time transformation, but don't we need to first solve the equations of motion to do this transformation? How does one understand this intuitively?

Canonical transformations preserve the Hamiltonian structure of a dynamical system, i.e., the Hamiltonian form of the equations of motion. Hence they are the natural class of transformations associated with these, in the same way as diffeomorphism are the natural class of transformations when studying manifolds and translations and Lorentz transformation are the natural class of transformations when studying Minkowski space.

They can be used, e.g., to find or classify symmetries of a system, or to simplify the form of the Hamiltonian. They also provide a short notation for the time shift operation, without having to solve the equation of motion. (Existence of the time shift as a canonical transformation is guaranteed if the Hamiltonian is twice continuously differentiable.)


Why do we need to a canonical transformation? Is it to simplify the form of Hamilton's equation?

It is just for our convenience. Mathematically, they preserve the Hamiltonian structure i.e. all the different forms represent the same dynamical system. When there are several choices to describe the system under consideration, one set of equations may be convenient over the rest. We tend to express our equations such that we find a maximum number of cyclic co-ordinates.

For example, to discuss the motion of a particle in a plane under the influence of a central force, we may use generalized co-ordinates as

  • Cartesian co-ordinates: $\quad q_1 = x, \quad q_2=y \quad$ both of which are not cyclic.

  • Plane polar co-ordinates: $\quad q_1 = r, \quad q_2=\theta \quad$ in which $\theta$ is cyclic.

The number of cyclic co-ordinates can thus depend on the choice of generalized co-ordinates. For each problem, there are particular choices in which many or even all co-ordinates are cyclic. Once we find the set, the calculations are trivial.

Why is it trivial? Suppose there exist a system with all its co-ordinates $q_i$ cyclic and $H$ is a constant of motion. Then, the conjugate momenta $p_i$ are all constants $alpha_i$ since, $\dot{p_i} = -\dfrac{\partial H}{\partial q_i}=0$. So Hamilton's equations of motion would give

$$\dot{q_i} = \dfrac{\partial H}{\partial \alpha_i} = \beta_i$$ with each $\beta_i$ is a function of only $\alpha_i$ and are independent of time. Hence, $q_i = \beta_i t + \delta_i$ as easier as it could get.


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