References: Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Section 2.8; Exercises 2.8.6 – 2.8.7.

Here we’ll examine an interesting relation between the action and the total energy of a system, as given by the Hamiltonian . Suppose a single particle moving in one dimension follows a classical path given by , and moves from an initial position at time of to a final position at time of . The action of this classical path is given by the integral of the Lagrangian

What can we say about the rate of change of the action with respect to the final time ? That is, we want to calculate , where all other parameters and are held constant. The situation can be illustrated as shown:

Since the only thing that is changing is , the particle starts at the same initial time (which we’ve taken to be in the diagram) and moves to the same location , but at a different time (in the diagram, later time). This means that the particle must follow a different path, possibly over its entire trajectory. This path, which we’ll call , is related to the original path by perturbing the original path by an amount :

In the diagram, the original path is shown in red and the perturbed path in blue. The amount is seen to be the vertical distance between these two curves at each time, and in the case of the paths shown in the diagram, .

The difference in the action between the two paths is due to two contributions: first, there is the contribution due to the extra time, from to , that the particle takes to complete its path. Second, there is the difference in the two actions over the path from to . The first contribution is entirely new and, for an infinitesimal extra time , it is given by

where is the Lagrangian evaluated at time . The other contribution can be obtained by varying the action over the path from to :

Since depends on and , we have

For infinitesimally different trajectories, we can see from the diagram above that at each point on the curve, so , so we get

In these equations, the derivatives of are evaluated on the original curve . To verify the second line, use the product rule on the second integrand and cancel terms to get the first line. The second term in the last is evaluated at only since we’re assuming that .

The quantity in brackets in the first integral is zero, because of the Euler-Lagrange equations which are valid on the original curve :

Putting everything together, we get for the total variation in the action:

Looking at the diagram above, the slope of the blue curve at the time is given by

From the definition 2 of we see that , so

This gives the final equation for the variation of the action:

where the second line follows from the definition of the canonical momentum .

The required derivative is

Using a similar technique, we can work out . In this case, the situation is as shown in this diagram:

The two trajectories now take the same time, but in the modified trajectory, the particle moves a distance further. Since both paths take the same time, there is no extra contribution . In this case , since the new (blue) curve is above the old (red) one . The derivation is the same as above up to 8, and the total variation in the action is now

At , , so we get

ExampleWe can verify 17 for the case of the one-dimensional harmonic oscillator. The general solution for the position is given by

The total energy is given by

where we just multiplied out the second line, cancelled terms and used .

To get the action, we need the Lagrangian:

The action for a trajectory from to is then

To proceed further, we need to specify and , since these depend on the boundary conditions (that is, on where we require the mass to be at and ). If we require and , then

Plugging these into 25 gives the energy as

Plugging and into 35, we get (using and , so that ):

Taking the derivative, we get

Thus the result is verified for the harmonic oscillator.

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