Relation between action and energy

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 {S} and the total energy of a system, as given by the Hamiltonian {H}. Suppose a single particle moving in one dimension follows a classical path given by {x_{cl}\left(t\right)}, and moves from an initial position at time {t_{i}} of {x_{cl}\left(t_{i}\right)=x_{i}} to a final position at time {t_{f}} of {x_{cl}\left(t_{f}\right)=x_{f}}. The action {S_{cl}} of this classical path is given by the integral of the Lagrangian

\displaystyle  S_{cl}=\int_{t_{i}}^{t_{f}}L\left(x,\dot{x}\right)dt \ \ \ \ \ (1)

What can we say about the rate of change of the action with respect to the final time {t_{f}}? That is, we want to calculate {\partial S_{cl}/\partial t_{f}}, where all other parameters {t_{i},x_{i}}and {x_{f}} are held constant. The situation can be illustrated as shown:

Since the only thing that is changing is {t_{f}}, the particle starts at the same initial time (which we’ve taken to be {t_{i}=0} in the diagram) and moves to the same location {x_{f}}, 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 {x\left(t\right)}, is related to the original path {x_{cl}\left(t\right)} by perturbing the original path by an amount {\eta\left(t\right)}:

\displaystyle  x\left(t\right)=x_{cl}\left(t\right)+\eta\left(t\right) \ \ \ \ \ (2)

In the diagram, the original path {x_{cl}} is shown in red and the perturbed path {x} in blue. The amount {\eta} 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, {\eta\left(t\right)<0}.

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 {t_{f}} to {t_{f}+\Delta t}, that the particle takes to complete its path. Second, there is the difference in the two actions over the path from {t_{i}} to {t_{f}}. The first contribution is entirely new and, for an infinitesimal extra time {\Delta t}, it is given by

\displaystyle  \delta S_{1}=L\left(t_{f}\right)\Delta t \ \ \ \ \ (3)

where {L\left(t_{f}\right)} is the Lagrangian evaluated at time {t_{f}}. The other contribution can be obtained by varying the action over the path from {t_{i}=0} to {t_{f}}:

\displaystyle  \delta S_{2}=\int_{0}^{t_{f}}\delta L\;dt \ \ \ \ \ (4)

Since {L} depends on {x} and {\dot{x}}, we have

\displaystyle  \delta L=\frac{\partial L}{\partial x}\delta x+\frac{\partial L}{\partial\dot{x}}\delta\dot{x} \ \ \ \ \ (5)

For infinitesimally different trajectories, we can see from the diagram above that {\delta x=\eta\left(t\right)} at each point on the curve, so {\delta\dot{x}=\dot{\eta}\left(t\right)}, so we get

\displaystyle   \delta S_{2} \displaystyle  = \displaystyle  \int_{0}^{t_{f}}\left[\frac{\partial L}{\partial x}\eta\left(t\right)+\frac{\partial L}{\partial\dot{x}}\dot{\eta}\left(t\right)\right]\;dt\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \int_{0}^{t_{f}}\left[-\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}+\frac{\partial L}{\partial x}\right]\eta\left(t\right)dt+\int_{0}^{t_{f}}\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right)dt\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  0+\left.\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right|_{t_{f}} \ \ \ \ \ (8)

In these equations, the derivatives of {L} are evaluated on the original curve {x_{cl}}. 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 {t=t_{f}} only since we’re assuming that {\eta\left(0\right)=0}.

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

\displaystyle  \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0 \ \ \ \ \ (9)

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

\displaystyle   \delta S_{cl} \displaystyle  = \displaystyle  \delta S_{1}+\delta S_{2}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \left[\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)+L\Delta t\right]_{t_{f}} \ \ \ \ \ (11)

Looking at the diagram above, the slope of the blue curve {x\left(t_{f}\right)} at the time {t_{f}} is given by

\displaystyle  \dot{x}\left(t_{f}\right)=\frac{\left|\eta\left(t_{f}\right)\right|}{\Delta t} \ \ \ \ \ (12)

From the definition 2 of {\eta} we see that {\eta\left(t_{f}\right)<0}, so

\displaystyle  \eta\left(t_{f}\right)=-\dot{x}\left(t_{f}\right)\Delta t \ \ \ \ \ (13)

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

\displaystyle   \delta S_{cl} \displaystyle  = \displaystyle  \left[-\frac{\partial L}{\partial\dot{x}}\dot{x}+L\right]_{t_{f}}\Delta t\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \left(-p\dot{x}+L\right)\Delta t\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  -H\Delta t \ \ \ \ \ (16)

where the second line follows from the definition of the canonical momentum {p=\partial L/\partial\dot{x}}.

The required derivative is

\displaystyle  \boxed{\frac{\partial S_{cl}}{\partial t_{f}}=-H\left(t_{f}\right)} \ \ \ \ \ (17)

Using a similar technique, we can work out {\partial S_{cl}/\partial x_{f}}. 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 {\Delta x} further. Since both paths take the same time, there is no extra contribution {L\Delta t}. In this case {\eta\left(t\right)>0}, since the new (blue) curve {x\left(t\right)} is above the old (red) one {x_{cl}\left(t\right)}. The derivation is the same as above up to 8, and the total variation in the action is now

\displaystyle  \delta S_{cl}=\left.\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right|_{t_{f}} \ \ \ \ \ (18)

At {t=t_{f}}, {\eta\left(t_{f}\right)=\Delta x}, so we get

\displaystyle   \delta S_{cl} \displaystyle  = \displaystyle  \left.\frac{\partial L}{\partial\dot{x}}\right|_{t_{f}}\Delta x\ \ \ \ \ (19)
\displaystyle  \frac{\partial S_{cl}}{\partial x_{f}} \displaystyle  = \displaystyle  \left.\frac{\partial L}{\partial\dot{x}}\right|_{t_{f}}=p\left(t_{f}\right) \ \ \ \ \ (20)

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

\displaystyle   x\left(t\right) \displaystyle  = \displaystyle  A\cos\omega t+B\sin\omega t\ \ \ \ \ (21)
\displaystyle  \dot{x}\left(t\right) \displaystyle  = \displaystyle  -A\omega\sin\omega t+B\omega\cos\omega t \ \ \ \ \ (22)

The total energy is given by

\displaystyle   E \displaystyle  = \displaystyle  \frac{1}{2}m\dot{x}^{2}+\frac{1}{2}m\omega^{2}x^{2}\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  \frac{m}{2}\left(\left(-A\omega\sin\omega t+B\omega\cos\omega t\right)^{2}+\omega^{2}\left(A\cos\omega t+B\sin\omega t\right)^{2}\right)\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2}\left(A^{2}+B^{2}\right) \ \ \ \ \ (25)

where we just multiplied out the second line, cancelled terms and used {\cos^{2}x+\sin^{2}x=1}.

To get the action, we need the Lagrangian:

\displaystyle   L \displaystyle  = \displaystyle  T-V\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}m\omega^{2}x^{2}\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{m}{2}\left(\left(-A\omega\sin\omega t+B\omega\cos\omega t\right)^{2}-\omega^{2}\left(A\cos\omega t+B\sin\omega t\right)^{2}\right)\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2}\left[A^{2}\left(\sin^{2}\omega t-\cos^{2}\omega t\right)+B^{2}\left(\cos^{2}\omega t-\sin^{2}\omega t\right)-4AB\sin\omega t\cos\omega t\right]\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2}\left(\left(B^{2}-A^{2}\right)\cos2\omega t-2AB\sin2\omega t\right) \ \ \ \ \ (30)

The action for a trajectory from {t=0} to {t=T} is then

\displaystyle   S \displaystyle  = \displaystyle  \int_{0}^{T}Ldt\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{4}\left[\left(B^{2}-A^{2}\right)\sin2\omega t+2AB\cos2\omega t\right]_{0}^{T}\ \ \ \ \ (32)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{4}\left[\left(B^{2}-A^{2}\right)\sin2\omega T+2AB\left(\cos2\omega T-1\right)\right]\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{2}\left[\left(B^{2}-A^{2}\right)\sin\omega T\cos\omega T+AB\left(\cos^{2}\omega T-\sin^{2}\omega T-1\right)\right]\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{2}\left[\left(B^{2}-A^{2}\right)\sin\omega T\cos\omega T-2AB\sin^{2}\omega T\right] \ \ \ \ \ (35)

To proceed further, we need to specify {A} and {B}, since these depend on the boundary conditions (that is, on where we require the mass to be at {t=0} and {t=T}). If we require {x\left(0\right)=x_{1}} and {x\left(T\right)=x_{2}}, then

\displaystyle   A \displaystyle  = \displaystyle  x_{1}\ \ \ \ \ (36)
\displaystyle  x_{1}\cos\omega T+B\sin\omega T \displaystyle  = \displaystyle  x_{2}\ \ \ \ \ (37)
\displaystyle  B \displaystyle  = \displaystyle  \frac{x_{2}-x_{1}\cos\omega T}{\sin\omega T} \ \ \ \ \ (38)

Plugging these into 25 gives the energy as

\displaystyle   E \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2}\left(x_{1}^{2}+\left(\frac{x_{2}-x_{1}\cos\omega T}{\sin\omega T}\right)^{2}\right)\ \ \ \ \ (39)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2\sin^{2}\omega T}\left(x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}\cos\omega T\right) \ \ \ \ \ (40)

Plugging {A} and {B} into 35, we get (using {c\equiv\cos\omega T} and {s\equiv\sin\omega T}, so that {s^{2}+c^{2}=1}):

\displaystyle   S \displaystyle  = \displaystyle  \frac{m\omega}{2s}\left[\left(x_{2}-x_{1}c\right)^{2}c-x_{1}s^{2}c-2x_{1}s^{2}\left(x_{2}-x_{1}c\right)\right]\ \ \ \ \ (41)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{2s}\left[\left(x_{2}^{2}-2x_{1}x_{2}c+x_{1}^{2}c^{2}\right)c-x_{1}^{2}s^{2}c-2x_{1}x_{2}s^{2}+2x_{1}s^{2}c\right]\ \ \ \ \ (42)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{2s}\left[\left(x_{1}^{2}+x_{2}^{2}\right)c-2x_{1}x_{2}\right]\ \ \ \ \ (43)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega}{2\sin\omega T}\left[\left(x_{1}^{2}+x_{2}^{2}\right)\cos\omega T-2x_{1}x_{2}\right] \ \ \ \ \ (44)

Taking the derivative, we get

\displaystyle   \frac{\partial S}{\partial T} \displaystyle  = \displaystyle  \frac{m\omega}{2s^{2}}\left[-\omega\left(x_{1}^{2}+x_{2}^{2}\right)s^{2}-\left(\left(x_{1}^{2}+x_{2}^{2}\right)c-2x_{1}x_{2}\right)\omega c\right]\ \ \ \ \ (45)
\displaystyle  \displaystyle  = \displaystyle  \frac{m\omega^{2}}{2s^{2}}\left[-\left(x_{1}^{2}+x_{2}^{2}\right)+2x_{1}x_{2}c\right]\ \ \ \ \ (46)
\displaystyle  \displaystyle  = \displaystyle  -\frac{m\omega^{2}}{2\sin^{2}\omega T}\left(x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}\cos\omega T\right)\ \ \ \ \ (47)
\displaystyle  \displaystyle  = \displaystyle  -E \ \ \ \ \ (48)

Thus the result is verified for the harmonic oscillator.

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