Tag Archives: action

Path integrals for special potentials; use of classical action

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercises 8.6.1 – 8.6.2.

We’ve seen that if we use the path integral formulation for a free particle, we get the exact propagator by considering only one path (the classical path) between the starting point {\left(x^{\prime},t^{\prime}\right)} and the end point {\left(x,t\right)}. In this case, the propagator has the form

\displaystyle  U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (1)

where {S_{cl}} is the classical action. It turns out that this form is true for a wider set of potentials, beyond just the free particle. The general form of the potential for which this is true is

\displaystyle  V=a+bx+cx^{2}+d\dot{x}+ex\dot{x} \ \ \ \ \ (2)

where {a,b,c,d} and {e} are constants. The general expression for the propagator is (where we’re taking the starting time to be {t^{\prime}=0}):

\displaystyle  U\left(x,t;x^{\prime}\right)=\int_{x^{\prime}}^{x}e^{iS\left[x\left(t^{\prime\prime}\right)\right]/\hbar}\mathfrak{D}\left[x\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (3)

where the notation {\mathfrak{D}\left[x\left(t^{\prime\prime}\right)\right]} means an integration over all possible paths from {x^{\prime}} to {x} in the given time interval.

For a given path, we can write the location of the particle {x\left(t^{\prime\prime}\right)} as composed of its position on the classical path {x_{cl}\left(t^{\prime\prime}\right)} plus the deviation {y\left(t^{\prime\prime}\right)} from the classical path:

\displaystyle  x\left(t^{\prime\prime}\right)=x_{cl}\left(t^{\prime\prime}\right)+y\left(t^{\prime\prime}\right) \ \ \ \ \ (4)

As the endpoints are fixed

\displaystyle  y\left(0\right)=y\left(t\right)=0 \ \ \ \ \ (5)

Also, since for any given potential and choice of endpoints, {x_{cl}\left(t^{\prime\prime}\right)} is fixed for all times, it is effectively a constant with regard to the path integration. Therefore

\displaystyle  dx=dy \ \ \ \ \ (6)

Making these substitutions into 3, we get, using Shankar’s slightly misleading notation:

\displaystyle  U\left(x,t;x^{\prime}\right)=\int_{0}^{0}e^{iS\left[x_{cl}\left(t^{\prime\prime}\right)+y\left(t^{\prime\prime}\right)\right]/\hbar}\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (7)

Usually, when the limits on an integral are the same, the integral evaluates to zero. However, in this case, the notation {\int_{0}^{0}\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right]} means that {y} starts and ends at zero, but covers all possible paths between these endpoints.

The action is the integral of the Lagrangian which, for the potential 2 is

\displaystyle   L \displaystyle  = \displaystyle  T-V\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\dot{x}^{2}-a-bx-cx^{2}-d\dot{x}-ex\dot{x} \ \ \ \ \ (9)

Because {L} is quadratic in both {x} and {\dot{x}}, we can expand it in a Taylor series up to second order without any approximation. That is

\displaystyle   L\left(x_{cl}+y,\dot{x}_{cl}+\dot{y}\right) \displaystyle  = \displaystyle  L\left(x_{cl},\dot{x}_{cl}\right)+\left.\frac{\partial L}{\partial x}\right|_{x_{cl}}y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}+\ \ \ \ \ (10)
\displaystyle  \displaystyle  \displaystyle  \frac{1}{2}\left(\left.\frac{\partial^{2}L}{\partial x^{2}}\right|_{x_{cl}}y^{2}+2\left.\frac{\partial^{2}L}{\partial x\partial\dot{x}}\right|_{x_{cl}}y\dot{y}+\left.\frac{\partial^{2}L}{\partial\dot{x}^{2}}\right|_{x_{cl}}\dot{y}^{2}\right) \ \ \ \ \ (11)

Look first at the last two terms on the RHS of the first line. Using the equations of motion, we have

\displaystyle  \left.\frac{\partial L}{\partial x}\right|_{x_{cl}}=\frac{d}{dt}\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right) \ \ \ \ \ (12)

To get the action, we need to integrate the Lagrangian over the time interval of interest. Integrating these two terms gives

\displaystyle   \int_{0}^{t}\left[\left.\frac{\partial L}{\partial x}\right|_{x_{cl}}y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}\right]dt^{\prime\prime} \displaystyle  = \displaystyle  \int_{0}^{t}\left[\frac{d}{dt}\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right)y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}\right]dt^{\prime\prime}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \left.\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right)y\right|_{0}^{t}-\int_{0}^{t}\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}dt^{\prime\prime}+\int_{0}^{t}\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}dt^{\prime\prime}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (15)

where we integrated the first term by parts. The integrated term in the second line is zero because {y=0} at both endpoints, and the last two terms cancel each other.

Returning to 11, we can calculate the three second derivatives explicitly:

\displaystyle   \frac{1}{2}\frac{\partial^{2}L}{\partial x^{2}} \displaystyle  = \displaystyle  -c\ \ \ \ \ (16)
\displaystyle  \left.\frac{\partial^{2}L}{\partial x\partial\dot{x}}\right|_{x_{cl}} \displaystyle  = \displaystyle  -e\ \ \ \ \ (17)
\displaystyle  \frac{1}{2}\frac{\partial^{2}L}{\partial\dot{x}^{2}} \displaystyle  = \displaystyle  m \ \ \ \ \ (18)

The integral of the first term on the RHS of 10 is just the classical action, so we get for the propagator 7:

\displaystyle  U\left(x,t;x^{\prime}\right)=e^{iS_{cl}/\hbar}\int_{0}^{0}\exp\left[\frac{i}{\hbar}\int_{0}^{t}\left(\frac{m\dot{y}^{2}}{2}-cy^{2}-ey\dot{y}\right)dt^{\prime\prime}\right]\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (19)

The remaining path integral can still be difficult to evaluate, but we can observe a few properties that it has. First, for any given path in the path integral, we must be able to express both {y} and {\dot{y}} as functions of time {t^{\prime\prime}}, so the complete path integral can depend only on the end time {t} (and, of course, on the constants {m}, {c} and {e}). That is, the propagator will always have the form 1:

\displaystyle  U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (20)

We have already evaluated the integral for the free particle where {c=e=0} and we found there that

\displaystyle  U\left(x,t;x^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar it}}e^{iS_{cl}/\hbar} \ \ \ \ \ (21)

Since the constant {b} doesn’t appear in 19, the propagator must have the same form for the more general case where {V=a+bx}. For more complex potentials, such as the harmonic oscillator, the function {A\left(t\right)} will in general have a different form and will have to be calculated explicitly in these cases.

As an example, we’ll consider the case of a particle subject to a constant force in the {x} direction, so that the potential is given by

\displaystyle  V\left(x\right)=-fx \ \ \ \ \ (22)

This gives a constant force of

\displaystyle  F=-\frac{dV}{dx}=f \ \ \ \ \ (23)

and thus a constant acceleration of {f/m}. For such a particle, its classical position is (from first year physics)

\displaystyle   x_{cl}\left(t^{\prime\prime}\right) \displaystyle  = \displaystyle  x_{0}+v_{0}t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\ \ \ \ \ (24)
\displaystyle  \dot{x}_{cl}\left(t^{\prime\prime}\right) \displaystyle  = \displaystyle  v_{0}+\frac{f}{m}t^{\prime\prime} \ \ \ \ \ (25)

To find {x_{0}} and {v_{0}}, we impose boundary conditions. At {t^{\prime\prime}=0}

\displaystyle  x_{cl}\left(0\right)=x_{0}=x^{\prime} \ \ \ \ \ (26)

At {t^{\prime\prime}=t}, its position is

\displaystyle   x_{cl}\left(t\right) \displaystyle  = \displaystyle  x=x^{\prime}+v_{0}t+\frac{f}{2m}t^{2} \ \ \ \ \ (27)

This gives

\displaystyle  v_{0}=\frac{x-x^{\prime}}{t}-\frac{f}{2m}t \ \ \ \ \ (28)

The classical Lagrangian is

\displaystyle   L \displaystyle  = \displaystyle  T-V\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\dot{x}_{cl}^{2}+fx_{cl}\ \ \ \ \ (30)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\left(v_{0}+\frac{f}{m}t^{\prime\prime}\right)^{2}+f\left(x_{0}+v_{0}t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\right)\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}m\left(\frac{x-x^{\prime}}{t}-\frac{f}{2m}t+\frac{f}{m}t^{\prime\prime}\right)^{2}+f\left(x^{\prime}+\left(\frac{x-x^{\prime}}{t}-\frac{f}{2m}t\right)t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\right) \ \ \ \ \ (32)

Note that {t} is a constant, as it is the time of the endpoint of the motion. To find the classical action, we must integrate this from {t^{\prime\prime}=0} to {t}. The integral is a straightforward integral of a quadratic in {t^{\prime\prime}}, although the algebra is tedious if done by hand, so is best done with Maple.

\displaystyle   S_{cl} \displaystyle  = \displaystyle  \int_{0}^{t}L\;dt^{\prime\prime}\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{3}{\frac{{f}^{2}{t}^{3}}{m}}+\left({\frac{x-x^{\prime}}{t}}-\frac{1}{2}{\frac{ft}{m}}\right)f{t}^{2}+\frac{1}{2}m\left({\frac{x-x^{\prime}}{t}}-\frac{1}{2}{\frac{ft}{m}}\right)^{2}t+fxt\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  -\frac{f^{2}t^{3}}{24m}+\frac{1}{2}ft\left(x+x^{\prime}\right)+\frac{m\left(x-x^{\prime}\right)^{2}}{2t} \ \ \ \ \ (35)

From 21, this gives a propagator of

\displaystyle  U\left(x,t;x^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar it}}\exp\left[\frac{i}{\hbar}\left(-\frac{f^{2}t^{3}}{24m}+\frac{1}{2}ft\left(x+x^{\prime}\right)+\frac{m\left(x-x^{\prime}\right)^{2}}{2t}\right)\right] \ \ \ \ \ (36)

This agrees with Shankar’s result in his equation 5.4.31.

As another example, consider the harmonic oscillator, where the potential is

\displaystyle  V=\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (37)

This potential is also of the form 2, so the propagator must have the form 20. This time, however, since {c\ne0}, the function {A\left(t\right)} will probably not have the form used in 21. The best we can say therefore is that

\displaystyle  U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (38)

where {A\left(t\right)} has the form (from 19):

\displaystyle  A\left(t\right)=\int_{0}^{0}\exp\left[\frac{i}{\hbar}\int_{0}^{t}\left(\frac{m\dot{y}^{2}}{2}-\frac{1}{2}m\omega^{2}y^{2}\right)dt^{\prime\prime}\right]\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (39)

We worked out the classical action for the harmonic oscillator earlier and found

\displaystyle  S_{cl}=\frac{m\omega}{2\sin\omega t}\left[\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right] \ \ \ \ \ (40)

where the particle is at {x^{\prime}} at {t^{\prime\prime}=0} and at {x} at {t^{\prime\prime}=t}. The propagator is therefore

\displaystyle  U\left(x,t;x^{\prime}\right)=A\left(t\right)\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right)\right] \ \ \ \ \ (41)

with {A\left(t\right)} given by 39.

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.