# 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.