Free-Particle Path Integral

The free-particle path integral is the first exact real-time path-integral calculation. It gives the same kernel derived by Fourier transform in Free-Particle Propagator:

$$K_0(x_f,t_f;x_i,t_i) = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} \exp\left[ \frac{im(x_f-x_i)^2}{2\hbar T} \right],$$

where

$$T=t_f-t_i.$$

The path-integral calculation shows why the exponent is the classical free-particle action and why the normalization is a fluctuation determinant.

Free-Particle Action

For a one-dimensional free particle,

$$L=\frac12m\dot x^2,$$

so the action is

$$S[x] = \int_{t_i}^{t_f}dt\,\frac12m\dot x^2.$$

The classical path connecting the endpoints is the straight line

$$x_{\rm cl}(t) = x_i+\frac{t-t_i}{T}(x_f-x_i).$$

Its action is

$$S_{\rm cl} = \frac{m(x_f-x_i)^2}{2T}.$$

The exact kernel will be a normalization factor times $e^{iS_{\rm cl}/\hbar}$.

Discretized Gaussian Integral

Slice time into $N$ intervals of size

$$\epsilon=\frac{T}{N},$$

and write $x_j=x(t_i+j\epsilon)$ with fixed endpoints

$$x_0=x_i, \qquad x_N=x_f.$$

The time-sliced free-particle expression is

$$\begin{aligned} K_N(x_f,T;x_i,0) &= \left( \frac{m}{2\pi i\hbar\epsilon} \right)^{N/2} \int dx_1\cdots dx_{N-1} \\ &\quad \times \exp\left[ \frac{im}{2\hbar\epsilon} \sum_{j=0}^{N-1} (x_{j+1}-x_j)^2 \right]. \end{aligned}$$

This is an $(N-1)$-dimensional oscillatory Gaussian integral.

Classical Path Plus Fluctuations

Decompose each sliced path as

$$x_j=x_j^{\rm cl}+\eta_j,$$

where

$$x_j^{\rm cl} = x_i+\frac{j}{N}(x_f-x_i), \qquad \eta_0=\eta_N=0.$$

Then

$$x_{j+1}-x_j = \frac{x_f-x_i}{N} + (\eta_{j+1}-\eta_j).$$

The discrete action separates:

$$\sum_{j=0}^{N-1}(x_{j+1}-x_j)^2 = \frac{(x_f-x_i)^2}{N} + \sum_{j=0}^{N-1}(\eta_{j+1}-\eta_j)^2.$$

The cross term vanishes because

$$\sum_{j=0}^{N-1}(\eta_{j+1}-\eta_j) = \eta_N-\eta_0 = 0.$$

Thus the endpoint dependence is entirely in the classical action, while the remaining integral is over endpoint-fixed fluctuations.

Fluctuation Determinant

The fluctuation part is a quadratic form in the $N-1$ variables $\eta_1,\ldots,\eta_{N-1}$:

$$\sum_{j=0}^{N-1}(\eta_{j+1}-\eta_j)^2 = \boldsymbol\eta^T A\boldsymbol\eta,$$

where

$$A= \begin{pmatrix} 2 & -1 & 0 & \cdots\\ -1 & 2 & -1 & \cdots\\ 0 & -1 & 2 & \cdots\\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$$

For this $(N-1)\times(N-1)$ Dirichlet discrete Laplacian,

$$\det A=N.$$

The Fresnel Gaussian integral gives

$$\int d^{N-1}\eta\, \exp\left[ \frac{im}{2\hbar\epsilon} \boldsymbol\eta^T A\boldsymbol\eta \right] = \left( \frac{2\pi i\hbar\epsilon}{m} \right)^{(N-1)/2} \frac{1}{\sqrt N},$$

with the usual real-time convergence prescription.

Resulting Propagator

Putting the classical and fluctuation factors together,

$$\begin{aligned} K_N &= \left( \frac{m}{2\pi i\hbar\epsilon} \right)^{N/2} \left( \frac{2\pi i\hbar\epsilon}{m} \right)^{(N-1)/2} \frac{1}{\sqrt N} \\ &\quad \times \exp\left[ \frac{im(x_f-x_i)^2}{2\hbar N\epsilon} \right]. \end{aligned}$$

Since $N\epsilon=T$,

$$K_N = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} \exp\left[ \frac{im(x_f-x_i)^2}{2\hbar T} \right].$$

The expression is independent of $N$ after the Gaussian integration. Therefore the continuum limit gives the same result:

$$K_0(x_f,t_f;x_i,t_i) = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} e^{iS_{\rm cl}/\hbar}.$$

Classical Action Form

For the free particle,

$$S_{\rm cl} = \frac{m(x_f-x_i)^2}{2T}.$$

The exact result

$$K_0 = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} e^{iS_{\rm cl}/\hbar}$$

is the simplest example of the semiclassical pattern:

• the phase is the classical action;
• the prefactor comes from fluctuations about the classical path;
• for quadratic actions, the semiclassical expression is exact.

For nonquadratic potentials, the same decomposition is generally an approximation or an organizing principle rather than an exact finite Gaussian integral.

Normalization

The factor

$$\left( \frac{m}{2\pi i\hbar T} \right)^{1/2}$$

has three jobs:

• it gives the kernel the correct dimension $1/\text{length}$;
• it makes the short-time limit a delta distribution;
• it makes the composition law work.

Dropping the normalization leaves the classical phase but destroys the propagator as an operator kernel.

Comparison with Momentum-Space Derivation

The momentum-space derivation begins from

$$K_0(x_f,T;x_i,0) = \int\frac{dp}{2\pi\hbar} \exp\left[ \frac{ip(x_f-x_i)}{\hbar} -\frac{ip^2T}{2m\hbar} \right].$$

That route is usually the shortest calculation. The path-integral route teaches a different lesson: the same answer can be seen as a sum over time-sliced histories, and for a quadratic action the infinite-dimensional oscillatory Gaussian can be evaluated exactly.

The two derivations agree because both compute the same matrix element of the same unitary operator.

Common Mistakes

• Dropping the determinant factor and keeping only $e^{iS_{\rm cl}/\hbar}$.
• Forgetting that the fluctuation endpoints are fixed: $\eta_0=\eta_N=0$.
• Treating the real-time Gaussian as an ordinary convergent integral without a prescription.
• Assuming the exact free-particle simplification carries over unchanged to nonquadratic potentials.
• Confusing the number of prefactors $N$ with the number of integrations $N-1$.
• Treating the path integral as a probability distribution over paths.

Cross-Links

• Time Slicing
• Free-Particle Propagator
• From Propagators to Path Integrals
• Why Path Integrals?
• Action and Phase
• Action Principles
• Gaussian Distributions

References

• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.
• J. Zinn-Justin, Path Integrals in Quantum Mechanics, Oxford University Press, 2005.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show that the cross term vanishes in the decomposition $x_j=x_j^{\rm cl}+\eta_j$.

Solution

The cross term is proportional to

$$\sum_{j=0}^{N-1} \frac{x_f-x_i}{N} (\eta_{j+1}-\eta_j).$$

The constant factor can be pulled out, leaving

$$\sum_{j=0}^{N-1}(\eta_{j+1}-\eta_j) = \eta_N-\eta_0.$$

Since the endpoints are fixed, $\eta_0=\eta_N=0$, so the cross term vanishes.

2. Verify $\det A=N$ for the Dirichlet discrete Laplacian of size $N-1$ using the recurrence $D_r=2D_{r-1}-D_{r-2}$.

Solution

Let $D_r$ be the determinant of the $r\times r$ tridiagonal matrix with $2$ on the diagonal and $-1$ on neighboring off-diagonals. Expanding along the last row gives

$$D_r=2D_{r-1}-D_{r-2}.$$

The initial values are

$$D_0=1, \qquad D_1=2.$$

The solution is

$$D_r=r+1.$$

For $r=N-1$, this gives $\det A=D_{N-1}=N$.

3. Work out the $N=2$ time-sliced free-particle integral explicitly and show that it gives the exact kernel for time $T=2\epsilon$.

Solution

For $N=2$, there is one intermediate position $x_1$:

$$K_2 = \left( \frac{m}{2\pi i\hbar\epsilon} \right) \int dx_1\, \exp\left[ \frac{im}{2\hbar\epsilon} \left( (x_1-x_i)^2+(x_f-x_1)^2 \right) \right].$$

Complete the square:

$$(x_1-x_i)^2+(x_f-x_1)^2 = 2\left( x_1-\frac{x_i+x_f}{2} \right)^2 + \frac{(x_f-x_i)^2}{2}.$$

The Fresnel integral gives

$$\int dx_1\, \exp\left[ \frac{im}{\hbar\epsilon} \left( x_1-\frac{x_i+x_f}{2} \right)^2 \right] = \left( \frac{\pi i\hbar\epsilon}{m} \right)^{1/2}.$$

Thus

$$K_2 = \left( \frac{m}{4\pi i\hbar\epsilon} \right)^{1/2} \exp\left[ \frac{im(x_f-x_i)^2}{4\hbar\epsilon} \right],$$

which is the free kernel with $T=2\epsilon$.

4. Why is the free-particle path integral exact while a generic potential path integral is not reduced to a single finite determinant?

Solution

The free action is quadratic in the path variables after time slicing. Therefore the finite-dimensional integrals are Gaussian and can be evaluated exactly. A generic potential produces nonquadratic terms in the sliced action, so the integrals are not Gaussian. One can use perturbation theory, stationary phase, numerical methods, or special exact techniques, but the simple determinant evaluation no longer applies.