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

$$T=t_f-t_i>0,\qquad m>0.$$

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

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

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 limit $T\mapsto T-i0^+$ taken from the lower half-plane. For $T>0$, this fixes $i^{-1/2}=e^{-i\pi/4}$ in the final kernel; the Gaussian integral carries the compensating Fresnel phase.

Resulting Propagator

Putting the classical and fluctuation factors together,

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

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:

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

Why the Gaussian answer is exact

The action contains no terms beyond quadratic order in the fluctuations. Integrating them therefore leaves exactly the determinant prefactor times $e^{iS_{\rm cl}/\hbar}$. For a nonquadratic potential, cubic and higher terms survive, and retaining only the determinant is generally a semiclassical approximation. Here independence of $N$ is also expected from exact composition of free kernels.

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.

References

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

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.