# 11.1.2: Doing the Path Integral - the Free Particle

• • Mark Tuckerman
• New York University
$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

The density matrix for the free particle

$H={P^2 \over 2m} \nonumber$

will be calculated by doing the discrete path integral explicitly and taking the limit $$P \rightarrow \infty$$ at the end.

The density matrix expression is

$\rho(x,x';\beta) = \lim_{P\rightarrow\infty }\left({mP \over 2\pi \beta \hbar^2} \right )^{P/2} \int dx_2 \cdots dx_P exp \left [ - {mP \over 2 \beta \hbar^2} \sum_{i=1}^P(x_{i+1}-x_i)^2\right] \vert _{x_1=x,x_{P+1}=x'} \nonumber$

Let us make a change of variables to

${u_1} = x_1 \nonumber$

${u_k} = x_k - \tilde{x}_k$

${\tilde {x}_k} = {(k-1)x_{k+1}+x_1 \over k} \nonumber$

The inverse of this transformation can be worked out explicitly, giving

${x_1} = u_1 \nonumber$

${x_k} = \sum_{l=1}^{P+1}{k-1 \over l-1}u_l + {P-k+1 \over P}u_1 \nonumber$

The Jacobian of the transformation is simply

$J = {\rm det}\left(\matrix{1 & -1/2 & 0 & 0 & \cdots \cr0 & 1 & -2/3 & 0 & \cdots \cr 0 & 0 & 1 & -3/4 & \cdots \cr 0 & 0 & 0 & 1 & \cdots \cr\cdot & \cdot & \cdot & \cdot & \cdot & \cdots}\right)=1 \nonumber$

Let us see what the effect of this transformation is for the case $$P = 3$$. For $$P = 3$$, one must evaluate

$(x_1-x_2)^2 + (x_2-x_3)^2 + (x_3-x_4)^2 = (x-x_2)^2 + (x_2-x_3)^2 + (x_3-x')^2 \nonumber$

According to the inverse formula,

${x_1} = {u_1}$

${x_2} = u_2 + {1 \over 2}u_3 + {1 \over 3}x' + {2 \over 3}x$

$x_3 = u_3 + {2 \over 3}x' + {1 \over 3}x$

Thus, the sum of squares becomes

${(x-x_2)^2 + (x_2-x_3)^2 + (x_3-x')^2 } = ( 2u_2^2 + {3 \over 2}u_3^2 + {1 \over 3}(x-x')^2 = {2 \over 2-1}u_2^2 + {3 \over 3-1}u_3^2 + {1 \over 3}(x-x')^2$

From this simple example, the general formula can be deduced:

$\sum_{i=1}^P(x_{i+1}-x_i)^2 = \sum_{k=2}^P {k \over k-1}u_k^2 +{1 \over P}(x-x')^2 \nonumber$

Thus, substituting this transformation into the integral gives

$\rho(x,x';\beta) = \left({m \over 2\pi\beta\hbar^2}\right)^{1/2} \prod _{k=2}^P \left ( {m_k P \over 2\pi \beta \hbar^2 } \right )^{1/2} \int du_2 \cdots du_P exp \left [ - \sum _{k=2}^P {m_kP \over 2\beta \hbar^2} u_k^2 \right] exp \left[-{m \over 2\beta\hbar^2}(x-x')^2\right] \nonumber$

where

$m_k = {k \over k-1}m \nonumber$

and the overall prefactor has been written as

$\left({mP \over 2\pi\beta\hbar^2}\right)^{P/2} =\left({m \over 2 \pi \beta \hbar^2} \right )^{1/2} \prod _{k=2}^P\left({m_k P \over 2\pi\beta\hbar^2}\right)^{1/2} \nonumber$

Now each of the integrals over the $$u$$ variables can be integrated over independently, yielding the final result

$\rho(x,x';\beta) = \left({m \over 2\pi\beta\hbar^2}\right)^{1/2}\exp\left[-{m \over 2\beta\hbar^2}(x-x')^2\right] \nonumber$

In order to make connection with classical statistical mechanics, we note that the prefactor is just $${1 \over \lambda}$$, where $$\lambda$$

$\lambda = \left({2\pi\beta\hbar^2 \over m}\right)^{1/2} =\left({\beta h^2 \over 2\pi m}\right)^{1/2} \nonumber$ is the kinetic prefactor that showed up also in the classical free particle case. In terms of $$\lambda$$, the free particle density matrix can be written as

$\rho(x,x';\beta) = {1 \over \lambda}e^{-\pi(x-x')^2/\lambda^2} \nonumber$

Thus, we see that $$\lambda$$ represents the spatial width of a free particle at finite temperature, and is called the "thermal de Broglie wavelength.''

This page titled 11.1.2: Doing the Path Integral - the Free Particle is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.