# Doing the path integral: the free particle

The density matrix for the free particle

\[H={P^2 \over 2m}\]

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

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

Let us make a change of variables to

\[\underline {u_1}\] | \(\underline {x_1}\) | ||

\[\underline {u_k}\] | \(\underline {x_k - \tilde{x}_k }\) | ||

\[\underline {\tilde {x}_k}\] | \(\underline { {(k-1)x_{k+1}+x_1 \over k} } \) |

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

\[\underline {x_1}\] | \(\underline {u_1} \) | ||

\[ \underline {x_k} \] | \( \sum_{l=1}^{P+1}{k-1 \over l-1}u_l + {P-k+1 \over P}u_1 \) |

The Jacobian of the transformation is simply

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

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

\[ (x_1-x_2)^2 + (x_2-x_3)^2 + (x_3-x_4)^2 = (x-x_2)^2 + (x_2-x_3)^2 + (x_3-x')^2\]

According to the inverse formula,

\[\underline {x_1}\] | \(\underline {u_1}\) | ||

\[ \underline {x_2}\] | \(u_2 + {1 \over 2}u_3 + {1 \over 3}x' + {2 \over 3}x \) | ||

\[\underline {x_3} \] | \(u_3 + {2 \over 3}x' + {1 \over 3}x \) |

Thus, the sum of squares becomes

\[\underline {(x-x_2)^2 + (x_2-x_3)^2 + (x_3-x')^2 } \] | \( 2u_2^2 + {3 \over 2}u_3^2 + {1 \over 3}(x-x')^2 \) | ||

\( {2 \over 2-1}u_2^2 + {3 \over 3-1}u_3^2 + {1 \over 3}(x-x')^2 \) |

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

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

Thus, substituting this transformation into the integral gives

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

where

\[ m_k = {k \over k-1}m \]

and the overall prefactor has been written as

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

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

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

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

\[ \lambda = \left({2\pi\beta\hbar^2 \over m}\right)^{1/2} =\left({\beta h^2 \over 2\pi m}\right)^{1/2}\]

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

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

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