# 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.''