Consider a classical particle moving in a one-dimensional potential well $u\left(x\right)$, as shown. The particle is in thermal equilibrium with a reservoir at temperature $T$, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

$\overline{x}=\left(\frac{\int x{e}^{-\beta u\left(x\right)}dx}{\int e^{-\beta u\left(x\right)}dx}\right)$

where each integral is over the entire $x$axis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(z) in a Taylor series about the equilibrium point

$$\begin{gathered} u\left(x\right)=u\left(x_0\right)+\left(x-x_0\right){\overline{)\frac{du}{dx}}}_{x_0}+\frac{1}{2}{\left(x-x_0\right)}^2{\overline{)\frac{d^{2}u}{d{x}^2}}}_{x_0}+\frac{1}{3!}{\left(x-x_0\right)}^3{\overline{)\frac{d^{3}u}{d{x}^3}}}_{x_0} \\ +........ \end{gathered}$$

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction $\overline{x}=x_0$.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for $\overline{x}$become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit $\overline{x}$ differs from by a term proportional to $kT$. Express the coefficient of this term in terms of the coefficients of the Taylor series $u\left(x\right)$

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

$u\left(x\right)=u_0\left[{\left(\frac{x_0}{x}\right)}^{12}-2{\left(\frac{x_0}{x}\right)}^6\right]$

Sketch this function, and show that the minimum of the potential well is at $x=x_0$, with depth $u_0$. For argon, $x_0=3.9\stackrel{\circ }{A}$and $u_0=0.010eV$. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of $u_0$. Evaluate the result numerically for argon, and compare to the measured value $\alpha =0.0007K^{-1}$

We have given that a classical particle moving in a one-dimensional potential well, as shown. The particle is in thermal equilibrium with a reservoir at temperature $T$, so the probabilities of its various states are determined by Boltzmann statistics

The average position of equal summation over $x$for the positions weighted by their probabilities,

$\overline{x}=\sum _{x}xP\left(x\right)$

The particle is inside one dimensional well, the probability:

$P\left(x\right)=\frac{e^{-\beta u\left(x\right)}}{Z}$

By substituting it in above equation

$$\begin{gathered} \overline{x}=\sum _{x}x\frac{e^{-\beta u\left(x\right)}}{Z} \\ Z=\sum _x{e}^{-\beta u\left(x\right)} \\ \overline{x}=\frac{\sum _{x}x{e}^{-\beta u\left(x\right)}}{\sum _x{e}^{-\beta u\left(x\right)}} \end{gathered}$$

Multiply and Divide RHS by $∆x$

$\overline{x}=\frac{{\displaystyle \sum _x}x{e}^{-\beta u\left(x\right)}∆x}{\sum _x{e}^{-\beta u\left(x\right)}∆x}$

As $\underset{∆x\to 0}{\lim }$the summation becomes integral from $-\infty to\infty$

$\overline{x}=\frac{\int x{e}^{-\beta u\left(x\right)}dx}{\int e^{-\beta u\left(x\right)}dx}$

Hence proved.

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

From figure we can say that the point $x_0$is local minimum,means slope is zero.

${\overline{)\frac{du}{dx}}}_{x_0}=0$

Therefore, Taylor series at point is

$$\begin{gathered} u\left(x\right)=u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3 \\ a=\frac{1}{2}{\overline{)\frac{d^{2}u}{d{x}^2}}}_{x_0}b=\frac{1}{6}{\overline{)\frac{d^{3}u}{d{x}^3}}}_{x_0} \end{gathered}$$

approximation,

$u\left(x\right)\approx u\left(x_0\right)+a{\left(x-x_0\right)}^2$

From part a

$$\begin{gathered} \overline{x}=\frac{\int x{e}^{-\beta \left|u\left(x_0\right)+a{\left(x-x_0\right)}^2\right|dx}}{\int e^{-\beta \left|u\left(x_0\right)+a{\left(x-x_0\right)}^2\right|dx}} \\ \overline{x}=\frac{\int x{e}^{-\beta \left|a{\left(x-x_0\right)}^2\right|dx}}{\int e^{-\beta \left|a{\left(x-x_0\right)}^2\right|dx}} \\ letx-x_0=y \\ dx=dy \\ \overline{x}=\frac{\int \left(y+x_0\right)e^{-\beta a{y}^2}dy}{\int e^{-\beta a{y}^2}dx} \\ \overline{x}=\frac{\int \left(y\right)e^{-\beta a{y}^2}dy+\int \left(x_0\right)e^{-\beta a{y}^2}dy}{\int e^{-\beta a{y}^2}dx} \\ \int \left(y\right)e^{-\beta a{y}^2}dy=0functionisodd \\ \overline{x}=\frac{\int \left(x_0\right)e^{-\beta a{y}^2}dy}{\int e^{-\beta a{y}^2}dx} \\ \overline{x}=x_0 \end{gathered}$$

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

If we keep cubic term in Taylor

$$\begin{gathered} \overline{x}=\frac{\int x{e}^{-\beta |u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3|dx}}{\int e^{-\beta |u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3|dx}} \\ letx-x_0=y \\ dx=dy \\ \overline{x}=\frac{\int \left(y+x_0\right)e^{-\beta |u\left(x_0\right)+a{y}^2+b{y}^3|dy}}{\int e^{-\beta |u\left(x_0\right)+a{y}^2+b{y}^3|dy}} \\ {e}^{-\beta y^3}=1-\beta b{y}^3 \\ \overline{x}=\frac{\int \left(x_0-\beta b{y}^4\right)e^{-\beta a{y}^2}dy}{\int e^{-\beta a{y}^2}dy} \\ {\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2}dy=\sqrt{\frac{\pi }{\beta a}} \\ {\int }_{-\infty }^{\infty }\beta b{y}^4{e}^{-\beta a{y}^2}dy=\frac{3b}{4\beta a^2}\sqrt{\frac{\pi }{\beta a}} \\ \overline{x}=x_0-\frac{3bkT}{4a^2} \end{gathered}$$

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

The Lennard Jones potential is given by$u\left(x\right)=u_0\left[{\left(\frac{x_0}{x}\right)}^{12}-2{\left(\frac{x_0}{x}\right)}^6\right]$

argon gas with

$u_0=0.010eV,x_0=3.9\times {10}^{-10}m$

Numerically,

$$\begin{gathered} \alpha =\frac{1}{\overline{x}}\frac{\partial \overline{x}}{\partial T}=\frac{1}{\overline{x}}\frac{\partial }{\partial T}\left(x_0-\frac{3bkT}{4a^2}\right) \\ \alpha =\frac{1}{\overline{x}}\left(-\frac{3bk}{4a^2}\right) \\ \alpha \approx -\frac{3bk}{4a^2{x}_0} \\ substitutewithaandb \\ \alpha =\frac{7k}{48u_0} \\ substitutewithgivenvaluesforargon \\ \alpha =\frac{7\left(8.62\times {10}^{-5}eV/K\right)}{48\left(0.010eV\right)} \\ =1.257\times {10}^{-3}{K}^{-1} \end{gathered}$$