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\begin{document}
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\heading{E0070: Ergodic density}
\auname{Alexander Elikashvili}
{\bf The problem:}
\Dn
Assuming a microcanonical ensemble with energy ${E}$ find an expression for probability density ${\rho\left(x\right)}$ of a particle which is
confined by a potential ${V\left(x\right)}$. Distinguish between ${1,2}$ and ${3}$ dimensional cases.
In particular show that in 2D case the probability density forms a step function. Contrast your
results with the canonical expression ${\rho\left(x\right)\propto \exp\left(-\beta V\left(x\right)\right)}$.
\Dn\Dn
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{\bf The solution:}
\[ \rho(x)=\frac{1}{(2\pi)^{D}}\int\rho(x,p)\delta(H(x,p)-E)d^{D}p \]
where D stands for dimension number and ${\rho(x,p)=\frac{1}{g_{D}(E)}}$,\; ${g_{D}(E)=\frac{1}{(2\pi)^{D}}\iint\delta(H(x,p)-E)d^{D}xd^{D}p}$.
Following mathematical relations will be helpful for solving the problem:
\[
(1)\;\;\;\delta(f(x))= \sum_{i}\frac{\delta(x-x_{i})}{|f'(x_{i})|},\;\;\;\;f(x_{i})=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\;\;\;\Theta(x)=\left\{\amatrix{0,\;\; x<0 \cr 1,\;\; x\geq 0}
\]
\Dn
{\bf 1D case:}
\[ \rho(x)=\frac{1}{2\pi g_{1}(E)}\int{\delta\left(\frac{p^{2}}{2m}+V(x)-E\right)dp} \]
One can distinguish between two possible solutions:
\Dn
${(\ast)}$ First case is when ${V(x)-E>0}$:
\[
\rho(x)=\frac{1}{2\pi g_{1}(E)}\int{\delta\left(\frac{p^{2}}{2m}+V(x)-E\right)dp}=0
\]
${(\ast\ast)}$ Second case is when ${V(x)-E<0}$, so with the help of relation (1) one gets:
\[
\rho(x)=\frac{1}{2\pi g_{1}(E)}\int_{-\infty}^{\infty}{\left(\frac{\delta\left(p-\sqrt{2m\left(E-V\left(x\right)\right)}\right)}{|\frac{p}{m}|}+\frac{\delta\left(p+\sqrt{2m\left(E-V\left(x\right)\right)}\right)}{|\frac{p}{m}|}\right)dp}=\frac{1}{2\pi g_{1}(E)}\sqrt{\frac{2m}{E-V\left(x\right)}}
\]
Using relation (2) one can combine solution ${(\ast)}$ with ${(\ast\ast)}$ and write:
\[
\rho(x)=\frac{1}{2\pi g_{1}\left(E\right)}\sqrt{\frac{2m}{E-V\left(x\right)}}\Theta\left(E-V\left(x\right)\right) \;\;\;\; g_{1}\left(E\right)=\frac{1}{2\pi}\int\sqrt{\frac{2m}{E-V\left(x\right)}}\Theta\left(E-V\left(x\right)\right)dx
\]
\Dn
\Dn
{\bf 2D case:}
\[ \rho(\vec{r})=\frac{1}{(2\pi)^2 g_{2}(E)}\int{\delta\left(\frac{\vec p^{2}}{2m}+V(\vec{r})-E\right)dp_xdp_y} \]
Just like in 1D case one has two possible solutions:
\Dn
${(\ast)}$ First one is when ${V(\vec{r})-E>0}$.
\[
\rho(\vec{r})=\frac{1}{(2\pi)^2g_{2}(E)}\iint\delta\left(\frac{\vec{p}^2}{2m}+V(\vec{r})-E\right)dp_xdp_y=0
\]
${(\ast\ast)}$ Second case is when ${V(\vec{r})-E<0}$, so with the help of relation (1) one gets:
\[
\rho(\vec{r})=\frac{1}{2\pi g_{2}(E)}\int_{0}^{\infty}{\left(\frac{\delta\left(p-\sqrt{2m\left(E-V\left(\vec{r}\right)\right)}\right)}{|\frac{p}{m}|}+\frac{\delta\left(p+\sqrt{2m\left(E-V\left(\vec{r}\right)\right)}\right)}{|\frac{p}{m}|}\right)pdp}=\frac{m}{2\pi g_{2}(E)}
\]
Using relation (2) one can combine solution ${(\ast)}$ with ${(\ast\ast)}$ and write:
\[
\rho(\vec{r})=\frac{m}{2\pi g_{2}(E)}\Theta \left(E-V\left(\vec{r} \right) \right) \;\;\;\;\; g_{2}(E)=\frac{m}{2\pi}\iint\Theta(E-V(\vec{r}))dxdy
\]
\Dn
\Dn
{\bf 3D case:}
\[
\rho(\vec{r})=\frac{1}{(2\pi)^3 g_{3}(E)}\int{\delta\left(\frac{\vec{p}^{2}}{2m}+V(\vec{r})-E\right)dp_xdp_ydp_z}
\]
In complete analogy with 1D and 2D cases one has two possible solutions:
\Dn
${(\ast)}$ First one is when ${V(\vec{r})-E>0}$.
\[
\rho(\vec{r})=\frac{1}{(2\pi)^3g_{3}(E)}\iiint\delta\left(\frac{\vec{p}^2}{2m}+V(\vec{r})-E\right)dp_xdp_ydp_z=0
\]
${(\ast\ast)}$ Second case is when ${V(\vec{r})-E<0}$, so with the help of relation (1) one gets:
\[
\rho(\vec{r})=\frac{1}{2\pi^2 g_{3}(E)}\int_{0}^{\infty}{\left(\frac{\delta\left(p-\sqrt{2m\left(E-V\left(\vec{r}\right)\right)}\right)}{|\frac{p}{m}|}+\frac{\delta\left(p+\sqrt{2m\left(E-V\left(\vec{r}\right)\right)}\right)}{|\frac{p}{m}|}\right)p^2dp}=\frac{m\sqrt{2m\left(E-V(\vec{r})\right)}}{2\pi^2 g_{3}(E)}
\]
Using relation (2) one can combine solution ${(\ast)}$ with ${(\ast\ast)}$ and write:
\[
\rho(\vec{r})=\frac{m\sqrt{2m\left(E-V(\vec{r})\right)}}{2\pi^2 g_{3}(E)}\Theta \left(E-V\left(\vec{r} \right) \right) \;\;\;\;\; g_{3}(E)=\iiint\frac{m\sqrt{2m\left(E-V(\vec{r})\right)}}{2\pi^2}\Theta \left(E-V\left(\vec{r} \right) \right)dxdydz
\]
\Dn
\Dn
In comparision with canonical ensemble microcanonical probability density has a wide range distribution, while canonical expression ${\rho}$ is concentrated around the minimum of the confining potential ${V}$.
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\end{document}