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\heading{E060: Oscillator in a microcanonical state} 
\auname{Elkana Porat}


{\bf The problem:}

\Dn

Assume that a harmonic oscillator with freqwency $\Omega$ and mass $m$ is prepared in a microcanonical state with energy E. 

\Dn

(1) Write down the probability distribution ${\rho\left(x,p\right)}$.

\Dn

(2) Find the projected probability distribution ${\rho\left(x\right)}$

\Dn\Dn

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{\bf The solution:}

\Dn

(1) 
The Hamiltonian: $H\left(x,p\right) = \frac{1}{2m}p^2 + \frac{m\Omega^2}{2}x^2$.

For a microcanonical state with energy E we get:

$${\rho\left(x,p\right)} = A\delta\left(H\left(x,p\right) - E\right)$$

where $A$ is a normalization factor we will find later.


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\Dn\Dn

(2)
The projected probability distribution is given by:

\begin{eqnarray}
	\rho\left(x\right)& = & A\int^{\infty}_{-\infty}{\delta(\frac{1}{2m}p^2 + \frac{m\Omega^2}{2}x^2 - E)\frac{dp}{2\pi}} \nonumber\\
	& = & \frac{A}{\pi}\sqrt{\frac{m}{2}}\int^{\infty}_{0}{\frac{\delta(u + \frac{m\Omega^2}{2}x^2 - E)}{\sqrt{u}}du} \nonumber\\
\nonumber\\
	& = & \left\{\begin{array}{ccc}
		\frac{A}{\pi\Omega}\frac{1}{\sqrt{\frac{2E}{m\Omega^2} - x^2}}& ;& x^2 < \frac{2E}{m\Omega^2}\\
		0&	;& o.w.
	\end{array}\right. \nonumber
\end{eqnarray}

when in the first step we used the simmetricity of $p^2$ and the change of variables: $p^2=2mu$.

\Dn

Finaly we find $A$:
$$1 = \int{\rho\left(x\right)dx} = \frac{A}{\pi\Omega} \int^{1}_{-1} {\frac{dx}{\sqrt{1 - x^2}}} = \frac{A}{\pi\Omega} \cdot \pi$$

$$A = \Omega$$


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