91Ó°ÊÓ

In this problem, our task is to calculate how high the temperature must be so that the probability of a system of distinguishable harmonic oscillators to occupy the ground state is less than $p=\frac{1}{2}$.

Similar to the notation used in the book, we also assume that the potential energy is shifted by$\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{Ä\S }{\mathit{Ó\neg }}_{\mathbf{0}}$so that the allowed energies of a harmonic oscillator is given by:

role="math" localid="1659762682190" ${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\mathbf{n}\mathit{Ä\S }{\mathit{Ó\neg }}_{\mathbf{0}}\mathbf{}\mathbf{where}\mathbf{}\mathbf{n}\mathbf{â‰\yen }\mathbf{0}\mathbf{,}$

and its ground state energy is${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$.

The key idea here lies from the fact that we are now dealing with an enormous amount of distinguishable harmonic oscillators, that is to say, the probability at which an oscillator is in an energy level $n$ is now given by the Boltzmann probability distribution:

$\mathrm{P}\left({\mathrm{E}}_{\mathrm{n}}\right)={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{n}}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Where ${\mathrm{k}}_{\mathrm{B}}$Boltzmann is’s constant, and is temperature.

Before we could utilize eq. (2), we first have to find the value of the normalization constant A . To do so, we plug in eq. (1) into (2), set the whole expression to l, and evaluate the integral over an interval dn from 0 to $∞$. This translates to:

$1={∫}_0^{∞}{\mathrm{Ae}}^{-{\mathrm{nÄ\S Ó\neg }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dn}$

Evaluating the above integral, and solving for A we obtain:

role="math" localid="1659763564607" $$\begin{gathered} 1=\mathrm{A}{∫}_0^{∞}{\mathrm{e}}^{-{\mathrm{nÄ\S Ó\neg }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ =\mathrm{A}{\left[-\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{Ä\S Ó\neg }}_0}{\mathrm{e}}^{-{\mathrm{nÄ\S Ó\neg }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\right]}_0^{∞} \\ =\mathrm{A}\left[-\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{Ä\S Ó\neg }}_0}\left(0-1\right)\right] \\ =\mathrm{A}\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{Ä\S Ó\neg }}_0}→\mathrm{A}=\frac{{\mathrm{Ä\S Ó\neg }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{Knowing},\mathrm{eq}.\left(2\right)\mathrm{now}\mathrm{becomes}: \\ \mathrm{P}\left(\mathrm{n}\right)=\frac{{\mathrm{Ä\S Ó\neg }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{e}}^{-{\mathrm{nÄ\S Ó\neg }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{temperature}\mathrm{T}\mathrm{where}\mathrm{the}\mathrm{probability}\mathrm{in}\mathrm{the}\mathrm{ground}\mathrm{state}\mathrm{is}\mathrm{less}\mathrm{than}1/2, \\ \mathrm{we}\mathrm{set}\mathrm{n}=0\mathrm{in}\mathrm{eq}.\left(3\right)\mathrm{and}\mathrm{transform}\mathrm{the}\mathrm{expression}\mathrm{into}\mathrm{an}\mathrm{inequality}: \\ \frac{1}{2}>\frac{{\mathrm{Ä\S Ó\neg }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{e}}^{-{\mathrm{nÄ\S Ó\neg }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{Solving}\mathrm{for}\mathrm{T}\mathrm{we}\mathrm{finally}\mathrm{get}: \\ \frac{1}{2}>\frac{{\mathrm{Ä\S Ó\neg }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}→\frac{2{\mathrm{Ä\S Ó\neg }}_0}{{\mathrm{k}}_{\mathrm{B}}} \end{gathered}$$