91影视

Like the notation used in the book, we also assume that the potential energy is shifted by$-\frac{1}{2}\mathrm{鈩�}{蝇}_0$so that the allowed energy of a harmonic oscillator is given by:

$E_n=n\mathrm{鈩�}{蝇}_0\text{where}n猢�0.............\left(1\right)$

And its ground state energy is$E_0=0$.

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

$P\left(E_n\right)=A{e}^{-E_n/k_{B}T}..........\left(2\right)$

where $k_B$is Boltzmann's constant, and T is temperature.

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

$1={鈭�}_0^{鈭�}A{e}^{-n\mathrm{鈩�}{蝇}_0/k_{B}T}dn$

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

$\begin{array}{rcl}1& =& A{鈭�}_0^{鈭�}{e}^{-n\mathrm{鈩�}{蝇}_0/k_{B}T}\\ & =& A{\left[-\frac{k_{B}T}{\mathrm{鈩�}{蝇}_0}{e}^{-n\mathrm{鈩�}{蝇}_0/k_{B}T}\right]}_0^{鈭�}\\ & =& A\left[-\frac{k_{B}T}{\mathrm{鈩�}{蝇}_0}(0-1)\right]\\ & =& A\frac{k_{B}T}{\mathrm{鈩�}{蝇}_0}\\ A& =& \frac{\mathrm{鈩�}{蝇}_0}{k_{B}T}\\ & & \end{array}$

Knowing A , eq. (2) now becomes:

$P\left(n\right)=\frac{\mathrm{鈩�}{蝇}_0}{k_{B}T}{e}^{-n\mathrm{鈩�}{蝇}_0/k_{B}T}...........\left(3\right)$

To find the temperature T where the probability in the ground state is less than$1/2$ , we set$n=0$ in eq. (3) and transform the expression into an inequality:

$\frac{1}{2}>\frac{\mathrm{鈩�}{蝇}_0}{k_{B}T}{e}^{-\left(0\right)\mathrm{鈩�}{蝇}_0/k_{B}T}$

Solving for T we finally get:

$$\begin{gathered} \frac{1}{2}>\frac{\mathrm{鈩�}{蝇}_0}{k_{B}T} \\ T>\frac{2\mathrm{鈩�}{蝇}_0}{k_B} \end{gathered}$$

Thus, the temperature is $T>\frac{2\mathrm{鈩�}{蝇}_0}{k_B}$.