91影视

The total energy Eof the particle can be written as:

$$\begin{gathered} \mathit{E}\mathbf{=}\left(n_a+n_b+n_c+n_d\right)\mathit{鈩�}{\mathit{蝇}}_{\mathbf{0}} \\ \mathbf{=}\mathbf{3}\mathit{鈩�}{\mathit{蝇}}_{\mathbf{0}} \end{gathered}$$

which follows that the sum of the quantum numbers corresponds to the macro-state:

${\mathit{n}}_{\mathbf{a}}\mathbf{+}{\mathit{n}}_{\mathbf{b}}\mathbf{+}{\mathit{n}}_{\mathbf{c}}\mathbf{+}{\mathit{n}}_{\mathbf{d}}\mathbf{=}\mathbf{3}$

The formula for the probability for one oscillator to be at state${\mathit{n}}_{\mathit{i}}$is:

where Mis the integer corresponding to the energy macro-state, ${\mathbf{n}}_{\mathbf{i}}$is the quantum number considered, and Nis the number of oscillators.

The total energy of the particle in the given microstate is $3\mathrm{鈩�}{蝇}_0$.

Let us now enumerate all possible microstates for the given microstate in which total energy is $3\mathrm{鈩�}{蝇}_o$:

$$\begin{gathered} (0,0,0,3);(0,2,0,1);(0,2,1,0);(1,2,0,0);(0,0,3,0);(2,0,0,1);(0,0,1,2);(0,1,1,1) \\ (0,3,0,0);(2,0,1,0);(0,1,0,2);(1,1,1,0);(3,0,0,0);(2,1,0,0);(1,0,0,2);(1,1,0,1) \\ (0,0,2,1);(0,1,2,0);(1,0,2,0);(1,0,1,1) \end{gathered}$$

In this problem, we have$M=3$, and$N=4$

Plugging in $n_i=0$, the probability at which an oscillator is at state $n_i=0$is thus:

$$\begin{gathered} P\left(n_i=0\right)=\raisebox{1ex}{${\displaystyle \frac{{\displaystyle (3-0+4-2)!}}{{\displaystyle (3-0)!(4-2)!}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{{\displaystyle (3+4-1)!}}{{\displaystyle 3!(4-1)!}}}$}\right. \\ =\raisebox{1ex}{${\displaystyle \frac{{\displaystyle 5!}}{{\displaystyle 3!2!}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{{\displaystyle 6!}}{{\displaystyle 3!3!}}}$}\right. \\ =\frac{10}{20} \\ =0.5 \end{gathered}$$

The remaining possible values for$n_i$are $1,2,\&3$.

For $n_i=1$the probability is:

$$\begin{gathered} P\left(n_i=1\right)=\raisebox{1ex}{$\frac{(3-1+4-2)!}{(3-1)!(4-2)!}$}\!\left/ \!\raisebox{-1ex}{$\frac{(3+4-1)!}{3!(4-1)!}$}\right. \\ =\raisebox{1ex}{$\frac{4!}{2!2!}$}\!\left/ \!\raisebox{-1ex}{$\frac{6!}{3!3!}$}\right. \\ =\frac{6}{20} \\ =0.3 \end{gathered}$$

For $n_i=2$we get:

$$\begin{gathered} P\left(n_i=2\right)=\raisebox{1ex}{$\frac{(3-2+4-2)!}{(3-2)!(4-2)!}$}\!\left/ \!\raisebox{-1ex}{$\frac{(3+4-1)!}{3!(4-1)!}$}\right. \\ =\raisebox{1ex}{$\frac{3!}{2!}$}\!\left/ \!\raisebox{-1ex}{$\frac{6!}{3!3!}$}\right. \\ =\frac{3}{20} \\ =0.15 \end{gathered}$$

For $n_i=3$we get:

$$\begin{gathered} P\left(n_i=3\right)=\raisebox{1ex}{${\displaystyle \frac{(3-3+4-2)!}{(3-3)!(4-2)!}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{(3+4-1)!}{3!(4-1)!}}$}\right. \\ =\raisebox{1ex}{${\displaystyle \frac{2!}{2!}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{6!}{3!3!}}$}\right. \\ =\frac{1}{20} \\ =0.05 \end{gathered}$$

Using our results in parts (b) and (c), the data are:

Plotting $n_i=1$as a function of n we thus have: