A Hint of Quantum Theory — Question 4

Question 4

An oscillator ladder has a geometric distribution of excitation numbers

A quantum harmonic mode has excitation energies $E_j=j\Delta$, $j=0,1,2,\ldots$, measured above its ground energy. The probability is proportional to $q^j$, where $q=e^{-\Theta/T}$ and $\Theta=\Delta/k_B=1000$ K. Use $\sum_{j=0}^\infty q^j=1/(1-q)$.

1. Normalize the probabilities and calculate the mean excitation number at 500 K by differentiating the geometric sum.

2. Find the probability of two or more quanta and compare the mean excitation energy with the classical value $k_BT$.

Question 4 - Solution

An oscillator ladder has a geometric distribution of excitation numbers

(a) Method and calculation

$$P_j=(1-q)q^j,\quad\langle j\rangle=(1-q)q\frac d{dq}\frac1{1-q}=\frac q{1-q}=\frac1{e^{\Theta/T}-1}.$$$$T=500\,\mathrm K:\quad\langle j\rangle=\frac1{e^2-1}\approx.156518.$$

(b) Interpretation and further calculation

$$P(j\ge2)=(1-q)\sum_{j=2}^\infty q^j=q^2=e^{-4}\approx.0183156.$$$$\langle E\rangle=.156518\Delta,\quad k_BT=.500\Delta,\quad\frac{\langle E\rangle}{k_BT}\approx.313035.$$Excitation energy is strongly suppressed when the spacing exceeds the thermal energy scale.

Physical check

The ground-state probability is nonzero, and the excitation number fluctuates; the mean need not be an integer.