Question

A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment with temperature 250 K. 1) What is the average internal energy?

Answer 1

To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.

Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by

$z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}$

Where,

$\epsilon_i =$ energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

$k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}$

$\epsilon_1=0J$

$\epsilon_2=1.6*10^{-21}J$

$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

$\beta = \frac{1}{(1.381*10^{23} m^2)(250)}$

$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

$\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}$

Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$

$\bar{\epsilon} = 0.9*10^{-22}J$

Therefore the average internal energy is $\bar{\epsilon} = 0.9*10^{-22}J$