The Exercise

A qubit has two levels, $|0\rangle$ with energy $E_0$ and $|1\rangle$ with energy $E_1$ . It is in a stationary state in which it is found in $|0\rangle$ with probability $p$ and in $|1\rangle$ with probability $1-p$ . The exercise assigns this state a temperature: the temperature $T$ at which these two probabilities are the Boltzmann populations of the two levels. It asks for $T$ .

Boltzmann Populations

Let $k_B$ be Boltzmann’s constant. The ground-state population $p_0(T)$ of the qubit at temperature $T$ is the Boltzmann factor of $|0\rangle$ divided by the sum of the two Boltzmann factors, $p_0(T) = \frac{e^{-E_0/(k_B T)}}{e^{-E_0/(k_B T)} + e^{-E_1/(k_B T)}},$ (1) and the excited-state population $p_1(T)$ of the qubit at temperature $T$ is the Boltzmann factor of $|1\rangle$ divided by the same sum, $p_1(T) = \frac{e^{-E_1/(k_B T)}}{e^{-E_0/(k_B T)} + e^{-E_1/(k_B T)}}.$ (2)

The temperature $T_p$ of a stationary state of the qubit with ground-state population $p$ is the real number $T_p = \frac{E_1 - E_0}{k_B \ln\bigl(p/(1-p)\bigr)}.$ (3)^✦ The logarithm is finite and nonzero exactly when $0 < p < 1$ and $p \neq 1/2$ . For $k_B > 0$ and $E_1 > E_0$ , $T_p$ is positive when the ground level is the more populated and negative under a population inversion, $p < 1/2$ .

The Claim

Theorem 1. Let $p$ satisfy $0 < p < 1$ and $p \neq 1/2$ , and let $k_B$ be nonzero. If the two levels have different energies, the Boltzmann populations at the temperature $T_p$ of (3) are $p$ and $1-p$ : $p_0(T_p) = p,$ (4)^✦ $p_1(T_p) = 1 - p.$ (5)^✦ If $T$ is a nonzero temperature with $p_0(T) = p$ , then $T = T_p.$ (6)^✦

What Is Checked

The claim is stated over the two Boltzmann populations (1) and (2), as functions of the temperature for given energies $E_0$ , $E_1$ and constant $k_B$ , and over the temperature (3) as a function of the population $p$ .