A three-state system with energies E = −ε0, 0, +ε0 is in a thermal equilibrium at a temperature T. If β ε0 = x, the probability of finding the system with energy E = 0 is [recall, cosh x = \(\frac{1}{2}\)(ex + e−x)]
This question asks about the probability of a three-state system being in a specific energy state when it is in thermal equilibrium at a temperature T. We are given the energies of the three states and a relationship involving the temperature and energy scale.
System Energy States and Thermal Equilibrium
The system has three distinct energy states with energies \(E_1 = -\epsilon_0\), \(E_2 = 0\), and \(E_3 = +\epsilon_0\). The system is in thermal equilibrium at temperature T. In thermal equilibrium, the probability of finding the system in a state with energy E is given by the Boltzmann distribution within the canonical ensemble:
\(P(E) = \frac{e^{-\beta E}}{Z}\)
Here, \(\beta = \frac{1}{k_B T}\) (where \(k_B\) is Boltzmann's constant) and Z is the partition function, which is the sum over all possible states i:
\(Z = \sum_{i} e^{-\beta E_i}\)
Partition Function Calculation
For our three-state system with energies \(-\epsilon_0\), \(0\), and \(+\epsilon_0\), the partition function Z is:
\(Z = e^{-\beta (-\epsilon_0)} + e^{-\beta (0)} + e^{-\beta (+\epsilon_0)}\)
\(Z = e^{\beta \epsilon_0} + e^0 + e^{-\beta \epsilon_0}\)
Since \(e^0 = 1\), we have:
\(Z = e^{\beta \epsilon_0} + 1 + e^{-\beta \epsilon_0}\)
Using the Given Relation
We are given that \(\beta \epsilon_0 = x\). Substituting this into the expression for the partition function:
\(Z = e^{x} + 1 + e^{-x}\)
\(Z = 1 + (e^{x} + e^{-x})\)
We are also reminded of the definition of the hyperbolic cosine function: \(\cosh x = \frac{1}{2}(e^x + e^{-x})\). From this, we can see that \(e^x + e^{-x} = 2 \cosh x\). Substituting this into the partition function expression:
\(Z = 1 + 2 \cosh x\)
Probability of Energy E=0
We need to find the probability of the system being in the state with energy \(E=0\). Using the Boltzmann distribution formula:
\(P(E=0) = \frac{e^{-\beta \cdot 0}}{Z}\)
\(P(E=0) = \frac{e^0}{Z}\)
\(P(E=0) = \frac{1}{Z}\)
Substituting the expression we found for Z:
\(P(E=0) = \frac{1}{1 + 2 \cosh x}\)
This can also be written using a negative exponent:
\(P(E=0) = (1 + 2 \cosh x)^{-1}\)
Let's compare this result with the given options:
- Option 1: \((2 \cosh x)^{-1}\) - This is incorrect.
- Option 2: \(\frac{1}{2}\)cosh x - This is incorrect.
- Option 3: \((1 + 2 \cosh x)^{-1}\) - This matches our result.
- Option 4: \(1 + 2 \cosh x\) - This is the partition function Z, not the probability.
The probability of finding the system with energy \(E=0\) is \((1 + 2 \cosh x)^{-1}\).
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