Let's break down **Problem 4.28** step by step: --- ### (a) Show for one oscillator: The given formulas are: \[ f = \frac{1}{2}\hbar\omega + kT \ln(1 - e^{-\beta\hbar\omega}) \tag{4.130} \] \[ s = k \left[ \frac{\beta\hbar\omega}{e^{\beta\hbar\omega} - 1} - \ln(1 - e^{-\beta\hbar\omega}) \right] \tag{4.131} \] \[ \bar{\epsilon} = \hbar\omega\left[\frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1}\right] \tag{4.132} \] --- ### (b) Mean energy of a system of \(N\) oscillators Given (4.132) for one oscillator, for \(N\) independent oscillators, the total mean energy is: \[ \boxed{ \bar{E}_N = N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1} \right] } \] --- ### (c) Compare with microcanonical ensemble (Problem 4.22) In the microcanonical ensemble, for a large number of oscillators, the mean energy at temperature \(T\) is found to be the same as above, i.e., \[ \bar{E}_N = N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1} \right] \] **Conclusion:** The canonical and microcanonical ensembles give the same mean energy for a system of harmonic oscillators in the thermodynamic limit (large \(N\)). This is a manifestation of the equivalence of ensembles. --- ### **Summary Table** | System | Mean Energy \(\bar{E}\) | |---------------------------------|---------------------------------------------------------------------------------| | 1 Harmonic Oscillator | \(\hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1} \right]\) | | \(N\) Harmonic Oscillators | \(N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1} \right]\) | | Microcanonical (large \(N\)) | \(N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta\hbar\omega} - 1} \right]\) | --- Let me know if you need to see derivations or more details about any of the steps!

Let's carefully analyze and solve **Problem 4.28**, based on the provided information and formulas. --- ### **Part (a): Show the expressions for a single oscillator** **Given formulas:** 1. **Free energy (Equation 4.130):** \[ f = \frac{1}{2}\hbar \omega + k T \ln (1 - e^{-\beta \hbar \omega}) \] where \(\beta = \frac{1}{k T}\). 2. **Entropy (Equation 4.131):** \[ s = k \left[ \frac{\beta \hbar \omega}{e^{\beta \hbar \omega} - 1} - \ln (1 - e^{-\beta \hbar \omega}) \right] \] 3. **Mean energy (Equation 4.132):** \[ \boxed{ \bar{\epsilon} = \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right] } \] --- ### **Part (b): Derive the mean energy of \(N\) oscillators** Since the oscillators are independent and identical, the total mean energy is simply: \[ \boxed{ \bar{E}_N = N \times \bar{\epsilon} = N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right] } \] This follows from the linearity of expectation: the total energy is the sum over all oscillators. --- ### **Part (c): Compare with microcanonical ensemble result and discuss ensemble equivalence** **Key point:** - **The canonical ensemble** calculates the average energy by summing over the probability distribution of states at temperature \(T\), leading to the result above. - **The microcanonical ensemble** considers all accessible microstates with a fixed total energy, typically applicable for large systems (thermodynamic limit). **Result:** - For the harmonic oscillator system, both ensembles give the **same average energy** in the large \(N\) limit: \[ \boxed{ \text{Mean energy per oscillator} = \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right] } \] - Therefore, **canonical and microcanonical ensembles are equivalent** in the thermodynamic limit for this system. --- ### **Summary of key points:** | Aspect | Result | |---------|---------| | Single oscillator mean energy | \(\bar{\epsilon} = \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right]\) | | \(N\) oscillators | \(\bar{E}_N = N \times \bar{\epsilon}\) | | Ensemble equivalence | Yes, both ensembles yield the same average energy in the large \(N\) limit | --- ### **Additional notes:** - The formula for the mean energy \(\bar{\epsilon}\) smoothly interpolates between quantum (low \(T\)) and classical (high \(T\)) regimes. - At **zero temperature** (\(T \to 0\)), \(\bar{\epsilon} \to \frac{1}{2} \hbar \omega\), the zero-point energy. - At **high temperature** (\(k T \gg \hbar \omega\)), \(\bar{\epsilon} \to k T\), recovering classical equipartition theorem (each degree of freedom contributes \(k T\)). --- **If you'd like, I can also provide detailed derivations of these formulas.**

Let's carefully analyze and solve **Problem 4.28** step by step. --- ### (a) Derive expressions for a single harmonic oscillator **Given:** - Energy levels: \[ E_n = \hbar \omega \left(n + \frac{1}{2}\right), \quad n=0,1,2,\dots \] - Partition function: \[ Z_1 = \sum_{n=0}^{\infty} e^{-\beta E_n} = e^{-\beta \hbar \omega/2} \sum_{n=0}^\infty \left( e^{-\beta \hbar \omega} \right)^n = \frac{e^{-\beta \hbar \omega/2}}{1 - e^{-\beta \hbar \omega}}. \] --- **1. Helmholtz free energy per oscillator:** \[ f = -k T \ln Z_1 = -k T \left[ -\frac{\beta \hbar \omega}{2} - \ln(1 - e^{-\beta \hbar \omega}) \right] \] \[ \Rightarrow \boxed{ f = \frac{1}{2} \hbar \omega + k T \ln (1 - e^{-\beta \hbar \omega}) } \] which matches **equation (4.130)**. --- **2. Entropy per oscillator:** \[ s = - \left( \frac{\partial f}{\partial T} \right)_{V} \] Express \(f\) in terms of \(x = \beta \hbar \omega = \frac{\hbar \omega}{k T}\): \[ f = \frac{1}{2} \hbar \omega + k T \ln (1 - e^{-x}) \] Since \(x = \frac{\hbar \omega}{k T}\), then: \[ \frac{\partial x}{\partial T} = - \frac{\hbar \omega}{k T^2} \] Calculating \(\frac{\partial f}{\partial T}\): \[ \frac{\partial f}{\partial T} = k \ln (1 - e^{-x}) + k T \cdot \frac{1}{1 - e^{-x}} \cdot e^{-x} \cdot \frac{\partial (-x)}{\partial T} \] \[ = k \ln (1 - e^{-x}) + k T \cdot \frac{e^{-x}}{1 - e^{-x}} \cdot \frac{\hbar \omega}{k T^2} \] \[ = k \ln (1 - e^{-x}) + \frac{\hbar \omega}{T} \cdot \frac{e^{-x}}{1 - e^{-x}} \] Note: \[ \frac{e^{-x}}{1 - e^{-x}} = \frac{1}{e^{x} - 1} \] Thus, \[ s = - \frac{\partial f}{\partial T} = -k \ln (1 - e^{-x}) - \frac{\hbar \omega}{T} \cdot \frac{1}{e^{x} - 1} \] But since \(x = \frac{\hbar \omega}{k T}\), \[ s = k \left[ - \ln (1 - e^{-\beta \hbar \omega}) + \frac{\beta \hbar \omega}{e^{\beta \hbar \omega} - 1} \right] \] which **matches equation (4.131)**. --- **3. Mean energy per oscillator:** \[ \overline{e} = - \frac{\partial \ln Z_1}{\partial \beta} \] Calculate: \[ \ln Z_1 = - \frac{\beta \hbar \omega}{2} - \ln (1 - e^{-\beta \hbar \omega}) \] Derivative: \[ \overline{e} = - \left[ - \frac{\hbar \omega}{2} - \frac{1}{1 - e^{-\beta \hbar \omega}} \cdot (- e^{-\beta \hbar \omega}) \cdot (- \hbar \omega) \right] \] Simplify: \[ \overline{e} = \frac{\hbar \omega}{2} + \frac{\hbar \omega e^{-\beta \hbar \omega}}{1 - e^{-\beta \hbar \omega}} = \frac{\hbar \omega}{2} + \hbar \omega \frac{1}{e^{\beta \hbar \omega} - 1} \] This is **Planck's mean energy formula** (equation 4.132): \[ \boxed{ \overline{e} = \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right] } \] --- ### (b) Total mean energy of \(N\) oscillators Since the oscillators are independent, \[ \boxed{ \overline{E}_N = N \times \overline{e} = N \hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right] \] --- ### (c) Comparing with microcanonical ensemble results In the thermodynamic limit (\(N \to \infty\)), the **average energy per oscillator** from the microcanonical ensemble matches the canonical result derived above. **Conclusion:** - Both ensembles give the same mean energy in the large \(N\) limit. - This demonstrates the **equivalence of ensembles** for large systems, which is a fundamental result in statistical mechanics. --- ### **Summary table:** | Quantity | Expression | |------------|------------| | Single oscillator mean energy | \(\hbar \omega \left[ \frac{1}{2} + \frac{1}{e^{\beta \hbar \omega} - 1} \right]\) | | \(N\) oscillators total energy | \(N \times \) above | | Ensemble equivalence | Yes, in the thermodynamic limit | --- ### Additional: Three-level system example Given microstates: | Energy level | Number of microstates \(\Omega\) | |----------------|---------------------------------| | 0 | 1 | | \(\epsilon\) | 3 | | \(2 \epsilon\) | 2 | Partition function: \[ Z_1 = 1 \times e^{-\beta \times 0} + 3 \times e^{-\beta \epsilon} + 2 \times e^{-2 \beta \epsilon} = 1 + 3 e^{-\beta \epsilon} + 2 e^{-2 \beta \epsilon} \] Mean energy: \[ \overline{e} = \frac{1}{Z_1} \left[ 0 \times 1 + \epsilon \times 3 e^{-\beta \epsilon} + 2 \epsilon \times 2 e^{-2 \beta \epsilon} \right] = \frac{\epsilon (3 e^{-\beta \epsilon} + 4 e^{-2 \beta \epsilon})}{Z_1} \] Total energy for \(N\) particles: \[ \boxed{ \overline{E} = N \times \overline{e} = N \epsilon \frac{3 e^{-\beta \epsilon} + 4 e^{-2 \beta \epsilon}}{1 + 3 e^{-\beta \epsilon} + 2 e^{-2 \beta \epsilon}} } \] --- **Final notes:** - The calculations demonstrate how statistical mechanics connects microscopic microstates to macroscopic thermodynamic quantities. - The equivalence of ensembles holds for large systems, reaffirmed by the identical energy predictions. --- If you need further derivations or clarifications, feel free to ask!