Let's break down the problem step by step as a helpful tutor! # **Step 1: Understanding the Problem** - **Energy spectrum:** \( E_1, E_2, E_3, E_4, \ldots = \epsilon, 2\epsilon, 3\epsilon, 4\epsilon, \ldots \) - **System:** \( N, V \)-system (single-particle energies as above) - **Supersystem:** Collection of \( M = 5 \) such systems - **Total energy of supersystem:** \( 10\epsilon \) - **Distributions:** \( \{n_j\} \) = number of subsystems (out of 5) with energy \( E_j \) - **Objective:** List all possible \( \{n_j\} \) consistent with total energy and system number, find their multiplicities \(\Omega\), the probability distributions, weights, averages, and normalization. --- # **Step 2: Listing All Possible Distributions \( \{ n_j \} \)** Each subsystem can have energy \( E_j = j\epsilon \) for integer \( j \geq 1 \). Let \( n_j \): Number of subsystems with energy \( E_j \). ### **Constraints** - Total number of subsystems: \( \sum_{j=1}^\infty n_j = M = 5 \) - Total energy: \( \sum_{j=1}^\infty n_j E_j = 10\epsilon \implies \sum_{j=1}^\infty j n_j = 10 \) We need **non-negative integers** \( n_j \) such that: \[ \sum_{j=1}^\infty n_j = 5, \quad \sum_{j=1}^\infty j n_j = 10 \] #### **Find all such integer solutions.** Let's systematically look for integer solutions. Since 5 systems and 10 units, maximum energy per particle is 10, minimum is 5 → energies per particle range from 1 to 10, but with only 5 particles. Try to assign \( n_j \) values for \( j = 1,2,3,4,5,6,7,8,9,10 \) such that their sum is 5 and the weighted sum is 10. But typically only \( j \leq 10 \) will be needed. Let's try all combinations for small \( j \): ### **Case 1: All at lowest energy** - \( n_1 = 5 \implies \) total energy = \( 5 \times 1 = 5 \) (too low) ### **Case 2: Four at 1, one at higher** - \( n_1 = 4, n_j = 1 \) for some \( j \) - Total energy: \( 4 \times 1 + j = 4 + j = 10 \implies j = 6 \) - Distribution: \( (n_1, n_2, n_3, n_4, n_5, n_6) = (4,,,,,1) \) ### **Case 3: Three at 1, two at higher** - \( n_1 = 3, n_k = 2 \) - \( 3 \times 1 + 2j = 3 + 2j = 10 \implies 2j = 7 \implies j = 3.5 \) (not integer) - \( n_1 = 3, n_j = 1, n_k = 1 \) - \( 3 \times 1 + j + k = 10 \implies j + k = 7 \) - Possible integer pairs: (2,5), (3,4), (4,3), (5,2) - Since \( j \geq k \), list unique combinations: - (2,5): \( (3,1,,,1,) \) - (3,4): \( (3,,1,1,,) \) ### **Case 4: Two at 1, three at higher** - \( n_1 = 2, n_j = 3 \) - \( 2 \times 1 + 3j = 2 + 3j = 10 \implies 3j = 8 \implies j = 8/3 \) (not integer) - \( n_1 = 2, n_j = 2, n_k = 1 \) - \( 2 \times 1 + 2j + k = 10 \implies 2j + k = 8 \) - Try \( j = 3 \Rightarrow 2 \times 3 + k = 8 \implies k = 2 \) - So, \( (2,2,1,,,) \) - \( j = 2 \Rightarrow 2 \times 2 + k = 8 \implies k = 4 \) - \( (2,2,,1,,) \) - \( j = 1 \Rightarrow 2 \times 1 + k = 8 \implies k = 6 \) - \( (2,,,,,1) \) - \( n_1 = 2, n_j = 1, n_k = 1, n_l = 1 \) - \( 2 \times 1 + j + k + l = 10 \implies j + k + l = 8 \) - Solutions for \( j, k, l \geq 2 \), sum 8, distinct ordering matters since they are indistinguishable. - Possible partitions: (2,2,4), (2,3,3) - (2,2,4): \( (2,2,,1,,) \) - (2,3,3): \( (2,,2,,,) \) ### **Case 5: One at 1, four at higher** - \( n_1 = 1, n_j = 4 \implies 1 + 4j = 10 \im 4j = 9, j = 2.25 \) (not integer) - \( n_1 = 1, n_j = 3, n_k = 1 \implies 1 + 3j + k = 10 \implies 3j + k = 9 \) - \( j = 3, k = \implies (1,,3,,,) \) - \( j = 2, k = 3 \implies (1,3,,1,,) \) - \( j = 2, k = 3 \implies (1,3,,1,,) \) - \( j = 1, k = 6 \implies (1,,,,,1) \) ### **Case 6: All at higher energies** - \( n_2 = 5 \implies 5 \times 2 = 10 \) - Distribution: \( (,5,,,,) \) ### **Summary of All Valid \( \{n_j\} \) Distributions:** Let's collect all unique distributions: | \( n_1 \) | \( n_2 \) | \( n_3 \) | \( n_4 \) | \( n_5 \) | \( n_6 \) | Total (should be 5) | Weighted Sum (should be 10) | |-----------|-----------|-----------|-----------|-----------|-----------|---------------------|------------------------------| | 4 | | | | | 1 | 5 | 4×1+1×6=10 | | 3 | 1 | | | 1 | | 5 | 3×1+1×2+1×5=10 | | 3 | | 1 | 1 | | | 5 | 3×1+1×3+1×4=10 | | 2 | 2 | 1 | | | | 5 | 2×1+2×2+1×3=10 | | 2 | | 2 | | | | 5 | 2×1+2×3=8 (skip) | | 2 | 1 | | 2 | | | 5 | 2×1+1×2+2×4=1+2+8=12 (skip) | | 2 | 2 | | 1 | | | 5 | 2×1+2×2+1×4=2+4+4=10 | | | 5 | | | | | 5 | 5×2=10 | | 1 | 3 | | 1 | | | 5 | 1×1+3×2+1×4=1+6+4=11 (skip) | | 1 | | 3 | | | | 5 | 1×1+3×3=1+9=10 | | 1 | 1 | 1 | 2 | | | 5 | 1×1+1×2+1×3+2×4=1+2+3+8=14 | | 1 | 2 | | | 2 | | 5 | 1×1+2×2+2×5=1+4+10=15 | | 2 | | | | | 3 | 5 | 2×1+3×6=2+18=20 (skip) | | | 4 | 1 | | | | 5 | 4×2+1×3=8+3=11 (skip) | After correcting and removing invalid ones, the **valid distributions** are: 1. (4, , , , , 1) 2. (3, 1, , , 1, ) 3. (3, , 1, 1, , ) 4. (2, 2, 1, , , ) 5. (2, 2, , 1, , ) 6. (, 5, , , , ) 7. (1, , 3, , , ) --- # **Step 3: Multiplicity \( \Omega(\{ n_j \}) \) for Each Distribution** Multiplicity for distributing \( M \) indistinguishable subsystems into distinguishable energy bins with \( n_j \) in each: \[ \Omega(\{ n_j \}) = \frac{M!}{\prod_j n_j!} \] Calculate for each distribution (use only non-zero \( n_j \)): 1. **(4, , , , , 1):** \( \Omega = \frac{5!}{4! \cdot 1!} = \frac{120}{24 \cdot 1} = 5 \) 2. **(3, 1, , , 1, ):** \( \Omega = \frac{5!}{3! \cdot 1! \cdot 1!} = \frac{120}{6 \cdot 1 \cdot 1} = 20 \) 3. **(3, , 1, 1, , ):** \( \Omega = \frac{5!}{3! \cdot 1! \cdot 1!} = 20 \) 4. **(2, 2, 1, , , ):** \( \Omega = \frac{5!}{2! \cdot 2! \cdot 1!} = \frac{120}{2 \cdot 2 \cdot 1} = 30 \) 5. **(2, 2, , 1, , ):** \( \Omega = \frac{5!}{2! \cdot 2! \cdot 1!} = 30 \) 6. **(, 5, , , , ):** \( \Omega = \frac{5!}{5!} = 1 \) 7. **(1, , 3,

# **Step 4: Probability Distribution \( p_i(\{ n_j \}) \)** The probability of each distribution is given by: \[ p_i(\{ n_j \}) = \frac{\Omega(\{ n_j \})}{\text{Total Configurations}} \] Where the total configurations for 5 indistinguishable systems can be calculated as: \[ \text{Total Configurations} = \sum_i \Omega(\{ n_j \}) \] ### **Calculating Total Configurations** Let's calculate \( \Omega \) for all valid distributions from the previous step: 1. **(4, 0, 0, 0, 1):** \( \Omega = 5 \) 2. **(3, 1, 0, 0, 1):** \( \Omega = 20 \) 3. **(3, 0, 1, 1, 0):** \( \Omega = 20 \) 4. **(2, 2, 1, 0, 0):** \( \Omega = 30 \) 5. **(2, 2, 0, 1, 0):** \( \Omega = 30 \) 6. **(0, 5, 0, 0, 0):** \( \Omega = 1 \) ### **Sum of Multiplicities** \[ \text{Total Configurations} = 5 + 20 + 20 + 30 + 30 + 1 = 106 \] ### **Calculating Probabilities for Each Distribution** Using the total configurations: 1. **(4, 0, 0, 0, 1):** \[ p_1 = \frac{5}{106} \approx 0.0472 \] 2. **(3, 1, 0, 0, 1):** \[ p_2 = \frac{20}{106} \approx 0.1887 \] 3. **(3, 0, 1, 1, 0):** \[ p_3 = \frac{20}{106} \approx 0.1887 \] 4. **(2, 2, 1, 0, 0):** \[ p_4 = \frac{30}{106} \approx 0.2830 \] 5. **(2, 2, 0, 1, 0):** \[ p_5 = \frac{30}{106} \approx 0.2830 \] 6. **(0, 5, 0, 0, 0):** \[ p_6 = \frac{1}{106} \approx 0.0094 \] --- # **Step 5: Relative Weights \( P_{rel} \)** The relative weight for each distribution can be calculated as: \[ P_{rel}(\{ n_j \}) = p_i(\{ n_j \}) \times \Omega(\{ n_j \}) \] ### **Calculating Relative Weights** 1. **(4, 0, 0, 0, 1):** \[ P_{rel} = 0.0472 \times 5 \approx 0.236 \] 2. **(3, 1, 0, 0, 1):** \[ P_{rel} = 0.1887 \times 20 \approx 3.774 \] 3. **(3, 0, 1, 1, 0):** \[ P_{rel} = 0.1887 \times 20 \approx 3.774 \] 4. **(2, 2, 1, 0, 0):** \[ P_{rel} = 0.2830 \times 30 \approx 8.490 \] 5. **(2, 2, 0, 1, 0):** \[ P_{rel} = 0.2830 \times 30 \approx 8.490 \] 6. **(0, 5, 0, 0, 0):** \[ P_{rel} = 0.0094 \times 1 \approx 0.0094 \] --- # **Step 6: Average of Probabilities \( \langle p \rangle \)** The average probability over all distributions is calculated as: \[ \langle p \rangle = \frac{1}{\text{Total Configurations}} \sum_i p_i(\{ n_j \}) \cdot \Omega(\{ n_j \}) \] ### **Calculating Average Probability** \[ \langle p \rangle = \frac{1}{106} \left( 5 \cdot 0.0472 + 20 \cdot 0.1887 + 20 \cdot 0.1887 + 30 \cdot 0.2830 + 30 \cdot 0.2830 + 1 \cdot 0.0094 \right) \] Calculating: \[ = \frac{1}{106} \left( 0.236 + 3.774 + 3.774 + 8.490 + 8.490 + 0.0094 \right) \] \[ = \frac{1}{106} \left( 24.7744 \right) \approx 0.2334 \] --- # **Step 7: Normalization of Probabilities** To check if the probabilities \( \{ p_i(\{ n_j \}) \} \) are normalized, we verify: \[ \sum_i p_i(\{ n_j \}) = 1 \] Calculating: \[ \sum_i p_i = \frac{5 + 20 + 20 + 30 + 30 + 1}{106} = \frac{106}{106} = 1 \] ### **Conclusion** - The probabilities \( \{ p_i(\{ n_j \}) \} \) are indeed normalized. - The distributions are consistent with the constraints provided. Thus, we have completed the solution!

# **Step 3: Multiplicity \( \Omega(\{ n_j \}) \) for Each Distribution** Multiplicity refers to the number of ways to arrange the indistinguishable particles (subsystems) into distinguishable energy states. The formula to calculate the multiplicity for a given distribution \( \{ n_j \} \) is: \[ \Omega(\{ n_j \}) = \frac{M!}{\prod_j n_j!} \] Where \( M \) is the total number of subsystems (which is 5 in this case), and \( n_j \) are the counts of subsystems at each energy level. ### **Calculating \( \Omega \) for Each Distribution** From the valid distributions we identified in the previous steps, we can calculate the multiplicity for each. 1. **Distribution: \( (4, 0, 0, 0, 1) \)** \[ \Omega = \frac{5!}{4! \times 1!} = \frac{120}{24 \times 1} = 5 \] 2. **Distribution: \( (3, 1, 0, 0, 1) \)** \[ \Omega = \frac{5!}{3! \times 1! \times 1!} = \frac{120}{6 \times 1 \times 1} = 20 \] 3. **Distribution: \( (3, 0, 1, 1, 0) \)** \[ \Omega = \frac{5!}{3! \times 1! \times 1!} = \frac{120}{6 \times 1 \times 1} = 20 \] 4. **Distribution: \( (2, 2, 1, 0, 0) \)** \[ \Omega = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = 30 \] 5. **Distribution: \( (2, 2, 0, 1, 0) \)** \[ \Omega = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = 30 \] 6. **Distribution: \( (0, 5, 0, 0, 0) \)** \[ \Omega = \frac{5!}{5!} = 1 \] ### **Summary of Multiplicities** | Distribution | \( n_1 \) | \( n_2 \) | \( n_3 \) | \( n_4 \) | \( n_5 \) | \( \Omega \) | |-----------------------------|-----------|-----------|-----------|-----------|-----------|--------------| | (4, 0, 0, 0, 1) | 4 | 0 | 0 | 0 | 1 | 5 | | (3, 1, 0, 0, 1) | 3 | 1 | 0 | 0 | 1 | 20 | | (3, 0, 1, 1, 0) | 3 | 0 | 1 | 1 | 0 | 20 | | (2, 2, 1, 0, 0) | 2 | 2 | 1 | 0 | 0 | 30 | | (2, 2, 0, 1, 0) | 2 | 2 | 0 | 1 | 0 | 30 | | (0, 5, 0, 0, 0) | 0 | 5 | 0 | 0 | 0 | 1 | --- This concludes Step 3. Next, we will move to Step 4, where we calculate the probability distribution \( p_i(\{ n_j \}) \) for each distribution.