Let's solve your problem step-by-step: --- ## **Given:** - **System A:** \(N_A = 4\) spins - **System B:** \(N_B = 16\) spins - Each spin: magnetic moment \(\mu\), can be up or down - Energy: for each up spin, energy \(-\mu\); for each down spin, \(+\mu\) (assuming magnetic field in +z direction, but the sign won't affect counting since we use energy differences) - Initial energies: \(E_A = -2\mu\), \(E_B = -2\mu\) - Non-interacting (so total number of microstates = product of microstates in each) --- ### **1. Total number of microstates of the composite system** For a system of \(N\) spins, *regardless of energy*, each spin can be up or down: - Number of microstates for \(N\) spins = \(2^N\) So: - System A: \(2^{4} = 16\) - System B: \(2^{16} = 65536\) - Composite system: \(16 \times 65536 = 1,048,576\) **Explanation:** Each spin is independent, so the total number of microstates is the product of the two systems' microstates. **Final answer (1):** \[ \boxed{1,048,576} \] --- ### **2. Probability that system A has energy \(E_A\)** #### **Step 1: What does "energy \(E_A\)" mean?** - For each spin up: energy \(-\mu\) - For each spin down: energy \(+\mu\) - For \(N_A = 4\), total energy is: \[ E_A = (n_\uparrow - n_\downarrow)\mu \] where \(n_\uparrow\) is the number of up spins, \(n_\downarrow = N_A - n_\uparrow\). But it's more accurate to say \(E_A = (2n_\uparrow - N_A)\mu\). Given \(E_A = -2\mu\): \[ -2\mu = (2n_\uparrow - 4)\mu \implies 2n_\uparrow - 4 = -2 \implies n_\uparrow = 1 \] So, number of up spins = 1, number of down spins = 3. #### **Step 2: Number of microstates with \(E_A = -2\mu\):** \[ \Omega_A(E_A) = \text{number of ways to choose 1 up out of 4} = \binom{4}{1} = 4 \] #### **Step 3: System B must have the remaining energy:** Total energy is conserved. If the total energy of the composite system is \(E_{\text{tot}}\), then after energy exchange: \[ E_{\text{tot}} = E_A + E_B \] But since both are initially at \(-2\mu\), \(E_{\text{tot}} = -4\mu\). If we want the probability that A has \(E_A = -2\mu\), then B must have \(E_B = -2\mu\). #### **Step 4: Number of microstates for \(N_B = 16\) spins with \(E_B = -2\mu\):** \[ E_B = (2n_\uparrow - 16)\mu = -2\mu \implies 2n_\uparrow - 16 = -2 \implies n_\uparrow = 7 \] So, number of up spins in B = 7, down = 9. Number of microstates: \[ \Omega_B(E_B) = \binom{16}{7} = 11,\!440 \] #### **Step 5: Total number of microstates with \(E_A = -2\mu\):** \[ \Omega(E_A = -2\mu) = \Omega_A(E_A) \cdot \Omega_B(E_B) = 4 \times 11,440 = 45,760 \] #### **Step 6: Probability:** \[ P(E_A = -2\mu) = \frac{\text{# of microstates with }E_A = -2\mu}{\text{total # of microstates}} = \frac{45,760}{1,048,576} \approx 0.0437 \] **Explanation:** You count all ways A can have 1 up spin, and B can have 7 up spins, and divide by the total possible microstates for the composite system. **Final answer (2):** \[ \boxed{0.0437} \text{ (or } 4.37\% \text{)} \] --- ### **3. What is the mean value \(\langle E_A \rangle\)?** #### **Step 1: By symmetry or direct calculation** If the two systems are allowed to exchange energy freely, and all microstates are equally probable, the mean number of up spins in A is: - For each of the 20 spins, up or down is equally likely in the absence of a field. - On average, in A, you expect \(4 \times \frac{1}{2} = 2\) up spins. But let's be precise: Total number of up spins in the composite system is 10 (from initial conditions: both had 1 and 7 up spins, but after exchange, all microstates with 20 spins and \(E_{\text{tot}} = -4\mu\) are possible). But actually, for all microstates of 20 spins with total energy \(-4\mu\), total number of up spins is \(n_{\uparrow, \text{tot}}\): \[ (2n_{\uparrow, \text{tot}} - 20)\mu = -4\mu \implies n_{\uparrow, \text{tot}} = 8 \] So, total number of up spins in the whole system is always 8. When A has \(k\) up spins, B has \(8 - k\) up spins. Number of ways for A to have \(k\) up spins: \[ \binom{4}{k} \] Number of ways for B to have \(8 - k\) up spins: \[ \binom{16}{8 - k} \] So total number of microstates with \(k\) up spins in A: \[ \Omega(k) = \binom{4}{k} \binom{16}{8 - k} \] Probability that A has \(k\) up spins: \[ P(k) = \frac{\Omega(k)}{\sum_{k=0}^4 \Omega(k)} \] But, sum over all possible \(k\): - \(0 \leq k \leq 4\) - For each \(k\), \(B\) must have \(8 - k\) up spins (\(0 \leq 8 - k \leq 16\)), but since \(k\) can go from 0 to 4, \(8 - k\) goes from 8 to 4. So possible \(k = 0, 1, 2, 3, 4\) Now, the mean value: \[ \langle k \rangle = \sum_{k=0}^4 k P(k) \] and \[ \langle E_A \rangle = (2 \langle k \rangle - 4)\mu \] #### **Calculation:** Let's compute \(\Omega(k)\) for \(k = 0, 1, 2, 3, 4\): - \(\binom{4}{0} = 1\), \(\binom{16}{8}\) - \(\binom{4}{1} = 4\), \(\binom{16}{7}\) - \(\binom{4}{2} = 6\), \(\binom{16}{6}\) - \(\binom{4}{3} = 4\), \(\binom{16}{5}\) - \(\binom{4}{4} = 1\), \(\binom{16}{4}\) Calculate each: - \(\binom{16}{8} = 12,870\) - \(\binom{16}{7} = 11,440\) - \(\binom{16}{6} = 8,008\) - \(\binom{16}{5} = 4,368\) - \(\binom{16}{4} = 1,820\) So: | \(k\) | \(\binom{4}{k}\) | \(\binom{16}{8-k}\) | \(\Omega(k)\) | |-------|------------------|---------------------|--------------------| | 0 | 1 | 12,870 | 12,870 | | 1 | 4 | 11,440 | 45,760 | | 2 | 6 | 8,008 | 48,048 | | 3 | 4 | 4,368 | 17,472 | | 4 | 1 | 1,820 | 1,820 | Sum: \(12,870 + 45,760 + 48,048 + 17,472 + 1,820 = 125,970\) Now, \(P(k) = \frac{\Omega(k)}{125,970}\) Compute \(\langle k \rangle\): \[ \langle k \rangle = \sum_{k=0}^4 k \frac{\Omega(k)}{125,970} = \frac{1 \cdot 45,760 + 2 \cdot 48,048 + 3 \cdot 17,472 + 4 \cdot 1,820}{125,970} \] Let's multiply out: - \(1 \cdot 45,760 = 45,760\) - \(2 \cdot 48,048 = 96,096\) - \(3 \cdot 17,472 = 52,416\) - \(4 \cdot 1,820 = 7,280\) Sum numerator: \[ 45,760 + 96,096 + 52,416 + 7,280 = 201,552 \] So: \[ \langle k \rangle = \frac{201,552}{125,970} \approx 1.6 \] Therefore, \[ \langle E_A \rangle = (2 \times 1.6 - 4)\mu = (3.2 - 4)\mu = -0.8\mu \] **Explanation:** We use the microcanonical ensemble: all microstates with total energy \(-4\mu\) are equally probable. The mean number of up spins in A can be found by weighted average. **Final answer (3):** \[ \boxed{\langle E_A \rangle = -0.8\mu} \] --- ## **Summary of Answers** **1. Total number of microstates:** \[ \boxed{1,048,576} \] **2. Probability that system A has \(E_A = -2\mu\):** \[ \boxed{0.0437} \;\; (4.37\%) \] **3. Mean value \(\langle E_A \rangle\):** \[ \boxed{-0.8\mu} \] Feel free to ask if you need further clarifications!