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A system of N non-interacting classical spins, where each spin can take values σ = - 1, 0, 1 , is placed in a magnetic field h. The single spin Hamilt
Question

A system of N non-interacting classical spins, where each spin can take values σ = - 1, 0, 1 , is placed in a magnetic field h. The single spin Hamiltonian is given by H=μBhσ+Δ(1σ2)H = -\mu_B h \sigma + \Delta (1 - \sigma^2)​​
where Delta are positive constants with appropriate dimensions. If M is the magnetization, the zero-field magnetic susceptibility per spin 1NMhh0\frac{1}{N} \frac{\partial M}{\partial h} \bigg|_{h \to 0}​.at a temperature T = 1//βkB is given by

A.

βμB2\beta \mu_B^2​​

B.

2βμB22+eβΔ\frac{2 \beta \mu_B^2}{2 + e^{-\beta \Delta}}​​

C.

βμB2eβΔ\beta \mu_B^2 e^{-\beta \Delta}​​

D.

βμB21+eβΔ\frac{\beta \mu_B^2}{1 + e^{-\beta \Delta}}​​

Correct option is B

Solution:

​σ = - 1, 0, 1

H=μBhσ+Δ(1σ2). Draw energy levels using σ=1,0,1 and H=μBhσ+Δ(1σ2) Z=eβμBh+eβμBh+eβΔF=kBTln(eβμBh+eβμBh+eβΔ)dF=SdTPdVμdh μ=(Fh)T,V=kBTeβμBheβμBheβμBh+eβμBh+eβΔβμB =kBTβμBeβμBheβμBheβμBh+eβμBh+eβΔ In the limit where h is small M=(1+βμBh1+βμBh)μB2+eβΔ =βμB22+eβΔH = -\mu_B h \sigma + \Delta (1 - \sigma^2) \quad .\\\ \\ \quad \text{Draw energy levels using } \sigma = -1, 0, 1 \text{ and } H = -\mu_B h \sigma + \Delta (1 - \sigma^2)\\\ \\Z = e^{\beta \mu_B h} + e^{-\beta \mu_B h} + e^{-\beta \Delta}\\F = -k_B T \ln \left( e^{\beta \mu_B h} + e^{-\beta \mu_B h} + e^{-\beta \Delta} \right)\\dF = -S dT - P dV - \mu dh\\\ \\\mu = \left( \frac{\partial F}{\partial h} \right)_{T, V} = k_B T \frac{e^{\beta \mu_B h} - e^{-\beta \mu_B h}}{e^{\beta \mu_B h} + e^{-\beta \mu_B h} + e^{-\beta \Delta}} \cdot \beta \mu_B\\\ \\ = k_B T \beta \mu_B \frac{e^{\beta \mu_B h} - e^{-\beta \mu_B h}}{e^{\beta \mu_B h} + e^{-\beta \mu_B h} + e^{-\beta \Delta}}\\\ \\\text{In the limit where h is small}\\\ \\M = \frac{\left(1 + \beta \mu_B h - 1 + \beta \mu_B h \right) \mu_B}{2 + e^{-\beta \Delta}}\\\ \\ = \frac{\beta \mu_B^2}{2 + e^{-\beta \Delta}}​​

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