Statistical mechanics: Particles with spin

AI Thread Summary
The discussion focuses on calculating the canonical partition function and related thermodynamic properties for a system of N non-interacting particles with spin in a magnetic field. For a single particle, the canonical partition function is derived as Z_1 = 2 cosh(βmB), leading to the probability of the spin-up state as P_+ = e^{-βmB}/(2 cosh(βmB)). The number of microstates for N particles is confirmed to be Ω(N) = 2^N. The Helmholtz free energy and average energy per particle are addressed, with the average energy formula provided as U = -∂ln(Z)/∂β. The calculations emphasize the importance of understanding the partition function in statistical mechanics.
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Homework Statement



We have N particles, each of which can either be spin-up (s_i = 1) or spin-down (s_i = -1) with i = 1, 2, 3...N. The particles are in fixed position, don't interact and because they are in a magnetic field with strength B, the energy of the system is given by:

E(s_1, ..., s_n) = -mB \sum_{i=1}^{N} s_i

with m > 0 the magnetic moment of the particles. The temperature is T.

a) Calculate the canonic partition function for N = 1 and the chance that this particle is in spin-up state P_+.

b) For any N, calculate the number of microstates \Omega(N), the Helmholtz free energy F(N,T) and the average energy per particle U(N, T)/N

The Attempt at a Solution



a) Z_1 = e^{-\beta m B} + e^{\beta m B} = 2 \cosh{\beta m B}
P_+ = \frac{e^{-\beta m B}}{2 \cosh{\beta m B}}

b) The number of possible microstates is \Omega(N) = 2^N, correct?

I know that U = -\frac{\partial \ln Z}{\partial \beta}, but I'm not sure how to calculate Z here.
 
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leave Z as the summation Z = Ʃ e-Eiβ where β = 1/KBTso ∂ln(Z)/dβ = (1/Z)(∂Z/∂β) = [-EiƩe-Eiβ]/Zi think b) is supposed to be (Z1)N sorry yeah your b) is right
 
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