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HW4 solutions - Physics 4120 Spring 2008 Homework#4...

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Physics 4120 Spring 2008 Homework #4 solutions 1) Reif 2.4 (a) The number of states with n 1 parallel and n 2 antiparallel spins is ! ! ! 2 1 n n N . Using our definition for energy H n n E μ ) ( 2 1 - - = and 2 1 n n N + = , we can rewrite our n’s in terms of energy: H E N H E N N n N n H E N n n N n N n n n H E μ μ μ μ 2 2 2 2 2 2 2 )) ( ( ) ( 1 2 1 1 1 1 2 1 + = - - = - = - = - = - - - = - - = So the number of states at a particular energy is ! 2 2 ! 2 2 ! + - H E N H E N N μ μ But we want the number of states within some interval δ E. If this interval is small, we can treat the number of states at a given energy as being constant within this interval. So the total number of states within the interval is just this number of states at a particular energy level multiplied by the number of different energy levels within our energy interval. The minimum difference in energy is 2 μ H (the energy to flip one spin), so the number of energy levels within our interval is δ E/2 μ H. So the total number of states within our interval is given by H E H E N H E N N μ δ μ μ 2 ! 2 2 ! 2 2 ! + - = (b) ( 29 + - - - + = ! 2 2 ln ! 2 2 ln 2 ln ! ln ln H E N H E N H E N μ μ μ δ Apply Sterling approximation: + + - + - + + - - - - + = H E N H E N N H E N H E N H E N H E N N N H E μ μ μ μ μ μ μ δ 2 2 2 2 2 2 ln 2 2 2 2 ln 2 2 ln 2 ln ln Note that the terms on the last line all cancel, so we have
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+ + - - - - + = H E N H E N H E N H E N N N H E μ μ μ μ μ δ 2 2 ln 2 2 2 2 ln 2 2 ln 2 ln ln It’s OK if you stop here, but I think it’s handy to go a bit further: + - - + + + - - +
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