Statistical mechanics problem of spin

athrun200
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Homework Statement


Consider a system consists of two subsystem A and A' in which A contains 3 spins and A' contains 2 spins. Suppose that, when the systems A and A' are initially separated from each other, measurements show the total magnetic moment of A to be - 3{\mu _0} and the total magnetic moment of A' to be + 4{\mu _0}. The systems are now placed in thermal contact with each other and are allowed to exchange energy until the final equilibrium situation has been reached. Under these conditions calculate:
(a) The probability P(M) that the total magnetic moment of A assumes anyone of its possible values M.

Homework Equations


Basic statistic and probability tools.

The Attempt at a Solution


The question implies that each of the spin in A has magnetic moment {\mu _0} while A' has 2{\mu _0}.
Originally, the system has the spin arrangement like this:
--- ++
where - means spin down and vice versa.

After they are in contact, they can have the following possibility.
--- ++ - 3{\mu _0}
--+ -+ - {\mu _0}
-+- -+ - {\mu _0}
+-- -+ - {\mu _0}
+-- +- - {\mu _0}
+-+ -- {\mu _0}
++- -- {\mu _0}
--+ +- - {\mu _0}
-++ -- {\mu _0}
-+- +- - {\mu _0}

So P( - 3{\mu _0}) = \frac{1}{{10}}, P( - {\mu _0}) = \frac{6}{{10}}, P({\mu _0}) = \frac{3}{{10}}.

But the answer is P({\mu _0}) = \frac{6}{7}, P( - 3{\mu _0}) = \frac{1}{7}.

What's wrong?
 
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Hi athrun200!

--+ -+ would have total moment -\mu_0, not +\mu_0, as is required by conservation of energy.
 
I know --+ -+ produces total moment - {\mu _0}
But how does it affect the answer?
 
The original configuration --- ++ had total moment +\mu_0, yes?

So all the possible configurations that you're counting should have total moment +\mu_0, not -\mu_0.
 
Oxvillian said:
The original configuration --- ++ had total moment +\mu_0, yes?

So all the possible configurations that you're counting should have total moment +\mu_0, not -\mu_0.

Oh! Now I get it, thanks a lot!
 
np :smile:
 
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