Probabilities for two spin systems interacting in isolation

  • Thread starter PhotonSSBM
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Homework Statement


Consider a system A consisting of a spin 1/2 having magnetic moment ##\mu_0##,
and another system A' consisting of 3 spins 1/2 each having magnetic moment ##\mu_0##.
Both systems are located in the same magnetic field B .
The systems are placed in contact with each other so that they are free to exchange energy.
Suppose that, when the moment of A points up (i.e., when A is in its + state), two o f the
moments o f A' point up and one of them points down.
Count the total number of states accessible to the combined system A + A ' w hen the moment o f A points up, and when it points down. Hence calculate the ratio P_/P+, where
P_ is the probability that the moment of A points down and P+ is the probability that it points up.
Assume that the total system A + A ' is isolated.

Homework Equations


Basic statistics.

The Attempt at a Solution


I feel like my answer this this is waaaay off just because the way I'm thinking about this makes it seem too easy.

It feels like the probability for A's moment to be + is 3/4, as just from reading previous examples in the book all they seem to note is that it would be the number of + spins over the total number of spins. The - probability would be similar.

What's telling me something is wrong about this is when I generalize to N number of spins in A', I don't get answers that seem right when using this logic.

Is it really this easy or am I missing something???
 

Answers and Replies

  • #2
kuruman
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You show a lot of feeling but you show no answers. What is your answer to the first question, count the total number of states accessible to the combined system A + A' when the moment of A points up? What about the second question, when the moment of A points down? Start from there.
 
  • #3
When A points up the total number of states is 3, as A always remains +, and there are three combinations of two + and 1 - in A'.

So P+ is 3/4

When A points down, we need to conserve energy, so all 3 of A' spins go to +. There is then only one configuration, (-|+++)

so P- is 1/4

I did offer those answers in my post, I guess I wasn't clear though.

Where am I going wrong here then???
 
  • #4
kuruman
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So far all this is correct and clear now. I assume you got the correct result for ratio P-/P+ that the problem is asking you to find. Now you say that when you generalize to N spins, what did you do? Did you assume that there is only one spin in A and N pins in A' or what? What assumptions did you make about the total energy? Bottom line, what makes you think that your generalization is not right?
 
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  • #5
Generalizing to N total spins in A' with n being up and N-n being down and finding P-/P+, while simplifying some things, just produced too neat an answer for me to be happy with given the rest of the homework.

I guess we got some easy problems then. No biggie.

I'm so used to working in Griffiths at this point I think that if an answer I get doesn't take me an hour I'm doing it wrong :)
 
  • #6
kuruman
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Yes, some problems in Griffiths are that easy and others are bears.
 

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