Statistical mechanics: particles in magnetic fields

[MatinSAR]

Homework Statement

We are examining a system that is composed of four electrons. Each of these electrons possesses a non-zero spin. These electrons each have a magnetic moment, denoted as $m$, which interacts with an external magnetic field, represented as $B$.
The spin of these electrons can align in two ways with respect to this external magnetic field. It can either be parallel (which we refer to as ‘up spin’) or anti-parallel (referred to as ‘down spin’).
When the magnetic field is present, the energy of the electrons changes depending on their spin. If an electron has an ‘up spin’, its energy is $-mB$. Conversely, if an electron has a ‘down spin’, its energy is $+mB$. Answer following questions:

Relevant Equations

Concepts in statistical mechanics.

Let’s consider that the total energy of this system is represented as $E=-2mB$.

Question 1: how many microstates correspond to this energy level?
We have $2^4=16$ microstates.

++++ Total magnetic moment: $4m$ Energy: $-4mB$
- - - - Total magnetic moment: $-4m$ Energy: $4mB$

- - -+ Total magnetic moment: $-2m$ Energy: $2mB$
- -+- Total magnetic moment: $-2m$ Energy: $2mB$
-+- - Total magnetic moment: $-2m$ Energy: $2mB$
+- - - Total magnetic moment: $-2m$ Energy: $2mB$
+++- Total magnetic moment: $2m$ Energy: $-2mB$
++-+ Total magnetic moment: $2m$ Energy: $-2mB$
+-++ Total magnetic moment: $2m$ Energy: $-2mB$
-+++ Total magnetic moment: $2m$ Energy: $-2mB$

++-- Total magnetic moment: $0$ Energy: $0$
+-+- Total magnetic moment: $0$ Energy: $0$
-++- Total magnetic moment: $0$ Energy: $0$
-+-+ Total magnetic moment: $0$ Energy: $0$
--++ Total magnetic moment: $0$ Energy: $0$
-++- Total magnetic moment: $0$ Energy: $0$

So the answer to first question is $4$.

Question 2: What is the probability that the system, when in equilibrium, is in one of its microstates?
$\dfrac {4}{16}$.

Question 3: What is the probability that a particle in this system has an up spin? Calculate the average magnetic moment.
Possible microstates are : +++-/++-+/+-++/-+++
The probability is $\dfrac 3 4$.
Average magnetic moment: $$m_{ave}=m(\dfrac 3 4)-m(\dfrac 1 4) = \dfrac m 2 $$ I have significant uncertainty about my answer to this part of the question.

Question 4: If the spin of particle one is up, what is the probability that the spin of particle two will also be up?
+++-/++-+/+-++ So I think $\dfrac 2 3$.

I'm aware that this question may seem straightforward, but I’m uncertain about the accuracy of my answers. Any ideas or corrections would be appreciated.

 

[Orodruin]

1: Significantly less writing to note that you need 3 spin up and one spin down to make net energy match. You have $4 \choose 3$ ways of selecting the 3 up spins.

2: Define equilibrium here. When all states are equally probable? Note that the energu is given! The problem is asking for the probability of a particular state.

3: what are you uncertain about?

4: Yes, basic conditional probability.

 

[MatinSAR]

1: Significantly less writing to note that you need 3 spin up and one spin down to make net energy match. You have $4 \choose 3$ ways of selecting the 3 up spins.

Yes, it is better to use your approach especially when we have many microstates.

2: Define equilibrium here. When all states are equally probable?

Yes. The energy corresponds to 4 microstates.
+++-
++-+
+-++
-+++
When the system is in equilibrium all states are equally probable, so each one has a probability of $1/4$.

3: what are you uncertain about?

The formula that I've used to calculate average of $m$. And I am not sure if I identified the probability of up and down spin($3/4$ and $1/4$) correctly.

4: Yes, basic conditional probability.

Many thanks.