Partition Function of a Single Magnetic Particle

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 7K views
M@B
Messages
3
Reaction score
0

Homework Statement



For a magnetic particle with an angular momentum "quantum number", j, the allowed values of the z component of a particles magnetic moment are:

µ = -jδ, (-j + 1)δ, ..., (j-1)δ, jδ

δ is a constant, and j is a multiple of 1/2

Show that the partition function of a single magnetic particle is

Z = sinh[βδB(j+1/2)] / sinh[(βδB)/2]


Homework Equations



in general, Z = Σ exp(β·E(s))

and for a magnetic particle: E(s) = -µB

1 + x + x2 + ... +xn = 1 - xn+1 / 1 - x


The Attempt at a Solution



If i did things correctly, I can get to an equation:

Z = [1 - exp(-βδB(j+1/2))] / [1 - exp(-βδB/2)]

I got this just by x = exp(-βδB/2) and noticing that the n in the finite sum is 2j. (if you add j to all µ to get a sequence from 0 to 2j instead of -j to j). Then I plugged into the mathematical identity I have above. The problem is converting this into the sinh term that the question asks for. Unless of course, it is completely wrong, in which case I'm rather lost on the subject.

Thanks for the help in advance,
M@

 
Physics news on Phys.org
I see two problems. First, if you have a sum of powers from -j to +j, e.g.
[tex]x^{-j} + x^{-j+1} + ... + x^j[/tex]
when you add j to each index, you're really multiplying each term by [tex]x^j[/tex]. In order to shift the indices, you need to divide the thing by [tex]x^j[/tex]. Second, for some reason you have a negative sign in your exponentials rather than a positive sign.
 
Thank you very much for your insight. I had completely overlooked the fact that adding j to the index was actually a multiplication. I've managed to make it work out properly by taking that into consideration.

Thanks again,
M@