I have to calculate the height and width of the multiplicity function for a two-state paramagnet. I'm supposed to do this analogously to the method in the text book, which I will try to outline below. The text book uses a lot of approximations: for instance, the first derivation considers a limit of $q\gg N$. Hopefully it's still clear what I'm asking.

Calculating the multiplicity of an Einstein solid for the limit $q\gg N$

First I'll consider the example from the text book for an Einstein solid of $q$ units of energy and $N$ oscillators.

Starting with the exact formula for multiplicity:


Taking the natural logarithm and applying Stirling's approximation:

$$\ln\Omega=\ln\left(\frac{\left(q+N\right)!}{q!N!}\right)$$ $$\ln\Omega=\ln\left(q+N\right)!-\ln q!-\ln N!$$ $$\ln\Omega\approx\left(q+N\right)\ln\left(q+N\right)-\left(q+N\right)-\left(q\ln q-q\right)-\left(N\ln N-N\right)$$ $$\ln\Omega=\left(q+N\right)\ln\left(q+N\right)-q\ln q-N\ln N$$

Using $q\gg N$, we can manipulate $\ln\left(q+N\right)$ using a Taylor series of only the first term:

$$\ln\left(q+N\right)=\ln\left[q\left(1+\frac{N}{q}\right)\right]$$ $$\ln\left(q+N\right)=\ln\left(1+\frac{N}{q}\right)$$ $$\ln\left(q+N\right)\approx\ln q+\frac{N}{q}$$

Plugging this result into the previous expression derived through Stirling's approximation, we find:

$$\ln\Omega\approx N\ln\frac{q}{N}+N+\frac{N^2}{q}$$

of which the last term is approximately zero in the limit $q\gg N$, to find a final expression of:

$$\Omega\left(N,q\right)\approx e^{N\ln(q/N)} e^N=\left(\frac{eq}{N}\right)^N$$

This is the example in the text book.

Calculating the multiplicity of a two-state paramagnet

The next step is to apply the same methods to a two-state paramagnet, for which the multiplicity function is given to be:


Taking the logarithm on both sides:

$$\ln\Omega=\ln\left(\frac{N!}{\left(N-N_\downarrow\right)!N_\downarrow!}\right)$$ $$\ln\Omega=\ln N! -\ln\left(N-N_\downarrow\right)!-\ln N_\downarrow!$$ $$\ln\Omega=N\ln N - \left(N-N_\downarrow\right)\ln\left(N-N_\downarrow\right)-N_\downarrow\ln N_\downarrow$$

Approximation of the logarithm $\ln\left(N-N_\downarrow\right)$ can be done by using the Taylor expansion about $a=0$:

$$\ln\left(1-x\right) \approx x + \frac{x^2}{2}+\frac{x^3}{3}+\dots$$

Manipulating the logarithm like so:

$$\ln\left(N\left(1-\frac{N_\downarrow}{N}\right)\right)=\ln N+\ln\left(1-\frac{N_\downarrow}{N}\right)$$

Allows to set $x=\frac{N_\downarrow}{N}$ (granted $\ln\left(1-x\right)\approx x$) to acquire a final expression (after cancelling some terms) of:

$$\ln\Omega\approx N_\downarrow\ln \frac{N}{N_\downarrow}+N_\downarrow+\frac{N_\downarrow^2}{N}.$$

Which consequently results in an expression for the multiplicity of:

$$\Omega\approx e^{N_\downarrow\ln \left(N/N_\downarrow\right)}e^{N_\downarrow}=\left(\frac{eN}{N_\downarrow}\right)^{N_\downarrow}$$

Similarly, the multiplicity as function of $N_\uparrow$ is:

$$\Omega\approx e^{N_\uparrow\ln \left(N/N_\uparrow\right)}e^{N_\uparrow}=\left(\frac{eN}{N_\uparrow}\right)^{N_\uparrow}$$

(In both cases the limit $q\gg N$ still applies, so as to take $\frac{N_\downarrow^2}{N}$ to be negligible in the final result.)

Intermezzo: the height of the peak of the multiplicity function for two large Einstein solids

Furthermore, in the text book example, the height of the peak in the multiplicity function of two large Einstein solids is calculated through:


And consequently setting $q_A=q/2$ to find:


Applying to the same methods to a two-state paramagnet

I've tried doing the same thing, but thing are starting to not add up now. Filling in $\Omega=\Omega_\uparrow\Omega_\downarrow$ (multiplicity of a combined system - using a combined system is a guess, but things are not adding up either if I calculate the multiplicity of a single solid), I find:


Filling in $N_\uparrow=N_\downarrow=N/2$, I find:


Even though I believe this is the wrong answer, the expression itself is not such a big issue until the next exercise, where I'm suppose to calculate the multiplicity function in the vicinity of the peak, by setting $N_\uparrow=\frac{N}{2}+x$ (from which automatically follows $N_\downarrow=\frac{N}{2}-x$).

So far so good, I plug that into the general equation I found for the multiplicity of the combined system. After rewriting a little, I find:


Apart from being able to plug in $x=0$ which checks out for my previous answer, the result seems very fishy, since I cannot work with it to estimate the height of the peak (third problem - follows now).

Calculating the width of the peak in the multiplicity function of combined system of two Einstein solids

To consider what the graph looks like near the peak, plug in the expressions for up and down-state $N$:


Manipulating the logarithm again:


After two steps this check out to be:


Plugging this back into the original equation for multiplicity, I find $\Omega$ to be:

$$\Omega=\left(\frac{e}{N}\right)^{2N}e^{N\ln(q/2)^2}e^{-N(2x/q)^2}=\Omega_\mathrm{max}\cdot e^{-N(2x/q)^2}.$$

The solution in the text book example is great! It allows to calculate an approximate peak width of $\Delta x=q/\sqrt{N}$ (at which point the multiplicity falls off by a factor of $1/e$). But my equation doesn't check out so nicely with my original expression of $\Omega_\mathrm{max}$, and I'm not sure what next step to take. Any suggestions welcome.


Perhaps I wrongly took the low temperature limit for my two-state paramagnet ? Can anyone clarify and explain why I should take the high temperature limit?


1 Answer 1


Think I found the answer:

First we must determine what value we must set for $N_\uparrow$, given is an expression for $x$, so we rewrite that first to be $N_\uparrow=\frac{N}{2}+x$.

Similarly to the answer to the previous exercise, we now calculate the multiplicity like so: $$\Omega=\frac{N!}{\left(\frac{N}{2}+x\right)!\left(N-\left(\frac{N}{2}+x\right)\right)!}. $$ We can rewrite one term a little: $N-\left(N/2+x\right)=N/2-x$, to then take the logarithm on both sides: $$\ln\Omega=\ln\left(\frac{N!}{\left(\frac{N}{2}+x\right)!\left(N-\left(\frac{N}{2}+x\right)\right)!}\right)$$ $$=\ln N!-\ln\left(\frac{N}{2}+x\right)!-\ln\left(\frac{N}{2}-x\right)!$$ $$=\left(N\ln N - N\right)-\left(\left(\frac{N}{2}+x\right)\ln\left(\frac{N}{2}+x\right)-\left(\frac{N}{2}+x\right)\right)-\left(\left(\frac{N}{2}-x\right)\ln\left(\frac{N}{2}-x\right)-\left(\frac{N}{2}-x\right)\right)$$ $$=N\ln N-\frac{N}{2}\left(\ln\left(\frac{N}{2}-x\right)+\ln\left(\frac{N}{2}+x\right)\right)+x\left(\ln\left(\frac{N}{2}-x\right)-\ln\left(\frac{N}{2}+x\right)\right).$$ Now, we have to manipulate the logarithms in the same fashion as (2.26):

$$\ln\left(\frac{N}{2}+x\right)=\ln\left(\frac{N}{2}\left(1+\frac{2x}{N}\right)\right)$$ $$=\ln\frac{N}{2}+\ln\left(1+\frac{2x}{N}\right)$$ $$\approx\ln\frac{N}{2}+\frac{2x}{N}$$ In the same fashion, we find $\ln\left(\frac{N}{2}-x\right)\approx\ln\frac{N}{2}-\frac{2x}{N}$. Substituting this result into equation \ref{eqn:ex2.24.5}:

$$\ln\Omega\approx N\ln N-\frac{N}{2}\left(\left(\ln\frac{N}{2}-\frac{2x}{N}\right)+\left(\ln\frac{N}{2}+\frac{2x}{N}\right)\right)+x\left(\left(\ln\frac{N}{2}-\frac{2x}{N}\right)-\left(\ln\frac{N}{2}+\frac{2x}{N}\right)\right)$$ $$= N\ln N-\frac{N}{2}\left(2\ln\frac{N}{2}\right)+x\left(-\frac{4x}{N}\right)$$ $$= N\left(\ln N-\ln\frac{N}{2}\right)-\frac{4x^2}{N}$$ $$= N\ln2-\frac{4x^2}{N}$$ Similar to equation (2.27), we find the following expression: $$\Omega=\Omega_\mathrm{max}\cdot e^{-4x^2/N}$$ Filling in $x=0$ shows $\Omega=\Omega_\mathrm{max}$ as desired.

  • $\begingroup$ If you can find a mistake in here, let me know. $\endgroup$
    – user55789
    Aug 20, 2014 at 18:01

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