3.3 The spin-half paramagnet

Take-home message: The paramagnet is the simplest illustration of statistical ideas

What is a spin- ${\textstyle \frac 1 2}$ paramagnet? A paramagnet is a substance which can be magnetised by an external magnetic field, and the magnetisation is aligned with the external field. Unlike a ferromagnet the response is weak and does not remain if the field is switched off.

A crystal of atoms with non-zero spin will act as a paramagnet, as the spins will tend to line up in an external field. From quantum mechanics we know the spin-projection along the field can only take certain discrete values. For simplicity we consider spin- ${\textstyle \frac 1 2}$, so that $s_z=\pm {\textstyle \frac 1 2}$. In an ideal paramagnet, the spins do not feel one another, but react independently to an external field.

Thus the ideal paramagnet is a lattice of $N$ sites at each of which the spin points either up or down. Each of these has a magnetic moment $\pm \mu$. Only the total magnetic moment is macroscopically measurable, and this is just the sum of the individual moments. If $n_\uparrow$ spins are pointing up and $n_\downarrow =N-n_\uparrow$ are pointing down, the total magnetic moment is

$$m=n_\uparrow \mu+n_\downarrow (-\mu)=\mu(2n_\uparrow -N).$$

In an external field, the spin-up atoms will have lower energy, $-\mu B$, and the spin-down atoms have higher energy, $\mu B$, so the the total energy is just

$$E=n_\uparrow (-\mu B)+n_\downarrow (\mu B)=-Bm$$

However we are going to start with zero external magnetic field so that all states have the same energy. The magnetisation is then an example of an extra degree of freedom not specified by the energy, as discussed in the previous section.

If you haven't already looked at the chequerboard examples, here and here, do so now. All the pictures carry over if for ``blue'' you read spin-up. (The chequerboard is 2-D and a crystal is 3-D, but in the absence of interaction the geometry is irrelevant; all that counts is the total number $N$ of atoms.)

So macrostates are characterised by their magnetic moment (or magnetisation, $M=m/V$), but microstates by the list of spins at each site. For $N=3$ there are four macrostates and nine microstates.

In general the number of microstates for a given macrostate, with $n_\uparrow = {\textstyle \frac 1 2}(N+m/\mu)$, is

$$\Omega(n_\uparrow )={N!\over n_\uparrow ! (N-n_\uparrow )!}$$

You should be able to prove that the sum of the $\Omega(N,n_\uparrow)$ over all $n_\uparrow$ is $2^N$.

Below we plot $\Omega(n)$, normalised to 1 at the peak, as a function of $n/N$, for different values of $N$.

As $N$ gets larger, the function is more and more sharply peaked, and it is more and more likely that in the absence of an external magnetic field there will be equal numbers of up and down spins, giving zero magnetisation.

For large $N$, the curve is very well approximated by a Gaussian,

$$\Omega(n)\propto e^{-{(n-N/2)^2\over N/2}}$$

with a mean of $N/2$ and a standard deviation $\sigma=\sqrt{N}/2$. Thus the fractional size of fluctuations goes as $1/\sqrt{N}$.

Since the probabilities of various sizes of fluctuations from the mean in a Gaussian are known, we can show that in a macroscopic system, $100\sigma$ deviations are vanishingly unlikely. Even they would be undetectable, so the macroscopic magnetisation is very well defined indeed.

References

• Mandl 2.1
• Bowley and Sánchez 4.3
• Kittel and Kroemer 1,2