Solid State Physics Homework 5: Assigned Mon, 13 Feb 2012; Due Wed, 22 Feb 2012

Dr. Colton, Winter 2012

1.  Spin polarization. The Zeeman effect describes an energy splitting that occurs between spin up and spin down states of electrons in a magnetic field. If you picture electrons as tiny bar magnets, the energy difference between the two states comes from the work you would need to do to take a magnet that’s aligned with the field, and rotate it 180° so that it’s aligned against the field. The equation is such:

In that equation, B is the magnetic field, mB is the “Bohr magneton” constant 9.274 ´ 10-24 J/T, and g is the electron’s “g-factor” for the material. For your reference, g is very close to 2 for free electrons and for silicon, but in some materials can be far from that. Electrons in GaAs, for example, have a g-factor of about 0.44.[*]

(a) One can often consider the spins of the electrons in a solid to be non-interacting, and therefore Boltzmann statistics apply. Use the Boltzmann factors for the upper and lower states to show that the spin polarization P at a certain field and a certain temperature is equal to:

The spin polarization P is defined to be the difference in spin populations, divided by the total spin population.

Hint: Although normally it’s easiest to call the lower state E = 0, in this case I think you’ll find it’s best to say E = 0 at the midpoint of the two states.

(b) What’s the spin polarization in GaAs at 7 T and 1.4 K? (That’s the highest field/lowest temperature that I can achieve in my lab.) What’s the spin polarization for Si under the same conditions?

2.  Density of states for 2D square lattice. In HW 4, problem 5, we found that for out-of-plane oscillations of a 2D square lattice, the dispersion is given by:

When plotted in 3D with C = M = a = 1, that equation looks like this, where the x- and y-axes cover the Brillouin zone:

(a) Note that the Brillouin zone can be divided into four equivalent squares. For simplicity let’s just consider the portion where kx is between 0 and p/a, and ky is also between 0 and p/a. Use a computer to divide that portion up into 400 smaller pieces (20 ´ 20; see the “Note” at end of problem). Calculate w for the center of each piece, then plot a histogram of the results. Use about 50 “bins”. This is the density of states.[*] (It is different from the 2D Debye model density of states that you solved in the problem above because the Debye model assumed a purely linear relationship between w and k. This problem uses the actual relationship between w and k, for this particular 2D situation.)

(b) You should see a cusp in the density of states. With reference to the above 3D plot, explain qualitatively why the cusp appears.

Note: In the instructions I said 400 pieces (20 ´ 20) and 50 bins so that you can do the problem with a spreadsheet such as Excel without too many headaches. However, if you can write a computer program to do this for you (I used Matlab), it’s easy to divide the Brillouin zone into many more pieces. The results are MUCH better if you divide things up into, say, 250,000 small pieces (500 ´ 500) and 1000 bins. It shouldn’t be a hard program to write if you have much programming experience; it took me fewer than 30 Matlab lines to get it to work.

Side note: This problem is based on one that I found in an older edition of Kittel (6th edition). This line from that problem made me chuckle—the instructions started with, “If you have access to a microcomputer…” How times have changed! J Incidentally, I did this as an extra-credit problem when I was a student and I believe it was the single-most helpful homework problem for me from the whole semester. It really allowed me to grasp the concept of density of states.

3.  Kittel 5.1. Singularity in density of states. Addition: in part (b), please do a quick sketch of your result, say from just below w0 to just above w0. Just a quick sketch on paper; no need to use Mathematica if you don’t want to. Note that this problem has an error in it; here’s what I wrote on the errata sheet:

p. 128: Problem 5-1, Singularity in density of states. In the last sentence, change the word “discontinuous” to “continuous, but has a kink.”

4.  Debye model: 1D. Calculate the specific heat due to phonons in the Debye model for the 1D case. Follow the steps I used in class (and in the handout) for the 3D case. To review, those steps were: (a) Write down the density of states. (b) Calculate the cutoff frequency, wD. (c) Write down the integral expression for the energy. Simplify as much as you can, and write the integral in terms of x (= w/kT) instead of w. (d) Do the integral for the low temperature regime. Take the derivative of U to get CV for this regime. Please leave this CV expression in terms of L. (e) Do the integral for the high temperature regime. Take the derivative of U to get CV for high temperatures. Please leave this CV expression in terms of N.

5.  Debye model: 2D. Repeat the previous problem, for the 2D situation. (For part (d), leave in terms of A.) By the way, these 1D and 2D problems are not just abstract mathematical constructs. Graphene, for example, acts in most respects as a 2D material. And there are other materials that act like 1D substances—for example, read through Kittel problem 5.4, Heat capacity of layer lattice. He describes a situation where you get a 2D-like heat capacity at low temperatures, CV ~ T2, and a 1D-like heat capacity at even lower temperatures, CV ~ T.

6.  Heat capacity of 1D linear chain. In Kittel problem 5.1 above, you solved for the density of states of the 1D linear chain situation—not a 1D Debye model, but rather the actual 1D monatomic balls & springs problem. Use that information to calculate the heat capacity for that chain (a) for low temperatures, and (b) for high temperatures. Note: I got it to work out by taking dU/dT before doing the low & high temperature approximations, and even before I substituted in for . So, to make the grader’s life easier, please do that also. Word of caution: When you do the high temperature approximation, you’ll have a term like sqrt(xm2 – x2). Since x is small in this approximation, you’ll be tempted to say that that expression equals sqrt(xm2 – small) = sqrt(xm2) = xm. Don’t do that! While it’s true that x is small, xm is also small, so you can’t do things that way. Just leave that term in the integral and use Mathematica (or other favorite method) to integrate it. Final hint: You should end up with answers that share similarities with the Debye model: 1D problem above: at low temperatures CV should be linear with temperature, and at high temperatures CV should be a constant (the same constant as found for the previous problem).

Phys 581 – HW 5 – pg 2

[*] To be precise, it’s actually –0.44. The negative sign just means that spin of the electrons tends to anti-align with the field, instead of align. (Obviously the bar magnet picture only works up to a certain point.) But since the g’s in the two equations should technically be |g|’s, don’t worry about the negative sign when doing the calculation in part (b).

[*] To be precise, the density of states is really the limit of this histogram as you use an infinite number of pieces and an infinite number of bins. And normalized, I suppose.