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Conduction electron ferromagnetism

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    We approximate the effect f exchange interactions among the conduction electrons if we assume that electrons with parallel spins interacti with each other with energy -V, and V is positive, while electrons with antiparallel spins do not interact with each other.

    (a) Show with the help of prblem 5 that the total energy of the spin-up band is:
    [tex] E^+=E_0(1+x)^(^5^/^3^)-(1/2)N \mu B(1+x) - (1/8)VN^2(1+x)^2[/tex]

    (b) Find a similar expression for E-. Minimise the total energy and solve for x in the limit x<<1. Show that the magnetisation is:

    [tex] M =(3N \mu B)/(2E_F-(3/2)VN) [/tex]

    (c) Show that with B =0 the total energy is unstable at x = 0 when [tex]V > (4E_f)/3N[/tex]

    2. Relevant equations
    [tex]M = n \mu x[/tex]


    3. The attempt at a solution
    This question is out of Kittel (8th edition) chapter 11. I easliy managed to do part (a).

    For (b) obviously E- was easy to find because it was just a couple of sign reversals. So for my total energy I just added E+ and E- to get:

    [tex] E_t_o_t = E_0(1+x)^(^5^/^3^)-(1/2)N \mu B(1+x) - (1/8)VN^2(1+x)^2 +E_0(1-x)^(^5^/^3^)+(1/2)N \mu B(1-x) - (1/8)VN^2(1-x)^2[/tex]

    Now I know I had to differentiate with respect to x to 'minimise' the total energy and then set that to 0 and solve for x. However, I don't get the right answer for the magnetisation. I found something on the web that said the following term is equivalent to ([tex](4/3)x[/tex]):

    [tex] (1+x)^(^2^/^3^) - (1-x)^(^2^/^3^)[/tex] *****

    If you're wondering where I get that term from it is part of the differential.

    [tex] dE_t_o_t/dx =(5/3)E_0[(1+x)^(^2^/^3^)-(1-x)^(^2^/^3^)] -N \mu B - (1/2)V(N^2)x [/tex]

    Now my main question with this part is why does that term ***** equal (4/3)x?? The question works out absolutely fine once I do this, but I don't get why???

    (c) For this last part the lecturer said we had to find the second derivative to show that it is unstable for V is greater than some value. But I don't quite understand this because there is no B or x in the second derivative. So what is the point of being given the conditions of B = 0 and x = 0???

    Thanks for your time.
    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 23, 2007 #2
    Part (b) specifies in the small x limit (ie close to the special x=0 point), so you use the usual expansion
    (1+x)^n ~ 1+nx+... and truncate the ... part. Voila: 4/3x.
    Part (c), the point is that the extrema of the energy are found by dE/dx=0; you are to focus on the extremum at B=0, x=0. The second derivative is independent of B and x as you say, but the extremum is still at those values of the parameters. You then notice that the energy is of the form "M-N", so you can show for one range of V the extremum is a minimum and thereore stable and for another of V is a local maximum and therefore unstable.
  4. Oct 23, 2007 #3
    Thanks for clearing that up!! I really appreciate it. Everything works fine now.
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