Why is the Term Equivalent to (4/3)x in Conduction Electron Ferromagnetism?

Click For Summary
SUMMARY

The discussion focuses on the derivation of the total energy expressions for conduction electron ferromagnetism, specifically addressing the interaction of electrons with parallel and antiparallel spins. The total energy expressions for the spin-up and spin-down bands are derived, leading to the magnetization formula M = (3N μ B) / (2E_F - (3/2)VN). The critical condition for stability is established, indicating that the total energy is unstable at x = 0 when V exceeds (4E_f) / (3N). The equivalence of the term ((4/3)x) is clarified through Taylor expansion techniques.

PREREQUISITES
  • Understanding of conduction electron ferromagnetism principles
  • Familiarity with energy expressions in quantum mechanics
  • Knowledge of Taylor series expansions
  • Basic concepts of magnetization and its dependence on energy
NEXT STEPS
  • Study the derivation of energy expressions in conduction electron systems
  • Learn about Taylor series expansions and their applications in physics
  • Explore the stability conditions in ferromagnetic systems
  • Investigate the role of exchange interactions in electron behavior
USEFUL FOR

Physicists, particularly those specializing in condensed matter physics, students studying quantum mechanics, and researchers focusing on magnetic materials and their properties.

big man
Messages
242
Reaction score
0

Homework Statement


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:
E^+=E_0(1+x)^(^5^/^3^)-(1/2)N \mu B(1+x) - (1/8)VN^2(1+x)^2

(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:

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

(c) Show that with B =0 the total energy is unstable at x = 0 when V &gt; (4E_f)/3N

Homework Equations


M = n \mu x

E_0=(3/10)NE_f

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:

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

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 ((4/3)x):

(1+x)^(^2^/^3^) - (1-x)^(^2^/^3^) *****

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

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

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:
Physics news on Phys.org
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.
 
Thanks for clearing that up! I really appreciate it. Everything works fine now.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
10
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K