Consider two simple bands in a 2-D square lattice with lattice constant a, one i

Question

Consider two simple bands in a 2-D square lattice with lattice constant a, one is given by: epsilon_1, k = - epsilon_0(cos k_x a + cos 2k_y a), another is given by: epsilon-2, k = - epsilon_0 (cos 2 k_x a + cos k_y a). a. Calculate the effective mass at k= 0 in the x-direction. Which band is heavier? b. Assume an electron is in the first bank at k = 0 at t = 0, an electric field E = E_0 x + 2 E_0 y is applied. We know that the motion of the electron will be periodic. Find the period of the motion.

Explanation / Answer

Solution :-

consider to simple bands in a 2-D square lattice with lattice constant a

One is given by : 1,k = -0 (Cos kxa + Cos 2Kya)

And another is given by 2,k = -0 (Cos 2kxa + Cos Kya)

a) Effective mass at k = 0

For first, 1=  -0 (Cos 0 + Cos 0)

1 =  0 ( Because Cos 0 = 1)

Hence mass =  1/ 0 = 1 units (Considering positive value)

And For second,  2,0 = -0 (Cos 0 + Cos 0)

Therefore mass = 2/-0 = 1 units (Considering positive value)

Hence both band is equal in mass

b) Given

Electric field, E= Eox + 2 Eoy

And we know that time for the electron to go to k = G state in extended Brillouin zone i.e. we limit ourselves to reduced wave vectors.

Therefore

dE/dk = d/dk (Eox + 2 Eoy)

Since k = G

dE/dG= d/dG (Eox + 2 Eoy)

At t = 0, x = y = 0

Hence the periof of the motion is as follows:-

dE/dG is approximate = 0

Hence period is Sin 0 becuse Sin0 is 0.

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