Parameters of electron dynamics can be directly related to...

Parameters of electron dynamics can be directly related to the band dispersion and band width. Consider a square lattice of atoms with the band energy \(\epsilon(\mathbf{q}) = \epsilon_0 + 2t\cos(q_xa) + 2t\cos(q_ya)\), where a is the lattice constant and \(t = \text{const}\) is the hopping parameter (interaction between the adjacent atoms). (a) Determine the band width. (b) Show that band dispersion is parabolic when \(|\mathbf{q}|a \ll 1\). (c) Use the parabolic dispersion to determine electron velocity and density of states at \(|\mathbf{q}|a \ll 1\). (d) Determine the effective mass in the same \(|\mathbf{q}|a \ll 1\) limit.

Transcript

00:01 Hello students, let's solve this question.

00:02 So the schrodinger equation is psi of r is equal to 1 by root of l cube into e raised to i k bar dot i bar now the born -gorman periodic boundary condition states that psi of r plus l is equal to psi of r so we can write 1 by root of l cube e raised to i into k bar into r bar plus l bar will be equal to 1 by root of l cube into e raised to i into k bar into r bar or we can say that e raised to i into k l will be equal to e raised to i 2 n r and here sorry 2 pi n or k can be written as 2 pi n divided by l where n value will be equal to 0 1 2 3 etc so we can say that the solution is quantized for the given value n now for the second part of the question the wave function of an electron is given by psi n is equal to c into exponential i into k n a minus omega t and therefore we can write psi n will be equal to c into exponential of i into k n minus 1 into a here.

01:52 This will be psi of n minus 1 into a minus omega t and psi of n plus 1 will be equal to c into exponential i into n plus 1 into a into k minus omega t so applying newton's law of motion we'll get f will be equal to minus beta psi n plus 1 plus psi n minus 1 minus 2 psi n therefore it can be written as minus k into dou psi n by dou t will be equal to minus of beta into a psi n plus 1 plus psi n minus 1 minus 2 into psi n now this can be written as a into b psi n minus a into psi n plus 1 plus psi n minus 1 now we can write minus i into t c exponential of i into k n a minus omega t into minus i omega minus b into sorry exponential of i into k n a minus omega t therefore it can be subtracted by minus a into c into exponential of i into k n a minus omega t plus k a plus c into exponential of i into k n a minus omega t minus k into a or it can be equal to k into omega is equal to b minus a e raised to i into k a plus e raised to minus i into k a which can be equal to b minus 2 a cos k a where k tends to 0 and cos k a is equal to 1 therefore omega will be equal to 1 by t b minus 2 a and if k tends to infinity omega will be equal to 1 by t into b minus 2 a cos k a therefore by knowing a and b we can find the exact energy of the electrons.

04:34 Let's plot that is from minus pi a here it is k tends to infinity and here it is pi by a we can draw the wave function...

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