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### homework4

Course: ECE 407, Fall 2009
School: Cornell
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of Department Electrical and Computer Engineering, Cornell University ECE 407: Physics of Semiconductor and Nanostructures Spring 2009 Homework 4 ` Due on Feb. 17, 2009 at 5:00 PM Suggested Readings: a) Lecture notes Problem 4.1 (1D lattice) Consider a 1D lattice of lattice constant a equal to 5 Angstroms. In this problem you will find an almost exact numerical solution to the problem of an electron in a...

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of Department Electrical and Computer Engineering, Cornell University ECE 407: Physics of Semiconductor and Nanostructures Spring 2009 Homework 4 ` Due on Feb. 17, 2009 at 5:00 PM Suggested Readings: a) Lecture notes Problem 4.1 (1D lattice) Consider a 1D lattice of lattice constant a equal to 5 Angstroms. In this problem you will find an almost exact numerical solution to the problem of an electron in a periodic 1D lattice and compare this solution with approximate analytical methods discussed in the lecture. Suppose the lattice potential can be written as: 2 V (x ) = 2 V1 cos x a Where V1 equals 0.2 eV. The exact solution for any wavevector k in the FBZ can be written as a superposition of plane waves in the form: k = Where: m = c k (Gm ) k + Gm = m = c (Gm ) 1 i (k + Gm ) x e L 2 a A good approximation to the exact can be obtained by terminating the series above at both ends, as follows: Gm = m k = m = N c k (Gm ) k + Gm N Where N is some large number, say 10. Now the solution looks more like a variational solution. a) Plug the assumed form of the solution in the Schrodinger equation and show that the resulting matrix equation looks like as shown below for the case when N=2. 1 O e(k + G 2 ) V1 V1 e(k + G 1 ) V1 V1 e(k ) V1 V1 e(k + G1 ) V1 V1 e(k + G2 ) O M ( c G 2 ) c (G 1 ) c (0 ) c (G 1 ) c (G 2 ) M M ( c G 2 ) c (G1 ) = E (k ) c (0 ) c (G1 ) c (G 2 ) M The advantage of this approach is that now one can solve the matrix eigenvectors and eigenvalues numerically, and in the limit N becomes large the solution obtained is pretty much the exact solution. For numerical solution, you will need to use software like Matlab or Mathematica. a) Assume N=10. For each value of k in the FBZ from 0 to /a, numerically solve for the smallest three eigenvalues of the matrix above. This will give you the energies, E1 (k ) , E 2 (k ) , and E 3 (k ) of the lowest three bands. Plot these energies vs the k-vector from 0 to to /a. NOTE: In Matlab the routine to use is eigs, and [V,D]=eigs(A,3,SM) will give the smallest three eigenvalues of the matrix A as diagonal components of the matrix D and the corresponding eigenvectors will be the column vectors of the matrix V. b) See if bandgaps open up anywhere in the plots you generated in part (a). What is the magnitude (in eV) of these bandgaps? How does the values of the bandgaps compare to the approximate theoretical values of the bandgaps? Now assume that the potential has an extra term: 2 4 V (x ) = 2V1 cos x + 2V2 cos x a a Where V2 equals 0.4 eV. c) Plug the assumed form of the solution in the Schrodinger equation and show the resulting matrix equation in the form given above for part (a). d) Assume N=10. For each value of k in the FBZ from 0 to /a, numerically solve for the smallest three eigenvalues of the matrix. This will give you the energies, E1 (k ) , E 2 (k ) , and E 3 (k ) of the lowest three bands. Plot these energies vs the k-vector from 0 to to /a. 2 e) See if bandgaps open up anywhere in the plots you generated in part (d). What is the magnitude (in eV) of these bandgaps? How does the values of the bandgaps compare to the approximate theoretical values of the bandgaps? Problem 4.2 (Energy bands for a hexagonal lattice in 2D) Consider the following 2D hexagonal lattice from your last homework along with the primitive lattice vectors: 3a r a1 a y a r a2 x The corresponding reciprocal lattice is shown below. 4 3a 4 a r b2 ky K r b1 M M kx Suppose the potential is: rr rr r r r r r r r V (r ) = V1 cos b1 . r + V1 cos b2 . r + V1 cos b1 + b2 . r + V2 cos b1 b2 . r Where in 2D: r r =x x+y y ( ) ( ) (( )) (( )) 3 The goal of this problem is to figure out the bandgap at the K-point in FBZ using a perturbative approach. a) As a result of the periodic potential, the free-electron state with k-vector equal to ( 0 , 4 3a ), which is the K-point in the first BZ, is coupled very strongly to 2 other degenerate free electron states. What are the k-vectors of these other free-electron states? Note that the K-point is located at the intersection of 2 different Bragg lines and therefore the free electron state here becomes strongly coupled to 2 other free electron states. b) Write the solution for the actual electron wavefunction at the K-point (0, 4 3a ) as a superposition of three degenerate free electron states, obtain a 3x3 matrix energy eigenvalue equation, and then solve it to obtain: i) the energies of the 3 energy bands at the K-point (0, 4 3a ) and, ii) the actual wavefunctions of the electron at the K-point (0, 4 3a ) corresponding to the 3 energy bands c) As a result of the periodic potential, the free-electron state with k-vector equal to ( 2 3 a ,0), which is the M-point in the first BZ, is coupled very strongly to other one degenerate free electron state. What is the k-vector of this other free-electron state? d) Write the solution for the actual electron wavefunction at the M-point ( 2 3 a ,0) as a superposition of two degenerate free electron states, obtain a 2x2 matrix energy eigenvalue equation, and then solve it to obtain: i) the energies of the 2 energy bands at the M-point ( 2 3 a ,0) and, ii) the actual wavefunctions of the electron at the M-point ( 2 energy bands 3 a ,0) corresponding to the 2 e) Starting from the following free-electron band diagram, sketch (on top of it, if you want to) the actual energy bands based on your calculations in parts (a), (b), (c) and (d). 4 If you did part (b) correctly you would have noticed that the 3-fold degeneracy of the lowest 3 freeelectron bands at the K-point (as found in problem 3.2(d) in homework 3) is only partially lifted when the periodic potential is turned on; two bands remain degenerate at the K-point but the third one has a higher energy. The hexagonal Bravais lattice under discussion describes graphene which is a single atomic layer of carbon atoms with two carbon atoms per unit cell, as shown below. The underlying hexagonal Bravais lattice is shown by the small black dots. The two carbon atoms per primitive cell are shown by the blue and red filled circles. y A graphene crystal r a1 r a2 3a a a x In part (b) you showed that two bands at the K-point remain degenerate even in the presence of the periodic atomic potential. This degeneracy has something to do with inversion symmetry of the graphene crystal. So we look at that next. Inversion Symmetry: a 2D crystal will have inversion symmetry with respect to y-axis if, V ( x, y ) = V (x, y ) . Similarly, a 2D crystal will have inversion symmetry with respect to the x-axis if, V (x, y ) = V (x, y ) . f) Verify that the potential: rr rr r r r r r r r V (r ) = V1 cos b1 . r + V1 cos b2 . r + V1 cos b1 + b2 . r + V2 cos b1 b2 . r has inversion symmetry with respect to the y-axis. ( ( ) ) ( ( ) ) (( (( )) )) (( (( )) )) g) Verify that the potential: rr rr r r r r r r r V (r ) = V1 cos b1 . r + V1 cos b2 . r + V1 cos b1 + b2 . r + V2 cos b1 b2 . r has inversion symmetry with respect to the x-axis. We will now break the inversion symmetry of the crystal and see what happens. In graphene, the blue and the red carbon atoms are exactly the same and therefore the atomic potential in graphene has inversion symmetry with respect to the y-axis. However, if the blue and the red atoms were not the same, say one was carbon and the other was silicon, then the inversion symmetry could be broken. In our model, we can break the inversion symmetry by assuming the following form of the potential: rr rr r r r r r r r V (r ) = V1 cos b1 . r + V1 cos b2 . r + V1 sin b1 + b2 . r + V2 cos b1 b2 . r ( ) ( ) (( )) (( )) h) Verify that the above potential does not ...

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