Consider a square wave potential with lattice constant a. In the first unit cell, this means:

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1 Consider a square wave potential with lattice constant a. In the first unit cell, this means: = for - a/4 < x < a/4, =0 for -a/ < x < - a/4 and a/4 < x < a/, This is repeated throuhout the crystal. 1. Find the two lowest enery eienvalues at the center and ede of the first Brillouin zone, as well as one point halfway between.. Find the wave function correspondin to each of these two bands at the three specified -values. 3. Graph the electron density as a function of position correspondin to each of these six wave functions. Note: I chaned the subscript on the constant in the amplitude, since I will use o for another purpose later. We are oin to use the central equation from the nearly free electron model to tacle this problem. The Central Equation states: 0 = E + G G. m G This defines not just one but a set of equations connectin each with every + G, for all G. Physically, this means connectin the coefficients of all the different bands correspondin to the same reduced-zone. That is in principle an infinite number of G- values, bands, and coefficients. It is a set of equations, not just 1, because I can substitute any value for my oriinal to et a different set of equations. In particular, if I choose a differin by a reciprocal lattice vector G, I will et a different equation involvin the very same set of wavefunction coefficients. I can enerate as many of these equations as I want just n by choosin different values of G =. a However, how many s we need to use depends on how many bands we are looin for and to a lesser extent how many components of G are sinificant. Low enery eienvectors typically have very little contribution from hiher G basis functions. Thus not all the s are connected appreciably. This is especially true if the G s fall off quicly. This problem is not the best to illustrate this, since square waves have pretty intense hih frequency components. Let s ive up some accuracy and just use five basis functions to

2 find five bands. Then the lowest enery two that we are solvin for should be pretty accurate. So we will consider only bands with G-values of {-, -1, 0, 1, or } times the primitive reciprocal lattice vector. That ives five bands. We will assume coefficients of all a hiher G-value basis functions are zero.

3 The central equation is 0= E + G G m G where G = n, for any inteer n and =, or a 0= E + n n m n We want to consider only five of our wavefunction frequency coeficients:,,,, We will assume all others are zero. So we need five equations. Our equations result from replacin with + j, j = inteer, and eepin only terms in the sum with coefficients of 0=,,,,. ( + j) m ( E) E + + m n + j n n ( + j) = 0 + E + m = λ +, + j + j n + j n n 0 + m n + j n n 0 ( + j) with λ+ j 0 +. m Once aain, we eep only terms includin,,,,. Then our equations are, for j = : = λ E for j = 1: = + λ E for j = 0 : = + + λ E + for j = 1: = λ E + for j = : = λ E

4 We can cast this in matrix-vector form: E 3 4 E 3 ( λ = 0. 3 λ E ( λ+ E) We can solve this set of equations a couple of ways. Tain the E terms to the other side of the equation maes this an eienvector/eienvalue problem. ltimately, this is what we want to do, because we want both the wavefunctions and the enery eienvalues. n = n as an eienvector problem, becomes With ± n sin, and 0=,the equation, rearraned λ λ λ 0 = E λ λ+ 3 See the accompanyin Maple spreadsheet for the solution to the remainder of the problem. Here are the resultin enery bands from the Maple results:

5 Althouh it loos lie the bands touch, close inspection of the eneries show that in every case we have a small band ap.

6 One additional note: If we only wanted the eneries, we could just note that a non-trivial solution requires that the determinant of the above equation be zero. 0 = ( ) λ E 3 4 E 3 E ( λ + E n = n ( λ ( λ E) = 0 ( λ ( λ+ E) ( λ+ 3 With ± n sin, and 0 =,the determinant equation becomes