So complex eigenvalues will never lead to bound

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getting small. So complex eigenvalues will never lead to bound states. In fact, complex eigenvaluescorrespond toscatteringstates, and the energy values for which eigenvalues are complex are theenergies at which the electron can move freely through the crystal.Suppose the eigenvalues are real. If|λ1|>1 then|λ2|= 1/|λ1|<1 and vice versa. So one of theproductsλn-11,λn-12will be growing large, and one will be getting small. So the only way thatxncan be getting small is if the coefficienta1ora2sitting in front of the growing product is zero.Now let us actually compute the eigenvalues. They areλ=-E±E2-42.178
If-2< E <2 then the eigenvalues are complex, so there are no bound states.The interval[-2,2] is the conduction band, where the electrons can move through the crystal.IfE=±2 then there is only one eigenvalue, namely 1. In this case there actually is only oneeigenvector, so our analysis doesn’t apply. However there are no bounds states in this case.Now let us consider the caseE <-2. Then the large eigenvalue isλ1= (-E+E2-4)/2 andthe corresponding eigenvector isv1=-1E+λ1. The small eigenvalue isλ2= (-E-E2-4)/2and the corresponding eigenvector isv2=-1E+λ2.We must now computea1and determinewhen it is zero. We have [v1|v2]a1a2=x1. This is 2×2 matrix equation that we can easily solvefora1a2. A short calculation givesa1= (λ1-λ2)-1(-(a+E)(E+λ2) + 1).Thus we see thata1= 0 whenever(a+E)(E-pE2-4)-2 = 0Let’s consider the casea= 5 and plot this function on the interval [-10,-2]. To see if it crossesthe axis, we also plot the function zero.>N=500;>E=linspace(-10,-2,N);>ONE=ones(1,N);>plot(E,(5*ONE+E).*(E-sqrt(E.^2 - 4*ONE)) - 2*ONE)>hold on>plot(E,zeros(1,N))Here is the result179
-20020406080100-10-9-8-7-6-5-4-3-2We can see that there is a single bound state in this interval, just below-5. In fact, the solutionisE=-5.2.The caseE >2 is similar. This time we end up with(a+E)(E+pE2-4)-2 = 0Whena= 5 this never has a solution forE >2. In fact the right side of this equation is biggerthan (5 + 2)(2 + 0)-2 = 12 and so can never equal zero.In conclusion, ifV1=-5, and all the otherVn’s are zero, then there is exactly one bound statewith energyE=-5.
2. Here is a diagram of the energy spectrum for this potential.-4-202-6EBound state energyConduction bandFor the bound state energy ofE=-5.2, the corresponding wave function Ψ, and thus the proba-bility that the electron is located at thenth lattice point can now also be computed. The evaluationof the infinite sum that gives the normalization constantN2can be done using a geometric series.
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IV.5.4. Conduction bands for a crystalThe atoms in a crystal are arranged in a periodic array. We can model a one dimensional crystalin the tight binding model by considering potential values that repeat a fixed pattern. Let’s focuson the case where the pattern is 1,2,3,4 so that the potential values areV1, V2, V3, . . .= 1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4, . . .In this case, if we start with the formulaxn=A(Vn-E)A(Vn-1-E)· · ·A(V1-E)10we can group the matrices into groups of four. The productT(E) =A(V4-E)A(V

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