The Dirac Comb: A First Glance at Periodic Potential Jed Brody Outline •Review of basic quantum mechanics •Solution to the Dirac comb problem •Some semiconductor devices I made 2 The Schrödinger Equation •Given V(x) and boundary conditions, find (x) and E that satisfy 2 d 2 ( x) V ( x) ( x) E ( x) 2 2m dx •We generally find an infinite number of solutions: 1(x) and E1 2(x) and E2 All energies except 3(x) and E3 . . . E1, E2, E3, etc. are forbidden!!! 3 Example: Infinite Square Well n En , 2 2ma n 1,2,3,... 2 2 Plot allowed energies in units 2 of 2 2ma 2 : First 10 allowed energies 2 100 0 4 Continuity of and d/dx • is continuous everywhere •d/dx is continuous everywhere except where V is infinite •If V(x)=-(x-a), d dx a d dx a 2m 2 (a) Outline of derivation: •Integrate Schrödinger equation from a-e to a+e •Let e0 5 Commuting Operators If operators A and B commute ([A,B]=AB-BA=0), then we can find eigenfunctions of A that are also eigenfunctions of B. Proof: Let [A,B] operate on a1, an eigenfunction of A with eigenvalue a1 (A a1 =a1 a1): AB BA 0 AB BA a1 0 AB a1 BA a1 0 AB a1 Ba1 a1 0 A( B a1 ) a1 ( B a1 ) 0 A( B a1 ) a1 ( B a1 ) B a1 ~ a1 B a1 b1 a1 6 The Dirac Comb •Consider an infinite, one-dimensional chain of regularly spaced atoms •Suppose each positively charged atomic core acts on electrons as a delta-function well (not realistic) V(x) -a a 2a 3a x Observe: V(x+a)=V(x) 7 Bloch’s Theorem If V(x+a)=V(x), then the solutions to the Schrödinger equation satisfy (x+a)=eiKa(x) for some constant K. (x+a) •Suppose we know (x) only between 0 and a: (x)=sin(x/a), 0<x<a (x) •Replace x with (x+a): (x+a)=sin[(x+a)/a], 0<x+a<a -a<x<0 •Bloch’s theorem gives (x)=e-iKa(x+a), (x)=e-iKasin[(x+a)/a], -a<x<0 -a a x 8 First Step of Proof: D and H commute •Define the displacement operator: Df(x)=f(x+a) •H is the Hamiltonian, (-ħ2/2m)d2/dx2+V(x), for a periodic potential: V(x+a)=V(x) •If D commutes with (-ħ2/2m)d2/dx2 and with V(x), then D commutes with H •D commutes with (-ħ2/2m)d2/dx2 because taking the second derivative of a function and then shifting it to the left is the same as shifting it to the left and then taking its second derivative •[D,V(x)]f(x)=[DV(x)-V(x)D]f(x) =DV(x)f(x)-V(x)f(x+a) =V(x+a)f(x+a)-V(x)f(x+a) =0 because V(x+a)=V(x) 9 Completion of Proof •D and H commute, so we can choose eigenfunctions of H that are also eigenfunctions of D •Let D act on one of these eigenfunctions: D(x)=(x), where is the eigenvalue of D •Thus since D(x)=(x+a), (x+a)=(x) •Note that (x+2a)=2(x), (x+3a)=3(x), etc. •To keep (x) from blowing up or vanishing upon repeated applications of D, ||=1 (x+a)=eiKa(x) 10 Restrictions on K •The electrical properties of a solid deep in its bulk are independent of edge effects. To avoid edge effects, use periodic boundary conditions •Wrap the x axis around on itself, forming a loop of length Na; N is the very large number of potential wells •After travelling a distance Na, you return to your starting point: (x+Na)=(x) •Bloch’s theorem gives (x+Na)=eiNKa(x), so eiNKa=1 NKa=2nK=2n/(Na), n=0,±1,±2,... 11 Solving the Dirac Comb Problem The potential is a series of delta functions: N 1 V ( x) ( x ja) j 0 In between delta functions (e.g.,0<x<a), V(x)=0: V(x) -a a 2a 3a x 2 d 2 E 2 2m dx 2 d 2 k 2 dx 2mE k 12 Conditions on (x) at x=0 ( x) A sin( kx) B cos( kx), 0 x a Replace x with (x+a): ( x a) A sin[ k ( x a)] B cos[ k ( x a)], a x 0 Apply Bloch’s theorem, (x)=e-iKa(x+a): ( x) e iKa { A sin[ k ( x a)] B cos[k ( x a)]},a x 0 (x) must be continuous at x=0: d dx 0 d dx B e iKa [ A sin( ka) B cos( ka)] 0 2m (0) : 2 kA e iKa 2m k[ A cos( ka) B sin( ka)] 2 B 13 Equation To Find Allowed Energies •Eliminating A and B from previous equations gives m cos( Ka) cos( ka) 2 sin( ka) k •On LHS, we know K=2n/(Na), where n=0,±1,±2,… •On RHS, define zka=a(2mE)1/2/ħ and ma/ħ2 sin z 2n cos f ( z) cos( z ) z N Example: N=8 n 2n/N cos(2n/N) 0 0 1 ±1 ±/4 21/2/2 ±2 ±/2 0 ±3 ±4 ±3/4 ± -21/2/2 -1 ±5 ±6 ±5/4 ±3 /2 -21/2/2 0 14 N=8 Solution sin z 2n cos f ( z) cos( z ) z N =5 15 N=50 Solution sin z 2n cos f ( z) cos( z ) z N 16