1. The Dirac Comb: A First Glance at Periodic Potential Jed Brody

2. 2 Outline Review of basic quantum mechanics Solution to the Dirac comb problem Some semiconductor devices I made

3. 3 The Schrödinger Equation Given V(x) and boundary conditions, find  (x) and E that satisfy We generally find an infinite number of solutions:  1 (x) and E 1  2 (x) and E 2  3 (x) and E 3. All energies except E 1, E 2, E 3, etc. are forbidden!!!

4. 4 Example: Infinite Square Well Plot allowed energies in units of :

5. 5 Continuity of  and d  /dx  is continuous everywhere d  /dx is continuous everywhere except where V is infinite If V(x)=-  (x-a), Outline of derivation: Integrate Schrödinger equation from a-  to a+  Let  0

6. 6 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 a 1 (A  a1 =a 1  a1 ):

7. 7 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) Observe: V(x+a)=V(x)

8. 8 Bloch’s Theorem If V(x+a)=V(x), then the solutions to the Schrödinger equation satisfy  (x+a)=e iKa  (x) for some constant K. Suppose we know  (x) only between 0 and a:  (x)=sin(  x/a),0<x<a 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 -iKa sin[  (x+a)/a],-a<x<0

9. 9 First Step of Proof: D and H commute Define the displacement operator: Df(x)=f(x+a) H is the Hamiltonian, (-ħ 2 /2m)d 2 /dx 2 +V(x), for a periodic potential: V(x+a)=V(x) If D commutes with (-ħ 2 /2m)d 2 /dx 2 and with V(x), then D commutes with H D commutes with (-ħ 2 /2m)d 2 /dx 2 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)

10. 10 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)=e iKa  (x)

11. 11 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)=e iNKa  (x), so e iNKa =1  NKa=2  n  K=2  n/(Na), n=0,±1,±2,...

12. 12 Solving the Dirac Comb Problem The potential is a series of delta functions: In between delta functions (e.g.,0<x<a), V(x)=0:

13. 13 Replace x with (x+a): Apply Bloch’s theorem,  (x)=e -iKa  (x+a):  (x) must be continuous at x=0: Conditions on  (x) at x=0

14. 14 Equation To Find Allowed Energies Eliminating A and B from previous equations gives On LHS, we know K=2  n/(Na), where n=0,±1,±2,… On RHS, define z  ka=a(2mE) 1/2 /ħ and  m  a/ħ 2 Example: N=8 n0±1 ±2 ±3 ±4 ±5 ±6 2  n/N 0±  /4 ±  /2 ±3  /4 ±  ±5  /4±3  /2 cos(2  n/N)12 1/2 /20-2 1/2 /2-1-2 1/2 /20

15. 15 N=8 Solution  =5

16. 16 N=50 Solution