# Kronig-Penney Model in Reciprocal Space

1. Apr 9, 2014

### irishhockey

1. The problem statement, all variables and given/known data
(a) For the delta-function potential and with P<<1, find at k=0 the energy of the lowest energy band.
(This is part A of Charles Kittel Solid State Physics problem 7.3)

2. Relevant equations

(P/Ka)sin(Ka)+cos(Ka)=cos(ka) (note, K and k are different variables)

ε=$\hbar$2K2/(2m)

3. The attempt at a solution

I have tried two different things.

attempt 1)

simply say (P/Ka)=0

so we get

cos(Ka)=cos(0)=1

or K=arcos(1)/a

plugging into energy equation

ε=$\hbar$2(arcos(1)/a)2/(2m)=0

which is obviously wrong

second attempt is to Taylor expand the trig functions, and assume the because P<<1, Ka<<1,
(in order to keep both left hand terms) and so (P/Ka)=1

after expanding and cancelling,

2cos(0)=1

which again stumps me

I'm not sure what to do. I have the sln manual, which says to expand the first equation, to find P≈(1/2)(Ka)2

Thank you for any help.

Last edited: Apr 9, 2014
2. Apr 11, 2014

### tman12321

We have P*sinc(Ka)+cos(Ka)=cos(ka)=1 at k=0, where I have used sinc(x)=sin(x)/x. Then P*sinc(Ka)=1-cos(Ka). We know P<<1, and that max(sinc(Ka))=1. This tells us that LHS=P*sinc(Ka)<=P<<1. If LHS<<1, then RHS<<1. Therefore 1-cos(Ka)<<1. What does this tell us about Ka? Recall the Taylor Expansion for cos(x) and sinc(x).