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Kronig-Penney Model in Reciprocal Space

  1. Apr 9, 2014 #1
    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=[itex]\hbar [/itex]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=[itex]\hbar [/itex]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. jcsd
  3. Apr 11, 2014 #2
    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).
     
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