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Finite square well potential energy

  1. Jun 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello.

    Imagine a particle bound in a square well potential of potential energy
    V0 if |x| > a
    0 if |x| < a

    The wave function of the particle is: (ignoring the time dependency)
    -A*exp(kx) if x<-a
    B*sin(3*pi*x/4a) if |x|<a
    A*exp(-kx) if x>a

    where k = sqrt(2mE)/ħ

    Find the energy of the particle.

    2. Relevant equations



    3. The attempt at a solution

    First of all I determined the value of A through the condition of continuity at the boundaries:
    A*exp(-k*a) = B*sin(3*pi*a/4a)
    A*exp(-k*a) = B/sqrt(2)
    A = B*exp(ka)/sqrt(2)

    Rewriting the wave function:

    -B*exp(k[x+a])/sqrt(2) if x<-a
    B*sin(3*pi*x/4a) if |x|<a
    B*exp(-k[x-a])/sqrt(2) if x>a

    After that I decided to use the normalization condition to find the value of k.

    [itex]\int[/itex]dx <ψ|ψ> = 1

    Separating the integral into three, one for each region, I concluded that:

    ∫dx A2 exp(2kx) = B2/4k
    ∫dx A2 exp(-2kx) = B2/4k
    ∫dx A2 sin2(3*pi*x/4a) = B2(a+1/2a)

    Therefore:
    B2/2k + B2(a+1/2a) = 1
    B2(1/2k + a + 1/2a) = 1

    However this doesn't help me much... I don't know what I should do next, or if what I'm doing is getting me closer to the answer.. The correct answer is:

    E = (9/32) π2ħ2/ma2

    Thanks for taking the time to read my problem.
     
  2. jcsd
  3. Jun 27, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    To answer the question, you don't need to worry about the normalization.

    Think about how E shows up in the Schrodinger equation.
     
  4. Jun 28, 2013 #3
    Thank you very much! I completely forgot about that!

    Cheers!
     
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