Quantam Griffiths 2.47 Finite Square Well

  1. Consider double finite square well with

    -Vo; -(a+b/2) < x < -b/2
    V(x) = 0; -b/2 < x < b/2
    -Vo; b/2 < x < a + b/2
    0; otherwise

    Sketch the ground wave function Psi(n=1) and the first excited state Psi(n=2) for
    1. b = 0
    2. b~a
    3. b >> a


    So...

    for b=0

    This is just the normal finite potential well. In particular, we can consider even and odd solutions.

    the transcendental equation is

    odd: tanz = ((zo/z)^2 - 1)^1/2
    even: -cotz = ((zo/z)^2 - 1)^1/2

    z=ka...zo=a(2mVo)^(1/2)/h_bar
    k=(2m(E+Vo))^(1/2)/h_bar

    As a result...for both even and odd...we get E_n + Vo ~ ((n*pi*h_bar)^2)/(2m(2a)^2)

    *Note* even though i included the above information, griffiths says "no computations!" "qualitative only."

    First Idea

    For x< -a...Ae^(kx) (even)...-Ae^(kx) (odd)
    For x > a....Ce^(-kx)

    i.e. => for these regions shape of Psi (n,x) is very simple, but If I was to graph for n=1 and n=2 on the same graph...would the n=2 state be higher? (i.e. higher y value for the same x values in this region)?

    For -a < x < a...

    I'm confused on the shape of the graph in this region...

    Do i construct for odd and even solutions?
    i.e. => like pics. on pg 32 (except for sin and cos as well)
    (the pics. on pg 32 are graphs of sin((n*pi/a)x) for n=1,2,3)

    then for b~a and b>>a....the results are similar since a well with width a and width 2a have the same physics

    Second Idea

    This is supposed to just mirror the results of the infinite square well.

    Particularly...we have e^(kx) (far left region) and e^(-kx) (far right region)

    Also, the graph inside the well would be of form sin((n*pi/2a)x)...
    I understand when we did the derivations of the infinite square well we got this result because of the B.C. #1...the continuity of Psi...as a result sin(2ka)=0...=> k=n*pi/2a

    I'm confused because griffiths says no computations. Because of that statement, I infer to look back to problems we have done to see how to pursue this problem. As a result, only time we have dealt with finite square well...we didn't even graph Psi...only got information for bound states and scattering states...in particular...no question in the text asks you to graph Psi for finite square well...as a result...only time we graphed Psi for something similar was delta function potential and infinite square well. Obviously, the infinite square well is more similar. So...should I do what my gut tells me and treat each well as Psi(n,x)=sin((n*pi/a)x)

    Thank you
     
  2. jcsd
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

0
Draft saved Draft deleted