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Transcendental equation from a finite square well potential

  1. Feb 26, 2014 #1
    if I have a transcendental equation such as this one: tan(l a) = -l / sqrt (64/a^2 - l^2 ) Where
    l=sqrt(2m(E+V) /hbar^2 ) and 'a' is the width of a finite square well, how can I solve this equation in terms of both l and a. I have successfully graphed the two sides of the equation together by assuming 'a' to be any constant, i.e 1 . but how can I graphically solve it and extract an exact solution for E that includes a.

    note: there are supposed to be three bound states:3 different solutions for E
     
  2. jcsd
  3. Feb 26, 2014 #2

    TSny

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    It might help to let ##z = a l## and express the transcendental equation in terms of ##z## alone.

    Note: For this type of problem you generally get two different transcendental equations. One for wave functions that are symmetric about the center of the well and one for anti-symmetric functions.
     
  4. Feb 26, 2014 #3
    letting x = la, I got this http://www.wolframalpha.com/input/?i=tan(x)+and+-x+/+sqrt(64+-+x^2)


    but i only have 2 bound states .


    here's the question, please tell me what I am doing wrong:

    V = infinity for x<0 and V= -Vo = -32hbar^2 / ma^2 for 0<x<a and V= 0 for x>a

    I let l = sqrt(2m(E+Vo)/hbar^2) and k = sqrt(-2mE/hbar^2)

    from the continuity of ψ at a: Be^(-ka) = Asin(la) (1)

    and from the continuity of dψ/dx at a : -kBe^(-ka) = lAcos(la) (2)

    divide (1) by (2) to get -l/k = tan(la) (3)

    but k^2 + l^2 = 2mVo / hbar^2 , so k = sqrt(64/a^2 - l^2)

    put the new k back in (3) to get tan(la) = -l / sqrt(64/a^2 - l^2)

    finally let x = la ----> tan(x) = -x / sqrt(64 -x^2)


    from which i get only two intersections on the positive y axis.


    "Note: For this type of problem you generally get two different transcendental equations. One for wave functions that are symmetric about the center of the well and one for anti-symmetric functions."

    TSny, the potential function is not even in this problem so I think it's only one transcendental. equation
     
  5. Feb 26, 2014 #4

    TSny

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    Oh, I didn't know that you were taking V = ∞ for x<0.

    Your work looks correct to me. There are in fact 3 solutions. The third one is difficult to pick up on your graph. Can you find it by regraphing?

    Here's a trick I've seen. By squaring both sides of your transcendental equation and using some trig identities, show that the solutions must satisfy

    ##|\sin (z)| = z/8##

    The graph of this is easier to analyze. However, you only want the solutions where you also satisfy the condition that ##\tan (z) <0##.
     
  6. Feb 26, 2014 #5
    "The graph of this is easier to analyze. However, you only want the solutions where you also satisfy the condition that tan(z)<0. "

    but where does this requirement come from?
     
  7. Feb 26, 2014 #6

    TSny

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    Look at your original transcendental equation.
     
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