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Homework Help: Finite/Infinite Potential Well Problems

  1. Apr 21, 2010 #1
    Hi, (first ever post)

    I have the following three problems

    1) A particle is in the ground state of an infinte well from 0<x<a. Find the probability that it is in the middle half of the well.

    I think I have managed this, its just the integral of the wave function squared then I put in the values 3a/4 and a/4 to get the probability. It all cancels out nicely to get 0.82 which seems about right

    psi = (2/a)^0.5 * sin(pi*x/a)
    psi^2 = (2/a) * sin^2(pi*x/a)
    Integral = ((2/a)((x/2)-((a/4pi)sin(2pix/a)
    then put in the limits 3a/4 and a/4 and get 0.82
    Is this the right method

    2) A particle is in second state n=3 of an infinite well from -a/2<x<a/2. Find the position of the two nodes.

    psi = (2/a)^0.5 * sin(pi*x*3/a)
    probability density = psi^2 = (2/a) * sin^2(pi*x*3/a)
    At the nodes i thought this must be equal to zero
    so (2/a) * sin^2(pi*x*3/a)=0
    therefore the nodes must be at x=a/3 and x=-a/3

    3) This is what I am stuck on. Consider a well of finite depth that is just deep enough to contain three states. Sketch n=3 wave function and give the two values of the two nodes in this case.

    I have sketched it from a textbook but I don't know how to find the nodes because I am confused about what the wave function is. Would it just be Asin(kx)+Bcos(kx). Any thoughts appreciated.

    Thanks
     
  2. jcsd
  3. Apr 21, 2010 #2
    for question no. (2):

    the wavefunction for the boundaries from x= -a/2 to x= a/2 (with n = 3) is sqrt(2/a) cos(npix/a) ..

    how will that change your answer?
     
  4. Apr 21, 2010 #3
    Really? I cant see why, looking at this diagram surely it describes a sine function at n=3?

    http://www.vectorsite.net/tpqm_02_05.png

    They all appear to be sine functions, because at the walls there are nodes. Maybe I am looking at it wrong actually, I can see how the symmetry of the n=3 graph would suggest it is a cos function.
     
  5. Apr 21, 2010 #4
    sqrt(2/a) cos(npix/a) if it is this then

    psi^2= (2/a)*cos^2 (3pix/a)
    for this to equal zero x=a/6 or -a/6

    so is that the correct answer for part 2?

    Does anyone have any bright ideas for part 3?

    thanks for the help by the way
     
  6. Apr 21, 2010 #5
    that is correct .. sorry for the late reply ..

    look .. when the boundaries run from -a/2 to a/2 .. then the function will alternate between being symmetric and antisymmetric ( cosine and sine ), then n=1 is cosine function, n=2 is a sine, n=3 is a cosine .. etc

    I can see that you are wondering why it is so ( I used to have the same question in my mind when I first studied it too ) ..

    you know that the function for the boundaries x = 0 to x= a is sqrt(2/a) sin(npix/a) quick comparison between it and the one between x= -a/2 to x= a/2 you would see that they have the same length (which is a) and the only difference is that the other one is shifted a/2to the left .. so change x to x-a/2 then the sine in the wavefunction would change from
    sin(npix/a) to sin(npix/a - npi/2) = sin(npix/a)cos(npi/2) - cos(npix/a)sin(npi/2)
    from there you can conclude that for n=1,3,5.. the solution would just be sqrt(2/a)cos(npix/a)..


    hopefully this is more clear now :) ..
     
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