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Particle in Spherical Well : Sudden Approximation

  1. Feb 19, 2014 #1
    1. The problem statement, all variables and given/known data

    In a spherical well in which..

    [tex]V=
    \begin{cases}
    0,\text{for }0 \le r < R \\
    ∞, \text{for } r > R
    \end{cases}
    [/tex]

    the s-wave eigenstates are

    [tex]\phi_n(r)=\frac{A}{r}\sin\left( \frac{n\pi r}{R} \right)[/tex]

    where A is a normalization constant. If a particle is in the ground state and R suddenly
    increases to ##R'=R/\epsilon##, where ##0 \le \epsilon \le 1##

    (i) Compute and plot the probability of finding the particle in the new ground state as a
    function of ##\epsilon##.
    (ii) Compute and plot the probability of finding the particle in a new state with quantum
    number n as a function of ##\epsilon## and n (this will be a 3D plot). What are the most important
    features you see in your plot?
    (iii) Sum all probabilities in (ii) for = 1=2 to show that they add to 1.
    Note: The relations

    [tex]\sum_{n=0}^{∞}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}[/tex]

    [tex]\sum_{n=0}^{∞}\frac{1}{(2n+3)^2(2n-1)^2}=\frac{\pi^2}{64}[/tex]

    may be useful here.



    2. Relevant equations

    Obviously the ones given in the problem.

    Also, this problem deals with the sudden approximation. Which says that when the system is changed very rapidly, the state remains unchanged initially and then is allowed to evolve.


    3. The attempt at a solution

    My solution is in the attached images. I know typing it out would have been ideal but it would have been too much and I could give more of my work this way. I scanned them to make sure they were high quality images.

    I don't know where my thought process went wrong, but what I am getting for part (iii) is obviously wrong since it diverges at n=2.
     

    Attached Files:

  2. jcsd
  3. Feb 19, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Generally, your work looks good. But,

    (1) Make sure you distinguish between probability amplitude and probability.

    (2) When R is switched to R/ε, the normalization constant for the new eigenstates will depend on ε.

    (3) It is not true that ##\frac{\sin(n\pi/2)}{n^2-4} = \frac{(-1)^{2n+1}}{n^2-4}##. For example, let n = 4. Also, n = 2 will need special treatment.
     
  4. Feb 20, 2014 #3
    Oh I forgot to replace n with 2n+1 in the denominator. After doing this, I get the second form of the second equation listed in the problem, except, It needs to be squared. In other words, I get.. (We will call it P)

    [tex]P=-\frac{8}{\pi}\frac{(-1)^{2n+1}}{(2n-1)(2n+3)}[/tex]

    So in order to get the summation, I need to first square this, THEN sum over n. I cannot determine if this makes sense or not. That means I am summing up the ##P^2## values, not the ##P## values.
     
  5. Feb 20, 2014 #4

    TSny

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    Homework Helper
    Gold Member

    Your expression for P is off by an overall numerical factor because you have not normalized the wave functions correctly for the case where the radius of the well is ##R/ \epsilon##.

    When you correct for this, you will have expressions for all of the transition amplitudes from the initial state to final states with odd quantum number ##\psi_{2n+1}##. You found that the transition amplitudes to states with even quantum number are zero. But there is an exception: the transition amplitude to ##\psi_2## is not zero.
     
  6. Feb 21, 2014 #5
    I believe I got it. Re-writing the normalization constant, I got a factor of 1/2. Meaning the summation I showed in my previous post is equal to 1/2. I then used L'Hopitals rule for when n=2 and got that term to be equal to 1/2. So 1/2+1/2=1.

    Hopefully, this is it. Thank you very much for your help!
     
  7. Feb 21, 2014 #6

    TSny

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    Homework Helper
    Gold Member

    Good.

    Another option for handling the n = 2 case is to set up the transition amplitude integral specifically for n = 2 and ##\epsilon = 1/2## and show the integral is ##\frac{1}{\sqrt{2}}##. So the probability for this transition is 1/2.
     
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