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Particle in the box

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Particle is put in into the infinity potential well. How the probability of finding particle between different maxima of probability density is changing by increasing of ##n##?

    2. Relevant equations
    ##\psi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}##
    ##E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}##


    3. The attempt at a solution
    ##\frac{d}{dx}|\psi_n(x)|^2=\frac{1}{a}\frac{2n\pi}{a}\sin\frac{2n\pi x}{a}=0##
    so
    ##\frac{2n\pi x}{a}=k\pi##
    From that ##x=\frac{ka}{2n}##.
    To bi maximum
    ##\cos \frac{2n\pi x}{a}<0##.
    Right?
    What is easiest way to solve this?
     
  2. jcsd
  3. Mar 8, 2014 #2

    Simon Bridge

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    Write out the probability calculation for n=1,2,3 ... and spot the pattern.
     
  4. Mar 9, 2014 #3
    For ##n=1## I get that ##\frac{d}{dx}\psi_1^2(x)=0## for ##x=\frac{ka}{2}##. How I now that ##k=1?##. If I take second derivative I get ##\frac{d}{dx}\psi_1^2(x)=\frac{4\pi^2}{a^3}\cos \frac{2\pi x}{a}##. If ##k=0##, ##x=0## and ##cos(0)=1>0##, so ##\psi_1(x=0)## is minimum. For ##k=1##, ##x=\frac{a}{2}## and ##cos(\pi)=-1<0## so for ##x=\frac{a}{2}## function has maximum. For ##k=2##, ##x=a## function again has minimum.

    For ##n=2## I get that ##\frac{d}{dx}\psi_2^2(x)=0## lead to ##x=\frac{k}{4}a##. So I have five possibilities ##k=0,1,2,3,4##. Two maxima are for ##k=1## and ##k=3##.

    For ##n=3## seven possibilities ##k=0,1,2,3,4,5,6##. Three maxima are for ##k=1,k=3,k=5##.

    But I still don't have real generalisation. I have only behaviour. I see that the all maxima are for odd number ##k##.

    Is there any other easier way to find this?
     
  5. Mar 9, 2014 #4

    Simon Bridge

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    You are overthinking it:
    - just sketch the probability densities for the first 3-4 energy levels, what sort of function are they?

    - mark in two adjacent maxima for each and shade the area between them that you need to calculate.
    Make a note of what the x values are (hint: try substituting ##\theta=kx## - what are the values of ##\theta## going to be for the first two maxima?)
     
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