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Schrodinger equation for one dimensional square well

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    the question as well as the hint is shown in the 3 attachments


    2. Relevant equations



    3. The attempt at a solution
    i know how to normalize an equation, however i do not understand what the hint is saying, or how to do these integrals, any guidance would be greatly appreciated
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2012 #2
    The hint is just telling you that each Sin(a pi x) is orthogonal to the other Sins
    I'll give you my hint - all you need to know is how to integrate Sin^2(x)

    Do you know how to do the other two problems?
     
  4. Mar 20, 2012 #3
    im not sure exactly what you mean by that, what is the integral i need to take?
     
  5. Mar 20, 2012 #4
    Well, do you know what it would mean for [itex]\psi(x)[/itex] to be normalized?
     
  6. Mar 20, 2012 #5
    normalizing ψ(x) means to determine if ∫ |Ψ(x)|^2 dx = 1 right?
     
  7. Mar 20, 2012 #6
    That is correct
    Do you know what it means if two functions are orthogonal?
     
  8. Mar 20, 2012 #7
    no i do not
     
  9. Mar 20, 2012 #8
    Okay, orthogonality is what the hint is describing
    two functions, f and g are orthogonal if [itex]\int f*g\ dx = 0[/itex]
    the hint is just restating that each of the Sins are orthogonal to the other ones, that is;
    [itex]\int Sin(\pi x)Sin(2 \pi x) dx = 0[/itex] etc

    Using this, how do you think you should proceed in determining A?
     
  10. Mar 20, 2012 #9
    do i expand the equation and cancel out the terms with Sin(πx)Sin(2πx) because they will integrate to 0?
     
  11. Mar 20, 2012 #10
    Yes, show me what you get
     
  12. Mar 20, 2012 #11
    does ∫Sin(πx)Sin(3πx)dx=0 and ∫Sin(2πx)Sin(3πx)dx=0 ?
     
  13. Mar 20, 2012 #12
    yes, if you work the integral out yourself you'll find that [itex]\int Sin[n \pi x] Sin[m \pi x] dx = 0[/itex] if [itex]m \ne n[/itex]
     
  14. Mar 20, 2012 #13
    alright so then in that case i will only be dealing with the sin2 functions and therefore |Ψ(x)|^2= (1/10a)sin2([itex]\pi[/itex]x/a)+(aA2/a)sin2(2[itex]\pi[/itex]x/a)+(9/5a))sin2(3[itex]\pi[/itex]x/a)
     
  15. Mar 20, 2012 #14
    That is not exactly true, the terms aren't zero on their own, what IS true is that

    [itex]\int |Ψ(x)|^2dx=\int (1/10a)sin2(πx/a)+(aA2/a)sin2(2πx/a)+(9/5a))sin2(3πx/a)dx[/itex]
     
  16. Mar 20, 2012 #15
    yes yes youre right i jumped the gun on that, now will the limits of my integral be from -a to a or from -[itex]\infty[/itex] to[itex]\infty[/itex]
     
  17. Mar 20, 2012 #16
    Since this is the particle in a box problem, the potential outside of the box is set to infinity and so we make ψ = 0 everywhere outside, so it doesn't matter where you set the limits of integration (as long as the box is contained in them of course) since we pick up exactly 0 from the outside region.
    If you look at the problem statement, you'll see that the box isn't -a<x<a, it's 0<x<a
     
  18. Mar 20, 2012 #17
    ah yes my mistake so using 0 to a as the limits of integration i get that it is equal to (1/10a)+(9/5a)+(2A2/a)
     
  19. Mar 20, 2012 #18
    Okay, so what are we going to do with this?
    What do we need to set this equal to and what do we then need to solve for?
     
  20. Mar 20, 2012 #19
    to normalize it it has to be set equal to 1 and we need to find A, also to fix my integral above, i did it out wrong and it is actualy equal to A^2 +(19/20) so that means if i set it equal to 1 that A = [itex]\sqrt{1/20}[/itex]
     
  21. Mar 20, 2012 #20
    with A solved for and pulgged back in the function is now normalized, so the next part of the question asks me to find the possible results of measures of the energy and what are the respective probabilities of obtaining each result
     
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