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Infinite well problem, not normal probability function(?)

  1. Apr 3, 2012 #1

    sz0

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    1. The problem statement, all variables and given/known data
    A particle with mass m is trapped inside the infinite potential well:

    0<x<L : U(x) = 0
    otherwise: U(x) = ∞

    ψ(x,t=0) = Ksin(3∏x/L)cos(∏x/L)

    What energies can be measured from this system, what are the probabilities for these energies ?

    2. Relevant equations

    Schrödinger equation, |ψ(x,t)|^2 = 1



    3. The attempt at a solution
    I can solve this type of problem normally when ψ = Asin(...) + Bcos(...), but the look of the ψ function confuses me. My guess i should use the Schrödinger equation to give me E like normally.. but maybe that's not how im supposed to do this. If it is, my math skills are not good enough. I've been trying quite hard to get something good out of (d/dx)^2(ψ) / ψ .

    So thanks in advance for any help i can get. :)
     
    Last edited: Apr 3, 2012
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  3. Apr 3, 2012 #2

    vela

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    You want to express ##\psi(x)## as a linear combinations of the eigenstates. Try using a trig identity to do this.
     
  4. Apr 6, 2012 #3

    sz0

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    Thank you, I will work on that this night and read up on how to do it, hopefully I will make it.
     
  5. Apr 9, 2012 #4

    sz0

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    I haven't solved it, but these are my new trys, am I doing anything right ?

    I start with what i think is a trigidentity,
    ψ(X) = Ksin([itex]\frac{3πx}{L}[/itex])cos([itex]\frac{πx}{L}[/itex]) = [itex]\frac{K}{2}[/itex][sin([itex]\frac{2πx}{L}[/itex])+sin([itex]\frac{4πx}{L}[/itex])] , solving K from this should be trivial, but I will wait with it for now.

    I have two approaches.
    1: If there is a function ψ'n = [itex]\frac{K}{2}[/itex]sin([itex]\frac{nπx}{L}[/itex]) then ψ'2 + ψ'4 = ψ, but ψn + ψm ≠ ψn+m right? so what is all this good for either way?

    2: [itex]\frac{-\hbar}{2m}[/itex][itex]\frac{dψ}{dx}[/itex]2 = Eψ with ψ =[itex]\frac{K}{2}[/itex][sin([itex]\frac{2πx}{L}[/itex])+sin([itex]\frac{4πx}{L}[/itex])] [itex]\Rightarrow[/itex] [itex]\frac{dψ}{dx}[/itex]2 = -[itex]\frac{K}{2}[/itex][sin([itex]\frac{2πx}{L}[/itex])+4sin([itex]\frac{4πx}{L}[/itex])]( [itex]\frac{4π}{L}[/itex])2 [itex]\Rightarrow[/itex] E = [itex]\frac{2\hbar}{mL^2}[/itex][itex]\frac{1+8cos(\frac{2πx}{L})}{1+2cos(\frac{2πx}{L})}[/itex]

    But what does this tell me about what values are allowed ? And was this a good way to go ?

    Well if you vela or someone else could help me out a bit more I would appreciate it, thanks.
     
  6. Apr 10, 2012 #5

    vela

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    Read up on the superposition of states.
     
  7. Apr 10, 2012 #6

    sz0

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    Ok I did, there was nothing about it in my textbook i think so becuse of different notation online it was a bit slow but i hope i got this right.

    ψ(X) = Ksin(3πx/L)cos(πx/L) = C[sin(2πx/L)+sin(4π/L)] is what I have after I solve 1= ∫|ψ|^2 dx [0,L] → K = 2/√L . For simplicity i introduce C = 1/√L = K/2 .

    I want to write ψ as a linear combination of different ψn that is
    ψ = Ʃ cnψn = C(c1sin(x/L)+c2sin(2x/L)+c3sin(3x/L)+c4sin(4x/L)+....) = C[sin(2πx/L)+sin(4π/L)]

    Another condition is Ʃcn2 = 1 [itex]\Rightarrow[/itex]
    c2 = c4 = 1/√2 [itex]\Rightarrow[/itex] Cc2=Cc4 = √(2/L) [itex]\Rightarrow[/itex]
    ψ2 = √(2/L)sin(2πx/L), ψ4 = √(2/L)sin(4πx/L)

    And now solving Schrödinger equation with ψ2
    E2 = [itex]\frac{2\hbarπ^2}{mL^2}[/itex]
    with ψ4
    E4 = [itex]\frac{8\hbarπ^2}{mL^2}[/itex]

    Is it correct to say now that these are the measurable energies, and that the constant cn shows the probability for En beeing measured, in this case they are the same so it's 50% each ?

    Thanks again.
     
  8. Apr 10, 2012 #7

    vela

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    Yup, you got it.
     
  9. Apr 11, 2012 #8

    sz0

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    Fantastic, thanks for your help.
     
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