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Find the normalisation constant using a trial wavefunction

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data

    a particle of mass m, confined to a one dimensional infinite potential of
    0[tex]\leq[/tex]x[tex]\leq[/tex]1 V(x) = 0
    elsewhere V(x) = [tex]\infty[/tex]

    2. Relevant equations

    Choose as a trial wavefunction

    [tex]\Psi[/tex](x) = Nx[1 - [tex]\alpha[/tex]x + ([tex]\alpha[/tex] - 1)x[tex]^{2}[/tex]]

    Verify that

    N[tex]^{2}[/tex] = [tex]\frac{K}{16 - 11\alpha + 2\alpha^{2}}[/tex]

    3. The attempt at a solution

    1 = <[tex]\Psi[/tex]|[tex]\Psi[/tex]>

    1 = [tex]\int^{1}_{0}[/tex]Nx[1 - [tex]\alpha[/tex]x + ([tex]\alpha[/tex] - 1)x[tex]^{2}[/tex]] Nx[1 - [tex]\alpha[/tex]x + ([tex]\alpha[/tex] - 1)x[tex]^{2}[/tex]] dx

    1 = N[tex]^{2}[/tex] [tex]\int^{1}_{0}[/tex] x[tex]^{2}[/tex][1 - [tex]\alpha[/tex]x + ([tex]\alpha[/tex] - 1)x[tex]^{2}[/tex]][tex]^{2}[/tex]

    Is this right so far?? I'm not sure how to carry on. Should I expand the brackets??
  2. jcsd
  3. Oct 23, 2009 #2
    Yes and Yes. Well, you could possibly integrate it as is, but it'd be easier to see after expanding the parenthesis.

    PS: Rather than typing {tex}code{/tex} for every variable, just write the whole thing in tex, it'll look a bit nicer.
  4. Oct 23, 2009 #3
    When I expand the brackets and multiply by x^2

    I get the following:

    x[tex]^{6} + x^{2} + 2 \alpha^{2} - \alpha^{4} - 2 \alpha x^{3} + \alpha^{2}x^{4} - \alpha x^{5}[/tex]

    This seems totally wrong when I look at what I am supposed to get for N^2...
  5. Oct 23, 2009 #4
    I think you expanded incorrectly. You should get

    \left(\alpha-1\right)^2x^6-2\alpha(\alpha-1)x^5+2(\alpha-1)x^4+\alpha^2x^4-2\alpha x^3+x^2

    When you integrate this over the range 0 to 1, you should get the answer.
  6. Oct 23, 2009 #5
    If it helps see a little bit more clearly, [tex]K=210[/tex]. Not sure why your question put [tex]K[/tex] rather than [tex]210[/tex], but it is what it is.

    Also, you could fully expand what I wrote down and then integrate, but you'll have a few extra terms to work with; it's usually easier to expand and regroup after integration.
  7. Oct 23, 2009 #6
    Ah, excellent! Thank you SO much for your help! :)
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