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How to normalize Schrodinger equation

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    The solution to the Schrodinger equation for a particular potential is psi = 0 for absolute x > a and psi = Asin(pi*x/a) for -a <= x <= a, where A and a are constants. In terms of a, what value of A is required to normalize psi?

    2. Relevant equations

    psi = 0 for absolute value of > a
    psi = Asin(pi*x/a) for -a<= x <= a

    3. The attempt at a solution

    The textbook gave the normalization as:

    [integral] psi^2 dx = 1. The integral has upper limit positive infinity and lower limit negative infinity.

    Can I just start by taking the integral of both sides of psi = Asin(pi*x/a)?

    which makes psi = A-cos(pi*x/a) * pi*x^2/(2a)?
     
    Last edited: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    SammyS

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    No.

    The integral of ψ(x) from -∞ to +∞ is zero. After all, ψ(x) is an odd function.

    (ψ(x))2 ≥ 0 for all x and it's greater than zero for some values of x, so its integral is definitely positive.
     
  4. Mar 13, 2013 #3
    How did you know the integral is zero. Why is psi an odd function?
     
  5. Mar 13, 2013 #4

    SammyS

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    [itex]\displaystyle \Psi(x)=A\sin\left(\frac{\pi x}{a}\right)[/itex]

    Therefore, [itex]\displaystyle \ \Psi(-x)=-\Psi(x)\,,\ [/itex] so [itex]\displaystyle \ \Psi(x)\ [/itex] is odd.

    That's pretty much beside the point except that it definitely shows that in general, [itex]\displaystyle \ \int \left(f(x)\right)^2\,dx\ne\left(\,\int f(x)\,dx\right)^2\ .[/itex]

    Solving your integral may be made easier by noting that [itex]\displaystyle \ \cos(2\theta)=1-2\sin^2(\theta) \ .\ \ [/itex] And solving for sin2(θ) gives [itex]\displaystyle \ \sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \ . \ [/itex]
     
    Last edited: Mar 13, 2013
  6. Mar 13, 2013 #5
    I know I stated -a<= x <= a. But x can be positive or negative. How did you know to choose x to be negative?

    How is cos(2θ)=1−2sin2(θ)? I don't remember this being a trig identity. I am weak in math.
     
  7. Mar 13, 2013 #6

    SammyS

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