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Integration help with normalizing wave function

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data
    ψ(x,t) = Ae-λxe-iwt

    Normalize this and solve for A


    2. Relevant equations
    [tex]\int_{-∞}^{∞}|ψ|2dx = 1


    3. The attempt at a solution
    I got to A2[itex]\int_{-∞}^{∞}e^{-2λx}dx[/itex]

    The solution manual multiplies it by 2 and only goes from 0 to ∞ instead of from -∞ to ∞. I'm trying to do it from -∞ to ∞ and I should get the same answer as the manual, but I'm not. They get a finite non-zero answer, but integrating what I got to, I have [itex]-\frac{A^{2}}{2λ}e^{-2λx}[/itex] and now, plugging in the limits of integration in for x, it will give me infinity as the answer.
    Anyone see where I made a mistake? Thanks.
     
  2. jcsd
  3. Jan 12, 2014 #2

    jbunniii

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    The integral as written above is infinite. It is not the same as integrating from ##0## to ##\infty## and multiplying by ##2##. Just plot the integrand to see this. If the integrand were ##e^{-2\lambda |x|}## (with absolute values) then it would be true.
     
  4. Jan 12, 2014 #3
    It is supposed to be absolute value. I keep forgetting about that. But now I'm getting 0 as the answer, since the ∞ in the exponent will make it look like this when I evaluate the integral over the limits:

    [tex]-\frac{A^{2}}{2λ}[e^{-2λ∞}-e^{-2λ∞}][/tex]

    Which is just

    [tex]-\frac{A^{2}}{2λ}[0 - 0][/tex]

    Which equals 0.

    I know I'm doing something wrong, though.

    Thanks for the response.
     
  5. Jan 12, 2014 #4

    jbunniii

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    What integral did you perform? What were the upper and lower limits?
     
  6. Jan 12, 2014 #5
    Upper limit was ∞ and lower was -∞. I plug them in for x, and since it's the absolute value of x, they're both just ∞, which means it turns to 0.
     
  7. Jan 13, 2014 #6

    jbunniii

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    So you are trying to perform this integral:
    $$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx$$
    Because of the absolute value, you can't evaluate the integral without breaking it into parts. (If you believe otherwise, then you need to produce an antiderivative for ##e^{-2\lambda|x|}##.) Try writing it this way:
    $$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx =
    \int_{-\infty}^{0} e^{-2 \lambda |x|} dx + \int_{0}^{\infty} e^{-2 \lambda |x|} dx$$
    Next step: simplify the integrand in each case by replacing ##|x|## with something simpler.
     
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