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Math Help - Normalise the Wave function

  1. #1
    Senior Member bugatti79's Avatar
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    Normalise the Wave function

    Hi All,

    I am going through notes our lecturer gave us, am having some problems with it
    Given the wave function \Psi(x,t)=Ae^{-\lambda \mid x \mid} e^{-i\omega t} where A, lambda and omega are psoitive real constants and

    \mid x \mid either (x greater and equal to 0) and (-x less than or equal to 0)

    1) Normalise Psi, 2) Determine the expectation values of x and x^2

    I can derive to far as  1=A^2(\int^{0}_{-\infty} e^{2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx) Then he changes the integral such that

    = -A^2\int^{0}_{-\infty} e^{-2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx = 2A^2\int^{\infty}_{0} e^{-2\lambda x} dx
    Where did the first termof the last expression come out of?

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post

    I can derive to far as  1=A^2(\int^{0}_{-\infty} e^{2\lambda x} dx + \int^{\infty}_{0} e^{-2\lambda x} dx) Then he changes the integral such that...
    Change the variable in the first integral to x'=-x to get to the final result.


    CB
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  3. #3
    Senior Member bugatti79's Avatar
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    ok, i get that, thanks. The next line becomes

    1=2A^2\frac{1}{-2\lambda}\int^{\infty}_{0} e^{-2\lambda x} d(-2\lambda x)= \frac{-A^2}{\lambda}e^{-2\lambda x} = \frac{A^2}{\lambda} upon substitution of the limits.
    He seems to have integrated the integral but then he changes the dx to d(-2\lambda x}) Why? I dont understand this?

    Thanks
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post
    ok, i get that, thanks. The next line becomes

    1=2A^2\frac{1}{-2\lambda}\int^{\infty}_{0} e^{-2\lambda x} d(-2\lambda x)= \frac{-A^2}{\lambda}e^{-2\lambda x} = \frac{A^2}{\lambda} upon substitution of the limits.
    He seems to have integrated the integral but then he changes the dx to d(-2\lambda x}) Why? I dont understand this?

    Thanks
    Just to be bloody minded and obscurantist, you can integrate it straight off without doing that (OK may be that is how he was taught to do such things, I would do it as shown below).

    \displaystyle 1= 2A^2\int_0^{\infty}e^{-2\lambda x}\;dx=2A^2 \left[\dfrac{1}{(-2\lambda)}e^{-2\lambda x} \right]_{x=0}^{\infty}=\dfrac{A^2}{\lambda}

    CB
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  5. #5
    Senior Member bugatti79's Avatar
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    yes, Im more comfortable with that too. Next I have a query wrt integration by parts using reduction formulae

    <x^2>=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx
    Setting up the formulas I get

    I_{n}=\int x^n e^{-ax}=-\frac{1}{a}x^n e^{-ax}+\frac{n}{a}\int x^{n-1}e^{-ax}dx =- \frac{1}{a}x^n e^{-ax}+\frac{n}{a}I_{n-1}

    Assuming the above is correct, do i proceed to calculate I2, I1 and I0? Im not sure where I put in the limits?

    Thanks
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post
    yes, Im more comfortable with that too. Next I have a query wrt integration by parts using reduction formulae

    <x^2>=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx
    Setting up the formulas I get

    I_{n}=\int x^n e^{-ax}=-\frac{1}{a}x^n e^{-ax}+\frac{n}{a}\int x^{n-1}e^{-ax}dx =- \frac{1}{a}x^n e^{-ax}+\frac{n}{a}I_{n-1}

    Assuming the above is correct, do i proceed to calculate I2, I1 and I0? Im not sure where I put in the limits?

    Thanks
    You should be able to calculate I_0 directly as an integral. The use the reduction formula to findI_1 then I_2

    CB
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You should be able to calculate I_0 directly as an integral. The use the reduction formula to findI_1 then I_2

    CB
    DO I sub in the limits in I_0 only and then find I_1 and I_2?
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post
    DO I sub in the limits in I_0 only and then find I_1 and I_2?
    No the recurence is for indefinite integrals so you have (check the algebra here please):

    I_0=-\frac{1}{a}e^{-ax}+c_0

    I_1=-\frac{1}{a}xe^{-ax}-\frac{n}{a^2}e^{-ax}+c_1

    (you can express $$ c_1 in terms of $$ c_0 if you like but there is no need)

    and so on for I_2, now apply the limits of integration.

    There is another way to evaluate the integral, which is to observe that for some quadratic ax^2+bx+c that:

     \dfrac{d}{dx}[(ax^2+bx+c)e^{-\lambda x}]=x^2e^{-\lambda x}

    doing the differentiation will then allow you to find the quadratic and so the required integral.

    CB
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  9. #9
    Senior Member bugatti79's Avatar
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    Thanks,

    I will respond to this when I get the chance
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  10. #10
    Senior Member bugatti79's Avatar
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    Hi CB. IM stil not sure how you work out the integral. Here is what I got so far.

    I_0=-\frac{1}{a}e^{-ax} +C_0
    I_1=-\frac{1}{a}xe^{-ax} + \frac{1}{a}(-\frac{1}{a}e^{-ax} +C_0)
    I_2=-\frac{1}{a}x^2e^{-ax}+\frac{2}{a}(-\frac{1}{a}xe^{-ax}-\frac{1}{a^2}e^{-ax}+\frac{c_0}{a})

    I have calcd I0 and I1 and substituted into I2. Assuming the above algebra is corect, what is the next step? I'm thinking that since its a definite integral we can eliminate the C0 therefore the above becomes

    I_2=\left[-\frac{1}{a}x^2e^{-ax}+\frac{2}{a}(-\frac{1}{a}xe^{-ax}-\frac{1}{a^2}e^{-ax}) \right]^{\infty}_{0}

    Thanks
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  11. #11
    Senior Member bugatti79's Avatar
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    Following on from last thread, I calculate I2 to be equal to \frac{1}{(2\lambda)^3}.
    Assuming this is correct, what do I do with this value in relation to 2A^2\int^{\infty}_{0} e^{-2\lambda x} dx

    My lecture notes goes on to let t=2\lambda x ??? I dont understand the overall methodology of integration by recursion with definite integrals.

    Any help appreciated.

    Thanks
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post
    Following on from last thread, I calculate I2 to be equal to \frac{1}{(2\lambda)^3}.
    Assuming this is correct, what do I do with this value in relation to 2A^2\int^{\infty}_{0} e^{-2\lambda x} dx

    My lecture notes goes on to let t=2\lambda x ??? I dont understand the overall methodology of integration by recursion with definite integrals.

    Any help appreciated.

    Thanks
    Goodness knows, I_2 comes from post #5 which is unrelated to the question in post #1.

    CB
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  13. #13
    Senior Member bugatti79's Avatar
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    CB,

    It turns out I was mislead by my notes and with some typos and apologies if I mislead you! :-) I just solved the following

    <x^2>=2\lambda\int^{\infty}_{0} x^2 e^{-2\lambda x} dx using integration by parts without any recursion. I dont know what the recursion calcs where for. If i find out I will inform you.

    Thanks
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