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Rose16

  • one year ago

I have another question, but please explain in more details. I do not have a good background, so even though you think it is not complicated, but actually I do not have the basic equation that I must use and the details. Therefore, I do not know how to solve this problem.

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  1. Rose16
    • one year ago
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    this is my question

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  2. IrishBoy123
    • one year ago
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    |dw:1441570821588:dw|

  3. Rose16
    • one year ago
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    Excuse me IrishBoy123 did you answer my question, I did not find the answer

  4. IrishBoy123
    • one year ago
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    so we're doing another fourier transform but the \(\sqrt{2 \pi}\)'s cancel out and its in the limit \(\bar k _x - \delta \lt k_x \lt \bar k _x + \delta \) can you process that?!

  5. IrishBoy123
    • one year ago
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    @Michele_Laino

  6. IrishBoy123
    • one year ago
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    so it's \[\huge \Psi(x,0) = \int_{\bar k _x - \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\]

  7. Rose16
    • one year ago
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    could you explain in more details? I do not know if you gave me the last result or not

  8. IrishBoy123
    • one year ago
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    |dw:1441575635884:dw| |dw:1441575674405:dw|

  9. IrishBoy123
    • one year ago
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    |dw:1441575769994:dw|

  10. IrishBoy123
    • one year ago
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    \[\huge \Psi(x,0) = \frac{1}{i \ x} \ \ e^{i \ k_x \ x} |_{k_x = \bar k _x - \delta }^{k_x = \bar k _x + \delta}\]

  11. Rose16
    • one year ago
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    what I wrote in ( NOTE ), just additional information and what I took in class and I do not know if I have to use it or not? I wrote it because I thought that might be helpful from you. but the question does not include the NOTE. ....... Note: the exercise in Quantum Mechanics by Merzbacher if you want to research about.

  12. IrishBoy123
    • one year ago
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    if you follow through with that integral, i reckon you will get somewhere

  13. IrishBoy123
    • one year ago
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    i will be back tomorrow .... i really hope we can take it forward you should post your ideas here.

  14. Rose16
    • one year ago
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    I hope so

  15. IrishBoy123
    • one year ago
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    This is it in long form, it is a inverse Fourier taken straight from the notes. I have no real idea what practical application you are looking at, but this makes sense just as a maths exercise. \[\ \Psi(x,0) = \frac{1}{\sqrt{2 \pi}}\int_{- \infty }^{ \bar k _x - \delta} \ 0 \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x - \delta }^{ \bar k _x + \delta} \sqrt{2 \pi} \ e^{i \ k_x \ x} \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x + \delta }^{ \infty} \ 0 \ dk_x\] \[\ = \int_{\bar k _x - \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\] \[ = \frac{1}{i \ x} \ \left[ e^{i \ k_x \ x} \right]_{k_x = \bar k _x - \delta }^{k_x = \bar k _x + \delta}\] \[\implies \Psi(x,0) = \frac{2}{x} e^{i \ \bar k_x \ x} sin( \delta . x)\] \[\ |\Psi(x,0) |^2 = \frac{4}{x^2} sin^2( \delta . x)\]

  16. Michele_Laino
    • one year ago
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    hint: we can write your function \( \large \phi \left( {{k_x}} \right)\), like below: \[\Large \phi \left( {{k_x}} \right) = \sqrt {2\pi } \left\{ {\theta \left( {{{\overline k }_x} + \delta - {k_x}} \right) + \theta \left( {{k_x} - {{\overline k }_x} + \delta } \right)} \right\}\] where \( \large \theta \left( {{k_x}} \right)\) is the function of \( \large Heaviside \) or the "step function"

  17. IrishBoy123
    • one year ago
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    mmmmm...... http://bueler.github.io/M611F05/M611heaviside.pdf i got the signum when i put my solution to the inverse fourier back into the fourier ie \[ \Psi(x,0) = \frac{2}{x} e^{i \ \bar k_x \ x} sin( \delta . x)\] \[\Phi = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \ \frac{2}{x} e^{i \ \bar k_x \ x} sin( \delta . x) \ e^{-i \ \bar k_x \ x} dx\] \[= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \ \frac{2}{x} sin( \delta . x) \ dx\] |dw:1441635116625:dw| this is viscious stuff.

  18. Rose16
    • one year ago
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    why do you think we use step function

  19. Michele_Laino
    • one year ago
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    because we can compute the requested function, applying the theory of distributions

  20. Rose16
    • one year ago
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    Is what you sent the function that I have to put in the Fourier function and then I do many steps to find it, or the final results.

  21. Michele_Laino
    • one year ago
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    yes! For example, using the theory of distribution, we get this: \[\Large F\left( {{\delta _ + }\left( x \right)} \right) = \theta \left( \omega \right)\] where at the right side we have the Fourier trasform operator F, and: \[\Large {\delta _ + }\left( x \right) = \frac{1}{2}\delta \left( x \right) + \frac{1}{{2\pi i}}P\left( {\frac{1}{x}} \right)\] where P is the principal value of 1/x

  22. Rose16
    • one year ago
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    it is so complicated for me

  23. Michele_Laino
    • one year ago
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    actually, the "delta" function, or the "Dirac" function is a generalized function or a distribution, here is why I have used some results of such theory are you familiar with the theory of distributions?

  24. Rose16
    • one year ago
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    Actually No, for this reason I do not know how to use it or complete the steps

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