## 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.

1. Rose16

this is my question

2. IrishBoy123

|dw:1441570821588:dw|

3. Rose16

Excuse me IrishBoy123 did you answer my question, I did not find the answer

4. IrishBoy123

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

@Michele_Laino

6. IrishBoy123

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

could you explain in more details? I do not know if you gave me the last result or not

8. IrishBoy123

|dw:1441575635884:dw| |dw:1441575674405:dw|

9. IrishBoy123

|dw:1441575769994:dw|

10. IrishBoy123

$\huge \Psi(x,0) = \frac{1}{i \ x} \ \ e^{i \ k_x \ x} |_{k_x = \bar k _x - \delta }^{k_x = \bar k _x + \delta}$

11. Rose16

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

if you follow through with that integral, i reckon you will get somewhere

13. IrishBoy123

i will be back tomorrow .... i really hope we can take it forward you should post your ideas here.

14. Rose16

I hope so

15. IrishBoy123

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

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

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

why do you think we use step function

19. Michele_Laino

because we can compute the requested function, applying the theory of distributions

20. Rose16

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

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

it is so complicated for me

23. Michele_Laino

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

Actually No, for this reason I do not know how to use it or complete the steps