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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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.

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

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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?!
so it's \[\huge \Psi(x,0) = \int_{\bar k _x - \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\]
could you explain in more details? I do not know if you gave me the last result or not
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\[\huge \Psi(x,0) = \frac{1}{i \ x} \ \ e^{i \ k_x \ x} |_{k_x = \bar k _x - \delta }^{k_x = \bar k _x + \delta}\]
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.
if you follow through with that integral, i reckon you will get somewhere
i will be back tomorrow .... i really hope we can take it forward you should post your ideas here.
I hope so
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. \[\ \Psi(x,0) = \frac{1}{\sqrt{2 \pi}}\int_{- \infty }^{ \bar k _x - \delta} \ 0 \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x - \delta }^{ \bar k _x + \delta} \sqrt{2 \pi} \ e^{i \ k_x \ x} \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x + \delta }^{ \infty} \ 0 \ dk_x\] \[\ = \int_{\bar k _x - \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\] \[ = \frac{1}{i \ x} \ \left[ e^{i \ k_x \ x} \right]_{k_x = \bar k _x - \delta }^{k_x = \bar k _x + \delta}\] \[\implies \Psi(x,0) = \frac{2}{x} e^{i \ \bar k_x \ x} sin( \delta . x)\] \[\ |\Psi(x,0) |^2 = \frac{4}{x^2} sin^2( \delta . x)\]
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"
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.
why do you think we use step function
because we can compute the requested function, applying the theory of distributions
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.
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
it is so complicated for me
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?
Actually No, for this reason I do not know how to use it or complete the steps

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