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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.
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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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441570821588:dw

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0Excuse me IrishBoy123 did you answer my question, I did not find the answer

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so 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?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so it's \[\huge \Psi(x,0) = \int_{\bar k _x  \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\]

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0could you explain in more details? I do not know if you gave me the last result or not

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441575635884:dw dw:1441575674405:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441575769994:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[\huge \Psi(x,0) = \frac{1}{i \ x} \ \ e^{i \ k_x \ x} _{k_x = \bar k _x  \delta }^{k_x = \bar k _x + \delta}\]

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0what 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2if you follow through with that integral, i reckon you will get somewhere

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i will be back tomorrow .... i really hope we can take it forward you should post your ideas here.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2This 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. \[\ \Psi(x,0) = \frac{1}{\sqrt{2 \pi}}\int_{ \infty }^{ \bar k _x  \delta} \ 0 \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x  \delta }^{ \bar k _x + \delta} \sqrt{2 \pi} \ e^{i \ k_x \ x} \ dk_x + \frac{1}{\sqrt{2 \pi}}\int_{\bar k _x + \delta }^{ \infty} \ 0 \ dk_x\] \[\ = \int_{\bar k _x  \delta }^{ \bar k _x + \delta} \ e^{i \ k_x \ x} \ dk_x\] \[ = \frac{1}{i \ x} \ \left[ e^{i \ k_x \ x} \right]_{k_x = \bar k _x  \delta }^{k_x = \bar k _x + \delta}\] \[\implies \Psi(x,0) = \frac{2}{x} e^{i \ \bar k_x \ x} sin( \delta . x)\] \[\ \Psi(x,0) ^2 = \frac{4}{x^2} sin^2( \delta . x)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: 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"

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2mmmmm...... 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.

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0why do you think we use step function

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0because we can compute the requested function, applying the theory of distributions

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0Is 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.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! 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

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0it is so complicated for me

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0actually, 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?

Rose16
 one year ago
Best ResponseYou've already chosen the best response.0Actually No, for this reason I do not know how to use it or complete the steps
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