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blue_leaf77

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What do you mean by "bounded by constants"?

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p is found from -ϒ+p_0 to ϒ+p_0 where ϒ and p_0 are positive constants.What do you mean by "bounded by constants"?

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blue_leaf77

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It is equal to another constant, C, between those bounds

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blue_leaf77

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I've normalized the wave function in momentum space and I've started taking the Fourier transform of the normalized function over the same bounds given above. It does not look like I'm going in the right direction.

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blue_leaf77

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What function do you get in position space? Is it even bound at all?

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It gives me 1/√(2ϒ)What function do you get in position space? Is it even bound at all?

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blue_leaf77

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Oh I'm sorry that was my answer for momentum space. In position space I haven't found an answer yet.

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blue_leaf77

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$$

\psi(x) = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty}\phi(p) e^{ipx/\hbar} dp = \frac{1}{\sqrt{2\pi \hbar}} \int_{p_0-Y}^{p_0+Y} \frac{1}{\sqrt{2Y}} e^{ipx/\hbar} dp

$$

How will you execute the next step?

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Yes that's what I have, I then compute the integral?

$$

\psi(x) = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty}\phi(p) e^{ipx/\hbar} dp = \frac{1}{\sqrt{2\pi \hbar}} \int_{p_0-Y}^{p_0+Y} \frac{1}{\sqrt{2Y}} e^{ipx/\hbar} dp

$$

How will you execute the next step?

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blue_leaf77

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Yes of course. In the end you should find that the wavefunction in position space is unbounded.Yes that's what I have, I then compute the integral?

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would it be useful to change my exponential into the form cos(theta)+isin(theta)?Yes of course. In the end you should find that the wavefunction in position space is unbounded.

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blue_leaf77

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It is easier to stay in exponential form, do you know how to integrate an exponential function?

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Yes. Thank you for your help!It is easier to stay in exponential form, do you know how to integrate an exponential function?

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