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In quantum physics we've defined: $$ \psi (x) = \sqrt{ \dfrac{1}{2 \pi \hbar} } \int^{ \infty }_{-\infty } \phi (p) \exp \left( i \dfrac{px}{ \hbar} \right) dp $$ Now, $$a(k) \equiv \sqrt{ \hbar } \phi (p)\quad {\rm and}\quad k = \dfrac{p}{ \hbar } $$ Where, $$ a(k) = \left\{ \begin{array}{cccc} 0 & k < - \dfrac{ \epsilon }{2} \\ \sigma + \dfrac{2 \sigma }{ \epsilon } k & - \dfrac{ \epsilon }{2} < k < 0 \\ \sigma - \dfrac{2 \sigma }{ \epsilon } k & 0 < k < \dfrac{ \epsilon }{2} \\ 0 & k > \dfrac{ \epsilon }{2} \\ \end{array} \right.$$

Normalizing $a(k)$ I get $\sigma$ to be:

$$ \sigma = \sqrt{ \dfrac{3}{ \epsilon } } $$ But I can't get anything reasonable from the Fourier Integral.

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closed as off-topic by tpg2114, Qmechanic Nov 3 '13 at 2:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

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I've replaced [tex] by double-dollar-signs and generally fixed the TeX... However, I don't understanding what you're asking. Could you please tell us what is the question? Do you want us to find the Fourier transform of a particular function, or is there some physics in your question? It's just completely unclear what you're after. Also, it would be helpful where your Ansatz $a(k)$ beneath "Where" comes from. What are you trying to do? – Luboš Motl Mar 27 '11 at 9:30

This isn't really physics, just math. Just perform straightforward piecewise integrations (I guess we can't really help you if you don't at least understand basics of integration).

Alternatively, you can observe that you can obtain your function (which is called triangular, by the way, for obvious reasons) as a convolution of the rectangular function. That means that its Fourier transform will be square of the Fourier transform of the rectangular function.

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If I understand well, you're asking what is the Fourier transform of the "triangle" function that is piecewise linear, continuous, nonzero in the interval $\{-\epsilon/2,+\epsilon/2\}$, and having maximum $\sigma$ at $k=0$.

Well, this is a purely mathematical question but the answer is $$ \psi(x) = \frac{\sigma\epsilon}{\sqrt{8\pi\hbar}} \, {\rm sinc}^2(\epsilon x/4\hbar) $$ where $${\rm sinc}(y) = \frac{\sin(y)}{y}$$ and you may substitute your value of $\sigma$. Marek suggests deriving the result as the product of the Fourier transform of two "rectangle" functions because the "triangle" function is the convolution of two "rectangle" functions and the Fourier transform maps convolution to a product. That's why we simply obtained the square of the ${\rm sinc}$ function.

The integrals may be calculated without any convolutions, directly, using the integration by parts. The required techniques are explained e.g. by Jacob Barnett, a 12-year-old student of Purdue University in Indianopolis, here:

And don't worry: Jake is here to help you with all your math phobias. The followup video by Jacob dealing with the trigonometric integrals may also be useful.

Just to be sure: if the original poster wanted to implicitly claim that the particular value of $a(k)$ is "canonical" or "special" or "more important than any other $L^2$-normalizable function $a(k)$", then he is wrong. It's just one function in the Hilbert space and there are infinitely many very different functions over there, too. The triangular function, in the $\epsilon\to 0$ limit, may be viewed as an approximation of the $\delta$-function, but even in this limited category, it is extremely far from being unique.

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