Quantum Mechanics and Momentum
Part 3′ (Mathematical Addendum): Fourier Transforms and Derivatives.
Again, we have our Fourier transform =\mathcal{F}[f(x)](k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-\imath{kx}}\,dx)
. Now, let us consider the transform of the derivative of a function, ](k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f'(x)e^{-\imath{kx}}\,dx)
. We can use integration by parts on this: with \,dx)
, 
; we have )
, and 
, and so
e^{-\imath{kx}}\,dx&=&\int_{-\infty}^{\infty}v\,du\\&=&\left[uv\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}u\,dv\\&=&\left[f(x)e^{-\imath{kx}}\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}f(x)\left(-\imath{k}e^{-\imath{kx}}\,dx\right)\end{eqnarray})
.
Now, if we require =0)
(as any meaningful signal, or quantum wavefunction, must be), the first term on the right side of the above vanishes, and thus:
e^{-\imath{kx}}\,dx&=&-\int_{-\infty}^{\infty}f(x)\left(-\imath{k}e^{-\imath{kx}}\,dx\right)\\&=&\imath{k}\int_{-\infty}^{\infty}f(x)e^{-\imath{kx}}\,dx\\\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f'(x)e^{-\imath{kx}}\,dx&=&\frac{\imath{k}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-\imath{kx}}\,dx\\\mathcal{F}[f'(x)](k)&=&\imath{k}\mathcal{F}[f(x)](k)\end{eqnarray})
.
This will prove important in later posts.
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Tags: Derivative, Fourier Transform, Integration by Parts, Math
This entry was posted on November 20, 2008 at 9:02 pm and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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