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Working with finite square wells, where $E < V_{0}$ I've derived the transmission coefficient for a rectangular barrier which seems to be correct. $$ T^{-1} = 1 + \frac{V_{0}^{2}}{4E(V_{0}-E)}sinh^{2}(\frac{2a}{\hbar}\sqrt{2m(V_{0}-E}) $$ We're now asked to discuss if this formula is better suited for narrow or broad wave packets. Any hints on how to tackle this problem?

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  • $\begingroup$ You sure about this formula? $E>V_0$, but $\sqrt{V_0-E}$? $\endgroup$ – JEB Apr 26 '18 at 17:52
  • $\begingroup$ The 2nd part: Is there a strong frequency dependency to $T$? That is, will a wide band signal be distorted? $\endgroup$ – JEB Apr 26 '18 at 17:54
  • $\begingroup$ @JEB my bad, edited now. The signal will most likely not be distorted since we haven't reviewed it in the course, this is introduction to quatum mechanics. $\endgroup$ – bullbo Apr 26 '18 at 18:00
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Actually wavepackets are not suitable with this equation because they do not have a constant wavelength. The less uncertainty in the wavenumber due to the less uncertainty in wavelength makes the broad wavepackets more suitable for the evaluation of the transmission coefficient through the equation.

Narrow wavepackets have a lesser number of waves and are confined to a small place. Uncertainty means that the function cannot be calculated by the experimenter but that doesn’t mean that the function does not exist or is not precise(has multiple solutions).The argument of $\sinh ^2 $ is $\frac{2a}{\hbar} \times \sqrt( {2mV_0} – {2mE})$ where $\frac{2mE}{\hbar}$ is the wavenumber.

Since there are more waves confined in a small area the uncertainty in wavenumber is high

Broad wavepackets are larger, so is their uncertainty of position but the uncertainty in wavenumber is very small and thus their energy .Because without knowing precisely the energy or the wavenumber the transmission coefficient is an approximation I think broad wavepackets are more suitable for this equation

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If you go back through the derivation you'll see that the fields were assumed to be time harmonic with a fixed frequency or wavenumber. So by definition the derivation is only for such waves. However, you can synthesize the response for a wave packet via a Fourier Integral. The easiest would be a Gaussian, where you can do things analytically. You already have T for each wavenumber. You'll see much more interesting behavior in the broad-band case.

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