# Transmission coefficient, best suitable for broad/narrow wave packets

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?

• You sure about this formula? $E>V_0$, but $\sqrt{V_0-E}$? – JEB Apr 26 '18 at 17:52
• The 2nd part: Is there a strong frequency dependency to $T$? That is, will a wide band signal be distorted? – JEB Apr 26 '18 at 17:54
• @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. – bullbo Apr 26 '18 at 18:00

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.