Looking at the functional form of the \(T(E)\), we can see that it becomes unity when \(\phi = 2n\pi\). Thus we have two cases :
Off Resonance : \(\phi \neq 2n \pi\).
Since the denominator is of the order 1, Clearly \(T(E) \propto exp(-4 \kappa l)\). Thus we have destructive interference in this case from the paths.
Resonance : \(\phi = 2n \pi\)
\[\begin{aligned} \Rightarrow & \frac{1}{\hbar} \oint p dx" = (n+ 1/2)\pi \end{aligned}\]
In this case, \(Z\) can be expanded as : \[\begin{split} Z = exp(-2\kappa l) [1 + [1-exp(-2 \kappa l)] + \\ [1-exp(-2 \kappa l)]^2 + [1-exp(-2 \kappa l)]^3 .... ] \end{split}\]
As discussed earlier, the \([1-exp(-2 \kappa l)]\) describes a single set of reflections from the barriers. Thus, it can be seen that the resonant condition is a result of the infinite series, which implies infinite reflection trajectories. This resonant behaviour can’t be obtained by only the single \(t_0\) contribution of \([1-exp(-2 \kappa l)]\), and hence sequential tunneling can not be responsible.
Similar to WKB, this \(T(E)\) expression can be plugged into the current expression to give us the I-V characteristics. Hence, we can safely conclude with sufficient mathematical backing that the negative resistance due to \(T(E) = 1\) arises from the resonant tunneling, not sequential tunneling.