# Transmission Coefficient of Finite Square Well

Now I was calculating the Transmission Coefficients of finite square well potential and found something weird

The transmission coefficient is given by $$T=\left[1+\frac{V_o^2\sin^2(2ka)}{4E(E+V_o)}\right]^{-1}$$ where $$2ka = 2\sqrt{\frac{2ma^2(E+V_o)}{\hbar^2}}=2z_o\sqrt{1+\frac{E}{V_o}}$$ where $$z_o=\sqrt{\frac{2ma^2V}{\hbar^2}}$$

Now special cases:
(i) At $$E\to \infty$$, $$T\to 1$$ which is very correct. Also when in some energy levels we see Resonant Transmission called the Ramsauer Townsend effect which is okay intuitively.

Now the problem parts
(ii) At $$E\to 0$$ , $$T\to 0$$
(iii) At $$E\to -1/2V_o$$, we get $$T=\left[1+\frac{V_o^2\sin^2(\sqrt{2}\,z_o)}{4(-1/2)(-V_o/2+V_o)}\right]^{-1}= \left[1-V_o\sin^2(\sqrt{2}\,z_o)\right]^{-1}$$ $$\bf T\geqslant 1$$ which is very weird because transmission coefficient can't more than 1. Am I doing something wrong here? Also in the second case where $$T\to 0$$, why transmission coefficient is zero?

• The transmission coefficient is undefined when $E<0$ as there is no incoming plane wave with negative energy. Commented Jun 13, 2021 at 18:44
• So for bound states there will be no tunneling? Without transmission coefficient how will I account for tunneling? Commented Jun 13, 2021 at 18:48
• This is a well not a barrier I assume? If it is indeed a well there is nothing to tunnel through surely? Commented Jun 13, 2021 at 19:19
• Okay. Suppose the particle is inside the well and $E<0$ , it is a bound state solution, will there be any possibility of the particle to tunnel through the well? Commented Jun 13, 2021 at 19:25
• hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html
– Gert
Commented Jun 13, 2021 at 19:39

Part (ii) has not yet been answered, so here goes. In general the result that $$T \to 0$$ as $$E \to 0$$ is true UNLESS the well supports half-bound states. In that case, you will find that a zero-energy incident wave is completely transmitted ($$T=1$$, a 'zero-energy transmission resonance') if the well is symmetric and $$0 if the well is asymmetric.