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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?

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  • $\begingroup$ The transmission coefficient is undefined when $E<0$ as there is no incoming plane wave with negative energy. $\endgroup$
    – mike stone
    Commented Jun 13, 2021 at 18:44
  • $\begingroup$ So for bound states there will be no tunneling? Without transmission coefficient how will I account for tunneling? $\endgroup$
    – Gandalf73
    Commented Jun 13, 2021 at 18:48
  • $\begingroup$ This is a well not a barrier I assume? If it is indeed a well there is nothing to tunnel through surely? $\endgroup$
    – mike stone
    Commented Jun 13, 2021 at 19:19
  • $\begingroup$ 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? $\endgroup$
    – Gandalf73
    Commented Jun 13, 2021 at 19:25
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html $\endgroup$
    – Gert
    Commented Jun 13, 2021 at 19:39

1 Answer 1

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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 $0<T<1$ if the well is asymmetric.

If you carefully take the zero-energy limit for the transmission coefficient at the top of the original post, you will find the conditions for when bound states appear in the square well.

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