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I stumbled on the question I can't quite grasp:

What is the meaning of poles for transmission probability $T(E)$?

$$ T(E) = \left( 1+\frac{1}{4}\frac{V_0^2}{E (E+V_0)} \sin^2 \left(\frac{2 a}{\hbar }\sqrt{2m (E + V_0)}\right) \right)^{-1} $$

$V(x)$ is a potential, $V(x) = -V_0$ for $-a<x<a$.

First of all, why this function would have any poles? Pole $z_0$ for a function $f(z)$ is by definition called a point where $f(z)$ goes to infinity (please, correct me if I am wrong. I used this link for a sanity check).

Here, on the contrary, $T(E)$ is a smooth function, for $E>V_0$ as well as for $E<V_0$ it is positive and less or equal to one.

Was this question incorrect? Did it implied other meaning for a pole?

Nel
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2 Answers2

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I think there is no pole in function $T(E)$. If $E>0$ and $V_0>0$, then denominator $$1+\frac{1}{4}\frac{V_0^2}{E (E+V_0)} sin^2\left(\frac{2 a}{\hbar }\sqrt{2m (E + V_0)}\right)\neq 0 $$ for any value of $E$ and $V_0$. However, if, $$\frac{2 a}{\hbar} \sqrt{2m (E + V_0)}=n\pi$$ then notice that the transmission coefficient is $1$. This means that there is perfect transmission and no reflection. The energies for this transmission is given as $$E_n+V_0=\frac{n^2\pi^2\hbar^2}{2m(2a)^2}$$ which is precisely the allowed energies for infinite square well.

Nel
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sslucifer
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  • Thanks for your reply.

    I had a typo in my initial question, so I've changed yours reply to remove my initial error)

     – Nel Jun 10 '20 at 02:09
  • You have already written that $T(E)$ is smooth function that is because it has no pole no matter what E is as long as it is greater than zero. So I am having a hard time understanding what your actual question is. – sslucifer Jun 10 '20 at 02:19
  • Let's say you have a function $f(x)=\frac{1}{1+x^2}$. This function will not have any poles as long as $x$ is real. That is why this function is smooth in real axis. But not smooth on complex plane. – sslucifer Jun 10 '20 at 02:21
  • Thanks for your reply, @sslucifer. I guess the initial question "What is the meaning of the poles of T(E)" (which was part of a homework) was just simply incorrect. – Nel Jun 10 '20 at 02:25
  • The initial question was whether this $T(E)$ may have any poles and if so what they are (which I probably should have formulated better). Which as you said is impossible for the given conditions. For other situations, like scattering, @Qmechanic posted some links which explains meaning of poles. – Nel Jun 10 '20 at 02:50
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For general 1D scattering (i.e. not just for the finite potential well) the poles in the transmission and reflection coefficients have a physical interpretation via bound state energies, cf. e.g. my Phys.SE answer here.

Qmechanic
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