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Analytic continuation to find scattering bound states

  1. Dec 9, 2012 #1

    I am trying to understand the idea of using analytic continuation to find bound states in a scattering problem. What do the poles of the reflection coefficent have to do with bound states? In a problem that my quantum professor did in class (from a previous final), we looked at the 1D potential

    $$ V(x) = \begin{cases} \infty & x \le 0 \\ V_0<0 & 0<x\le a \\ 0 & x > a \end{cases} $$

    After some work, we found that the transmission coefficent near $x=a$ has to look like

    $$ S(k) = e^{-2ika} \frac{1+i\tan(k'x)\frac{k}{k'}}{1-i\tan(k'a)\frac{k}{k'}} $$

    $$k'=\frac{\sqrt{|V_0|-E}}{\hbar} \quad\text{and}\quad k=\frac{\sqrt{2mE}}{\hbar}$$ .

    (Notice that k' is real but k is imaginary.) He then said that to find the bound states we want to analytically continue and then look for poles in S(k) (why?!), so we take $$k\to i\kappa$$ (why?!) where $$\kappa = \frac{\sqrt{2m|E|}}{\hbar}$$, so the denominator of S(k) (at the simple pole) becomes

    $$ 0=1-i\tan (k'a)\frac{(i\kappa)}{k'} = 1+\tan (k'a)\frac{\kappa}{k'} $$

    Therefore (why?!) the bound states are given by solutions to $$\tan(k'a) = -\frac{k'}{\kappa}$$.

    This is the last part of a four-part problem, so it could be that I'm not including a critical detail. The full problem statement is problem 3 here:

    Last edited: Dec 9, 2012
  2. jcsd
  3. Dec 12, 2012 #2


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    Look at it like this: A true bound state is a scattered wave without incident wave. Hence it must coincide with a pole of the scattering matrix as the latter is the quotient of the scattered and the incident radiation.
    There are other poles with complex energy and momentum. These are called "resonances".
  4. Dec 12, 2012 #3
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