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Bound states and scattering length

+ 1 like - 0 dislike
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  1. What is the relationship between bound states and scattering length?

  2. What is the relationship between scattering states and scattering length?

  3. When we say, potential is 'like' repulsive for positive scattering length and viceversa, are we talking with respect to scattering states or bound states? (though the answer should be scattering states, but the literature everywhere assumes you to know it on your own.)

All these concepts are found in the theory of BCS-BEC crossover.

This post imported from StackExchange Physics at 2015-06-15 19:36 (UTC), posted by SE-user cleanplay
asked Jul 12, 2013 in Theoretical Physics by cleanplay (80 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

Consider non-relativistic quantum mechanics with a finite range potential. The spectrum of the Hamiltonian has a discrete sector (possibly empty), corresponding to bound states, and a continuous sector, corresponding to scattering states. The asymptotic wave function of the scattering states can be characterized by phase shifts. The small momentum limit of the phase shifts determines the scattering lengths (and scattering volumes etc.).

By the magic of analyticity there are some relations between binding energies and phase shifts. One example is Levinson's theorem (http://ajp.aapt.org/resource/1/ajpias/v32/i10/p787_s1). Another important example has to do with shallow bound states. If there is a shallow bound state with energy $E=-E_B$, then the s-wave scattering length a satisfies $E_B=1/(ma^2)$. This is explained in most text books on QM (I was perusing Weinberg's book earlier this year, and he has a whole section devoted to shallow bound states.)

With regard to your questions:

1) For shallow bound states, $E_B=1/(ma^2)$. In general, no direct relation except for ``global'' statements like Levinson's theorem.

2) The scatttering length is defined through the low-momentum limit of the scattering phase shift; phase shifts determine asymptotic behavior of scattering states.

3) We call negative $a$ attractive and positive $a$ repulsive because the asymptotic wave functions are pulled in or pushed out, respectively. Also, the simplest mechanism for small negative/positive $a$ is a weak repulsive/attractive potential. If there is a shallow bound state then $a$ is positive. This means that the scattering wave is pushed out (``repulsive'') even though the underlying potential is obviously attractive. This means that at low energy one cannot distinguish scattering from weakly repulsive potentials and strongly attractive ones with a shallow bound state.

This post imported from StackExchange Physics at 2015-06-15 19:36 (UTC), posted by SE-user Thomas
answered Jul 13, 2013 by tmschaefer (705 points) [ no revision ]
thanks, nice answer

This post imported from StackExchange Physics at 2015-06-15 19:36 (UTC), posted by SE-user cleanplay

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