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When we calculate bound states for finite square well we use symmetry of the potential and take the solutions as combination of even and odd functions . But we do not use the symmetry at the time of calculating scattering state solutions. What is the exact basis of this specific difference of choosing the method of solution? What if we use the symmetry to make the scattering state solution as the combination of even and odd solutions?

Thanks in advance.

Note: An answer which is stated in a footnote of 'Introduction to Quantum Mechanics'- Griffiths:2nd ed. Page 93 is "scattering problem is inherently asymmetric,since the waves come in from one side"; If so, then, similarly, for bound state, from where did the particle came in? Isn't it true that the particle for bound state has to come in from any of the sides of the potential well? Otherwise how did the particle get inside the finite square well in bound state?

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  • $\begingroup$ "similarly, for bound state, from where did the particle came in?" - is a particle 'coming in' in a bound state? $\endgroup$ – Alfred Centauri Sep 29 '17 at 0:18
  • $\begingroup$ Actually I meant how did the particle get in there if we do not say the same for bound state as was stated for scattering state. Because the author said "scattering problem is inherently asymmetric,since the waves come in from one side only" - then what will be the equivalent statement for bound state? Otherwise symmetry ought to be equally applied on scattering. Let me know if I have any dubious language on it. :) @Alfred Centauri $\endgroup$ – Deepayan Turja Sep 29 '17 at 1:31
  • $\begingroup$ Related: physics.stackexchange.com/q/44003/2451 and links therein. $\endgroup$ – Qmechanic Sep 29 '17 at 5:39

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