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A question is given in Griffiths. Consider the "step" potential V(x) which rises to Vo (at x=0) which is 0 for x<0. For the scattering states and E less than Vo, the reflection coefficient comes out to be 1. It means that the particle can't cross the barrier from the left-hand side.

Doesn't scattering state mean that the particle can cross every (finite) barrier with some non-zero probability? If not, then what is meant by a scattering state?

I want to state humbly that I'm not asking the definition of scattering state on the basis of potential at infinity.

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I think you are asking the difference between scattering state and bound state.

You are solving the Schrodinger with a potential that goes to zero at large distances.

When E<0, ψ goes to zero at large distance. That is a bound state.

When E>0, ψ goes to e^ikr. That is a scattering state. These particles have high probability to be far away. The particles start far, go to the scattering center and go far.

But what you are asking is that when the energy of the particle is smaller then the magnitude of the potential,the particle is bounded to the well that is generating the potential. It is in a bound state.

But if the energy is greater then the magnitude of the potential, the particle will cross the well.

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  • $\begingroup$ I have a doubt in your last statement. I should give an example. For V(x)={ Vo, for -a<x<a; 0, for |x|>a} all the three cases {0<E<Vo, E=V0, E>Vo} where V0>0, have non-zero transmission coefficient. That means the particle can cross the well even when 0<E<V0. Then what's the problem in going inside the well in the above case? $\endgroup$ – Asit Srivastava Jun 3 '18 at 6:12

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