Why can't I add the energies in this WKB approximation example to get the allowed energies for the given potential?
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Thanks. For #2, do you mean that the form that you solve is ∫r002m(E+U0)−−−−−−−−−−√dr=πℏ(n+1/2)∫0r02m(E+U0)dr=πℏ(n+1/2) where Ueff=−U0+ℏ2ℓ(ℓ+1)2mr2=−U0Ueff=−U0+ℏ2ℓ(ℓ+1)2mr2=−U0 since ℓ=0ℓ=0? Also, it seems strange that the Coulomb potential isn't used here at all. – user2561523 Dec 15 '13 at 3:25
Yes, I think so(with minus sign in the expression E−UE−U, of course. Physical reason is that this phenomenological nuclear physics problem. Wave-function correspond to αα-particle, that has positive charge sign. Nucleus has same sign of charge, therefore Coulomb potential here is repulsive! But, there some extra nuclear forces. We can approximate them by simple −U0−U0 potential well. They attract particle to nucleus. –
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Thanks. For #2, do you mean that the form that you solve is ∫r002m(E+U0)−−−−−−−−−−√dr=πℏ(n+1/2)∫0r02m(E+U0)dr=πℏ(n+1/2) where Ueff=−U0+ℏ2ℓ(ℓ+1)2mr2=−U0Ueff=−U0+ℏ2ℓ(ℓ+1)2mr2=−U0 since ℓ=0ℓ=0? Also, it seems strange that the Coulomb potential isn't used here at all. – user2561523 Dec 15 '13 at 3:25
Yes, I think so(with minus sign in the expression E−UE−U, of course. Physical reason is that this phenomenological nuclear physics problem. Wave-function correspond to αα-particle, that has positive charge sign. Nucleus has same sign of charge, therefore Coulomb potential here is repulsive! But, there some extra nuclear forces. We can approximate them by simple −U0−U0 potential well. They attract particle to nucleus. –
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To keep them moving straight, charges are necessary to establish an electric field in the y-direction to counterbalance the magnetic force, and these charges are shown by the small + and - symbols. ... For most metals, the Hall coefficient is negative, as expected if the charge carriers are electrons.
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