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For the delta function potential, U0 → ∞, b → 0, U0b → V0. Then the above reduces to Q2 − K 2 sinh Qb sin Ka = cos k(a + b) 2QK Taking the limit as b → 0, this becomes Q2 − K 2 Qb sin Ka = cos k(a + b) 2QK Dividing through by Qb and taking th

Author

uday singh

Language

EN