Explanation The wave function in region ψ x < 0 is the combination of incident particle’s wave function ψ i n c and reflected particle’s wave function ψ r e f . ψ x < 0 = ψ i n c + ψ r e f In order to find wave function in the region, the Schrödinger equation should be solved using the potential energy for that region. d ψ 2 d x 2 = − 2 m ℏ 2 ( E − U ( x < 0 ) ) ψ x < 0 = − 2 m ℏ 2 ( E − 0 ) ψ x < 0 = − 2 m E ℏ 2 ψ x < 0 The coefficients of wave function ψ x < 0 are grouped into variable k . k = 2 m E ℏ 2 Substitute values of variable k in Schrödinger equation. d ψ 2 d x 2 = − 2 m E ℏ 2 ψ x < 0 = − k 2 ψ x < 0 The solution to the Schrödinger equation in the region is the combination of positive and negative imaginary exponentials. ψ x < 0 = A e i k x + B e − i k x In order to find wave function in region, the Schrödinger equation should be solved using the potential energy for that region. d ψ 2 d x 2 = − 2 m ℏ 2 ( E − U 0 ) ψ 0 < x < L The coefficients of wave function ψ x < 0 are grouped into variable k ' . k ' = 2 m ( E − U 0 ) ℏ 2 Substitute the value of variable k ' in Schrödinger equation. d ψ 2 d x 2 = − 2 m ( E − U 0 ) ℏ 2 ψ 0 < x < L = − k ' 2 ψ 0 < x < L The solution to the Schrödinger equation in the region before the step is a combination of positive and negative imaginary exponentials. ψ x < 0 = C e i k ' x + D e − i k ' x To determine the wave function after the step, solve the Schrödinger equation, using the potential energy in that region 0 . d ψ 2 d x 2 = − 2 m ℏ 2 ( E − U ( x > L ) ) ψ x > L = − 2 m ℏ 2 ( E − 0 ) ψ x > L = − 2 m E ℏ ψ x > L The coefficients of the wave function ψ x > L are the same as the coefficients in the region before the step. Regroup them into k 2 . d ψ 2 d x 2 = − 2 m E ℏ 2 ψ x > L = − k 2 ψ x > L The solution to me Schrödinger equation in this region is a combination of positive and negative exponentials. ψ x > L = F e i k x + G e − i k x There is no reason to have a wave travelling in the negative direction in this region, so the second coefficient G must be 0 . This simplifies the resulting wave. ψ x > L = F e i k x In order to be a valid solution, the wave functions on both sides of the step up must meet in the middle, meaning their limits must be the same. lim x → 0 ( ψ x < 0 ) = lim x → 0 ( ψ 0 < x < L ) A + B = C + D The other condition defining a valid solution isthe derivatives of the wave functions on each side of the step up must approach the same value. lim x → 0 ( d ψ x < 0 d x ) = lim x → 0 ( d ψ 0 < x < L d x ) k A − k B = k ' C − k ' D On the second side of the step, the same continuity conditions apply. The limits of the wave functions must be the same from both sides. lim x → L ( ψ 0 < x < L ) = lim x → L ( ψ x > L ) C e i k ' L + D e − i k ' L = F e i k L The limits of the derivatives must be the same as well

2nd Edition

ISBN: 9780805303087

Author: Randy Harris

Publisher: Addison Wesley

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Connection formulae for WKB approximation

In quantum mechanics, Schrödinger’s equation for certain systems with varying parameters cannot be solved directly. In such a case, an approximation method is used to obtain the approximate solution. WKB method is a special technique used to obtain the a…

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Solution Summary: The author explains that the Schrödinger equation should be solved using the potential energy for the region.

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