- #1
hnicholls
- 49
- 1
Is it the case that there is no general solution for the step potential with E = V0?
This seems as though it must be the case as there are only certain energy eigenvalues which would be allowed and, therefore, we cannot for all possible values for the potential barrier find allowed energy eigenvalues.
This would be appear to be reflected in the inconsistent solutions for the wave function in interval 2 (right of the step potential x > 0).
Wave equation for interval 2: Ψ2 = A2eik2x + B2e-ik2x 0 < x < ∞
Wave number for interval 2
k2 = [2m/ħ2 (E - V2)]1/2 Interval 2 ; x > 0 V2 = E
k2 = [2m/ħ2 (0)]1/2 = 0
The angular frequency in this eigenfunction corresponds to the wave number k2 and is:
ω2 = 0 = 0
Thus, the terms in the eigenfunction correspond to the following:
A2 – incident wave (momentum 0)
B2 – reflected wave (momentum 0)
Producing the following wave equation for interval 2: Ψ2 = A2 + B2
0 < x < ∞
This result is inconsistent with the solution to the TISE for interval 2
(-ħ2/2m)d2/dx2Ψ2(x) + v(x)Ψ(x)2 = EΨ(x)
where v(x)Ψ(x)2 = EΨ(x)
So, d2/dx2Ψ2(x) = 0
In order for d2/dx2Ψ2(x) to equal 0
d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2
This seems as though it must be the case as there are only certain energy eigenvalues which would be allowed and, therefore, we cannot for all possible values for the potential barrier find allowed energy eigenvalues.
This would be appear to be reflected in the inconsistent solutions for the wave function in interval 2 (right of the step potential x > 0).
Wave equation for interval 2: Ψ2 = A2eik2x + B2e-ik2x 0 < x < ∞
Wave number for interval 2
k2 = [2m/ħ2 (E - V2)]1/2 Interval 2 ; x > 0 V2 = E
k2 = [2m/ħ2 (0)]1/2 = 0
The angular frequency in this eigenfunction corresponds to the wave number k2 and is:
ω2 = 0 = 0
Thus, the terms in the eigenfunction correspond to the following:
A2 – incident wave (momentum 0)
B2 – reflected wave (momentum 0)
Producing the following wave equation for interval 2: Ψ2 = A2 + B2
0 < x < ∞
This result is inconsistent with the solution to the TISE for interval 2
(-ħ2/2m)d2/dx2Ψ2(x) + v(x)Ψ(x)2 = EΨ(x)
where v(x)Ψ(x)2 = EΨ(x)
So, d2/dx2Ψ2(x) = 0
In order for d2/dx2Ψ2(x) to equal 0
d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2