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[Quantum mechanics] Step barrier R^2 and T^2 
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#1
Mar912, 05:07 PM

P: 14

1. The problem statement, all variables and given/known data
There is a step barrier at x=0, V_0 > E I am given: ψi = e^i(kx−ωt) > ψr = R e^i(−kx−ωt) < ψt = T e^(−αx−iωt) > Part of question I am confused about: State the two boundary conditions satisﬁed by the wave function at x = 0 and hence find expressions for R^2 and T^2 2. Relevant equations N/A 3. The attempt at a solution I have already worked out an equation for alpha in the previous part. ψ1 = ψi + ψr ψ2 = ψt I started to apply the conditions ψ1(0) = ψ2(0) to get 1 + R = T and d/dx ψ1(0) = d/dx ψ2(0) to get ik  iRk = αT I solved the above to get R = (ik +αT) / ik with T = 1 + R etc I understand that R^2 = R x R* and T^2 = T x T* However I am told that T = 1  R since these are probability coefficients. Do I solve with T = 1R (probability coefficients) or T = 1+R (using ψ1(0) = ψ2(0))? I am confused which one to use and why. Many thanks. 


#2
Mar912, 09:39 PM

HW Helper
P: 2,327

The probability interpretation only makes since with respect to a normalization, usually the L2 norm for wavefunctions. Unfortunately, the friendly plane wave that makes naive calculations so simple wreaks havoc on the L2 norm. (Plane waves for free particles actually live in a different kind of quantum space than normalizable/quantized wavefunctions.) I don't know what you mean by calling R and T "probability coefficients", but I believe that your equations for R and T (the ones that you derived) are correct (assuming that the wavefunction obeys the Schrodinger equation, which T = 1  R would actually violate at x = 0). I think it doesn't even make sense to compare R^{2} to T^{2} as probabilities, because I think that the probability of transmission is zero, but I will have to think back to my QM (and that is a long way back). 


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