- #1
mjordan2nd
- 173
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This is not a homework question per se, but it's a simple enough computation question from some notes I'm reading that I think it belongs in this section. I will do my best to keep to you guys' formatThis is not a homework question per se, but it's a simple enough computation question from some notes I'm reading that I think it belongs in this section. I will do my best to keep to you guys' format.
1. Homework Statement
We are given the following Schrodinger equation
-\frac{1}{2} \psi "(x) + c_0 \delta(x) \psi(x) = E \psi(x)
and we want to calculate the transmission and reflection coefficients. The answers are given in the book, but based on my calculations the answers should be different. So I'm either misunderstanding how to do the calculation or the book is wrong, and I would appreciate if you guys could help me understand which.
The solution to the equation is (if x<0)
\psi(x) = Ae^{ipx} + Be^{-ipx}
\psi'(x) = ip \left( Ae^{ipx} - Be^{-ipx} \right)
and if x>0
\psi(x) = Ce^{ipx}.
\psi'(x) = ipCe^{ipx}
[/B]
Integrating the Schrodinger equation over a small interval (-\epsilon, \epsilon) gives us
-\frac{1}{2} \left[ \psi'(\epsilon) - \psi'(-\epsilon) \right] + c_0 \psi(0) = E \int_{-\epsilon}^\epsilon \psi(x) dx.
Letting \epsilon \rightarrow 0 gives us the following condition according to my calculation:
\frac{ip}{2} \left[-C+A-B \right] +c_0 C = 0.
However, the notes say that this condition should be
\frac{ip}{2} \left[C-A+B \right] c_0 C = 0.
Who is right? This leads to different transmission and reflection amplitudes. According to the notes they are
T=\frac{C}{A} = \frac{p}{p+ic_0}
R = =\frac{B}{A} = -\frac{ic_0}{p+ic_0}
whereas according to my calculations they are
T=\frac{2c_0-ip}{c_0-ip}
and I didn't really calculate R because it didn't seem worth it until I knew whether or not I was correct or not. Anyway, your assistance would be appreciated in clarifying this matter. Thanks you.
1. Homework Statement
We are given the following Schrodinger equation
-\frac{1}{2} \psi "(x) + c_0 \delta(x) \psi(x) = E \psi(x)
and we want to calculate the transmission and reflection coefficients. The answers are given in the book, but based on my calculations the answers should be different. So I'm either misunderstanding how to do the calculation or the book is wrong, and I would appreciate if you guys could help me understand which.
Homework Equations
The solution to the equation is (if x<0)
\psi(x) = Ae^{ipx} + Be^{-ipx}
\psi'(x) = ip \left( Ae^{ipx} - Be^{-ipx} \right)
and if x>0
\psi(x) = Ce^{ipx}.
\psi'(x) = ipCe^{ipx}
The Attempt at a Solution
[/B]
Integrating the Schrodinger equation over a small interval (-\epsilon, \epsilon) gives us
-\frac{1}{2} \left[ \psi'(\epsilon) - \psi'(-\epsilon) \right] + c_0 \psi(0) = E \int_{-\epsilon}^\epsilon \psi(x) dx.
Letting \epsilon \rightarrow 0 gives us the following condition according to my calculation:
\frac{ip}{2} \left[-C+A-B \right] +c_0 C = 0.
However, the notes say that this condition should be
\frac{ip}{2} \left[C-A+B \right] c_0 C = 0.
Who is right? This leads to different transmission and reflection amplitudes. According to the notes they are
T=\frac{C}{A} = \frac{p}{p+ic_0}
R = =\frac{B}{A} = -\frac{ic_0}{p+ic_0}
whereas according to my calculations they are
T=\frac{2c_0-ip}{c_0-ip}
and I didn't really calculate R because it didn't seem worth it until I knew whether or not I was correct or not. Anyway, your assistance would be appreciated in clarifying this matter. Thanks you.
