Calculating Reflection Coefficient for a Potential Well

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Homework Help Overview

The problem involves calculating the reflection coefficient for a particle encountering a potential well, where the potential is zero for x<0 and Vo for x>0. The particle's energy is greater than Vo, and it is incident from the left side.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the time-independent Schrödinger equation to both regions and derive expressions for the wave functions. They raise questions about the validity of their solutions and the resulting reflection coefficient.
  • Participants question specific aspects of the mathematical derivation, including the presence of terms and signs in the equations.

Discussion Status

The discussion is ongoing, with participants actively questioning the mathematical steps taken and exploring potential errors in the derivation. There is no explicit consensus on the correctness of the approach or the results obtained so far.

Contextual Notes

Participants are grappling with the implications of their mathematical results, particularly regarding the reflection coefficient being equal to 1, which seems counterintuitive given the problem's physical context.

captainjack2000
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Homework Statement


A potential well with V(x)=0 for x<0 and V(x) = Vo for x>0. The particle has energy E greater than Vo and is incident from the left side. Calculate the reflection coefficient.

Homework Equations



Relfection coefficient is given by R= (B*B)/(A*A)

The Attempt at a Solution


I think that the time independent Schrödinger equation for the two regions are
x<0 (-hbar/2m)dphisquared/dphi =Ephi(x)
and
x>0 ((-hbar/2m)dphisquared/dphi +Vo phi = Ephi(x)

the solutions to this
x<0 Aexp(ik1x) +Bexp(-ik1x) = phi(x)
x>0 Cexp(ik2x)

is this right
carrying this through and equating two solutions at x=0
A+b = C
and equating differentials at x=0 gives
ik1(A-B) = -ik2C/k1

but when I solve for A and B i get
A=(c/2)(1-ik2/k1)
B=(c/2)(1+ik2/k1)

when I find R i get R=1 which must wrong because particle is not always reflected it should pass over the potential barrier?
 
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Things started to break down when you evaluated the derivatives. Why is there a k1 on the left side? And why didn't the i's cancel out in the final solution?
 
ik1(A-B) = -ik2C/k1
Apart from the extra k1 on the left side which Dick pointed out, why is there a minus sign on the right side?
 
sorry should the differentials be (A-B)=-ik2C/k1?
in which case i still get the same solutions for A and B and a coefficient of 1
 
Why is there a minus sign on the left hand side?
 

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