Quantum mechanics, potential barrier

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Homework Help Overview

The discussion revolves around a quantum mechanics problem involving a particle encountering a potential barrier. The original poster presents a scenario where the potential is defined as V(x)=0 for certain regions and V(x)=V_0 for the barrier. The tasks include writing the Schrödinger equation and calculating the transmission coefficient under different energy conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the Schrödinger equation and derive the transmission coefficient, expressing confusion over the algebraic steps and the interpretation of the results from a textbook reference.

Discussion Status

Participants are actively engaging with the original poster's calculations, questioning the setup of the wave functions and the implications of reflections at the boundary. There is a recognition of the need to clarify the relationship between the ratio F/A and the transmission coefficient, with some participants indicating a shift in understanding regarding the modulus operation.

Contextual Notes

Participants are discussing the potential for reflections at the boundary, which may affect the wave function in the second region. There is also mention of reliance on a specific textbook for guidance, indicating a shared resource for understanding the problem.

fluidistic
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Homework Statement


A particle of mass m goes toward the unidimensional barrier potential of the form V(x)=0 for x\leq 0 and a\leq x and V(x)=V_0 for 0<x<a.
1)Write the corresponding Schrödinger's equation.
2)Calculate the transmission coefficient for the cases 0<E<V_0 and E>V_0. Hint: Check out quantum mechanics books.

Homework Equations


Schrödinger's equation.

The Attempt at a Solution


I solved the Schrödinger's equation for Psi.
The result I have is
\Psi _I (x)=Ae^{ik_1 x}+Be^{-i _k1x}
\Psi _{II}(x)=De^{-k_2x}
\Psi _{III}(x)=Fe^{ik_1x}.
The continuity conditions give:
(1) A+B=D.
(2) i k_1 A-i k_1 B=-k_2 D.
(3) De^{-k_2a}=Fe^{ik_1a}
(4) -k_2 De^{-ak_2}=ik_1 Fe^{ik_1a}.

I isolated B in function of A. I reached B=A \cdot \frac{\left (1 +\frac{ik_1}{k_2} \right ) }{ \left ( \frac{ik_1}{k_2} -1\right )}.
Since D=A+B, I got D in function of A only.
And since F=De^{-a(k_2+ik_1)}, I reached F in function of A only.
Now the coefficient of transmission is \frac{F}{A}. This gives me \left [ 1+ \frac{\left ( 1+ \frac{ik_1}{k_2} \right ) }{\left ( \frac{ik_1}{k_2}-1 \right ) } \right ] e^{-a(k_2+i_k1)}.
I checked out in Cohen-Tanoudji's book about this and... he says that F/A represent the transmission coeffient just like me but then in the algebra, he does T= \big | \frac{F}{A} \big | ^2. He reaches T=\frac{4k_1^2k_2^2}{4k_1^2k_2^2+(k_1^2-k_2^2)^2 \sin k_2 a}.
So I don't understand why he says F/A but does the modulus of it to the second power. And he seems to get a totally different answer from mine.
By the way my k_1=\sqrt {\frac{-2mE}{\hbar ^2}} and my k_2=\sqrt {\frac{2m(V_0-E)}{\hbar ^2}}.
What am I doing wrong?!
 
Last edited:
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It's possible to get reflections at x=a, so the region II wave function needs to be modified.

The transmission coefficient is |F/A|2.
 
fluidistic said:
I checked out in Cohen-Tanoudji's book about this and... he says that F/A represent the transmission coeffient just like me but then in the algebra
Where did he say this? I can never find anything in that encyclopedia of a textbook.

The closest thing I saw was that he said you could get the transmission coefficient from the ratio, but he never actually said the ratio itself was the transmission coefficient.
 
vela said:
It's possible to get reflections at x=a, so the region II wave function needs to be modified.

The transmission coefficient is |F/A|2.
Ah whoops, I totally missed this.
vela said:
Where did he say this? I can never find anything in that encyclopedia of a textbook.

The closest thing I saw was that he said you could get the transmission coefficient from the ratio, but he never actually said the ratio itself was the transmission coefficient.
Again oops once again. You're right, he says that the ratio allows us to determine the coefficients.
Ok I understand now why it's the modulus to the second power. I'll rework on that.
 

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