Quantum mechanics, potential barrier

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fluidistic
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Homework Statement


A particle of mass m goes toward the unidimensional barrier potential of the form [itex]V(x)=0[/itex] for [itex]x\leq 0[/itex] and [itex]a\leq x[/itex] and [itex]V(x)=V_0[/itex] for [itex]0<x<a[/itex].
1)Write the corresponding Schrödinger's equation.
2)Calculate the transmission coefficient for the cases [itex]0<E<V_0[/itex] and [itex]E>V_0[/itex]. 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
[itex]\Psi _I (x)=Ae^{ik_1 x}+Be^{-i _k1x}[/itex]
[itex]\Psi _{II}(x)=De^{-k_2x}[/itex]
[itex]\Psi _{III}(x)=Fe^{ik_1x}[/itex].
The continuity conditions give:
(1) [itex]A+B=D[/itex].
(2) [itex]i k_1 A-i k_1 B=-k_2 D[/itex].
(3) [itex]De^{-k_2a}=Fe^{ik_1a}[/itex]
(4) [itex]-k_2 De^{-ak_2}=ik_1 Fe^{ik_1a}[/itex].

I isolated B in function of A. I reached [itex]B=A \cdot \frac{\left (1 +\frac{ik_1}{k_2} \right ) }{ \left ( \frac{ik_1}{k_2} -1\right )}[/itex].
Since [itex]D=A+B[/itex], I got D in function of A only.
And since [itex]F=De^{-a(k_2+ik_1)}[/itex], I reached F in function of A only.
Now the coefficient of transmission is [itex]\frac{F}{A}[/itex]. This gives me [itex]\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)}[/itex].
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 [itex]T= \big | \frac{F}{A} \big | ^2[/itex]. He reaches [itex]T=\frac{4k_1^2k_2^2}{4k_1^2k_2^2+(k_1^2-k_2^2)^2 \sin k_2 a}[/itex].
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 [itex]k_1=\sqrt {\frac{-2mE}{\hbar ^2}}[/itex] and my [itex]k_2=\sqrt {\frac{2m(V_0-E)}{\hbar ^2}}[/itex].
What am I doing wrong?!
 
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Answers and Replies

  • #2
vela
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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.
 
  • #3
vela
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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.
 
  • #4
fluidistic
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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.
Ah whoops, I totally missed this.
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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