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QM - one-dimensional barrier - a simple step

analyzing the simple-step scattering problem for E<V, we find that the solution to the schroedinger equation is:

PHI(left) = Aexp(ikx)+Bexp(-ikx)
PHI(right) = Cexp(-qx)

Continuity of the function and it's derivative at x=0 gives the relations between the parameters A, B and C.

Solving the appropriate equations we obtain

R = |B\A|^2 = 1
meaning that there is total reflection; hence the transmission must be zero.

But if we calculate the probability of finding a particle, say in the interval a<x<2a, we get a non-zero probability (because PHI(right) does not venish).
How is it possible, if every particle must be reflected?
 

Answers and Replies

Hootenanny
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Solving the appropriate equations we obtain

R = |B\A|^2 = 1
meaning that there is total reflection; hence the transmission must be zero.

But if we calculate the probability of finding a particle, say in the interval a<x<2a, we get a non-zero probability (because PHI(right) does not venish).
How is it possible, if every particle must be reflected?
If you obtain a reflection value of one then you have incorrectly determined the coefficients and/or the R value.
 
The calculation of R is taken from a book, there shouldn't be a mistake there.

And this is also a quotation from the book:
"In classical physics region 2 is a "forbidden" domain. In quantum mechanics, however, it is possible for particles to penetrate the barrier."

Region 2 is the region of the step (where V=V0).
 
Hootenanny
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The calculation of R is taken from a book, there shouldn't be a mistake there.
Then you must have incorrectly determined the relationships between the coefficients and/or wave numbers. What is the formula you have for the R value?
And this is also a quotation from the book:
"In classical physics region 2 is a "forbidden" domain. In quantum mechanics, however, it is possible for particles to penetrate the barrier."

Region 2 is the region of the step (where V=V0).
That is correct...
 
R = |B\A|^2 = |(1-iq/k)/(1+iq/k)| = 1

Anyway, if the particles can penetrate the barrier, it's logical that the probability to find a particle in a certain interval on the "other side" is non-zero;
But still R=1. Does it mean that the particles are actually reflected from the inside of the barrier?
 
Hootenanny
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R = |B\A|^2 = |(1-iq/k)/(1+iq/k)| = 1
Sorry my bad, I thought you were talking about the case where E>V. In the case where E<V, then yes, the R value is unity.
Anyway, if the particles can penetrate the barrier, it's logical that the probability to find a particle in a certain interval on the "other side" is non-zero;
But still R=1. Does it mean that the particles are actually reflected from the inside of the barrier?
So yes, although the wavefunction is 'totally reflected' at the boundary, if we calculate the probability density inside the potential step, we find it to be non-zero. Therefore, even though the entire wavefunction is reflected, there is still some non-zero probability to find the particles inside the potential. This is what is referred to as tunnelling.
 
thanks:smile:
 
George Jones
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What Hootenanny said in different words:

Very roughly, unlike classical reflection, the position at which quantum reflection occurs is uncertain; the particle can sometimes penetrate into the forbidden region before being reflected.
 

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