Finite potential well transmission coefficient

[lys04]

In the last page of this image, the formula for the transmission coefficient, i'm not sure exactly what it means.
The page says there is no reflection when the sine term is 0 cuz T=1), but for scattering states E>0 anyways? So won't it always pass through? Or is there a chance for a particle with E>0 to not pass through the barrier?

 

[renormalize]

In the last page of this image, the formula for the transmission coefficient, i'm not sure exactly what it means.
The page says there is no reflection when the sine term is 0 cuz T=1), but for scattering states E>0 anyways? So won't it always pass through? Or is there a chance for a particle with E>0 to not pass through the barrier?

Attachment is WAY too small to read.

 

[Nugatory]

The attachment is not readable. Please post the specific formula you are asking about, using the forum’s Latex feature: https://www.physicsforums.com/help/latexhelp/

 

[lys04]

Attachment is WAY too small to read.

I have put bigger images in, sorry

 

[Nugatory]

but for scattering states E>0 anyways? So won't it always pass through? Or is there a chance for a particle with E>0 to not pass through the barrier?

Look at that expression for $T^{-1}$: If the sine term is non-zero then $T^{-1}\gt 1$, meaning that $T\lt 1$, so the particle will not always pass through the barrier under those conditions.

 

[DrClaude]

The page says there is no reflection when the sine term is 0 cuz T=1), but for scattering states E>0 anyways? So won't it always pass through? Or is there a chance for a particle with E>0 to not pass through the barrier?

This is a potential well, not a potential barrier.

But yes, there is a non-zero chance that the particle will be reflected even though E>0. This is purely a quantum mechanical phenomenon.

 

[lys04]

This is a potential well, not a potential barrier.

A potential barrier is like this image right? Whereas a potential well is the opposite.

Ok so a particle with E>0 may be reflected in this case that makes sense, but how does transmission coefficient work for a well? Like a particle with E<0 might pass through the well or smth?

 

[Nugatory]

Ok so a particle with E>0 may be reflected in this case that makes sense, but how does transmission coefficient work for a well? Like a particle with E<0 might pass through the well or smth?

A particle with $E\lt 0$ is bound, trapped in the well and cannot escape to infinity in either direction.. However, the solutions to Schrodinger's equation do have exponentially decaying tails outside of the well (if the well is not of infinite depth) so there is some non-zero probability that a position measurement will find the particle outside of the well.
@DrClaude already mentioned that the non-zero probability of reflection when $E\gt 0$ is a uniquely quantum mechanical phenomenon; this non-zero probability of finding a bound particle somewhere outside the well is another such.

Even more entertaining is quantum tunnelling: consider two wells side by side. Classically there is no way that a particle with $E\lt 0$ can ever move from one well to the other, but in fact they can.

 

[lys04]

A particle with E<0 is bound, trapped in the well and cannot escape to infinity in either direction.. However, these solutions do have exponentially decaying tails outside of the well (if the well is not of infinite depth) so there is some non-zero probability that a position measurement will find the particle outside of the well.

Thats cool!
If I rearrange the equation 2a*sqrt(2m(E+V_0))/hbar = npi (This is when there is 100% transmission), I get V_0+E=n^2hbar^2pi^2/8ma^2, does that mean when E=n^2hbar^2pi^2/8ma^2-V_0 then the particle can escape the well 100%?

 

[DrClaude]

Thats cool!
If I rearrange the equation 2a*sqrt(2m(E+V_0))/hbar = npi (This is when there is 100% transmission), I get V_0+E=n^2hbar^2pi^2/8ma^2, does that mean when E=n^2hbar^2pi^2/8ma^2-V_0 then the particle can escape the well 100%?

There are no bound states when E>0.

 

[lys04]

There are no bound states when E>0.

does that mean 2a*sqrt(2m(E+V_0))/hbar = npi only apply for bound states? And E>0 scattering states will always "pass" the well?

 

[DrClaude]

does that mean 2a*sqrt(2m(E+V_0))/hbar = npi only apply for bound states? And E>0 scattering states will always "pass" the well?

No, it applies to scattering states. I think this is pretty clear in the text you posted.

When E>0, you will always get transmission, i.e., there is always a non-zero probability that the particle will go from one side of the well to the other. What is surprising from a classical point of view is that there can be a non-zero probability of the particle being reflected by the well, which would never happen classically. What is surprising from a quantum point of view is that for certain scattering energies, there is no reflection (as if the potential was not there).