# Qm 2

• Apr 16th 2008, 08:42 PM
DiscreteW
Qm 2
1a.) Calculate the $\displaystyle |T|^2$ using the equation below at $\displaystyle V_0 = 5eV, 10eV, 15eV, 20eV$ respectively, assuming that particle has an energy of $\displaystyle E = 2eV$, and has a width of the barrier $\displaystyle a=20~ nm$.

Equation: $\displaystyle |T|^2 = \frac{(2k\alpha)^2}{(k^2 + \alpha^2)^2\sinh^2{2\alpha a} + (2k\alpha)^2}$

b.) Using the same equation, calculate $\displaystyle |T|^2$ at $\displaystyle a = 10,20,40,80~nm$, respectively, if $\displaystyle E = 2eV$ and $\displaystyle V_0 = 5eV$.

c.) Explain why the above happens in a.) and b.). Explain the physical meaning of $\displaystyle |T|^2$.

d.) Suppose that it's a larger system (that is, it's a calssical system where $\displaystyle a$ and $\displaystyle V_0$ are large) with $\displaystyle E < V_0$ - explain if it's possible to locate the particle in region $\displaystyle x > a$. Why is this the case?

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So I know $\displaystyle \alpha^2 = \frac{2m(V_0 - E)}{\hbar^2} ~\mbox{and} ~k^2 = \frac{2mE}{\hbar^2}$ but I have to get the units correct for a and b and I'm awful at units in physics. Then I guess I'll be able to see a trend if we have the ratio of transmitted current to incident current.

For d.), I'm going with "no". I say this, because in my book it says "there is a finite probability for the particle to be inside the classically forbidden barrier region where its kinetic energy is zero, but the point is that nobody can "see" a particle actually go through a classically forbidden region. Particle detectors can detected only objects of kinetic energy greater than zero; if you insert a detector inside the barrioer to see the particle, you are not only making a hole in the potential, but also in your ojbective, because the ojbect will no longer belong to a classically forbidden region where you wanted to find it."

Thanks!
• Apr 17th 2008, 04:10 AM
topsquark
I will get to this one later, as well.

-Dan
• Apr 17th 2008, 10:35 AM
topsquark
Quote:

Originally Posted by DiscreteW
1a.) Calculate the $\displaystyle |T|^2$ using the equation below at $\displaystyle V_0 = 5eV, 10eV, 15eV, 20eV$ respectively, assuming that particle has an energy of $\displaystyle E = 2eV$, and has a width of the barrier $\displaystyle a=20~ nm$.

Equation: $\displaystyle |T|^2 = \frac{(2k\alpha)^2}{(k^2 + \alpha^2)^2\sinh^2{2\alpha a} + (2k\alpha)^2}$

b.) Using the same equation, calculate $\displaystyle |T|^2$ at $\displaystyle a = 10,20,40,80~nm$, respectively, if $\displaystyle E = 2eV$ and $\displaystyle V_0 = 5eV$.

c.) Explain why the above happens in a.) and b.). Explain the physical meaning of $\displaystyle |T|^2$.

d.) Suppose that it's a larger system (that is, it's a calssical system where $\displaystyle a$ and $\displaystyle V_0$ are large) with $\displaystyle E < V_0$ - explain if it's possible to locate the particle in region $\displaystyle x > a$. Why is this the case?

------------------------------------------

So I know $\displaystyle \alpha^2 = \frac{2m(V_0 - E)}{\hbar^2} ~\mbox{and} ~k^2 = \frac{2mE}{\hbar^2}$ but I have to get the units correct for a and b and I'm awful at units in physics. Then I guess I'll be able to see a trend if we have the ratio of transmitted current to incident current.

For d.), I'm going with "no". I say this, because in my book it says "there is a finite probability for the particle to be inside the classically forbidden barrier region where its kinetic energy is zero, but the point is that nobody can "see" a particle actually go through a classically forbidden region. Particle detectors can detected only objects of kinetic energy greater than zero; if you insert a detector inside the barrioer to see the particle, you are not only making a hole in the potential, but also in your ojbective, because the ojbect will no longer belong to a classically forbidden region where you wanted to find it."

Thanks!

If I'm recognizing this $\displaystyle \left | T^2 \right |$ is the transmission probablity for a particle of mass m out of a finite square well potential. I'd convert all of your energies to J and your distances to m. Then k and alpha have units of 1/m. (Use the value of hbar in Js.)

As to part d) you are correct. You can't quite take any sort of meaningful limit to this equation to get that result. (The typical attempt at this is to let $\displaystyle \hbar \to 0$.) If you want to see how this happens you need to look up the WKB approximation for the Schrodinger equation. This gives a rough, but doable, combination of Classical and Quantum Physics.

-Dan