Qm 3

• Apr 16th 2008, 10:19 PM
DiscreteW
Qm 3
For a system, a particle that has a square-well potential looks like:

Code:

      V(x)       |    ---  |  -------> x  -a|  |  |a   |  |  |   |__|___| -V_0
Tried drawing as best as I could (the center line is an arrow going up that is $\displaystyle V(x)$, and the corners where the horizontal line meets the vertical line is $\displaystyle -a$ and $\displaystyle a$ (left and right).) The base is $\displaystyle -V_0$.

Questions:

a.) Explain how different the particle w/ $\displaystyle E>0$ or $\displaystyle E<0$ acts.

b.) Determine which of the $\displaystyle E$ situations in part a.) has restriction on the allowed energies. Explain what mathematics is the cause of the restriction.

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

This question is kind of tricky. We know E > 0 usually has reflection and dominates at low energy. The current is same everywhere..

$\displaystyle \frac{\hbar k}{m}(1 - |R|^2) = \frac{\hbar k}{m}|T|^2$

I'm assuming there's no transmission/reflection?

Eigenvalues (can't be divided, unit has no path). They are discrete.

The bigger $\displaystyle V_0$ is, the more boundary conditions? Always has 1 bound state... that's pretty much where I'm at.

b.) Not sure.
• Apr 17th 2008, 04:08 AM
topsquark
I will get to this later today, if someone else doesn't get here first.

-Dan
• Apr 17th 2008, 10:44 AM
topsquark
Quote:

Originally Posted by DiscreteW
For a system, a particle that has a square-well potential looks like:

Code:

      V(x)       |    ---  |  -------> x  -a|  |  |a   |  |  |   |__|___| -V_0
Tried drawing as best as I could (the center line is an arrow going up that is $\displaystyle V(x)$, and the corners where the horizontal line meets the vertical line is $\displaystyle -a$ and $\displaystyle a$ (left and right).) The base is $\displaystyle -V_0$.

Questions:

a.) Explain how different the particle w/ $\displaystyle E>0$ or $\displaystyle E<0$ acts.

For E < 0 the particle is trapped within the potential well. However the probability for the particle to be inside the well is not quite 1. For this reason we have to assume that the wavefunction exists outside the well, which is classically forbidden. (You may think of this as a result of the position-time uncertainty relation.) The wavefunction in the classically fobidden region is modeled on a decaying exponential. This particle will end up having discrete energy eigenvalues.

For E > 0 there are no restrictions on where the particle might be. In this case the particle will have a continuous spectrum of energies. Note, though, that the particle is not quite free while it is "over" the potential well, so the wavefunction is slightly modified here. (Rather like water waves traveling moving from shallow water to deep water abruiptly: the amplitude of the wave gets smaller all of the sudden.)

Note that we have a slight problem for wave energies near E = 0. One can interpret this problem as being akin to the possiblility the wave with E > 0 might "clip" the edge of the potential well. Similarly for E < 0 the particle might "oscillate" into the free particle region. Most analyses overlook this problem as it is difficult to solve the problem exactly.

-Dan
• Apr 17th 2008, 10:46 AM
topsquark
Quote:

Originally Posted by DiscreteW
For a system, a particle that has a square-well potential looks like:

Code:

      V(x)       |    ---  |  -------> x  -a|  |  |a   |  |  |   |__|___| -V_0
Tried drawing as best as I could (the center line is an arrow going up that is $\displaystyle V(x)$, and the corners where the horizontal line meets the vertical line is $\displaystyle -a$ and $\displaystyle a$ (left and right).) The base is $\displaystyle -V_0$.

Questions:

b.) Determine which of the $\displaystyle E$ situations in part a.) has restriction on the allowed energies. Explain what mathematics is the cause of the restriction.

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

This question is kind of tricky. We know E > 0 usually has reflection and dominates at low energy. The current is same everywhere..

$\displaystyle \frac{\hbar k}{m}(1 - |R|^2) = \frac{\hbar k}{m}|T|^2$

I'm assuming there's no transmission/reflection?

Eigenvalues (can't be divided, unit has no path). They are discrete.

The bigger $\displaystyle V_0$ is, the more boundary conditions? Always has 1 bound state... that's pretty much where I'm at.

Qualitatively what is happening in the E < 0 region is the wave is mostly reflecting off the boundary of the potential well, setting up a standing wave inside the well. This is the origin of the discrete energy spectrum. (You can see the same thing happening in a standing wave on a string: Only specific values of the frequency are allowed on the string.)

-Dan