Another potential barrier question

In summary, the particle will be reflected back if its energy is greater than the potential drop (V0). R, the reflectivity constant, is found by solving the Schrodinger equation to the right of the potential drop and applying boundary conditions.
  • #1
holden
30
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Since these seem to be popular around here these days, I thought I'd another to the mix :)

"A particle of mass m and energy E > 0 approaches a potential drop -V0 from the left. What is the probability that it will be reflected back if E = V0/3? The potential energy is V(x) = 0 for x<0 and V(x) = -V0 for x>=0."

So what we're looking for is R, the reflectivity constant. My book tells me a formula for 1/T for this case (E > 0, V < E), but that's for a rectangular well and includes the width of the well in the formula.. and this one is infinite.

What I've tried to do so far is using [tex]\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0[/tex] for the Schrodinger's equation to the left of the well. I'm not really sure why I'm using this other than it's in my notes (cause V(x) is zero?). So from this I'm getting [tex]\psi(x) = Ae^{\frac{i\sqrt{2mE}x}{\hbar}} + Be^{\frac{-i\sqrt{2mE}x}{\hbar}}[/tex].

I know R = abs(B/A)^2.. but I'm not sure what to do next. I think apply the boundary conditions..? I'm really lost on how to do that.. I think I'm missing something fundamental here.

Thanks in advance for the help!
 
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  • #2
holden said:
Since these seem to be popular around here these days, I thought I'd another to the mix :)

"A particle of mass m and energy E > 0 approaches a potential drop -V0 from the left. What is the probability that it will be reflected back if E = V0/3? The potential energy is V(x) = 0 for x<0 and V(x) = -V0 for x>=0."

So what we're looking for is R, the reflectivity constant. My book tells me a formula for 1/T for this case (E > 0, V < E), but that's for a rectangular well and includes the width of the well in the formula.. and this one is infinite.

What I've tried to do so far is using [tex]\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0[/tex] for the Schrodinger's equation to the left of the well. I'm not really sure why I'm using this other than it's in my notes (cause V(x) is zero?). So from this I'm getting [tex]\psi(x) = Ae^{\frac{i\sqrt{2mE}x}{\hbar}} + Be^{\frac{-i\sqrt{2mE}x}{\hbar}}[/tex].

I know R = abs(B/A)^2.. but I'm not sure what to do next. I think apply the boundary conditions..? I'm really lost on how to do that.. I think I'm missing something fundamental here.

Thanks in advance for the help!
To the right of the drop, the solution will be
[tex] C e^{+i k_2 x} [/tex]
where [tex] k_2 = { {\sqrt {2m(E+ V_0)}} \over \hbar} [/tex]
Now impose that the wavefunction is continuous at x=0. This gives obviously [itex] A+B =C [/itex]. Now impose that the derivative is continuous too. That will give a second relation relation the three coefficients and the energy. Using those two equations, you should be able to write B in terms of A and then you can get R.

Hope this helps

Patrick
 
  • #3
It does! Thanks! One question, just for my own understanding: why is there only a C term to the right, instead of a C and a D term? This is where I get confused on these.
 
  • #4
why is there only a C term to the right, instead of a C and a D term?
I think it will help if you think about the conditions at +/-infinity.

Also -- physically:
To the left of the barrier you have a forward propagating part (exp(+number)) and a reflected part (exp(-number))... once you hit the barrier, you have just forward propagating part that tunneled into the material (this is kinda like optics).
 
  • #5
holden said:
It does! Thanks! One question, just for my own understanding: why is there only a C term to the right, instead of a C and a D term? This is where I get confused on these.
A wave of the form [itex]e^{i kx}[/itex] is a traveling wave moving to the right (assuming k is a real positive constant). In that problem, the inicident wave comes from the left. There is a reflected wave and a transmitted wave to the right of the origin. So to the right of the origin, there can be no wave moving to the left (this could only happen if there was a second incident wave coming from x=+infinity...but if the only incident wave is coming from x=-infinity, there can be no wave traveling to the left in the x>0 region)

Glad I could help a little
 
  • #6
That makes perfect sense. Thanks to all for the help.
 

Related to Another potential barrier question

1. What is a potential barrier?

A potential barrier is an energy barrier that prevents particles from moving freely across a boundary. It can be in the form of a physical barrier or an energy barrier, and it can affect the movement of particles such as electrons, ions, or molecules.

2. How does a potential barrier affect particle movement?

A potential barrier acts as a resistance to the movement of particles, as they need to overcome the energy barrier in order to cross it. This can slow down or completely block the movement of particles, depending on the strength of the barrier.

3. What are the types of potential barriers?

There are two main types of potential barriers: physical and energy barriers. Physical barriers are physical objects that physically block the movement of particles, such as a wall or a cell membrane. Energy barriers, on the other hand, are differences in energy levels that particles need to overcome in order to move across a boundary.

4. How can potential barriers be overcome?

Potential barriers can be overcome by providing enough energy to the particles to help them overcome the barrier. This can be done by applying a strong enough electric field or by increasing the temperature of the system, which can provide the particles with more energy to cross the barrier.

5. What are some real-world examples of potential barriers?

There are many examples of potential barriers in our daily lives. Some examples include cell membranes in biology, electric fences in agriculture, and energy barriers in electronic devices such as diodes and transistors. In physics, potential barriers are also important for understanding quantum mechanics and the behavior of particles at the atomic and subatomic level.

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