Another potential barrier question

[holden]

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 $$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0$$ for the Schrödinger'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 $$\psi(x) = Ae^{\frac{i\sqrt{2mE}x}{\hbar}} + Be^{\frac{-i\sqrt{2mE}x}{\hbar}}$$.

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!

 

[nrqed]

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 $$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0$$ for the Schrödinger'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 $$\psi(x) = Ae^{\frac{i\sqrt{2mE}x}{\hbar}} + Be^{\frac{-i\sqrt{2mE}x}{\hbar}}$$.

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
$$C e^{+i k_2 x}$$
where $$k_2 = { {\sqrt {2m(E+ V_0)}} \over \hbar}$$
Now impose that the wavefunction is continuous at x=0. This gives obviously $A+B =C$. 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

 

[holden]

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.

 

[physics girl phd]

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).

 

[nrqed]

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 $e^{i kx}$ 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

 

[holden]

That makes perfect sense. Thanks to all for the help.