Rectangular Potential and Constraints

[Kreizhn]

Hey all,

A friend asked me for help the other day on his QM homework. The problem regards a rectangular potential
[tex] U(x) = \begin{cases} V_0 & -a \leq x \leq a \\ 0 & \text{otherwise} \end{cases}, \qquad E<V_0 [/tex]
I thought about this for a while and checked a few textbooks. If we solve this in a piecewise form, we get
[tex] \psi(x) = \begin{cases} L_1 e^{ik_Lx} + L_2 e^{-ik_Lx} & x < -a \\ C_1 e^{kx} + C_2 e^{-kx} & -a < x < a \\ R_1 e^{ik_Rx} + R_2 e^{-ik_Rx} & x > a \end{cases} [/tex]
Now by demanding that [itex] \psi(x) [/itex] be continuous and have continuous derivatives, we get 2 conditions from [itex] x=-a [/itex] and 2 from [itex] x=a [/itex] for a total of 4 conditions. Normalization gives us a 5th condition, but we need 6 in total. Now according to the textbooks, we can just set [itex] R_2 = 0 [/itex]. My question is, what is the motivation that allows us to set [itex] R_2 = 0[/itex]?

Edit: Sorry, I perhaps should have been more explicit. I hope it's clear I'm talking about solving the time dependent Schrodinger equation, and [itex] k_L, k_R [/itex] are appropriately defined constants. I didn't think they were important but meant to include them originally.

 

[LostConjugate]

Edited out my first paragraph, I noticed it was incorrect.

I suppose if in the book they are assuming the incoming particles are only coming from the negative x side, then you could set [tex]R_2[/tex] to 0. It is the probability amplitude a particle will reflect off the potential coming from the positive side. If there is no particles heading in that direction, there is no probability.

 

[dextercioby]

Actually a full analysis discusses the solutions including the time dependency. That analysis will include the notions of incoming probability wave and outgoing probability wave. When these 2 are properly defined, 2 of the complex exponentials terms will vanish.

 

[LostConjugate]

which 2?

 

[dextercioby]

Both for the incoming wave and for the outgoing one the complex exponentials in t and x must have opposite signs (I assumed incoming wave from the left (-infty -> x) going to the right (x->+\infty)).

 

[Avodyne]

The condition [itex]R_2=0[/itex] is motivated by saying that the [itex]L_1[/itex] term corresponds to a wave incident from the left, the [itex]L_2[/itex] term corresponds to a wave that was reflected from the potential, and the [itex]R_1[/itex] term corresponds to a wave that was transmitted through the potential. We would want the [itex]R_2[/itex] term only if we had an incident wave from the right as well as the left.

With [itex]R_2=0[/itex], the reflection probability is [itex]|L_2/L_1|^2[/itex], and the transmission probability is [itex]|R_1/L_1|^2[/itex]. These should sum to 1, and will if you solve the problem correctly.

 

[LostConjugate]

Doesn't the transmission probability have to account for the difference in wave number k (difference in momentum)?

 

[Kreizhn]

Thanks for the replies everyone. Sorry I'm so late in replying myself, my email screwed up :S