- #1
gnome
- 1,041
- 1
Given:
in region 1, x < 0, U = u0 > 0
in region 2, 0≤x≤L, U = 0
in region 3, L < x, U = u0 (same as in region 1)
total energy E constant everyplace, E > u0, and a particle is moving towards the right beginning at some x<0.
So, in region 1:
[tex]\frac{d^2\psi_1}{dx^2} = \frac{-2m}{\hbar^2}\left(E-u_0\right)\psi[/tex]
therefore
[tex]\psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}[/tex]
in region 2, U = 0 so
[tex]\frac{d^2\psi_2}{dx^2} = \frac{-2m}{\hbar^2}E\psi[/tex]
therefore
[tex]\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x}\; \textrm{where}\; k_2 = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
in region 3, U = u0 again, but here there is no edge to produce a wave propagating towards the left, so
[tex]\psi_3(x) = Fe^{ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}[/tex]
Matching conditions give me
at x=0:
[tex]\psi_1(0) = \psi_2(0) \implies A +B = C+D[/tex]
[tex]\frac{d\psi_1}{dx}(x=0) = \frac{d\psi_2}{dx}(x=0)\implies k_1A-k_1B=k_2C-k_2D [/tex]
at x=L:
[tex]\psi_2(L) = \psi_3(L) \implies Ce^{ik_2L} +De^{-ik_2L} = Fe^{ik_1L}[/tex]
[tex]\frac{d\psi_2}{dx}(x=L) = \frac{d\psi_3}{dx}(x=L)\implies k_2Ce^{ik_2L}-k_2De^{-ik_2L}=k_1Fe^{ik_1L} [/tex]
I am to show that if 2L = the de Broglie wavelength of the particle in region 2 there will be NO reflected wave in region 1. I understand that this means that, since the path distance L represents 1/2 wavelength, waves reflected by the well edge at x=L will interfere destructively with waves reflected by the well edge at x=0.
But I don't see how to rearrange the equations to show that explicitly. I don't even know exactly what I'm looking for. How would equations in exponential form depict destructive interference? Or do I have to switch to equations in terms of sine and cosine?
I notice that since [tex]k=\frac{2\pi}{\lambda}[/tex],
[tex]\lambda_2 = 2L \implies k_2 = \frac{2\pi}{2L} = \frac{pi}{L}[/tex]
and that then leads to
[tex]k_1=\sqrt{\frac{\pi^2}{L^2} - \frac{2mu_0}{\hbar^2}}[/tex]
but I can't see that that leads to anything useful.
Any hints will be appreciated.
in region 1, x < 0, U = u0 > 0
in region 2, 0≤x≤L, U = 0
in region 3, L < x, U = u0 (same as in region 1)
total energy E constant everyplace, E > u0, and a particle is moving towards the right beginning at some x<0.
So, in region 1:
[tex]\frac{d^2\psi_1}{dx^2} = \frac{-2m}{\hbar^2}\left(E-u_0\right)\psi[/tex]
therefore
[tex]\psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}[/tex]
in region 2, U = 0 so
[tex]\frac{d^2\psi_2}{dx^2} = \frac{-2m}{\hbar^2}E\psi[/tex]
therefore
[tex]\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x}\; \textrm{where}\; k_2 = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
in region 3, U = u0 again, but here there is no edge to produce a wave propagating towards the left, so
[tex]\psi_3(x) = Fe^{ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}[/tex]
Matching conditions give me
at x=0:
[tex]\psi_1(0) = \psi_2(0) \implies A +B = C+D[/tex]
[tex]\frac{d\psi_1}{dx}(x=0) = \frac{d\psi_2}{dx}(x=0)\implies k_1A-k_1B=k_2C-k_2D [/tex]
at x=L:
[tex]\psi_2(L) = \psi_3(L) \implies Ce^{ik_2L} +De^{-ik_2L} = Fe^{ik_1L}[/tex]
[tex]\frac{d\psi_2}{dx}(x=L) = \frac{d\psi_3}{dx}(x=L)\implies k_2Ce^{ik_2L}-k_2De^{-ik_2L}=k_1Fe^{ik_1L} [/tex]
I am to show that if 2L = the de Broglie wavelength of the particle in region 2 there will be NO reflected wave in region 1. I understand that this means that, since the path distance L represents 1/2 wavelength, waves reflected by the well edge at x=L will interfere destructively with waves reflected by the well edge at x=0.
But I don't see how to rearrange the equations to show that explicitly. I don't even know exactly what I'm looking for. How would equations in exponential form depict destructive interference? Or do I have to switch to equations in terms of sine and cosine?
I notice that since [tex]k=\frac{2\pi}{\lambda}[/tex],
[tex]\lambda_2 = 2L \implies k_2 = \frac{2\pi}{2L} = \frac{pi}{L}[/tex]
and that then leads to
[tex]k_1=\sqrt{\frac{\pi^2}{L^2} - \frac{2mu_0}{\hbar^2}}[/tex]
but I can't see that that leads to anything useful.
Any hints will be appreciated.