- #1
ALime88
- 12
- 0
Hello,
Damn Griffiths 3-star problems... The setup for the potential is a 2D cross basically, so there's a column of zero potential 2a wide (straddling the y-axis) going to +/- infinity, and there's the same thing on the x-axis, so they cross over the origin. Everywhere else the potential is infinity. It's Griffith's 7.20 in the 2nd Edition. The goal is to find the lowest energy which propagates to infinity (doesn't decay to zero along the "arms" of the cross). The provided hint is to "Go way out one arm (say, x>>a) and solve the Schroedinger equation through Sep of Vars; if the wave function goes out to infinity the dependence on x must take the form e^ikx, k>0". Here's what I did:
\frac{-\hbar^2}{2m}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\psi=E\psi
sub in \psi=X(x)Y(y)
\frac{X''}{X}=-\frac{Y''}{Y}-\frac{2mE}{\hbar^2}=-K^2
X=Ae^{iKx}+Be^{-iKx}, Y=Ce^{iTy}+De^{-iTy}
where T^2=-K^2+\frac{2mE}{\hbar^2}
Now I'm having trouble with the boundary conditions, as well as what Griffiths means by "Go way out one arm". Assuming we're already down this arm, Y(a)=Y(-a)=0 due to the infinite potential. That makes C=D, so
Y=\frac{C}{2}\cos(Ta)=0
Therefore T=\frac{(n+1/2)\pi}{a}
That's pretty much as far as I can get. I don't know what the boundary conditions for X would be, and that's the only way I can think of proceeding. I'm really stumped on this one, so any pointers would be nice!
- AL
Damn Griffiths 3-star problems... The setup for the potential is a 2D cross basically, so there's a column of zero potential 2a wide (straddling the y-axis) going to +/- infinity, and there's the same thing on the x-axis, so they cross over the origin. Everywhere else the potential is infinity. It's Griffith's 7.20 in the 2nd Edition. The goal is to find the lowest energy which propagates to infinity (doesn't decay to zero along the "arms" of the cross). The provided hint is to "Go way out one arm (say, x>>a) and solve the Schroedinger equation through Sep of Vars; if the wave function goes out to infinity the dependence on x must take the form e^ikx, k>0". Here's what I did:
\frac{-\hbar^2}{2m}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\psi=E\psi
sub in \psi=X(x)Y(y)
\frac{X''}{X}=-\frac{Y''}{Y}-\frac{2mE}{\hbar^2}=-K^2
X=Ae^{iKx}+Be^{-iKx}, Y=Ce^{iTy}+De^{-iTy}
where T^2=-K^2+\frac{2mE}{\hbar^2}
Now I'm having trouble with the boundary conditions, as well as what Griffiths means by "Go way out one arm". Assuming we're already down this arm, Y(a)=Y(-a)=0 due to the infinite potential. That makes C=D, so
Y=\frac{C}{2}\cos(Ta)=0
Therefore T=\frac{(n+1/2)\pi}{a}
That's pretty much as far as I can get. I don't know what the boundary conditions for X would be, and that's the only way I can think of proceeding. I'm really stumped on this one, so any pointers would be nice!
- AL