- #1
bowlbase
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- 2
Homework Statement
Consider the Hamiltonian ##H=-\frac{d^2}{dx^2}+V(x)##
##V(x) =\begin{cases} \infty & x < 0 ,x>\pi \\V_0 & \frac{\pi}{2}-\frac{a}{2} \geq x \leq \frac{\pi}{2}+\frac{a}{2} \\ 0 &elsewhere\end{cases}##
Use the boundary conditions at x=0 and x=##\pi## to set two of the constants and then use the boundary conditions of continuity of the wave functiond and the first derivative at ##x=\frac{\pi}{2}-\frac{a}{2}## and when ##x=\frac{\pi}{2}+\frac{a}{2}## as well as normalization to solve for the lowest 4 eigenstates.
Homework Equations
The Attempt at a Solution
So when Schrodinger's is zero: ##\frac{d^2\psi(x)}{dx^2}=E\psi(x)##
So I found that ##\psi_1(x)=Asin(\sqrt{E}x)+Bcos(\sqrt{E}x)##
Then when ##V(x)=V_0##:##\frac{d^2\psi(x)}{dx^2}+V_0\psi(x)=E\psi(x)##
Then if E>V: ##\psi_2(x)=Csin(\sqrt{E-V_0}x)+Dcos(\sqrt{E-V_0}x)##
Or if V>E: ##\psi_3(x)=Le^{\sqrt{V_0-E}x}+Fe^{-\sqrt{V_0-E}x}##
I know that if x=0 then A must be zero. I'm not sure what to do with pi though. Another student showed me that they had an additional equation for ##\psi## but I'm not sure where it came from:
##\psi(x)_4=Gsin(\sqrt{V_0-E}(x-\pi))+Hcos(\sqrt{V_0-E}(x-\pi))##
Before I can even give the second part of the question a shot I think I need to reduce the number of coefficients first. If the above equation is accurate (though I have no clue where it came from at the moment) then G would need to be 0 as well. That's 6 coefficients left.
From here I believe that ##\psi_1=\psi_2=\psi_3## (I can't guess about the fourth). I'm just not sure what my next steps should be to get rid of so many variables.