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I posted a small question about part of this problem earlier and got some tips that kept me going for a while. However, I'm still having trouble and I fear that my approach to the problem may be fundamentally wrong, since I'm not too confident with this stuff yet. So I thought I'd post the entire problem and my attempt at solving it so far:
Given the following one-dimensional potential \inline{ V(x) = -g\delta(x) -g\delta(x-a)} where \inline g and \inline a are positive constansts:
a) Graph the potential
This seemed easy enough. The difference between the two delta functions would be 0-0 everywhere except at x = 0, where the potential would equal \inline{ -\infty}, and at x = a, where it would also be \inline{ -\infty}. So the graph of the potential would consist of two vertical lines starting at the x-axis at those two points and going down to negative infinity. Correct?
b) Solve the time-independent Schrodinger equation for E < 0
My approach was as follows. We derived the time independent equation in class. Assuming the wave function has the following form:
\Psi(x,t) = \psi(x)T(t)
then: -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi
I then divided the area being considered into three regions:
i) x < 0
E < 0 \therefore E = -|E|
So: \frac{d^2\psi}{dx^2} = -\frac{2m}{\hbar^2}(-|E|)\psi
\frac{d^2\psi}{dx^2} = \frac{2m|E|}{\hbar^2}\psi
Solving this differential equation:
\psi(x) = Ae^{\kappa x} + Be^{-\kappa x}
where \inline{\kappa^2 = \frac{2m|E|}{\hbar^2}}
As far as I know, in the other regions, the differential equation ends up being the same, so we have:
ii) 0 < x < a
\psi(x) = Ce^{\kappa x} + De^{-\kappa x}
iii) x > a
\psi(x) = Fe^{\kappa x} + Ge^{-\kappa x}
It also seemed to me that as \inline{ x \rightarrow -\infty, Be^{-\kappa x} \rightarrow \infty }
and as \inline{ x \rightarrow \infty, Fe^{\kappa x} \rightarrow \infty }
Neither of these situations is physically possible, so \inline{ B = F = 0 }
One continuity condition is \inline{ \psi(0^+) = \psi(0^-) }
Ae^0 = Ce^0 + De^0 \Rightarrow A = C + D
Also: \inline{ \psi(a^+) = \psi(a^-) }
Ce^{\kappa a} + De^{-\kappa a} = Ge^{-\kappa a}
Ce^{2\kappa a} + D = G
and so Ge^{-\kappa x} = Ce^{\kappa(2a-x)} + De^{-\kappa x}
To summarise, my solution is:
\psi(x) = \left\{\begin{array}{cc}Ae^{\kappa x} & x < 0 \\Ce^{\kappa x} + De^{-\kappa x} & 0 < x < a\\Ce^{\kappa(2a-x)} + De^{-\kappa x} & x > a \end{array}
(A = C + D)
This may or may not satisfy my prof's requirements for part (b), but I decided to continue in the same way as the example problems in my notes did by integrating Schrodinger's equation:
-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}{\frac{d^2\psi}{dx^2}dx} -g\int_{-\epsilon}^{\epsilon}{\delta (x)\psi(x)dx} -g\int_{-\epsilon}^{\epsilon}{\delta(x-a)\psi(x)dx} = E\int_{-\epsilon}^{\epsilon}{\psi dx}
-\frac{\hbar^2}{2m}[\psi'(\epsilon) - \psi'(-\epsilon)] - g\psi(0) - g\psi(a) = 0
Take the limit as \inline{ \epsilon \rightarrow 0} :
-\frac{\hbar^2}{2m}[\psi'(0^+) - \psi'(0^-)] - g\psi(0) - g\psi(a) = 0
\frac{d\psi}{dx} = \left\{\begin{array}{cc}A\kappa e^{\kappa x} & x < 0 \\C\kappa e^{\kappa x} - D\kappa e^{-\kappa x} & 0 < x < a\end{array}
And so -\frac{\hbar^2}{2m}[(C\kappa - D\kappa) - (C\kappa + D\kappa)] - g\psi(0) - g\psi(a) = 0
-\frac{\hbar^2}{2m}(-2D\kappa) = g(C+D) + g(Ce^{\kappa a} + De^{\kappa a})
As it must be clear by now, I'm not exactly sure what I'm doing, and I'm at a loss how to proceed from here. If I had some inkling, then I wouldn't have bothered posting all of this. I'd really appreciate some help with this problem. The last two parts of the question are:
c) Find the energy of the bound states (i.e. E < 0).
d) Find normalized eigen-functions for the energies in part (c).
Given the following one-dimensional potential \inline{ V(x) = -g\delta(x) -g\delta(x-a)} where \inline g and \inline a are positive constansts:
a) Graph the potential
This seemed easy enough. The difference between the two delta functions would be 0-0 everywhere except at x = 0, where the potential would equal \inline{ -\infty}, and at x = a, where it would also be \inline{ -\infty}. So the graph of the potential would consist of two vertical lines starting at the x-axis at those two points and going down to negative infinity. Correct?
b) Solve the time-independent Schrodinger equation for E < 0
My approach was as follows. We derived the time independent equation in class. Assuming the wave function has the following form:
\Psi(x,t) = \psi(x)T(t)
then: -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi
I then divided the area being considered into three regions:
i) x < 0
E < 0 \therefore E = -|E|
So: \frac{d^2\psi}{dx^2} = -\frac{2m}{\hbar^2}(-|E|)\psi
\frac{d^2\psi}{dx^2} = \frac{2m|E|}{\hbar^2}\psi
Solving this differential equation:
\psi(x) = Ae^{\kappa x} + Be^{-\kappa x}
where \inline{\kappa^2 = \frac{2m|E|}{\hbar^2}}
As far as I know, in the other regions, the differential equation ends up being the same, so we have:
ii) 0 < x < a
\psi(x) = Ce^{\kappa x} + De^{-\kappa x}
iii) x > a
\psi(x) = Fe^{\kappa x} + Ge^{-\kappa x}
It also seemed to me that as \inline{ x \rightarrow -\infty, Be^{-\kappa x} \rightarrow \infty }
and as \inline{ x \rightarrow \infty, Fe^{\kappa x} \rightarrow \infty }
Neither of these situations is physically possible, so \inline{ B = F = 0 }
One continuity condition is \inline{ \psi(0^+) = \psi(0^-) }
Ae^0 = Ce^0 + De^0 \Rightarrow A = C + D
Also: \inline{ \psi(a^+) = \psi(a^-) }
Ce^{\kappa a} + De^{-\kappa a} = Ge^{-\kappa a}
Ce^{2\kappa a} + D = G
and so Ge^{-\kappa x} = Ce^{\kappa(2a-x)} + De^{-\kappa x}
To summarise, my solution is:
\psi(x) = \left\{\begin{array}{cc}Ae^{\kappa x} & x < 0 \\Ce^{\kappa x} + De^{-\kappa x} & 0 < x < a\\Ce^{\kappa(2a-x)} + De^{-\kappa x} & x > a \end{array}
(A = C + D)
This may or may not satisfy my prof's requirements for part (b), but I decided to continue in the same way as the example problems in my notes did by integrating Schrodinger's equation:
-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}{\frac{d^2\psi}{dx^2}dx} -g\int_{-\epsilon}^{\epsilon}{\delta (x)\psi(x)dx} -g\int_{-\epsilon}^{\epsilon}{\delta(x-a)\psi(x)dx} = E\int_{-\epsilon}^{\epsilon}{\psi dx}
-\frac{\hbar^2}{2m}[\psi'(\epsilon) - \psi'(-\epsilon)] - g\psi(0) - g\psi(a) = 0
Take the limit as \inline{ \epsilon \rightarrow 0} :
-\frac{\hbar^2}{2m}[\psi'(0^+) - \psi'(0^-)] - g\psi(0) - g\psi(a) = 0
\frac{d\psi}{dx} = \left\{\begin{array}{cc}A\kappa e^{\kappa x} & x < 0 \\C\kappa e^{\kappa x} - D\kappa e^{-\kappa x} & 0 < x < a\end{array}
And so -\frac{\hbar^2}{2m}[(C\kappa - D\kappa) - (C\kappa + D\kappa)] - g\psi(0) - g\psi(a) = 0
-\frac{\hbar^2}{2m}(-2D\kappa) = g(C+D) + g(Ce^{\kappa a} + De^{\kappa a})
As it must be clear by now, I'm not exactly sure what I'm doing, and I'm at a loss how to proceed from here. If I had some inkling, then I wouldn't have bothered posting all of this. I'd really appreciate some help with this problem. The last two parts of the question are:
c) Find the energy of the bound states (i.e. E < 0).
d) Find normalized eigen-functions for the energies in part (c).
Any suggestions?!?!
