#### cepheid

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*Given the following one-dimensional potential [tex] \inline{ V(x) = -g\delta(x) -g\delta(x-a)} [/tex] where [tex] \inline g [/tex] and [tex] \inline a [/tex] 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 [tex] \inline{ -\infty} [/tex], and at x = a, where it would also be [tex] \inline{ -\infty} [/tex]. 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:

[tex] \Psi(x,t) = \psi(x)T(t) [/tex]

then: [tex] -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi [/tex]

I then divided the area being considered into three regions:

i) x < 0

[tex] E < 0 \therefore E = -|E| [/tex]

So: [tex] \frac{d^2\psi}{dx^2} = -\frac{2m}{\hbar^2}(-|E|)\psi [/tex]

[tex] \frac{d^2\psi}{dx^2} = \frac{2m|E|}{\hbar^2}\psi [/tex]

Solving this differential equation:

[tex] \psi(x) = Ae^{\kappa x} + Be^{-\kappa x} [/tex]

where [tex] \inline{\kappa^2 = \frac{2m|E|}{\hbar^2}} [/tex]

As far as I know, in the other regions, the differential equation ends up being the same, so we have:

ii) 0 < x < a

[tex] \psi(x) = Ce^{\kappa x} + De^{-\kappa x} [/tex]

iii) x > a

[tex] \psi(x) = Fe^{\kappa x} + Ge^{-\kappa x} [/tex]

It also seemed to me that as [tex] \inline{ x \rightarrow -\infty, Be^{-\kappa x} \rightarrow \infty }[/tex]

and as [tex] \inline{ x \rightarrow \infty, Fe^{\kappa x} \rightarrow \infty }[/tex]

Neither of these situations is physically possible, so [tex] \inline{ B = F = 0 }[/tex]

One continuity condition is [tex] \inline{ \psi(0^+) = \psi(0^-) }[/tex]

[tex] Ae^0 = Ce^0 + De^0 \Rightarrow A = C + D [/tex]

Also: [tex] \inline{ \psi(a^+) = \psi(a^-) }[/tex]

[tex] Ce^{\kappa a} + De^{-\kappa a} = Ge^{-\kappa a} [/tex]

[tex] Ce^{2\kappa a} + D = G [/tex]

and so [tex] Ge^{-\kappa x} = Ce^{\kappa(2a-x)} + De^{-\kappa x} [/tex]

To summarise, my solution is:

[tex]\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} [/tex]

[tex] (A = C + D) [/tex]

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:

[tex]-\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}[/tex]

[tex] -\frac{\hbar^2}{2m}[\psi'(\epsilon) - \psi'(-\epsilon)] - g\psi(0) - g\psi(a) = 0 [/tex]

Take the limit as [tex] \inline{ \epsilon \rightarrow 0} [/tex] :

[tex] -\frac{\hbar^2}{2m}[\psi'(0^+) - \psi'(0^-)] - g\psi(0) - g\psi(a) = 0 [/tex]

[tex] \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} [/tex]

And so [tex] -\frac{\hbar^2}{2m}[(C\kappa - D\kappa) - (C\kappa + D\kappa)] - g\psi(0) - g\psi(a) = 0 [/tex]

[tex] -\frac{\hbar^2}{2m}(-2D\kappa) = g(C+D) + g(Ce^{\kappa a} + De^{\kappa a}) [/tex]

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).

d) Find normalized eigen-functions for the energies in part (c).