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I am facing a problem of Quantum Mechanics, and I gently need your help in continuing to solve it.

The problem is the old usual problem of a particle subject to a potential, which this time has the form

$$V(x) = \alpha \delta(x^3+2ax^2-a^2x - 2a^3)$$

And we need to find the energies and the wave function normalization.

So first of all I used the well known identity for the Dirac Delta Distribution in order to write the potential as

$$V(x) = \alpha \left(\frac{1}{6a^2}\delta(x-a) + \frac{1}{2a^2}\delta(x+a) + \frac{1}{19a^2}\delta(x+2a)\right)$$

By the way, we can take $\alpha = 1$ in case.

From here, a simple sketch of the potential highlights $4$ regions:

$$\begin{cases} x < -a \\ -a < x < +a\\ a < x < 2a \\ x > 2a \end{cases} $$

But my first doubt is: shall I split the second region into two other regions like

$$\begin{cases} -a < x < 0\\ 0< x < +a \end{cases} $$

or not?

Also, I attempted to write down the general solution fo the EVEN wave function case, and I got stuck also because of the previous regions question. I think I shall go for

$$\psi_e(x) = \begin{cases} A e^{-kx} ~~~~~ x > 2a \\ A e^{kx} ~~~~~ x < -a \\ \ldots \end{cases} $$

Where the $\ldots$ represent my doubts about how to write the general solution in those cases...

I would really be grateful for any help or clarification about this!

Qmechanic
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Les Adieux
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  • I forgot the right constants from the potential, in the wave solutions, but this can be fixed. – Les Adieux Jun 17 '18 at 15:25
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    Isn't the "old usual problem" about bound states in an attractive potential? If $\alpha >0$, this is a repulsive potential, and there wouldn't be any bound states because the spectrum of the Hamiltonian is bounded below by 0. Or are you thinking about scattering instead? – Michael Seifert Jun 17 '18 at 15:45
  • The answer here, shows how to approach such things in general: 1. the ansatz 2. how to obtain the correct boundary conditions for such problems. The double delta potential problem, to be read afterwards, is here. – Raphael J.F. Berger Sep 30 '20 at 06:08

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From here, a simple sketch of the potential highlights $4$ regions: $$\begin{cases} x < -a \\ -a < x < +a\\ a < x < 2a \\ x > 2a \end{cases} $$

This is wrong. A delta function $\delta(x + x_0)$ has a peak at $-x_0$, not at $x_0$. You've flipped the sign of $x$.

But my first doubt is: shall I split the second region into two other regions like $$\begin{cases} -a < x < 0\\ 0< x < +a \end{cases} $$ or not?

No, you don't. I guess you're doing this out of habit, because you've seen that done in other problems, but think about it: what is special about $x = 0$? Why should something happen there at all? Why not also split at $x = a/2$ or $x = \pi^\pi a$?

You only need to split the solution at $x = 0$ if the potential actually changes there. In many problems, the potential is set up this way for convenience, but it doesn't happen here.

Also, I attempted to write down the general solution fo the EVEN wave function case, and I got stuck also because of the previous regions question. I think I shall go for $$\psi_e(x) = \begin{cases} A e^{-kx} ~~~~~ x > 2a \\ A e^{kx} ~~~~~ x < -a \\ \ldots \end{cases} $$ Where the $\ldots$ represent my doubts about how to write the general solution in those cases...

Again, I guess you're trying an even wavefunction out of habit, but this is not right. If the potential is even or odd, it can be shown that your energy eigenstates to be chosen to be even or odd. But the potential you're dealing with here is neither. If you demand your solution is even, you will get no solution at all.

I would really be grateful for any help or clarification about this!

To the left of the first delta function, take a growing exponential. To the right of the last delta function, take a decaying exponential. In both of the two intermediate regions, take superpositions of growing and decaying exponentials.

knzhou
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