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Let an infinite square well have its left corner at origin and right corner at $x=L$. We then have the following boundary conditions for a wavefunction $\Psi(0)=\Psi(L)=0$. Now for a different system let $V$ be the potential $V(x)=\lambda \delta (x)$ where $\delta(x)$ is the dirac delta function and $\lambda >0$. The general solution (according to my physics professor) is for a particle coming from the left

$$\Psi(x)=\begin{cases} Ae^{ikx}+Be^{−ikx} &\text{ for }x \le 0 \\ Ce^{ikx} &\text{ for }x \ge 0 \end{cases}$$

and he states "the physical solution must be continous at $x=0$. I think I agree with that however when I asked him "Why don't we have the boundary condition in this case $\Psi(0)=0$ due to $\delta(0)=\infty$ just as in the case with the infinite square well. He answered something but I didn't really understand. And could the solution be as following?

$$\Psi(x)=\begin{cases} Ae^{ikx}+Be^{−ikx} &\text{ for }x < 0 \\ Ce^{ikx} &\text{ for }x > 0 \\ 0 &\text{ for }x = 0 \end{cases}$$

So my question still remains. Thanks in advance.

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Keep in mind that the Dirac delta function is not actually a function. It is typically introduced to physics students as an "infinite spike" such that for if $f(x)=\delta(x)$ then $f(0)=\infty$. This is not very rigorous, and it leads to confusions like the one you are facing now.

Instead, consider the Dirac delta "function" as the limit of a function that has a spike of finite height and width at $x=0$ as we cause the spike to get taller and thinner while still requiring the area under the spike to be constant. Then you will see that the spike is just part of the potential; it is not a boundary condition.

Remember, the reason we don't want a non-zero wavefunction in the infinite potential region of the square well is because it is a non-infinitesimal region with infinite potential. This is not the case for the above limit, as the potential will only approach infinity at a single point, so we have no issues with not setting $\psi(0)=0$ where the Dirac delta function is centered.

As for your proposed solution, if you were to work out the rest of the problem you would find $A=B=C=0$, so the only way $\psi(0)=0$ is if you have no particle at all.

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  • $\begingroup$ Would it be a problem though to set $\Psi(0)=0$ as I did at the bottom in my question? And if so, what problem would occur? $\endgroup$
    – ludz
    May 6 at 17:34
  • $\begingroup$ @ludz See my edit. I'll leave the work of that to you. $\endgroup$ May 7 at 1:26
  • $\begingroup$ Great answer. +1 from me $\endgroup$
    – Sarthak
    May 10 at 3:41
  • $\begingroup$ Did you conclude $A=B=0$ by continuity of first derivative of $\psi$ at $x=0$? Won't the particle be 100% reflected with B=-A in the above problem? $\endgroup$
    – Sarthak
    May 10 at 3:45
  • $\begingroup$ @LoneAcademic No, and No. $\endgroup$ May 10 at 9:20
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Physical wave functions must have finite energy, but it is not always the case that $V(x) = \infty \Rightarrow \psi(x) =0$

For the infinit square potential, if $\psi$ is non zero outside the well, then $\int \psi^* H\psi$ will be infinite.

For the $\delta$ potential (or the Coulomb potential, for that matter), there are wave functions $\psi$ with $\psi(0) \neq 0$ and $\int \psi^* H \psi < \infty$.

This is because $\int {\rm{d}} x \psi^* (x) \delta(x) \psi(x) = |\psi(0)|^2$.

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