2
$\begingroup$

I am attempting to solve the Schrödinger equation with the potential $V = - \delta (x)$. This leads to a differential equation

$$ \alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0 $$

where

$$ \alpha \equiv \frac{\hbar^2}{2m} $$

My instinct to solve the equation was to use an integral transform as I didn't know what else to do with the delta function. Since I am concerned with all space, I chose the Fourier transform. To get the Fourier transform, I transform each term, $\alpha \psi''(x)$, $E \psi(x)$, $\delta(x) \psi(x)$, and $0$. I get these transforms:

\begin{align} &\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\ &\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\ &\mathcal{F}\{ \delta(x)\psi(x) \} = \hat{\psi}(p) \\ &\mathcal{F} \{0 \} = 0 \end{align}

Where $\hat{\psi}(p)$ is the Fourier transform of $\psi(x)$. Unfortunately, when I combine this with the differential equation above, I get

$$ (1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = 0 $$

which results in the $\hat{\psi}(p)$ term going away and leaving me with a function only of $p$, so I am not able to solve for $\psi(x)$ via an inverse Fourier transform like I expected. I did replace the right hand side with an arbitrary function and work through it, but that results in $\psi(x) = 0$ is that function is $0$.

Is there something that I have done incorrectly in my approach? Is there a better approach to solving this equation? My professor and book both solved this question by solving on $(-\infty, 0)$ and $(0, \infty)$ and determining the condition at the origin by integrating the Schrödinger equation over $(-\epsilon, \epsilon)$, but I was hoping to be able to solve it in a more general way and plan to move to a Dirac comb next, which should be very similar using Fourier transformations.

$\endgroup$
  • $\begingroup$ Note that $\mathcal{F}[\delta(x)f(x)] \not= \mathcal{F}[f(x)]$. $\endgroup$ – Winther Sep 3 '15 at 0:58
  • $\begingroup$ I was under the impression that $\mathcal{F}\{\delta(x) f(x)\} = \mathcal{F}\{\delta(x)\} \ast \mathcal{F} \{f(x)\} = \mathcal{F} \{f(x)\}$ since $\delta(x) \ast f(x) = f(x)$, but I forgot to use $\mathcal{F} \{\delta(x)\}$ instead of $\delta(x)$. $\endgroup$ – danielunderwood Sep 3 '15 at 1:04
1
$\begingroup$

Fourier transforms is a perfectly fine tool to solve the system. The advantage of using it is that you avoid having to solve two equations, one for $x<0$ and one for $x>0$, and then matching at $x=0$. The drawback is that you will have to compute the inverse transform which is usually a bit more work than solving and performing the matching if you have not done it before. However if you can use Fourier transform tables then this is imo the simplest way to solve it. The only mistake in your approach so far is not using the correct formula for the Fourier transform of $\delta(x)\psi(x)$. If you correct this then the derivation is straight forward and below I show an example of such a derivation for completeness.


For a general delta-function sink, $V(x) = -\lambda\delta(x)$, the Schrödinger equation reads

$$-\alpha\frac{d^2\psi(x)}{dx^2} - \lambda \delta(x)\psi(x) = E\psi(x)$$

where $E<0$ for a bound state (which is the case we are interested in). Taking the Fourier transform, using the convention $\int {\rm d}x\, e^{ipx}$ for simplicity, we obtain

$$\alpha p^2\hat{\psi}(p) - \lambda\psi(0) = E\hat{\psi}(p) \implies \hat{\psi}(p) = \frac{A}{p^2 + p_0^2}$$

where $p_0 = \sqrt{\frac{-E}{\alpha}}$ and $A= \frac{\lambda\psi(0)}{\alpha}$. Transforming back to real space we get the wavefunction

$$\psi(x) = \frac{A}{2\pi}\int_{-\infty}^\infty\frac{e^{-ipx}}{p^2+p_0^2}{\rm d}x = \frac{A}{2 p_0} e^{-p_0|x|}$$

where I simply used a Fourier transform table to compute the integral. If you want to know how to solve it analytically see this answer. The final thing left is to compute the energy which is determined by imposing wave-function normalization

$$\int_{-\infty}^\infty |\psi(x)|^2{\rm d}x = 1\implies \psi(0) = \sqrt{p_0} \implies E = -\frac{\lambda\alpha}{2}p_0 = -\frac{\lambda^2}{4\alpha}$$

so the full solution can be written

$$\psi(x) = \sqrt{p_0}e^{-p_0|x|}~~~\text{where}~~~p_0 = \frac{\lambda}{2\alpha}$$

$\endgroup$
  • $\begingroup$ A couple questions. First, I notice that you don't get a $4\pi^2$ term -- is this because you use the transform $\int e^{ipx} dx$ instead of $\int e^{2 \pi i \omega x} dx$? Does this affect anything else? Second, should you not use the $\frac{2a^3}{\omega^4 + a^4}$ transform since your $p_0^2$ term will be negative since $E < 0$? If you used $p_0^4$, this would not be an issue. I know we're looking at $E < 0$ anyway, so the $p_0$ term will be positive in our case, but it still seems a bit odd. $\endgroup$ – danielunderwood Sep 3 '15 at 23:51
  • $\begingroup$ @danielu13 Yes, that is correct. The only thing it affects is that by choosing $e^{ipx}$ one must remember to use $\frac{1}{2\pi} e^{-ipx}$ for the inverse transform (which I used). Note that $p_0^2 = \frac{-E}{\alpha} > 0$ so this term is positive. What do you mean by "the ... transform" and "If you used $p_0^4$"? I choose to name the constant $p_0^2$ just for simplicity. It does not matter what I name it the result is the same no matter what. $\endgroup$ – Winther Sep 3 '15 at 23:59
-1
$\begingroup$

From the Schrodinger equation $$\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0$$ and \begin{align} &\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\ &\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\ &\mathcal{F}\{ \delta(x)\psi(x) \} = \psi(0) \\ &\mathcal{F} \{0 \} = 0 \end{align} then it is seen that: $$(1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = \psi(0).$$ Now, since $\hat{\psi}$ is on $(-\infty, \infty)$ then the only way to determine $\psi(0)$ is to consider the two regions, $p < 0$ and $p > 0$, and use $\psi(0)$ as the matching condition. In general the delta function acts like a discontinuity and needs to be treated as a discontinuity.

$\endgroup$
  • $\begingroup$ Note that it should be $\psi(0)$ not $\hat{\psi}(0)$. This constant does not really matter as this can be fixed by the requirement that the wave-function is normalized. Also considering two regions in Fourier space makes no sense; there should be no need for any matching condition (this is only an issue when working in real space). $\endgroup$ – Winther Sep 3 '15 at 2:03
  • $\begingroup$ Pointing out mistakes in your answers is a waste of time as you never correct them, but for OPs sake let me restate it: the final part is still after the edit dead wrong (plus the big typo). No matching is needed. The delta function is a discontinuity in real space and this does not translate into a fourier space discontinuity. $\endgroup$ – Winther Sep 3 '15 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.