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If we consider the potential $V= - a V_0 \delta (x)$, it satisfies the bound state energy $$E= - \frac{m a^2 V_0^2 }{2 \hbar^2}.$$ I have done this solving Schrodinger's equation outside the potential for $E<0$. Only wave-function is continuous here, not the potential.

My main goal right now is applying a limiting procedure to a square well, so it will become a delta potential like this case. What approximation should I consider to make this happen? How?

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    $\begingroup$ Hint: Can you think of a sequence of square well potentials that converges to the delta potential? $\endgroup$ – Qmechanic Nov 8 '16 at 0:36
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Consider the square well potential $$ V(x) = \begin{cases}0 & \mathrm{if}\ x<-a/2 \\ -V_0 & \mathrm{if}\ -a/2<x<a/2 \\ 0 & \mathrm{if}\ x>a/2, \end{cases} \tag 1 $$ and look for the ground-state eigenfunction at energy $E=-\frac12 \kappa^2$ (in atomic units, so the Schrödinger equation reads $[-\frac12 \partial_x^2 +V(x)]\psi = E\psi$).

The outside regions are easy to solve for, giving you $\psi(x) = Ae^{-\kappa |x|}$ for $|x|>a/2$. Similarly, the inner region is relatively simple, since we require even parity, so we have $$\psi(x)=B\cos\left(\sqrt{2V_0-\kappa^2}\,x\right)$$ for $|x|<a/2$. To get the quantization condition, you essentially want the vectors $(\psi(a/2),\psi'(a/2))$ from the two regions to be parallel, which with some fiddling can be made into the requirement that the Wronskian of the two solutions vanish: \begin{align} W & = \det\begin{pmatrix}\psi_\mathrm{in}(a/2) & \psi_\mathrm{out}(a/2) \\ \psi_\mathrm{in}'(a/2) & \psi_\mathrm{out}'(a/2)\end{pmatrix} \\ & = \det\begin{pmatrix}B\cos\left(\sqrt{2V_0-\kappa^2}\,a/2\right) & Ae^{-\kappa a/2} \\ -B\sqrt{2V_0-\kappa^2}\sin\left(\sqrt{2V_0-\kappa^2}\,a/2\right) & -\kappa Ae^{-\kappa a/2}\end{pmatrix} \\ & = AB e^{-\kappa a/2} \left[ \sqrt{2V_0-\kappa^2}\sin\left(\sqrt{2V_0-\kappa^2}\,a/2\right) -\kappa \cos\left(\sqrt{2V_0-\kappa^2}\,a/2\right) \right] \\ & = 0, \end{align} or in other words $$ \tan\left(\sqrt{2V_0-\kappa^2}\,a/2\right) = \frac{\kappa}{\sqrt{2V_0-\kappa^2}}. \tag 2 $$


That's the main stage. To get a delta potential out of this, you want to take the limit where the well becomes deeper and deeper, while also squeezing the width $a$ to be smaller and smaller as $V_0$ grows, in such a way that the ground state energy stays constant, and the ground state converges to the single bound state of a delta potential.

There's a few things that you can see need to happen here:

  • First of all, you look at the right-hand side of $(2)$, which is independent of $a$ and can therefore be analyzed using only the limit of $V_0\to\infty$ (or, more accurately, $V_0\gg \kappa^2$). In this limit the right-hand side goes to zero.

  • This means that the left-hand side will also go to zero, which puts the tangent in the tangent regime. This tells you that $a$ must go to zero faster than $1/\sqrt{V_0}$.

  • In this limit, you can simply linearize the equation, which gives you $$ \sqrt{2V_0-\kappa^2}\,a/2 \approx \frac{\kappa}{\sqrt{2V_0-\kappa^2}}, $$ or in other words $$ V_0 a\approx (2V_0-\kappa^2)\,a/2 \approx \kappa. $$

With that you're essentially done, but there's still two distinct approaches you can take. One is to take the limit on the $V_0 a=\mathrm{const}$ curve, and accept the fact that the ground-state energy will move around a bit and then finally converge; the other is to take a slightly more complicated expression for $a$ as a function of $V_0$ by inverting expression $(2)$.

Either way, you're left with two ways of obtaining a delta-spike potential out of a finite square well by a limiting procedure:

  • The potential in $(1)$ converges to the delta potential $V(x) = -V_0 a\delta(x)$, and its ground state energy converges to $-\frac12 a^2V_0^2$, if you take the limit of $V_0\to\infty$ whilst making the width tend to zero as $a=\kappa/V_0$.

Alternatively,

  • The potential in $(1)$ converges to the delta potential $V(x) = -V_0 a\delta(x)$, and its ground state energy $-\frac12 \kappa^2$ stays constant, if you take the limit of $V_0\to\infty$ whilst making the width tend to zero as $$ a = \frac{2/\kappa}{\sqrt{2V_0/\kappa^2-1}} \arctan\left(\frac{1}{\sqrt{2V_0/\kappa^2-1}}\right) $$ (which of course reduces to $a\approx \kappa/V_0$ as $V_0\to\infty$).

    Mathematica graphics

Interestingly enough, when you do this, you can look at what happens to the cosine: the position $a$ gets closer and closer to zero, but the frequency $\sqrt{2V_0-\kappa^2}$ keeps increasing, so you could hope for the switchover at $a$ to keep a constant relative position on the cosine's shoulder.

However, this is not the case. Instead, you can rephrase the tangent-based quantization condition $(2)$ in terms of the cosine, and it gives you $$ \cos\left(\sqrt{2V_0-\kappa^2}\,a/2\right) = \sqrt{1-\frac{\kappa^2}{2V_0}}, $$ identically for all $V_0$. This means, therefore, that the switchover from the inner to the outer region is increasingly at the top of the cosine crest, with barely any time to turn down from the flat top into the inclined sides. This is a bit counter-intuitive, since the slope at the switchover needs to be $-\kappa$, at least in the limit, and the top is relatively flat; the paradox is broken by the fact that the inner length scale is very sharp and therefore the cosine slopes, even if you barely get into them, are also very steep.

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  • $\begingroup$ I understand your approach, but I'm confused about why "Wronskian of the two solutions vanish". Is it due to get an infinite spike potential? Can't we find the equation 2 any other way? $\endgroup$ – user58143 Nov 8 '16 at 2:41
  • $\begingroup$ You can - it's simply the cleanest way to ensure the continuity of $\psi$ and $\psi'$ simultaneously. $\endgroup$ – Emilio Pisanty Nov 8 '16 at 11:53

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