Solutions for the wavefunction in an infinite square well with a delta function barrier in the middle are easily found online (see here for an example). I am wondering what the wavefunction is for an infinite square well with a delta function well in the middle. The setup is the bottom of the infinite square well is defined to be zero energy. I realize that there will be two situations, one where the particle's energy is less than zero and will therefore be bound to the delta function well and one where the particle's energy is greater than zero and is bound to the infinite square well. What are the wavefunctions for these two situations?

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    $\begingroup$ Try to work it out (use $-\alpha \delta (x)$)! It might turn out not to be all that different. $\endgroup$ – Danu Oct 9 '13 at 8:50
  • $\begingroup$ I'm curious; since this is a PHYSICS site, why are questions permitted that are totally mathematical, and hypothetical, and unrelated to any physical reality ? I'm just asking, since I'm a novice member. $\endgroup$ – user26165 Oct 9 '13 at 19:50
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    $\begingroup$ @GeorgeE.Smith, what you describe is basically the quantum mechanics classroom experience, although potentials have been created in labs that come close to these hypothetical potentials. As I understand it, many physical situations can be expressed as "perturbations" of these idealized situations, and so these totally mathematical questions may have physical application after all. However, I am not that far yet in my coursework and so can't say much more. Additionally, your question would be better asked in the Meta section. $\endgroup$ – NeutronStar Oct 9 '13 at 22:01
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    $\begingroup$ @GeorgeE.Smith: if solving Schrödinger's equation for toy potentials ain't physics, I don't know what is. $\endgroup$ – Jerry Schirmer Oct 10 '13 at 2:28

Consider an infinite square with free region $[0, L]$. Place a delta function potential at $L/2$ with strength $\alpha$. Then, Schrödinger's equation is

$$E\psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x ^{2}} + \alpha \delta(x- \frac{L}{2})\psi$$

First, note that the delta function is zero everywhere except at the center, which means that the energy eigenstates everywhere else are given by $\psi = A \sin (kx + \phi)$. Furthermore, we know that $\psi(0) = \psi(L) = 0$. Thus, we know that on the left hand side of the delta, we have $\phi = 0$, and on the right hand side, we have $\phi = - kL$. Thus, for $0<x<L/2$, we have $\psi = A \sin(kx)$, while for $L/2 < x < L$, we have $\psi = B \sin (k(x-L))$. The wave function must be continuous at $L/2$, so this guarantees that $A=-B$.

We can derive a restriction on $k$ by integrating Schrödinger's equation over an arbitrarily small region around $x = L/2$. Since the wave function is continuous, the left hand side drops off, and we're left with:

$$0 = -\frac{\hbar^{2}}{2m}(\psi^{\prime}_{+} - \psi{\prime}_{-}) + \alpha\psi(L/2)$$

Or, more concretely,

$$\psi^{\prime}_{+} = \frac{2m\alpha}{\hbar^{2}}\psi(L/2) + \psi^{\prime}_{-} $$

Taking the derivative and substituting, we find:

$$-Ak\cos(kL/2) = \frac{2m\alpha}{\hbar^{2}}A\sin(kL/2)+Ak\cos(kL/2)$$

Finally, this gives us the trancendental equation

$$\tan(kL/2) = - \frac{\hbar^{2}k}{4m\alpha}$$

Noting that we still have, as in the finite square well case, $E = \frac{\hbar^{2}k^{2}}{2m}$, all solutions have $E > 0$, irrespective of the sign of $\alpha$, although simple graping shows that there are infinite solutions to this equation, independently of the sign of $\alpha$ (though the sign of and value of $\alpha$ will shift where those solutions are quite dramatically).

  • $\begingroup$ I'm sure we can still have E<0, the wavefunction will just have a different form than what you proposed here. $\endgroup$ – NeutronStar Oct 11 '13 at 1:37
  • $\begingroup$ @Joshua: what I gave here was a complete eigenbasis of the Hilbert space. All square integrable functions on $[0,L]$ can be expressed as a sum of these functions. $\endgroup$ – Jerry Schirmer Oct 11 '13 at 3:24
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    $\begingroup$ @Jerry Schimer: transcendental equation should be $tan(kL/2$) = - $\dfrac{\hbar^2k}{m\alpha}$ Or am I missing something? $\endgroup$ – user73508 Feb 19 '15 at 10:21
  • $\begingroup$ @JerrySchirmer I have read your answer, and I think that it's very interesting. But can we have an explicit expression for the energy levels? $\endgroup$ – sunrise Sep 6 '15 at 22:00
  • $\begingroup$ @JerrySchirmer There is also a non-trivial solution for E=0 which exists only under a certain condtion for a(the "height" of the delta) $\endgroup$ – TheQuantumMan May 1 '16 at 18:26

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