From Griffiths, Introduction to Quantum Mechanics, pg. 73:

Evidently, the delta-function well, regardless of its "strength" $\alpha$, has exactly one bound state

$$\psi(x) = \frac{\sqrt{m \alpha}}{\hbar} e^{-m \alpha |x| / \hbar^2} ; \qquad E = - \frac{m \alpha^2}{2 \hbar^2} \, .$$

I don't understand how the author concludes that the delta-function well has one bound state. I understood every part of the derivation that led to the 2 equations above, but I don't see how all of these conclusions point to the delta-function well having one bound state. Also, I don't understand how the delta-function can be a "well" in the first place. It's a spike!

I understand a bound state as being a state that the particle is in where it's energy is less than the potential that bounds it, so because it doesn't have the energy required to surpass the potential, it can never leave the region of potential.

The parameters of the problem: \begin{align} \psi(x)=&Be^{kx} \quad\text{for } x \leq 0 \\ \psi(x)=&Be^{-kx} \quad \text{for } x \geq 0 \\ V=&- \alpha\ \delta(x) \\ k=&\frac{\sqrt{-2mE}}{\hbar} \, . \end{align}

  • $\begingroup$ A good exercise would be to consider a square well and then try to scale it such that it become deeper and narrower but such that you retain more than one solution in the scaling limit. $\endgroup$ – Count Iblis Oct 2 '16 at 18:19
  • $\begingroup$ If you check pages 78 thru 80 of Griffiths, you'll find the finite potential well problem and the resulting transcendental equation solved graphically. If you study that graphical solution on page 80 a bit, you'll literally see why there is just one bound state in the limit as the finite well goes to the delta well. In fact, problem 2.31 asks that you show this mathematically. $\endgroup$ – Alfred Centauri Oct 2 '16 at 23:24

Here is an argument.

  1. On one hand, in 1D the $n$th bound state has $n\!-\!1$ nodes. But a bound state in the delta function well is in a classically forbidden region (and hence exponentially decaying) for $x\in \mathbb{R}\backslash\{0\}$, so there cannot be any nodes. Hence $n\leq 1$.

  2. On the other hand, any attractive potential in 1D has a bound state, so $n\geq 1$.

  • $\begingroup$ yeah this is better formulated than my comment. $\endgroup$ – ZeroTheHero Feb 21 '19 at 18:59

I don't understand how the delta-function can be a "well" in the first place. It's a spike!

One way to think of the "delta-function potential" is as the limit of the finite potential well. We can consider a potential of the form $$ V(x) = \begin{cases} - V_0 & |x| < L/2 \\ 0 & |x| > L/2 \end{cases} $$ (Note that this is an overall shift in the energy from the Wikipedia article linked above.) The delta-function potential is then the limit of this potential as $V_0 \to \infty$ and $L \to 0$ simultaneously. Since $\int V(x) dx = -V_0 L$ for a finite potential well, and the integral of the delta-function potential is $-\alpha$, we will have to take this limit in such a way that $V_0 L = \alpha$. In other words, as the well gets deeper, it simultaneously gets narrower so that $V_0 L = \alpha$ at all times.

In the general case for the finite well (arbitrary $V_0$ and $L$), explicitly solving the Schrödinger equation in the three regions of the real axis ($x < -L/2$, $-L/2 < x < L/2$, $x > L/2$) yields a set of boundary conditions at the interfaces. These boundary conditions can only be satisfied for certain discrete values of the energy, given by the solutions to $$ \sqrt{u_0^2 - v^2} = \begin{cases} v \tan v \\ -v \cot v \end{cases} $$ depending on whether the wavefunction is even or odd, respectively. Here, the parameter $u_0$ is defined as $$ u_0^2 = \frac{m L^2 V_0}{2 \hbar^2} $$ and $v = \sqrt{2mE} L/{2 \hbar}$.

Depending on the values of $L$ and $V_0$, this equation will have a certain number of discrete solutions; they can be found via a nice graphical technique described in the article linked above. In particular, one of the important findings is that the number of bound states in a finite potential well is always equal to $$ N = \left\lfloor \frac{2 u_0}{\pi} \right\rfloor + 1. $$

Now return to the delta-function limit. If $V_0 L = \alpha$, we have $u_0^2 = m \alpha L/2 \hbar^2$; and as $L \to 0$, we will have $u_0 \to 0$ as well. Thus, as we take the limit of a finite potential well becoming very narrow and very deep, we see that there is only going to be one bound state.


I like Michael Seifert's approach of taking the limit of a square potential well, but I think it can be simplified quite a bit to get at the essence of what's going on. A standing-wave solution to the Schrodinger equation, for a square potential well, is going to have three pieces: an oscillating part in the middle, and tails on the left and right, in the classically forbidden region.

Let the well have width $L$ and depth $\propto 1/L$. Now consider a solution that has $n$ humps in the oscillating region. For the higher values of $n$, it's a good approximation to say that this oscillating wave has $n$ half-wavelengths. Its momentum goes like $n/L$, so its kinetic energy goes like $(n/L)^2$. That means that as you contract $L$, the "price" of having a higher $n$ goes like $L^{-2}$, and this outstrips the additional binding energy, which only goes like $-1/L$.

So given this analysis, the mystery is not why there is only one bound state, the mystery is why there are any at all. Why is $n=1$ bound? The idea here is that it's only approximately true that $n$ humps are $n$ half-wavelengths. We're splicing the sine wave on to the two exponential tails, and this means that we don't actually complete $n$ half-wavelengths within the well. We do complete $n-1$ full half-oscillations. For $n=1$ there are no full half-oscillations. For small $L$, the $n=1$ wavefunction just looks like two exponentials with a little curve joining them, and this curve is much less than a full wavelength.

  • $\begingroup$ One can also argue that the well is so tight there is no place for any solution that crosses the axis, as this implies very high curvature and thus very high (i.e. too high) kinetic energy. This precludes anything but $n=1$. $\endgroup$ – ZeroTheHero Feb 21 '19 at 16:49

The delta function in one dimension has one bound state because the equations admit one solution. That does sound circular, but you can gain a physical intuition about the problem by looking at it from a classical point of view. Solving the time invariant Schrodinger equation is no different than looking for the normal modes in a wave equation for a string. In the case of a delta function potential, you've attached an anti-spring (spring with negative spring constant) to the string. The lowest energy state of this string is to let the spring push it off to one side.

Things get interesting when the number of dimensions gets higher, though. I don't recall all of the details, but I do recall that more than one solution can exist (depending on the number of solutions there are to a transcendental equation), and the solutions involve modified Bessel functions of the second kind.

Also, delta functions are always limits that are meant to be inside of integrals. So you can think of a delta function potential well as just an approximation of an ordinary finite square well that is very narrow and very deep compared to the other scales in the problem.

  • $\begingroup$ "...because the equations admit one solution. " - one solution for the energy? $\endgroup$ – whatwhatwhat Oct 2 '16 at 18:07
  • $\begingroup$ Nope. One solution that has a particular energy will match the boundary conditions: the wave function is continuous, the wave function goes to zero as $|x|\rightarrow \infty$, and the derivative of the wave function changes in the prescribed way across the delta function. $\endgroup$ – Sean E. Lake Oct 2 '16 at 18:09

After one has solved the scrodinger equation , there are a few steps to get the actual wave function for the corresponding system. Usually one makes the wave function continuous and it's first derivative continuous . For cases where the potential is infinite the first derivative of the wave function is not continuous , but one can solve for the arbitrary constants. In case of Dirac Delta potential while treating the discontinuity of the wave function at the origin one gets the acceptable energy for the particle in the Dirac Delta potential. And if you see carefully in the final expression of Energy Eugene value there are just constants. Hence D.J Griffith comments that hence E is not discreet and not continuous and has only and only a constant value hence there can be only one bound state.


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