# $\delta$ potential has highest probability for highest potential

I can't understand this intuitively. Figure 2.9 in Griffith's QM says that the wavefunction at $$x=0$$ for a delta function potential is $$\sqrt{\kappa}$$, and to the right it decays like $$\psi_+=\sqrt{\kappa}e^{-\kappa x}$$, and it's symmetric about $$x=0$$. But the potential is infinitely high at $$x=0$$, so shouldn't the probability of observing the particle there be zero?

• The potential at $x=0$ is not infinity as if it were an infinite well. You can think of the delta function as a limit of a square potential where the heigh increases as the width decreases, but the area remains constant. While it is true that the potential is infinite at $x=0$, it is infinite just there. – TheAverageHijano Feb 21 at 12:15

But the potential is infinitely high at $$x=0$$

You've misunderstood the configuration. To the extent that this kind of language makes sense, at $$x=0$$ the potential is infinitely low.

The delta-function potential well is best thought of as the limit of a sequence of finite square wells whose depth goes to infinity as they become increasingly narrow, while keeping the ground-state energy constant.

This is distinct from a delta-function potential barrier, where it's the barrier height that goes to infinity. The delta-function potential barrier does not admit bound states.

• Hum, this looks like a pretty humongous oversight on my part, getting confused about the distribution being talked about. I was thinking about what you call a barrier in my head, probably because it resembles what I have always thought of as a delta function. I'm very sorry, I should be more meticulous while reading. I checked Griffiths and he appears to be talking about bound states for a delta well, as you said. – JumpingChimp Feb 21 at 13:41
• @JumpingChimp No worries - it happens to all of us, believe me. – Emilio Pisanty Feb 21 at 14:53
• If you feel like doing penance, though, this is probably an excellent opportunity for you to figure out what happens if you have two delta barriers next to each other at some finite separation $L$. Intuition suggests that there should be some form of resonance phenomenon between them if you can fit an integer number of half-wavelengths between the two, but what's the nature of that resonance? Does it admit bound states? If not, then what happens? – Emilio Pisanty Feb 21 at 14:55
• In some video's discussion of the infinite square well, the prof said that if you have two infinite wells next to each other, you can think of them as separate situations which can be solved in the same manner. So should I be able to model that system as one bound infinite square well-like function inside and something else outside? But that gets weird with normalization and I'm not sure what the number of bodies in that situation would be, since I'm thinking of one whole particle in the 'square well' between the barriers. – JumpingChimp Feb 21 at 16:00
• Perhaps this is an opportunity to employ multiple 'step' potentials and throw in some limits? – JumpingChimp Feb 21 at 16:00

Eigenstates satisfy the Schrödinger equation:

$$\hat H \psi(x)=E\psi(x)$$

Multiplying by $$\psi^*(x)$$ and integrating over all the space:

$$\int\psi^*(x) \left(\frac{p^2}{2m} +V(x)\right)\psi(x)=E$$

If we had an infinite potential well, $$\int\psi^*(x)V(x)\psi(x)$$ would be infinite if $$\psi$$ weren't zero, so we would have an infinite energy. We can't allow that to happen, so we take $$\psi=0$$ where the potential is infinite.

The delta function however is integrable, so we don't need to force the wave function to be zero at $$x=0$$.