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In this question I asked about the possibilities of what a boundary might be like, with emphasis on the storytelling.

Now I'd like to investigate what a hard boundary would mean in quantum mechanics, in a more “hard SF” manner.

Imagine a bubble of walled-off spacetime that occurred in the lab so it could be examined up close an personal. Whether the inside is in stasis, destroyed, censored, or whatever is not important here. The interesting thing is that a boundary exists and quantum-mechanical wave functions are prohibited from entering the region contained by the boundary.

Imagine, perhaps, that it's an energy well of arbitrary height. Or, I'm intrigued by @Beta’s remark, “The mirror is 100% legal; all fields and space-time curvature are symmetrical at the boundary, or equivalently the boundary condition allows no normal components of anything.” Or, it just somehow prevents a wavefunction collapse from ever choosing that position.

The phenomena resulting from this should be benign. It needs to interact with normal matter! The bubble won’t just fall through the Earth like a neutrino, or fly off at the speed of light. Rather, it needs to be massive so it can stay put in the room, follow the standard path through spacetime like a massive object, and be held and pushed by normal matter. E.g. it could be placed in a stand and stay put in the lab, even as the Earth turns under it.

I'm thinking that something like Pauli Exclusion could be made to work: the electrons of matter would feel the excluded bubble nearby and distort the shape of the wave function, requiring higher energy. Making it a hard solid object is the main issue!

Second, what might be happening very close to the edge? If it's pushed and must move like a billiard ball, but it is not a point-mass, the force must somehow be communicated around the entire space. If it's infinity rigid there would be problems with motion due to relativistic effects.

Anybody up to communing with the Hamiltonian?

JDługosz
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    I don't think I have enough for an answer for now, but wouldn't it just be the case of the infinite well, except the rest of the universe is in the well? That is, instead of a deep (energy) hole, you get an infinitely tall plateau? – PipperChip Dec 08 '15 at 16:26
  • An energy plateau spanning an area of a macroscopic object, existing inside an otherwise normal universe. I don't have a link to an infinite-well description. – JDługosz Dec 08 '15 at 16:34
  • No rush... I'll be in a long drive this weekend, and can ponder the plot elements that are giving me trouble, unrelated to the interesting physics of the artifact. – JDługosz Dec 08 '15 at 16:36

1 Answers1

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I'm a physicist, but an experimental one. This means I don't deal with hypothetical stuff like this very often, and when I do, I feel the need to figure out a real-world example or test for it. Also, this post contains a fair amount of math and is a little lengthy, so this is your warning. The bullet points at the bottom show my summary and conclusions from this.

I'm afraid I have no idea how to develop such a thing; an object or region of space that is an infinitely high energy wall to particles. I know normal matter and fields can be energy barriers, but I don't think any of them can be infinite in height. (They can be, however, practically infinite, which is something else entirely!) That second question, though, is something I can tell you about.

I'm going to assume some nice symmetry (like a sphere) so that the math is easy, but we also get to really focus on what it means to have an infinite plateau. I'm also going to assume, since this is a macroscopic object, that I don't have to worry about electron tunneling (that changes the problem setup and answer).

As I understand it, your potential energy is a discontinuous, piece-wise function which looks like:

$U(x) = 0$ outside the object ($x\geq0$)

$U(x) = \infty$ inside and on the border of the object ($x<0$)

Therefore I'm going to use the one dimensional time independent Schrödinger Wave equation.

$$\hat{H}\Psi = E\Psi$$ $$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x) }{{\partial x}^2}+U(x)\Psi(x) = E\Psi(x)$$ Inside the object (on on the border) $\Psi(x) = 0$ is the only solution. That's boring, and we don't really care about that, until we start applying boundary conditions for the other part. Now, just outside the object is where things could get interesting: $$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x) }{{\partial x}^2}+ 0\Psi(x) = E\Psi(x)$$ $$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x) }{{\partial x}^2} = E\Psi(x)$$ Since the wavefunction must be 0 at the boundary (x=0), I'm going to say: $$\Psi(x) = A\sin{(\alpha x)}$$ If we solve for $\alpha$, we get an equation like: $$\alpha = \sqrt{{\frac{2mE}{\hbar^2}}}$$

and after normalizing it, your wave function looks like:

$$\Psi(x) = \Bigg(\frac{2\sqrt[4]{{\frac{2mE}{\hbar^2}}}}{\sqrt{2\sqrt{{\frac{2mE}{\hbar^2}}} x - 2\sin(2\sqrt{{\frac{2mE}{\hbar^2}}} x)}}\Bigg) \sin{\Big(\sqrt{{\frac{2mE}{\hbar^2}}} x\Big)}$$

Which is pretty ugly. However, there are some takeaways from this:

  • $\Psi^2$, which shows the odds of finding a particle interacting with the plateau at any particular point in space, is a decaying sine wave. (See here for a generalized, not-entirely-accurate plot where $\alpha = 1$.) This means that what happens near the edge also happens in set distances away from the edge.
  • VERY close to the edge and at some particular distances, the wavefunction goes to zero. This means near the very edge of this object and at some particular distances, there is a "true vacuum." (This also makes me think of band structures, but it's not the same!)
  • It doesn't really matter what the energy of the particle is; the energy just acts as a scaling factor in this equation. All particles approaching this object behave in the same way.
  • A particle near this behaves like it's in an infinite well, but also a little bit like a free particle. It can have any energy level, but still goes to zero at some points in space.
  • This also assumes there is only one particle interacting with the object. Multiple particles, especially ones that interact with each other, can change this wavefunction. This isn't entirely useless, though! It can act as a guide for our intuition for *real world*$^{TM}$ situations.
PipperChip
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  • I wish I understood this answer, as the question reminded me of the Dyson Pair in the Starflyer War trilogy and the Dark Fortress force fields used to isolate the systems, which were absolute. – Draco18s no longer trusts SE Dec 11 '15 at 16:08
  • @Draco18s Is it the math, the physics, or the presentation? I want this to be understandable to the asker, but also random internet folks, too. – PipperChip Dec 11 '15 at 16:13
  • The math and physics. :D Its over my head. The presentation is fine. – Draco18s no longer trusts SE Dec 11 '15 at 16:18
  • Practically infinite: perhaps "as high as needed" so the boundary never is breached but some skin effects are present, maybe it can use the incoming energy against itself like how a superconductor expells a magnetic field. – JDługosz Dec 12 '15 at 03:22
  • Can you elucidate what the decaying sin wave means by explaining what happens to a particle that approaches this anomaly? – JDługosz Dec 12 '15 at 03:32
  • @JDługosz A particle gets close, really close, and then would reflect (it's a wave, after all). The decaying sine wave really means it's more likely to be near the anomaly, and less so the further you get. – PipperChip Dec 13 '15 at 23:56
  • It seems you are saying that there are multiple distances where the amplitude is zero. Does that effect ever stop, and does it mean that photons can't cross such a radius? – JDługosz Dec 13 '15 at 23:59
  • @JDługosz At those places where psi^2 is 0, you will not find the photon/particle in question. They can cross those distances just fine; that's tunneling for you. And no, this wavefunction never stops, it just stops being an appreciable! – PipperChip Dec 14 '15 at 02:54
  • "Being an appreciable"? What about a solid object brought near the surface? What happens to parts of the solid that are at these distances? And what scale are we talking about? – JDługosz Dec 14 '15 at 04:08
  • @JDługosz sorry, that was meant to say "being appreciable" (I'm not always the best at reviewing my posts!) Solid objects, or objects that are large enough for "solid" to have useful meaning, don't get QM applied to them. The largest objects which get QM applied to them are smaller molecules or effects which arise from QM effects, like bonding in quantum chemistry. – PipperChip Dec 14 '15 at 06:34
  • So the surrounding atoms and molecular bonds dominate and wash out the effects of the psi from the anomaly? – JDługosz Dec 14 '15 at 07:36
  • No, The effects of wavefunctions (psi) simply get so small that they behave like normal, macroscopic particles. If you had, say, 1 or 2 particles approaching the anomaly, QM would be super important. A few hundred, and maybe QM will be important. A few million? QM wouldn't be super important; newtonian models would be OK. – PipperChip Dec 14 '15 at 18:28