What would a hard boundary mean in physics on the quantum-mechanical level?

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 Schrodinger 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.

posted over 8 years ago

4y ago by HDE 226868‭

PipperChip‭

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