You might think I’m about to tell a joke. But no, not me. This is serious physics, folks!

Suppose a particle falls into a hole, and, as in a nightmare *(or as in a certain 1970s movie featuring light sabers)*, the walls of the hole start closing in. The particle will just stay there, awaiting the end. But if the same thing happens to a wavicle, the outcome is very different. Like a magician, the wavicle will escape!

Today I’ll explain why.

As I described last time, stationary particles, waves and wavicles differ in their basic properties.

stationary particle | standing wave | standing wavicle | |

location | definite | indefinite | indefinite |

energy | definite, container-independent | adjustable | definite, fixed by frequency |

frequency | none | container-dependent | container-dependent |

amplitude | none | adjustable | fixed by frequency & container |

*A stationary particle, standing wave, and standing wavicle, placed in an identical constrained space and with the lowest possible energy that they can have, exhibit quite different properties.*

Stationary particles can have a fixed position and energy. Stationary (i.e. standing) waves have a definite frequency, but variable amplitude and energy. And standing wavicles are somewhere in between, with no fixed position, but with a definite frequency and energy. Let’s explore an important consequence of these differences.

The Collapsing Well

Imagine that we place a tiny object at the bottom of a deep but wide well. Then we bring the walls of the well together, making it narrower and narrower. What happens?

A particle will just sit patiently at the bottom of the well, even as the walls close in, seemingly unaware of or unconcerned about its impending doom (Fig. 1).

A wavicle, by contrast, can’t bear this situation (Fig. 2). Inevitably, as the walls approach each other, the wavicle will always leap bodily out of the well, avoiding ever being trapped.

The only way to keep a wavicle inside a collapsing well would be to extend the walls of the well upward, making them infinitely high.

This difference between particles and wavicles has a big impact on our world, and on the atoms and subatomic “particles” out of which we are made.

## Energy Between Walls

In my last post I discussed what happens to a particle located between two walls that are separated by a distance ** L**, as in Fig. 3. If the particle has rest mass

**and is stationary, it will have energy**

*m***, no matter what**

*E=mc*^{2}**is.**

*L*A wavicle standing between two walls is different, because the energy of the wavicle grows when ** L** decreases, and vice versa. A wavicle of mass

**will therefore have energy**

*m***larger**than

**.**

*mc*^{2}The energy will be just a *little* larger than ** mc^{2}** if the distance

**is long, specifically much longer than the wavicle’s “Compton wavelength”**

*L***(where**

*h / m c***, Planck’s constant, is a constant of nature, like**

*h***). But if**

*c***is much shorter than the Compton wavelength, then the wavicle’s energy can greatly exceed**

*L***.**

*mc*^{2}## Click here for a math formula showing some details.

*Throughout this post I’m going to be quantitatively imprecise, keeping only those conceptual and mathematical features which are needed for the conceptual lessons. A complete mathematical treatment is possible, but I think it would be less instructive and more confusing.*

Roughly speaking, the formula for the wavicle’s energy is

Notice that

- if is infinite,
- if is large, , just slightly larger than
- if is small, , much larger than

## Energy in a Well

Now let’s imagine digging a hole that has a width ** W** and a depth

**, as shown in Fig. 5. Let’s first imagine the well is quite wide, so**

*D***is relatively large.**

*W*If we put a stationary particle of mass ** m** in the well, it just sits at the bottom. How much energy does it have?

If a particle placed at ground level, outside the well, has energy ** E=mc^{2}**, then a particle below ground level at a depth

**has energy**

*D*- ,

where ** g** is the acceleration of objects due to the Earth’s gravity. The energy has been reduced by the lowering of the particle’s altitude; the larger is

**, the greater the reduction.**

*D**(This reflects the fact that in a hole of greater depth, it would take more energy to lift the particle out of the hole.)*But the particle’s energy shows no dependence on

**.**

*W*Suppose we instead put a stationary wavicle of mass ** m** in a wide well. It too sits inside the well, but it’s different in detail. It vibrates as a standing wave whose length is set by

**, and whose frequency**

*W***therefore also depends on**

*f***. Since a wavicle’s energy and frequency are proportional, via the Planck-Einstein quantum formula**

*W***, that means its energy depends on**

*E=fh***too.**

*W*Being in the well, at a reduced altitude, reduces its energy by the same factor ** m g D** that applies for the particle. As before, the larger is

**, the greater the reduction.**

*D*But the wavicle’s energy gets a boost, relative to that of the particle in the well, because the finite width of the well increases its frequency. The smaller is ** W**, the larger the energy boost from the narrowness of the well.

Thus there is a competition between the well’s depth, which lowers the wavicle’s energy as it does the particle’s, and the well’s width, which raises the wavicle’s energy relative to the particle.

## Click here for a math formula showing how this works

Specifically, inside a well of width ** W** and depth

**, its energy is approximately**

*D*So what happens, now, if the well starts collapsing and the walls close in? What do the particle and wavicle do as ** W** decreases?

## Leaping Out of the Well?

As the well becomes narrower, the particle does nothing. Its energy doesn’t depend on the width ** W** of the well, and so the particle doesn’t care how far apart the walls are until they actually come in contact with it. The particle’s energy is always less than the

**energy of a particle sitting on the ground outside the hole. That means that the particle never has sufficient energy to leave the hole on its own.**

*mc*^{2}The wavicle is quite another matter. As ** W** decreases, the energy of the wavicle

**. And at some point, when**

*increases**is small enough,*

**W****the energy of the wavicle**

*in*the well becomes greater than the energy of the wavicle that extends*outside*the well.To keep things simple, let’s imagine ** L** to be very large, so large that when outside the well, both the particle’s and wavicle’s energy are almost exactly equal to

**. In that context, consider Fig. 7, where I show the**

*mc*^{2}**-dependence of the energy of four objects:**

*W*- the wavicle and particle outside the well (blue), whose energy is independent of
and*D*,*W* - the particle in the well (orange), whose energy depends on
but not*D*,*W* - the wavicle in the well (green), whose energy depends on both
and*D*.*W*

Only the last of these depends appreciably on ** W**, which is why the blue and orange lines are straight.

When ** W** is large (at right in the plot,) then the fact that the well is deep assures that the energy of the wavicle in the well (green) is lower than the energy of the wavicle that sits on the ground (blue), filling the whole space. That implies that the wavicle will remain within the well, just as a particle in the well would.

But inevitably, when ** W** is small enough (at left in the plot,) the situation reverses!

*(How small exactly? It depends on details, specifically on both the mass*The wavicle in the narrow well, unlike a wavicle in a wide well, has energy

**m**and the hole’s depth**D**.)**greater**than a wavicle outside the well. That means that the wavicle in the narrow well has sufficient energy to leave the well entirely, and to become a standing wave that sits on the ground, occupying the whole region between the outer walls.

Notice this is completely general! No matter how deep we make the well, as long as the depth is finite, there is always a small enough width for which the wavicle’s escape becomes possible.

## Wavicles Push Back; Particles Don’t

To say this another way, a wavicle in a narrow well has more energy than one in a wide well. Therefore, squeezing a well with a wavicle in it costs energy, whereas to squeeze a well with a mere particle inside costs none. As we shrink ** W**, adding more and more energy to the system, there will always come a point where the wavicle will have enough energy to pop out of the hole. It’s almost as though the wavicle is springy and resists being compressed. A particle in a well, by contrast, is completely inert.

This remarkable property of wavicles, related to Heisenberg’s uncertainty principle, has enormous implications for atomic physics and subatomic physics. In my next post, we’ll see examples of these implications, ones of central importance in human existence.

———-

## Aside: We can also compare wavicles with ordinary, familiar waves. We’ve seen how important it is that shortening a wavicle increases its energy. What about the waves on a guitar string? Can they, too, hop out of a hole?

A vibrating guitar string has a standing wave on it that produces sound waves at a particular frequency, heard as a particular musical note. A guitar player, by shortening the string with one finger, can make the frequency of the wave increase, which makes the musical note higher. But doing so need not increase the energy of the string’s vibration! There is no relation between energy and frequency for an ordinary wave, because the number of wavicles that makes up that wave can change. The frequency might increase, but if the number of wavicles decreases, then the energy could stay the same, or even decrease.

It’s only when the vibrating string’s standing wave consists of a **single** wavicle (or an **unchangeable number** of wavicles) that energy must be added to increase frequency.

For this same reason, a large wave in a well need not pop out of the well as its walls contract, because shrinking the well’s size, which may increase the wave’s frequency, need not increase its energy.

## 24 Responses

Dr.Strassler:

I have often thought of your description above, the particle in a well, as also pertaining to degeneracy pressure. For instance, in highly compressed “matter” like a neutron star, the width (the space) that the neutrons are confined to, is so small, that the particles have so much kinetic energy, that they exert an “outward” pressure. This would not be a force resisting the inward crush of gravity, but the particles having so much energy, in their extremely small confines, would be flying out of the gravity well….so to speak. Does that have any merit?

I have often thought this would apply to the structure of atoms. If you already had two electrons, in a lower orbital, a third electron, trying to push its way in, would effectively make the spacing so small, that the electron would fly back out (for lack of a better word) because it would have so much kinetic energy. In other words, a third electron trying to push its way in, would make the spacing so small, that’s its kinetic energy would remain too high to stay in the lower orbital….giving resistance, in a sense to the crushing of matter.

This isn’t unreasonable thinking, but it turns out to be wrong.

There is a key difference (not at all obvious in anything I’ve said about wavicles here, though it is covered in the book’s last chapter) between fermionic wavicles and bosonic wavicles.

Bosonic wavicles don’t have degeneracy pressure, while fermionic wavicles do. That’s because bosonic wavicles are happy all to sit in the same state, while two fermionic wavicles cannot be in the same state; fermionic fields simply don’t allow it. That’s the Pauli Exclusion principle.

The sort of “uncertainty pressure” that I’ve described in this post, by contrast, is true for both fermions and bosons. For this reason, a hydrogen atom with a bosonic electron would be almost the same as a hydrogen atom with a fermionic electron.

But an oxygen atom would be completely different, because all the electrons in oxygen would be in the same, low-energy state if they were bosons, and wouldn’t form the structure that we see in nature, which is enforced by the Pauli Exclusion Principle.

Because fermionic wavicles cannot be in the same state, each one you add (well, each two, because there are two spin states which typically have almost equal energy) has more energy than the last one. Thus, while the energy of N photons in a box can simply be N times that of a single photon in the lowest possible state, the energy of N neutrinos in the same box is much, much larger than that. The dependence of this latter effect on the box’s size is degeneracy pressure.

Re: Compton wavelength

When equating energies – mc²=hf=h(c/L) – I get the following formula for the Compton wavelength: L=h/(mc), same as on Wikipedia https://en.wikipedia.org/wiki/Compton_wavelength.

This article and the previous one use L=hc/m in several places.

Yes, that’s a typo (you can see it because h/mc has units of length, but hc/m does not.) Thanks for pointing it out, will fix it now.

Here we have a canicule, a heatwave.

Anyway, if a wavicle is used as a vehicle, a medium to express or to achieve something, I wonder what that something is supposed to be in functional terms.

Is a wavicle comparable with a vehicle?

Maybe in pilot wave versions of quantum field theory?

Maybe as carrier of energy?

Nah — anything can carry energy…

While I used to take the concept of a “particle” as seriously as anyone, my reaction now about a these days is that the word itself is a classical-first, energy-indifferent math abstraction that has no meaning in the physical universe. For example, any attempt to shrink an atomic orbital electron down to the scale of, say, a proton, would require use of confinement gamma rays with wavelengths comparable to protons. You would utterly obliterating your experiment before it begins.

That said, alas, Dr. Strassler, I still don’t get how you distinguish a “wavicle” from an otherwise fully standard probability wave. All I see in the equations are waves with wave fronts confined by known emission times and the speed of light. Those two constraints partially localize the waves in space, but in no way make them particle-like, which is how I think most folks read “wavicle.” They are merely expanding probability waves, very much like what you get if you toss a rock (known initiation time) into a pond (known propagation speed).

If that kind of initiation-plus-velocity-limits is all you mean by “wavicle,” that’s fine. However, most folks just call those probability waves. If you mean waves confined by potentials such as protons (orbitals) or electron-reflecting walls, again, fine. However, in that case an atomic orbital _is_ a wavicle. All such wavicles expand if released, not as particle-like entities, but as ordinary fast-expanding wave fronts. It’s just that the wave fronts of light particles expand faster.

Here’s a more specific question: If a photon is released on the other side of the universe and encounters nothing more than gravitational potentials to guide its path to an earth telescope, does it travel as “wavicle” with finite dimensions, or as a probability wave that occupies a large chunk of the known universe before arriving at earth? If it’s the latter — which is what I see in the equations — why bring up the wavicle concept at all?

First things first, Terry.

Do you agree that a system of six electrons consists of six (6!) wavicles moving in

the three-dimensional space that you and I live in, while there is a single (only 1!) probability wave describing this system, and that wave moves about in a high-dimensional space that is calledthe configuration space of the system, with a minimum of 18 dimensions and in a sense many more?If we don’t agree on that, then there’s no point in trying to answer your other questions… because I’m going to tell you that your question about a photon doesn’t make any sense. A photon is not a probability wave.

https://profmattstrassler.com/2024/04/25/why-a-wave-function-cant-hurt-you

Dr. Strassler,

Wavicle is not a standard physics term, and to the best of my knowledge, it doesn’t have any a standard mathematical definition, so let’s toss that one right back: What is a wavicle in QED?

Also, are you seriously suggesting that a photon coming out of a pinhole has no wavelike behaviors in experimental physics? I didn’t ask how you modeled in configuration space — that’s the math. You are an experimentalist. Is a single photon coming out of a dim pinhole best modeled as a a probability wave, or not? That’s an experimentally verifiable question.

Cheers,

Terry

You’re being absurd, Terry. A wavicle in free QED is that object created by a field raising operator; there’s one such operator that creates photons, one that creates electrons, and one that creates positrons. In interacting QED, it is much more complex, but that’s because electrons are much more complex due to interacttions; in that case, a better definition of the electron and positron is that they are any member of a hyperbola of energy and charge eigenstates that has no additional free parameters.

Of course I am not suggesting that a photon has no wavelike behaviors — I’m calling it a “wavicle”, for goodness sakes. Why do you think “wav-” appears in the name?

But it is NOT a probability wave. A probability wave is something entirely different. AS AN EXPERIMENTALIST. A single photon coming out of a dim pinhole is not a probability wave. This is most obvious if two photons come out. Then there are two wavicles, but only one probability wave describing them.

This is most definitely experimentally viable, and my colleagues check it every day in their experiments.

The universe has huge numbers of photons and electrons and all sorts of other objects. It has exactly one (ONE) (uno) (1) (I) probability wave.

I can think of two examples of an energy well for an electron: (a) ‘orbiting’ the nucleus of a Hydrogen atom; (b) In a mutually assured destructive embrace with a positron.

Why doesn’t the Heisenberg uncertainty principle ensure that in (b), the electron and positron orbit one another with a minimum energy in a manner similar to (a)?

As far as the energetics is concerned, the orbits are exactly the same; the wavicle physics (captured in the Heiserberg uncertainty principle) DOES ensure that “the electron and positron orbit one another with a minimum energy in a manner similar to (a)”. The positronium spectrum is *exactly* the same as the hydrogen spectrum — except for an overall factor of 2 for all energies, coming from the fact that the positron has the same mass as the electron. while the proton’s mass is much larger (known as the “Reduced Mass” effect.)

The positronium “atom” has a overall frequency of about 10^21 cycles per second, and even the internal frequency (which, in classical physics, would correspond to the orbital time) is 10^16 cycles per second. The lifetime of positronium is 10^-10 seconds. That means that the electron and positron “orbit” each other

a million timesbefore they annihilate.A million orbits. For the Earth and Moon system, that’s almost 100,000 years. The destruction inherent in the “embrace” is a very, very slow one; positronium lasts for eons, when understood in terms of its natural timescales.

In other words, annihilation is a tiny afterthought… an effect that a mini-person living a hundred orbits on a positron planet as it co-orbits the electron would never even know was coming. And except for that tiny though destructive effect, the physics of the positronium “atom” is the same as that of hydrogen, except for the reduced mass.

There also exist electron-antimuon atoms. They last for a microsecond.

It seems like the equation below Fig 5 should be changed so that you are not subtracting the gravitational energy, mgD, from the wavicle’s intrinsic energy, mc^2, but subtracting it from the ground level gravitational energy, mg, would that be correct?

E=mc^2 + mg – mgD

Maybe you trying to keep things simple.

That’s not entirely wrong but there are mistakes too.

First, “m g” is a force, not an energy, and cannot appear in the equation that you have written.

If you want to follow your logic, then what should instead appear is mc^2 + m g h_1 – m g h_2 where h_1 is the height of the floor and h_2 is the height of the bottom of the well. But the difference in those two heights is simply the depth: (h_1 – h_2) = D. So even from your logic, when you do it correctly you will get my answer.

One other way to see that this is true is to set D = 0; then there is no well, and the energy has to be mc^2.

There is an important point about energy, though, an issue which often comes up already in first-year physics. We are always free to reset the zero of energy when we are not considering a local process and not worrying about the global universe, because resetting the zero is a shift by a constant, and that has no effect on how things behave locally and in the absence of strong gravity. Really the energy of the object should be mc^2 – GMm/R, where G is Newton’s constant, M is Earth’s mass, and R is the radius of the Earth. In the well, it is mc^2 – GMm/(R+D) However, R is vastly greater than D, so if we included this, it would simply shift all of the discussion by GMm/R everywhere, which is a constant, and will only shift my discussion and my energy plot down by some constant factor. It therefore has no effect on the arguments or on the conclusion.

Fig 6 and the last of the three in Fig 2 are not exactly right. The other parts of 2 are close enough visually. Fig 6 will have parts of the wavefunction outside the well (as the corresponding part of 2 does).

The last fig 2 part will have a sharper Gaussian bump on it, over the deeper well.

Yes, correct. However, I viewed these as details that were not essential to the point I was wanting to convey. (1) the leakage of the wave functions across and through the walls may be extremely tiny, depending on the height and thickness and type of the walls, which I did not want to specify, and so would be a distraction. (2) the effect of the deeper well on the shape of the wavicle that extends beyond the well is a complicated detail that again depends on D, W and L, which I did not want to get into. It doesn’t change the energetics very much, especially if L is very large — and it’s the dependence of the energy on W that is really the main point. You could correctly complain that the energy of the wavicle in the full space depends slightly on W — it isn’t exactly mc^2 — but again, that’s a very minor effect if L is always much greater than W. (Imagine the space is a meter long and the well is a millimeter across — this is not going to be important in the escape of the wavicle.)

Is Figure 6 supposed to be interactive? It looks as if I could grab one of those handles and change it, but it seems to be just an image.

Thanks, just an errant screen-shot; fixed.

What would happen if a wavicle fell into a well followed by a particle on top of the wavicle?

Would the wavicle rise as the walls closed in and the particle remaining at the bottom or would the particle attach itself to the wavicle and rise together?

Unfortunately your question has no answer, as you haven’t been specific enough. The answer depends on how the two objects interact.

For example, I can’t answer the question as to what happens if one wavicle falls on top of another wavicle in a well without much more specificity about the two wavicles’ masses and interactions. The behavior of two electrons would be very different from the behavior of an electron and a muon, which in turn would be very different from the behavior of an electron and an anti-muon. As I emphasized in this post, the key to answerng the question is to understand how the energy of the system depends on the distance between the walls of the well — and once there are two objects in the well, that energy depends on exactly how those objects interact and what their two masses are.

Do you want to leave it at that, or do you want to ask a more specific question? Try to keep it simple, since if it’s complicated and I have to do a real calculation, I may not have time for it.

I think that your question might be approximated with a hydrogen atom. On the scale of the electron’s orbital, the proton is very particle-like and the entire system is bound. (If wavicle and particle don’t interact then by definition they’ll do their own seperate thing. (Though the atom seen as a single unified system means…))

So what happens when you squeeze an atom? If the energy needed to escape the well for both particle and wavicle is less than the binding energy between the two then the wavicle will ‘lift’ the particle, the entire system will be ejected from the well.

Alternatively, if the binding energy is less than that needed for ejection of the bound system, then the wavicle will be separated from the particle and escape the well, leaving the particle behind. Attempting to ‘pin’ a wavicle ‘beneath’ a particle is just setting an arbitarily high binding energy for the two, in which case both must be ejected or not leave at all.

I hope that gets at the spirit of your question without having to dig into too much actual physics. It’s a very broad-strokes answer.