Of Particular Significance

# Zero-Point Motion

First, the basic idea; then, for those who want more details, I’ll show you the simple math.

## Overview of Zero-Point Motion

The principle of uncertainty, discovered by Werner Heisenberg in 1927, states that an object’s position and speed (more precisely, its momentum, to which its speed is related) cannot both be precisely known. This means that it is impossible for any object to be at a known position and absolutely still, because then we would know more than quantum physics permits us.

Instead, we could know its position perfectly, but nothing about its speed, or vice versa. Or we can possess some partial knowledge of both. This last is true of a ball on a spring or at the end of a pendulum; there’s some uncertainty in the ball’s position and some uncertainty in its speed.

It’s almost as though the ball were moving around somewhat randomly. However, although this image of random motion has some merits, it’s not really right. To move randomly with a definite trajectory would imply an object’s position and speed could be known by someone who knew that trajectory better than we do. The point of quantum physics is that the trajectory is not even meaningful.

Because of this, an object such as a ball on a spring or a pendulum never can truly stop in one place. This is not to say that its location is completely undetermined. The spring or the pendulum’s string to which the ball is connected make it exceedingly unlikely that it will be found twenty miles, or even twenty meters, away. Thus its position is partially known. Because of this, its speed cannot be perfectly known. It can at best be partially known, and thus it cannot be exactly zero.

The ball, therefore, must in some sense always be in motion. This is its zero-point motion.

Since it is never motionless, it always has motion energy, even when it possesses the lowest possible energy that it can have. This (along with energy stored in the spring or pendulum as the ball moves around) is called its zero-point energy. If an object can vibrate simply and would do so with frequency $f$, then its zero-point energy — the energy it has even when it is not vibrating — written in terms of Planck’s constant $h$ (the cosmic certainty limit), is

• $E_{{\rm zero-point}} = \frac{1}{2} f h$

(Notice this differs from the Planck-Einstein quantum formula, which states that the energy of a quantum of vibration is $f h$. The difference of 1/2 is real. Zero-point energy, the energy of “the ground state”, the lowest possible energy that a system can have, must not be confused with the additional $f h$ energy of a quantum of vibration, which puts the system in the “first excited state”; the total energy of the first excited state, then, is the sum of the zero-point energy plus the energy of that one quantum of vibration, and thus its total energy is $\frac{3}{2}f h$. There’s a bit more on this point in this page’s last section.)

## Details of Zero-Point Motion and its Energy

To understand, using simple math, how quantum physics creates zero-point “motion” and its associated energy, we can start making a few observations about a simple vibrating ball on a spring in the absence of quantum physics. (If you would like a more advanced treatment, try this page and its sequel.)

### Vibration Without Quantum Physics

Imagine placing the ball at its “equilibrium position” $x=0$, where the spring is neither compressed nor stretched. If we ignore quantum physics, the ball will remain at $x=0$ forever, or until someone disturbs it.

If you disturb the ball, it will begin to bounce back and forth, its position $x(t)$ vibrating over time $t$ around the equilibrium position. The vibrations occur at the resonance frequency $f$:

• $x(t) = x_{{\rm max}} \sin( 2 \pi f t)$

where $x_{{\rm max}}$ is the ball’s furthest distance from equilibrium.

As it bounces back and forth, the ball’s position is sometimes to the right of the equilibrium point ($x>0$) and sometimes to the left ($x<0$), and so its average position is zero:

• ${\rm average}(x) = 0$

But the average of the square of the ball’s position is always positive or zero, and so its average is not zero. Instead,

• ${\rm average}(x^2) = \frac12 x_{{\rm max}}^2$
Here’s a proof:

The ball traverses one back-and-forth cycle in a time $T=1/f$. The average of the position is

• ${\rm average}(x^2) =\frac{1}{T} \int_{t=0}^{t=1/ f} dt\ x_{{\rm max}}^2\sin^2( 2 \pi f t) = f (x_{{\rm max}}^2/2 f) = \frac12 x_{{\rm max}}^2$

Similarly, the average velocity $v$ of the motion is zero — on average, the ball isn’t going anywhere — but its velocity-squared is always positive or zero, so the average velocity-squared is not zero. It turns out to be:

• ${\rm average}(v^2) = \frac12 (2\pi f)^2 x_{{\rm max}}^2$

[Side note: Except for the numerical factors out front, you could guess this from dimensional analysis.]

Because the motion-energy of the ball is $m v^2/2$, its average motion-energy is

• ${\rm average\ motion \ energy} = \frac12\times m \times \frac12 (2\pi f)^2 x_{{\rm max}}^2 = \frac14 m (2\pi f x_{{\rm max}})^2$

The average energy stored in the spring turns out to be exactly the same as average motion-energy of the ball, with the result that the total energy of the vibration is twice the average motion energy:

• ${\rm total \ energy} = \frac12 m (2\pi f x_{{\rm max}})^2$

### With Quantum Physics, but Without Vibration

In quantum physics, if we place the ball at its equilibrium position; it won’t stay there, even though it is not vibrating, thanks to quantum uncertainty. Instead our uncertainty in its position $\delta x$ and the uncertainty in its speed $\delta v$ satisfy Heisenberg’s uncertainty relation

• $m\ \delta x \ \delta v \geq \frac{h}{4\pi}$

where $h$ is Planck’s constant (i.e., the cosmic certainty limit.) The uncertainty of the position is defined to be the square root of the average value of the square of its position, and similarly for the square root of the velocity:

• $\delta x = \sqrt{{\rm average}(x^2)} = \sqrt{1/2} x_{{\rm max}}$
• $\delta v = \sqrt{{\rm average}(v^2)} = \sqrt{1/2}(2 \pi f) x_{{\rm max}}$

where I have used our earlier formulas for vibration in deriving this formula. (This last step is illegal, because the earlier formulas were for true, non-quantum, back-and-forth vibration, whereas here we are trying to capture the effects of random zero-point motion. A logically correct derivation does in the end require the use of quantum physics techniques, beyond our current scope.)

For the lowest energy arrangement of a simple vibrating quantum system near equilibrium (known as the “ground state”), the total uncertainty is minimized, so the uncertainty relation becomes an equality:

• $m\ \delta x \ \delta v = \frac{1}{2} m (2 \pi f) x_{{\rm max}}^2 = \frac{h}{4\pi}$

From here, we find

• $x_{{\rm max}}^2 = \frac{h}{4 \pi^2 m f}$

So even when the ball is at equilibrium, the fact that it is a quantum object gives it a non-zero position-squared, as though it were oscillating or moving around in a random way, over a distance that increases with h and decreases with the mass of the object. This is the zero-point motion. It is exceedingly minuscule even for a ball whose mass is only a gram (less than a sugar cube’s mass.) But for atoms, and for subatomic particles, it’s significant.

### The Zero-Point Energy

There is energy associated with this uncertainty. How much? As noted earlier, half of it comes from its motion energy $m v^2/2$, which from our earlier equation for motion energy is

• $\frac14 m (2\pi f x_{{\rm max}})^2 = \frac14 m (2\pi f)^2 (\frac{h}{4 \pi^2 m f}) = \frac{1}{4} f h$

And since the average stored energy in a spring is the same as the average kinetic energy, the total zero-point energy of the object is

• $E_{{\rm zero-point}} = \frac{1}{2} f h$

This looks as though we have discovered the quantum formula, but with a mistake of 1/2. However, this is the energy associated with having no quantum of vibration at all. The quantum formula will tell us that adding each quantum of vibration shifts the energy upward by $f h$, not by $f h/2$.

Instead, what this means is that the total energy of having a vibrating ball with $n$ quanta of vibration is

• $E_{n} = \left(n+\frac{1}{2}\right) f h$

one $f h$ for each quantum, plus an extra $f h/2$ from the zero-point motion that has nothing to do with vibration itself. That extra 1/2 is the signal that nothing in quantum physics can ever truly stop. There is no such thing as a completely stationary object, at least not in the conventional sense that we imagine — no dot-like object ever sits still.

(But then again, an electron is not a dot.)

That’s all we will need to know about a ball on a spring. (If you are seeking a more advanced treatment of what I’ve just discussed, try this page and its sequel.) In a later note we’ll see that the same ideas affect fields, to even greater effect.

Click here for the next article in this series.

### 8 Responses

1. Daniel Du Prie says:

So what *would* actually happen if “h” were, say, 100x larger than it is, or 1000x or even bigger than that? Would everything dissolve? Or blow up? Would we have Schrodinger cats that are *literally* both alive & dead simultaneously? Would it be impossible to form stable matter? I’d love to find out more…!

1. This question has to be posed more carefully; what are you holding fixed as you increase h? This is actually quite tricky, because “h” appears in the definition of various quantities.

I think the best answer is that greater uncertainty would also lead to less stable atoms and molecules, so that they would mostly be lost at room temperature. But of course, you might want to change room temperature to compensate for this… after all, the earth’s history will be different if h is different, and so will that of the whole universe. And this is just the first set of issues that complicate giving a complete answer.

Physicists often ask questions such as yours, but find them really tricky to answer because of all these niggling questions about what to change and what not to change, and why.

But Schrodinger cats would be no more alive or dead than they are. This is not an issue of hbar’s size but of a basic conceptual issue that doesn’t really depend on what one uses.

2. Peter B says:

“So even when the ball is at equilibrium, the fact that it is a quantum object gives it a non-zero position-squared, as though it were oscillating or moving around in a random way, over a distance that increases with h and decreases with the mass of the object. This is the zero-point motion.”

How can increase with h if h is a constant?

1. Good question: I probably didn’t write this as clearly as I could have. I don’t literally mean that one could change h in our universe. The point is that h controls the size of the effect; if you compared our universe to one in which everything was the same except that h was larger, then the quantum effects would be larger.

Physicists often imagine changing constants — i.e., imagining universes in which the constants were different — as a powerful tool for getting a conceptual handle on how our universe works. You can do the same with the speed of light — hold certain quantities fixed but imagine the speed of light were ten times faster — even though there is no way to change that quantity in our own universe. Or with the electron’s mass — what if it were a hundred times smaller?

[BTW, I discuss (and use) this classic technique in my book; it’s central to how physicists grapple with the universe.]

3. Maccasa Wuuuud says:

Well, your initial proposition involves “objects” but that includes cricket balls, ice-breakers and Egyptian pyramids.

I think you meant sub-atomic particles.

The difference is important.

Heisenberg et Al were postulating about the behavior of electrons, mesons, Quarks, Strangeness and their interactions. Let’s call that Charm

None of which have been proven, it’s all still a theory.
Ask Foucault about balls on the end of strings.

The math obfuscates the reality, and vice versa, in measurable terms none of this argument about the Uncertainty principle relates to what Newtonian physics has successfully defined.

Out on a limb, I’d say it’s guess work, at best. I defy you to prove me wrong.

I’m just putting on my troll-proof armour

1. Torbjörn Larsson says:

There is a deep irony that you write this on devices that use transistors, which can only be understood – in the sense of being improved – by using quantum physics.

4. Jason Stanidge says:

From what I’ve read of the history behind the discovery of Heisenberg’s uncertainty principle, Heisenberg saw it as a consequence of the measuring apparatus unavoidably disturbing what was being measured whereas Bohr saw it as a fundamental physical property of our world, analogous to the relativity of simultaneity in special relativity. But now it seems to me that some authors claim the uncertainty principle can be ‘derived’ from the axioms of quantum mechanics. So it’s reassuring for me when you state above: “where h is Planck’s constant (i.e., the cosmic certainty limit.)” giving it the same level of importance as c, the cosmic speed limit which I think ends up hidden by the way QM is presented nowadays.

What strikes me about Heisenberg’s uncertainty principle the way you’ve presented it on your blog, is how it relates the time-like component of the position and energy-momentum four vectors to one another as one relation, likewise for the space-like components as another mathematical relationship. I also know that:

1. Energy and momentum are defined via Noether’s theorem: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/mass-and-energy/

2. Space-time is affected by energy and momentum: https://profmattstrassler.com/2023/10/26/the-impossible-commentary-is-gravity-a-force-is-it-an-illusion/

3. Keeping track of the conservation of energy and momentum where gravity is significant over long distances can be difficult: https://profmattstrassler.com/2023/10/26/the-impossible-commentary-is-gravity-a-force-is-it-an-illusion/#comment-449565

Once energy-momentum within a space-time region becomes large enough at a quantum level to affect space-time; this will lead to ambiguities in how energy, momentum, space-interval and time-intervals can be defined. I believe the latter has something to do with ‘space-time foliation’ which, until recently, was completely new to me. So my thoughts above finally leads me to ask:

Is the Heisenberg uncertainty principle related to the ambiguity in being able to define energy, momentum, space and time over a region of space-time; where the energy-momentum there significantly affects the flatness of space-time?

1. The answer to your final question is no, although the previous paragraph is correct, more or less. You are right that things become complicated when spacetime itself becomes quantum (space-time foam, as it is sometimes called) because that’s where quantum gravity kicks in and we don’t entirely know what we are doing.

However, the Heisenberg uncertainty principle is much simpler than that. The space and the time may be perfectly well-defined. The uncertainty principle refers not to the space and time themselves but to the locations of objects within that space and time, and to the energy and momentum of those objects, which are typically far too small to affect the space and time.

Another way to say this is that the uncertainty principle is perfectly well-defined in a world without gravity — that is, in a world with space and time but no spacetime curvature.

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