# Are bubble chamber tracks inconsistent with quantum mechanics?

I am reading the book How Is Quantum Field Theory Possible? by Sunny Auyang, and he raises an interesting point in chapter 4 (p. 23):

L. E. Ballentine argued that the projection postulate leads to wrong results. Even when the quantum system somehow triggers its environment to produce a measurable eigenvalue, its state does not collapse. Consider the track left by a charged particle in a cloud chamber. The incoming particle is usually represented by a momentum amplitude. It encounters the first cloud-chamber atom and ionizes it, leaving the tiny droplet that we observe. This process is sometimes construed as a position measurement that collapses the particle's amplitude into a position eigenstate. The interpretation is untenable. A position eigenstate is a spherical wave that spreads out in all directions. Hence it would be impossible for the particle to ionize subsequent atoms to form a track that indicates the direction of the original momentum, which is allegedly destroyed in the first ionization.

In other words, the projection postulate of QM is inconsistent with bubble chamber tracks. Is there an accepted resolution to this?

I can think of a few ideas:

1. The projection postulate is wrong.
2. Droplets in bubble chambers do not count as position measurements.
3. The droplets are position measurements, but only localize the position to a finite region of space, and this allows some of the "momentum" part of of the wavefunction to remain intact upon collapse.

But all of these seem to have issues and conflict with other principles of QM. Curious if there is a standard resolution, or if this necessarily gets into the contentious realm of quantum interpretations.

• This is sometimes called the Mott problem. See i.e. the Wikipedia article: en.m.wikipedia.org/wiki/Mott_problem – Bob Jacobsen May 12 '19 at 3:27
• @Bob: Sorry if this is a stupid question but in the article, it is said they solve the problem by using the configuration space and the related article mentions it is a classical space, while the QM one is the state space. Does that mean they just shift the representation independently of QM wavefunctions? – Exocytosis May 12 '19 at 6:43
• – Stéphane Rollandin May 12 '19 at 8:54
• “Measurement” is complicated in condensed matter. In some cases, like this one, the interaction is with more than one atom. So you have to think about the entire configuration of atoms. Another example is talking about phonons, the coherent motion of multiple atoms, instead of the mechanical motion of just one. It’s still a quantum approach, because you’re using the quantum formalism. It’s just about more complicated configurations. – Bob Jacobsen May 12 '19 at 20:11

I am replying to the title question:

Are bubble chamber tracks inconsistent with quantum mechanics?

I worked with bubble chamber data for years and never encountered these esoteric interpretations.

Here is a bubble chamber event and a charged pion decaying into a muon and an electron:

The main interaction happens at the vertex on the top. That has the specific wave function which the experiment is studying, i.e. measuring multiplicity, and finding energy and momentum by using the imposed magnetic field.

Each little dot is a mesurement of another wavefunction solution " atom +pion" (the magnetic field is the cherry on the pie that allows for momentum measurment, using the particle manifestation) scattering , a completely different wavefunction than the initial one. It has a probability of getting a pion with an unmeasurably smaller momentum + an electron as a dot, getting the momentum balance. And so on, with innumerable tiny scatters and innumerable new wavefunctions. The little curl up at the many track vertex is where the probability of getting an electron with a measurable momentum won, and the electron momentum could be measured.

In my opinion there is no paradox but a misunderstanding of what a wavefunction solution is: it depends on the boundary conditions and the potentials which are continuously changing with small interactions along the track. Each dot is a different wavefunction manifestation for the pion.

The answer to the title question is : there is no inconsistency .

As any form of higher level mathematical models of quantum mechanics are based on the solutions of the basic equations and the postulates governing them, my opinion is that there is something wrong with this "projection" business, either in interpretation or in definition.

Here is what I find for the projection postulate:

The postulate in quantum mechanics that observation of a physical system, by determining the value of an observable, results in the transition of the quantum state of the system to a particular eigenstate corresponding to the eigenvalue of the observed quantity.

From the discussion above, I conclude that the confusion comes by not realizing that there is a continuous series of interactions on the trail of the track, and continuously new wavefunctions/states. These interactions are of the same mathematical form as the main vertex interaction, but governed by different potentials in the scattering( different Feynman diagrams too) , at each dot.

The bubble-track phenomenon is not in conflict with the projection postulate, as long as we use the projection postulate appropriately. Applying the projection postulate directly to the particle's position observable $$\hat X$$ (the one defined by $$\hat X\psi(x)=x\psi(x)$$) is not appropriate. Real measurements have finite resolution, and applying the projection postulate directly to $$\hat X$$ amounts to assuming that the measurement has infinite resolution.

To naturally account for the finite resolution of the real measurement, we can use a model in which the molecules comprising the bubble chamber (and atmosphere, etc) are included as part of the quantum system, along with their interaction with the quantum electromagnetic field. In this model, the formation of bubbles, the reflection of light by the bubbles, the dissipation of heat, and so on, are all described as quantum phenomena at the microscopic level. Doing the calculations explicitly would be too difficult, but based on experience with less-daunting models, we know what will happen: the particle's position will become practically irreversibly entangled with the rest of the system, including with the light that reflected from the bubbles. Then, instead of applying the projection postulate to an observable $${\hat X}$$ associated directly with the particle's position, we can apply it to an observable $${\hat M}$$ associated with the reflected light, such as an observable corresponding to a two-dimensional array of photon-counters, which has a discrete set of eigenspaces.

Let $$|\psi\rangle$$ denote the state after a bubble has formed and scattered some light, but before applying the projection postulate. We can write this state as a sum of eigenstates $$|\psi_m\rangle$$ of the observable $${\hat M}$$: $$|\psi\rangle=\sum_m|\psi_m\rangle,$$ When applied to the observable $${\hat M}$$, the projection postulate says that after the formation of a bubble and the reflection of light, we might as well replace the state of the whole system (the particle, the bubbles, the light, the air) with one of the eigenstates $$|\psi_m\rangle$$. As usual, the relative frequencies of these various possible outcomes are given by Born's rule $$\frac{\langle\psi_m|\psi_m\rangle}{\langle\psi|\psi\rangle}.$$ Thanks to the entanglement that developed between the light and the particle's position in the original state $$|\psi\rangle$$, each of the eigenstates $$|\psi_m\rangle$$ is a state in which the particle's position is concentrated in a small region determined by the resolution of the bubble-chamber system, as described in Ryan Thorngren's answer. The important point is that the particle's position is concentrated only in a small region, not at a point. This finite resolution comes naturally when we extend the model to include the physical processes involved in the measurement.

To see how this finite resolution can fix the problem described in the OP, suppose that the bubble-chamber system resolves the particle's position to $$\sim 1$$ micrometer. This means that in each of the eigenstates $$|\psi_m\rangle$$, the particle's position is concentrated in $$\sim 1$$-micrometer-wide neighborhood of some point $$\mathbf{x}_0$$, with momentum concentrated in a neighborhood of $$\mathbf{p}_0$$. Let $$\Delta x$$ and $$\Delta p$$ denote the widths of these neighborhoods. We must have $$\Delta x\,\Delta p\gtrsim\hbar$$, but if $$\Delta x\sim 1$$ micrometer, then $$\Delta p$$ can still be as small as $$\Delta p\sim \frac{\hbar}{\Delta x} \sim 10^{-28}\frac{\text{ kg}\cdot\text{m}}{\text{s}}.$$ That's small enough to allow for the formation of a long bubble-track.

The key is that real measurements have finite resolution, and we can account for this naturally by applying the projection postulate to an observable that is farther "downstream" in the cascade of effects caused by the passage of a particle through the bubble chamber, such as an observable associated with the light reflected from the bubbles.

By the way, this is how so-called "weak measurements" can be treated in quantum theory using only the usual projection postulate.

• It's the fluid which does the "measurement" in this problem, not a light source. – Ryan Thorngren May 14 '19 at 11:20
• @RyanThorngren The fluid is sufficient, yes. I included the light source only because it's easier to intuitively identify an appropriate observable associated with the light source than with the fluid. Any observable sufficiently far "downstream" in the cascade of effects will suffice, and yes, the fluid is already sufficiently far downstream, because the particle's effect on the fluid is already practically irreversible, even without the light source. – Chiral Anomaly May 14 '19 at 12:11
• @RyanThorngren Clarification: In the previous comment, "any observable" means any observable that is (indirectly) sensitive to the particle's position, of course. – Chiral Anomaly May 14 '19 at 12:37

I think the measurement in the bubble chamber is more closely modelled as a weak measurement, which does not collapse the wavefunction into an eigenstate, but which "squeezes" it in position space around a particular point. You can read more about it here.

The result is that in any small time window between scattering events, the wavefunction looks something like a Gaussian wave packet, with small $$\Delta x$$ but also small $$\Delta p$$. These wave packets have linear trajectories and if you repeatedly (weakly) measure them along their trajectory (ie. all scattering is with a vanishingly small momentum transfer), then you can do so without disturbing their shape. In fact the squeezing helps to mitigate the dispersing of the packet due to uncertainty, similar to the quantum Zeno effect, resulting in what looks like a classical trajectory.