Why aren't particles constantly “measured” by the whole universe?

Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?

Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?

• Very closely related question here. It's basically the same idea, but specialized to a particular apparatus. – knzhou Feb 14 at 21:02
• Measurement doesn't do anything. See related questions physics.stackexchange.com/q/453410 and physics.stackexchange.com/q/459754 – R.. Feb 15 at 0:53
• "in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?" Yeah that's pretty much the entire field of quantum computing. – DanielSank Feb 15 at 8:40
• Particles are constantly measured, that's why for the longest time we thought we lived in a classical universe. – Aron Oct 9 at 5:22

Seems like the whole universe is receiving information about the electron's position.

Yes, the influence that an electron exerts on the rest of the universe does depend on the location of the electron, but that's not enough to constitute a measurement of the electron's location. We need to consider the degree to which the electron's influence on the rest of the universe depends on its location.

Consider something analogous to but simpler than a double slit experiment: consider an electron in deep space, in a superposition of two different locations $$A$$ and $$B$$. Even in deep space, the electron is not alone, because space is filled with cosmic microwave background (CMB) radiation. CMB radiation has a typical wavelength of about $$1$$ millimeter. When CMB radiation is scattered by an electron, the resulting state of the radiation depends on the electron's location, but the key question is how much it depends on the electron's location. If the locations $$A$$ and $$B$$ differ from each other by $$\gg 1$$ millimeter, then the CMB radiation will measure the electron's location very effectively, because an electron in location $$A$$ will have a very different effect on the CMB radiation than an electron in location $$B$$ would have. But if locations $$A$$ and $$B$$ differ from each other by $$\ll 1$$ millimeter, then an electron in location $$A$$ will not have a very different effect on the CMB radiation than an electron in location $$B$$ would have. Sure, the electron has a significant effect on the CMB radiation regardless of its location, but that key is whether the effect differs significantly when the location is $$A$$ versus $$B$$. The CMB radiation measures the electron's location, but it does so with limited resolution. Widely-spaced locations will be measured very effectively, but closely-separated locations will not.

For this to really make sense, words are not enough. We need to consider the math. So here's a version that includes a smidgen of math.

Let $$|a\rangle$$ denote the state of the universe (including the electron) that would result if the electron's location were $$A$$, and let $$|b\rangle$$ denote the state of the universe that would result if the electron's location were $$B$$. If the electron started in some superposition of locations $$A$$ and $$B$$, then the resulting state of the universe will be something like $$|a\rangle+|b\rangle$$. Whether or not the electon's location is effectively measured, these two terms will be essentially orthogonal to each other, $$\langle a|b\rangle\approx 0$$, simply because they differ significantly in the location of the electron itself. So the fact that the final state is $$|a\rangle+|b\rangle$$ with $$\langle a|b\rangle\approx 0$$ doesn't tell us anything about whether or not the electron's location was actually measured. For that, we need a principle like this:

• The electron's location has been effectively measured if and only if the states $$|a\rangle$$ and $$|b\rangle$$ are such that $$\langle a|\hat O|b\rangle\approx 0$$ for all feasibly-measurable future observables $$\hat O$$. (Quantifying "$$\approx 0$$" requires some care, but I won't go into those details here.)

For an operator $$\hat O$$ to be "feasibly measurable", it must be sufficiently simple, which loosely means that it does not require determining too many details over too large a region of space. This is a fuzzy definition, of course, as is the definition of measurement itself, but this fuzziness doesn't cause any problems in practice. (The fact that it doesn't cause any problems in practice is frustrating, because this makes the measurement process itself very difficult to study experimentally!)

In the example described above, the suggested condition is satisfied if locations $$A$$ and $$B$$ differ by $$\gg 1$$ millimeter, because after enough CMB radiation has been scattered by the electron, the states $$|a\rangle$$ and $$|b\rangle$$ differ significantly from each other everywhere, and no operator $$\hat O$$ that is simple enough to represent a feasibly-measurable observable can possibly un-do the orthogonality of the states $$|a\rangle$$ and $$|b\rangle$$. Loosely speaking, the state $$|a\rangle$$ and $$|b\rangle$$ aren't just orthogonal; they're prolifically orthogonal, in a way that can't be un-done by any simple operator. In contrast, if locations $$A$$ and $$B$$ differ by $$\ll 1$$ millimeter, then we can choose an operator $$\hat O$$ that acts just on the electron (and is therefore relatively simple) to obtain $$\hat O|a\rangle\approx |b\rangle$$, thus violating the condition $$\langle a|\hat O|b\rangle\approx 0$$. So in this case, the electron's location has not been effectively measured at all. The states $$|a\rangle$$ and $$|b\rangle$$ are orthogonal simply because they differ in the location of the electron itself, but they are not prolifically orthogonal because the effect on the rest of the universe doesn't depend significantly on whether the electron's location was $$A$$ versus $$B$$.

What I'm doing here is describing "decoherence" in a different way than it is usually described. The way I'm describing it here doesn't rely on any factorization of the Hilbert space into the "system of interest" and "everything else." The way I'm describing it here (after quantifying some of my loose statements more carefully) can be applied more generally. It doesn't solve the infamous measurement problem (which has to do with the impossibility of deriving Born's rule within quantum theory), but it does allow us to determine how effectively a given observable has been measured.

Some quantitative calculations — including quantitative results for the specific example I used here — are described in Tegmark's paper "Apparent wave function collapse caused by scattering" (https://arxiv.org/abs/gr-qc/9310032), which is briefly reviewed in https://physics.stackexchange.com/a/442464. Those calculations use the more traditional description of decoherence, but the results are equally applicable to the way I described things here.

• Am I understanding correctly that the coherence of electron's superposition in my example is shared with the universe and is lost over time, but not fast enough to be completely lost? Also I do not seem to understand the difference between decoherence and wavefunction collapse. Can it be "partially" collapsed? In the work by Tegmark you mentioned there is this equation that describes speed of decoherence in case of radiation scattering, which is parsmetrized by particle flux. What can be particle flux analogy in my example? Aren't intersction rates, exerted by electron, infinite? – FunkyLoiso Feb 15 at 20:02
• @FunkyLoiso Decoherence is a physical process in which the electron's location becomes entangled with the rest of the system in a way that is practically impossible to undo. Decoherence can be slight or extreme, and it can be slow or fast. In a typical deliberate measurement, it's extreme and very fast. In contrast, "collapse" is something we do on paper when we throw away all but one of the terms in that special entangled state that decoherence produces. We might as well do this because we only experience one of those terms. This is part of the "measurement problem." It's weird. – Chiral Anomaly Feb 15 at 21:31
• @FunkyLoiso I'm not sure what you mean by the coherence of the electron's superposition being shared with the universe, so I can't address that part of your comment. In the 2-slit example, there are several things analogous to the decoherence-by-particle-flux in Tegmark's analysis. Particle flux could be one of them. Another is the interaction between the electron's charge and the atoms in the double-slit material. The fact that the material is electrically neutral overall can help subdue this one, but it can still be important. The electron's gravitational effect is present but negligible. – Chiral Anomaly Feb 15 at 21:35

There are time-scales related to interactions, or, equivalently, interaction rates. These interaction rates are often calculated in lowest order based on Fermi’s Golden Rule. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.

In interference experiments, we therefore define a coherence time for the interfering particles.

In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).

• Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself? – FunkyLoiso Feb 14 at 21:50
• Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing. – Ben Crowell Feb 14 at 22:23
• But isn‘t a weak interaction with gravity still a measurement? How weak must an interaction be so that it doesn‘t lead to a collapse of the wave function? – asmaier Feb 15 at 9:06
• Interaction rates are a red herring here. The rate at which an accelerating charge emits photons or gravitons is infinity. These infinitely many very low-energy particles do not cause a measurement as explained in Dan Yand's answer. – knzhou Feb 15 at 12:52
• @asmaier: But isn‘t a weak interaction with gravity still a measurement? How weak must an interaction be so that it doesn‘t lead to a collapse of the wave function? Take a look at Joos and Zeh, Z Phys B 59 (1985) 223. They do the case of interactions with photons. The rate of decoherence $\Lambda$ is proportional to the cross-section and to the number density of the photons. So in the case of a very weak interaction, you don't get a Copenhagen-style sudden collapse, you get a slow decoherence. "Measurement" and "collapse" are approximate notions specific to Copenhagen. – Ben Crowell Feb 15 at 16:51

The other answers are great, but not very useful for experimentally minded people. I will try to address this question from the practical point of view.

A nice heuristic way of thinking about measurement is through the energy level shifts that actually cause decoherence (and ultimately measurement). This is the way quantum computing folk often thinks about qubits.

An energy perturbation will cause roughly a $$e^{i\, \delta E \, t}$$ phase shift to the quantum state that normally has a phase $$e^{iE_0t}$$. If these are big phase shifts and occur randomly, then the quantum state will evolve in a way that is unrelated to the Hamiltonian you think it obeys. However, if $$\delta E$$ is small compared to the energy spacing (like gravity compared to electromagnetism) and the timescale you are considering, then the phase slips are negligible and no "measurement" has occurred in the practical sense.

To give a concrete example, think about the reflection of a photon with energy $$\hbar \omega$$ from a mirror. A naive thought is that the reflection of the photon counts as a "measurement" of the photon because the photon transfers momentum to the mirror, and thus will "collapse" the wave function. Let's see if that's true.

The photon's momentum changes from $$+\hbar \mathbf{k}$$ to $$-\hbar \mathbf{k}$$, giving $$2\hbar \mathbf{k}$$ momentum to the mirror. This change in momentum doesn't come for free, naturally some energy has to be transferred to the kinetic energy of the mirror. Let's assume the light is visible light with a wavelength of 500nm, and the mirror weighs 100 grams. Then the energy transferred to the mirror is:

$$E_\textrm{mirror}=\frac{p^2}{2m}=\frac{4\hbar^2 k^2}{2m} \approx 2\cdot10^{-34} \textrm{eV}=9\cdot10^{-34} \hbar \omega_{\textrm{photon}} \implies 10^{13} \,\textrm{years}$$

This means that the "measurement" process of the mirror will cause a single phase slip by $$2 \pi$$ on a timescale of roughly 10 billion years. You can imagine that under normal circumstances, this is not a "measurement", so the photon maintains its quantum state.

• I think a natural follow-up question is why complex structures like molecules aren’t constantly “measuring” their own constituents - with interference effects being demonstrated with larger and larger structures, it seems strange that the entire composite can be placed in superposition despite the rigid geometric/bond relationships within. – JPattarini Feb 16 at 14:53
• @JPattarini, in fact larger molecules etc. do measure themselves so to speak. They lose the phase coherence I described, and ultimately stop being in pure quantum states. That's why to get quantum properties of larger molecules to manifest they must be cooled down, otherwise they look classical. The article "More is different" by P. Anderson discusses this exact example. – KF Gauss Feb 16 at 17:29

Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.

Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.

So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.

As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.