Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Suppose I have the following double split experiment set up:

  • a monochromatic electron source of low intensity, which we can model as emitting a single electron at a time with energy $T$.
  • a diffraction screen at a distance $A$ with two macroscopic slits equidistant from the source with width $w$ and distance from the center of the screen $\frac{d}{2}$.
  • a detector screen sitting parallel to the diffraction screen at a distance $B$ which consists essentially of a lattice of ammeters each connected with a square surface of side $r$.

                                 Double slit experiment

Classical quantum mechanics, and experiments, tell us that the following behaviour is experienced:

  • Only one random ammeter is disturbed each time an electron is fired
  • The probability distribution of the random ammeters being fired is as given by QM.

The interaction of the electron with the detector is not modeled, the electron is essentially modeled as a wave with interference and which ammeter is fired, essentially an effect of particle behaviour, cannot be discussed within this picture.

So, the measurement or decoherence of the electron wave function corresponds to this switch between wave-like and particle-like models of the electron - which is clearly unsatisfying, intellectually, because QM is vague about this - it merely associates mathematical operations with the procedure, without giving any particular physical or mathematical justification of it.

The questions:

Do further refinements of QM (QED or QFT or ST) give any better explanation of what justifies the measurement "recipe" of classical QM by correctly modelling the electron/detector interaction?

How, in layman terms, is the change of picture from wave to particle modeled (if it is at all)?

share|cite|improve this question
Try to look at one dot on the detector screen as at a bit of information about the whole interference pattern. And look at the interference picture as at a picture of something complicated, non trivial, non reducible to one dot. The interference picture (photo) describes the whole experimental set, not one electron. – Vladimir Kalitvianski Jan 30 '11 at 14:25
Sklivvz, To tease out more of the kind of answer I think you might be looking for, consider a relativistic version of the two-slit. Here if D is a distance in the experiment (c=light speed), then what about opening and closing shutters in times small compared to D/c, versus large compared to D/c? The answers might reveal how this is thought to work in a relativistic QM. – Roy Simpson Jan 30 '11 at 16:06
@Roy, maybe you care to elaborate in an answer? – Sklivvz Jan 30 '11 at 19:39
Sklivvz, the dots on the display are OK, they're classical scale thermodynamic events in the detectors, but the dots "traveling" from the source to the detector are prejudicial to the question of which interpretations of QM we might be discussing. While only when the state preparation device is present and powered up do we see events at the screen, nonetheless there might be no particles or nothing at all traveling from one to the other (or both ways in some interpretations). You could draw and think about the experiment without those intermediate dots. – Peter Morgan Mar 14 '11 at 13:06
If you don't have the Feynman Lectures On Physics, volumes 1-3, you might want to get them, though I'm sure there are other equally good sources I don't know about. They're very easy to follow. Even in the first volume, he does a beautiful job of explicating the wave-particle duality. – Mike Dunlavey Dec 4 '11 at 23:27

6 Answers 6

up vote 2 down vote accepted

An important contemporary approach to the two-slit situation you ask about is decoherence. Many contributors to this forum are fans of decoherence, I am not, but it is very important and worth attention. I couldn't help noticing you still had queries about it even after the explanations of these «fans».

Since one should always be able to include more of the outside world in the box of quantum analysis, always be able to push the boundary out, you are basically asking if any progress has been made by including the slits, and detectors, in the unitary Quantum picture, or if instead any progress has been made by changing the unitary picture even a little. The decoherence approach does not change the unitarity of the evolution, and neither will QFT. (Some have wondered if gravity or other non-linearities will, but we will not go into that right now.)

Brief sketch of decoherence

For simplicity, assume there are only two ammeters behind the slits. As always with unitary evolutions, the electron, after being diffracted by the slits, and when it reaches the plane of the ammeters, is in a superposition of two states belonging to the different locations on that plane of the two ammeters (normally it will be in a superposition of many states but we can simplify here, too). $c_1\psi_1 + c_2\psi_2$. (We will simply neglect the occurrences where neither ammeter fires.) The state space is effectively C$^2$. Now, the ammeters are being modelled quantum-ly, too, so the set of two ammeters has a Hamiltonian and a Hilbert Space for the state vectors (wave functions) of the system of two ammeters, $H_{amm}$. The state space for the combined system of electron--being--measured and measurement apparatus (two ammeters) is then C$^2\otimes H_{amm}$. The decoherence approach makes more or less the same key assumption Mott, London, Wigner, and many others make down to today (but which I query, elsewhere, see my other posts and links), which is that whatever dictionary or correspondence there is between the macroscopic idea of «ammeter 1 fires» or, parallellly, «ammeter 2 fired», is to be modelled by a collection of quantum states in $H_{amm}$ (this is what I usually complain about, but not here), or, since we can pass to their closed span and simplify to assume there is one state, label them $\phi_1$ and $\phi_2$. (A key point in this regard will be expanded upon in a minute...) Now since this means that $\psi_1 \otimes \phi_o$ evolves unitarily to $\psi_1 \otimes \phi_1$ (Here, $\phi_o$ is the initial state of the ammeter-lattice while it waits to discharge or fire) but that $\psi_2 \otimes \phi_o$ would evolve to $\psi_2 \otimes \phi_2$, by linearity (no progress has been made, IMHO, by tinkering with the assumption of linearity), what actually happens is that our electron and ammeter as a combined system evolve to $$c_1 \psi_1 \otimes \phi_1 + c_2 \psi_2 \otimes \phi_2,$$ which is an entangled state: neither the ammeter nor the electron can be considered as a separate system anymore.

At this point, decoherence, depending on which flavour is administered, points out something undeniable: in fact, there are many more degrees of freedom within the ammeters: our simplification (which is the same as that of Wigner, EPR, and many others) overlooks something important. It is, according to this theory, important that the ammeters are at least weakly coupled to the environment. It is not quite a closed system, so the analysis above is only approximate.

All decoherence approaches (as far as I know) use density matrices. (I would be interested in a current reference to one that works only with pure states and not density matrices.) It can be shown in theory, rigorously, that this coupling with the environment leads to a further thermodynamic-like evolution to a density matrix which is very nearly diagonal and so can be regarded as a classical (or Bayesian) probability distribution on the two states, which are each obviously separable: $$\psi_1 \otimes \phi_1$$ and $$\psi_2 \otimes \phi_2.$$

The Coleman--Hepp model and Bell's response

In my opinion, the grand-daddy of all decoherence theories is the so-called Coleman--Hepp model. I learned it in Bell's famous paper, a freely available copy of it is here: in which he tranlsates it from the language of QFT in C*algebras to the Schroedinger picture and Heisenberg picture. Not on Los Alamos archive. Of course Coleman and Hepp are two most distinguished physicists. Briefly, the criticism, which I agree with, is that a density matrix is not a state, so this is really no better from a logical or foundational point of view than the open system approach and suffers from the same question-begging. (Which is more than thiry years old now, so it's not progress.)

The Physics of these models

These models all use a kind of thermodynamics, and this is surely right, as Peter Morgan (hi peter) said in a previous post, the possible detections are thermodynamic events...and the way to make that precise is to take some sort of limit as the number of degrees of freedom goes to infinity. But none of these models reflects in the model that measurement is a kind of amplification, none of the thermodynamic limits involved uses negative temperature, and this is surely wrong. Feynman's opinion was that this was decisive, see Feynman and Hibbs, Quantum Mechanics and Path Integrals, New York, 1965, p. 22. So these models do not incorporate Feynamn's insight. The models of Balian et al. cited above do incorporate this, and study phase transitions induced by tiny disturbances from an unstable equilibrium which is indeed the physics of bubble chambers and photographic emulsions. Bohr thought that the apparatus had to be classical, and decoherence models do not use a limiting procedure that introduces a classical approximation. So they do not incorporate Bohr's insight.

There is experimental evidence that interaction with the environment induces decoherence, but not yet in a way relevant to measurement. There is also experimental evidence , the so-called spin echoes, that mesoscopic systems can recover their coherence after losing it, which is what Wigner always assumed.

share|cite|improve this answer
Wow - this is exactly the kind of answer I wanted. Thanks. – Sklivvz Aug 21 '13 at 20:06
This came up recently on… about the Balian et al. approach – joseph f. johnson Aug 29 '13 at 13:54

Dear sklivvz, the very same question was asked a few days ago. Quantum field theory, string theory, or any other viable theory that may supersede quantum mechanics directly reduces to non-relativistic quantum mechanics in the non-relativistic limit and changes nothing about the basic postulates of quantum mechanics.

It means that you may find the corresponding low-energy, low-speed (multi-)particle states in the Hilbert space of QFT or string theory or anything else - essentially creation operators acting on the relevant vacuum - and you may prove that the QFT or stringy Hamiltonian acts on these states exactly as the non-relativistic Hamiltonian does, plus corrections that go like positive powers of $1/c$ and may be neglected in the non-relativistic limit.

So nothing changes about quantum mechanics and the double-slit experiment in QFT or string theory and it's likely that those things will never change. Cheers, LM

share|cite|improve this answer
Hi Lubos, this does not address the question - it's a given that more advanced theories must reduce to QM, as QM models successful experiments. This is just like saying that any gravitational theory must reduce to Newton. However, for example, QED lets us model the detectors so, in theory, one should be able model the measurement in a way that reduces to classical QM. What is this model? How does it reduce to classical QM? – Sklivvz Jan 30 '11 at 12:15
Dear Sklivvz, if you're asking how a silicon plate or iron or glass plate is represented in QED, then you're not asking a question about the double slit experiment; instead, you're asking a question about condensed matter physics. The precise description of a material depends on the material. At any rate, QED or SM or string theory won't say anything substantially different than non-relativistic QM. If you're assuming that relativistic QFT or string theory change anything about the interpretation of QM or the measurement, then your assumption is incorrect. They don't change anything. – Luboš Motl Jan 30 '11 at 16:46
Dear Lubos, I am asking whether modelling the detectors gives us a better understanding on why we observe a particle-like effect (one ammeter spikes per electron), instead of a wave like effect (all ammeters have spikes distributed in interference patterns). Classical QM doesn't explain this at all (or for that matter why we should use a position operator when we are measuring current). – Sklivvz Jan 30 '11 at 19:32

The answer is "yes and no". Important research is being done on modelling, with a Hamiltonian, the joint interaction between a microscopic particle being measured and the macroscopic measurement apparatus doing the measuring. Allahaverdyan, Balian, et al. have done the best, latest work. H.S. Green did a very stylised model long ago. Hannabuss has been doing important work on this.
Bibliographic information for their papers, and others by Collet, Milburn, et al., in Quantum Optics, there is also C. Gardiner and P. Zoller, Quantum Noise, can be found in the bibliography to my Thermodynamic Limits, Non-commutative Probability, and Quantum Entanglement, published, available for free at and my longer paper at

That was the "yes" part. The cited authors are mainstream important researchers.

But. It doesn't do much to change the big picture.

In particular, it does not resolve the controversy about decoherence. The work of these authors is valid whether or not there is decoherence, nor do they address the question of what would produce it and when.
Instead, they address directly the measurement process as a unitary, quantum mechanical process. The electron is still entangled with the slits and ammeter and everything, but they can say something about the precise way in which that entanglement is "negligible for all practical purposes". So it is important work.

More precisely, to answer the last parts of your question, it is not even a good idea to think of the "particle" as sometimes being examined in a a wave picture and sometimes in a a particle mode. That is just sloppy undergraduate thinking, you won't find that in the axioms of QM and you won't find it in these papers. There is no "this switch between wave-like and particle-like models of the electron" in their careful analyses of the measurement process, nor in the axioms of QM. In the axioms there is a switch from using the unitary evolution axioms of the wave function to the axioms for a measurement process, but this has nothing to do with modelling the electron. These cited papers succeed in modelling the measurement process as a unitary deterministic evolution of a quantum combined system. The resulting entangled superposition of states is not the same thing as what the measurement axioms predict, because the entangled state is a superposition of quantum states and the "result of a measurement process" is supposed to be a probability distribution on separable states. But they can show that the difference between these two different things is practically negligible because the "coherence" in the superposition is very very low...If now at this step you, dear reader, take the point of view of the "decoherence" crowd and apply it and go a little further than these results do, you can then say that since the coherence is negligible, this superposition of entangled states will appear the same as a mixed state, and a mixed state is, « as we all know », a probability distribution on its different components.

So yes, progress has been made, but J.S. Bell would never have accepted going from a quantum superposition of states to even a diagonal density matrix since logically these are distinct conceptions, so the controversy continues.

share|cite|improve this answer

So, the measurement or decoherence of the electron wave function corresponds to this switch between wave-like and particle-like models of the electron - which is clearly unsatisfying, intellectually, because QM is vague about this - it merely associates mathematical operations with the procedure, without giving any particular physical or mathematical justification of it.

Quantum mechanics doesn't have two models of the electron. For intuitive purposes it can be useful to visualize very broad wave packets (that are almost a momentum eigenstate) as "waves" and very localized wave packets (that are almost a position eigenstate) as "particles", but the QM description is unique.

The Copenhagen interpretation assumes that a measurement of position "collapses" the (position) wavefunction to a very localized wave packet with a width in the order of the measurement uncertainty. Decoherence explains the apparent collapse of the wavefunction by an entanglement between the system and environmental degrees of freedom due to the measurement process.

share|cite|improve this answer
"Decoherence explains the apparent collapse of the wavefunction by an entanglement between the system and environmental degrees of freedom due to the measurement process.": Ok can you be more clear about this by avoiding jargon? I don't really understand what you are saying. – Sklivvz Jan 30 '11 at 13:48
If you are measuring a system, you can't avoid entangling the system with macroscopic objects (measurement device, observer, ...) because the state of this objects must be modified by the measurement. So you won't observe a superposition of measurements because your state will be entangled with the measurement (in the MWI the different measurement outcomes are in "different branches"). – mmc Jan 30 '11 at 14:15
All "mainstream" interpretations of quantum mechanics assume that the results of measurements will be non-deterministic. In the Copenhagen Interpretation it's just a postulate, while in the Many Wolrds Interpretation it comes from you being unable to know the "branch" where you will end. I don't know enough about the Consistent Histories Interpretation to explain it in detail, but I know it doesn't deterministically select an outcome (I think that it just discards unobservable entanglements to get a "classical superposition".) – mmc Jan 30 '11 at 14:34
My question is not about which interpretation of QM is correct, but about why QM is correct as-is. – Sklivvz Jan 30 '11 at 19:38
@Skli Because it agrees with experiments – user1708 Jan 30 '11 at 20:41

I find it strange that even after 85 years since its inception, quantum theory has not been fully appreciated! I don't know what you meant by the term "classical quantum mechanics". I guess, it reflects your desire to "explain" quantum theory in terms of classical intuition. But what one should do is really the opposite. As far as your question is concerned, quantum theory in its current formulation has no problem. No agonizing scrutiny of the basic postulates are at all required. "Measurement" in QT has always confused some people and doubtless will continue to do so. But once decoherence is understood, it is not such a big deal.

share|cite|improve this answer
By classical QM I mean QM as it is taught in 99% of all courses (e.g. Copenhagen, not relativistic, Schrödinger, etc). – Sklivvz Jan 30 '11 at 13:44
Exactly, sb1! 85 years is a pretty long time. People often humiliate historical figures who didn't accept heliocentrism or something like that 20 years after they were established in science, but sometimes the very same people deny key insights of physics that were found 85 years ago. – Luboš Motl Jan 30 '11 at 16:48
I fully appreciate QM. That said, it doesn't mean that it explains everything or that it is a complete theory or that I can't have a question on it - comparing having doubts about something and not believing in heliocentrism is quite frankly demeaning and laughable. Especially because what is the correct interpretation of QM is still an open problem in physics. – Sklivvz Jan 30 '11 at 19:37
@skli not in physics, in philosophy – user1708 Jan 30 '11 at 20:43
@kake, that is completely debatable. – Sklivvz Jan 30 '11 at 22:35

I am always amazed that no one ever points out the obvious in these discussion: it is impossible experimentally to verify a one-to-one correspondence between the emission of an electron and the firing of a detector. The reason is that we do not yet have a pea-shooter for electrons. We never quite know when one has been fired. All we can say experimentally is that the rate of firing detectors is in some way proportional to the average flux of electrons.

share|cite|improve this answer
A thin layer of a weak beta-emitter on a wire or as a dot on a metal sheet is such a "one electron at a time gun. A multiplier arrangement hit by such an electron is able to detect it clearly. – Georg Dec 5 '11 at 10:22
1 "A new type of electron sources has emerged which permits to inject particles in a controllable manner, one at a time, into an electronic circuit." – anna v Dec 5 '11 at 14:39
It is interesting that the authors of the paper argue that until March of this year, my position was the correct one: that we did not yet have a pea-shooter for electrons. The authors claim to have recently gotten around this constraint. I'm betting that they haven't. – Marty Green Dec 5 '11 at 15:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.