Are tracks in collision experiments proof of particles?

[vanhees71]

''not enough detail'' is a strong exaggeration - he is completely silent about decoherence or measurement!

He just needs Born's rule for interpreting the final outcome. This makes it an exemplary contribution to the foundations. He explains without reference to anything outside the quantum formalism.

Moreover, there is no reference to the $\alpha$ particle! This makes his analysis very close to a field theoretical treatment. It is consistent with the possibility (implicitly indicated in the formulation of the thread title) that particles do not exist but are just a way of visualizing invisible happenings in the microscopic domain.

But we cannot use Mott's analysis directly in a QFT treatment since there is a mismatch between the statistical view of a train of many temporally separated particles in a beam (as an ensemble in the QM1 sense) and the temporally resolved view of many-particle QFT, where everything happening in space and time is described by correlation functions only.

Of course he (and in my opinion all other theoreticians after him) only need Born's rule to interpret the meaning of the state. There's nothing else in the formalism. Sometimes you find attempts to derive Born's rule from the other postulates of quantum theory. I think that Weinberg has given a convincing line of arguments that this is not possible (in his newest textbook: Lectures on Quantum Mechanics, Cambridge University Press). Of course, to follow such (mostly mathematical) endeavers is very interesting and sometimes fruitful. A famous example is the attempt to derive the parallel postulate of Euclidean geometry from the other axioms, which lead Gauß et al to the discovery of non-Euclidean geometry.

Within quantum theory Mott's analysis is fully sufficient to explain the observation of tracks in matter from quantum theory. It's of course an interesting question to investigate, how to generalize the non-relativstic wave-function treatment in the "first-quantization formalism" to QFT.

 

[A. Neumaier]

Indeed I meant classical field; I wouldn't expect any such thing from a quantum field, as it is an abstract concept devoid of simple visualization. Still, how do you explain with quantum field that charge on oil drops occurs in multiples of $e$?

It is because in QED, the total charge operator has a discrete spectrum consisting of integral multiples of $e$. Thus the analogy with angular momentum is complete; one doesn't need a particle concept.

 

[A. Neumaier]

The rest is a matter of terminology, some people call it wave-particle duality, others just call it a quantum particle. I just don't think there is a distinction to be made here at all in what seems to be universal behaviour.

The behavior is indeed universal, but the terminology matters a lot for the intuition and the resulting apparent weirdness.

The tension between particles and waves goes back in the case of light to the times of Huygens (1690) and Newton (1704). Through the interference experiments of Young (1801) it was settled (conclusively for more than a century) in favor of waves. The particle picture was later explained through the paraxial approximation that leads to geometric optics. Thus one can get particles in some limit as an approximation of waves; but there is no way to get waves from particles. Thus the waves are fundamental. This is also the reason why elementary particle physics is based on quantum field theory and not on quantum particle theory! Nomen est omen.

There are only fields in Maxwell's equations but the particle picture is a useful approximation for many optical phenomena (not involving destructive interference). In the same way, there are only correlation functions in quantum field theory, the fundamental theory in modern physics, but the particle picture is a useful approximation for many microscopic phenomena, as long as one acknowledges its limitations and intrinsic approximate nature.

 

[strangerep]

It is because in QED, the total charge operator has a discrete spectrum consisting of integral multiples of $e$. [...]

Ouch! To my shame, I was not aware of this. Where can I find a derivation?

 

[vanhees71]

For free electrons and positrons you have
$$\hat{Q}=-e \int \mathrm{d}^3 \vec{x} :\hat{\bar{\psi}}(t,\vec{x}) \gamma^0 \psi(t,\vec{x}): = -e \sum_{\sigma=\pm 1/2} \int \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} [\hat{a}^{\dagger}(\vec{p},\sigma)\hat{a}(\vec{p},\sigma)-\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma)].$$
A complete basis are the Fock states, which are the eigenstates of occupation numbers, which are either 0 or 1.

 

[A. Neumaier]

Ouch! To my shame, I was not aware of this. Where can I find a derivation?

In the free case, it is a consequence of the fact that the total charge operator is $e$ times the difference of the number operators for positrons and for electrons. Since the latter have a nonnegative integral spectrum and commute, the statement follows. In the interacting case, one has the same situation in naive perturbation theory. Then one has to argue that nothing bad happens to this fact during renormalization.
I don't think the limits involved change anything in the conclusion.

A complete basis are the Fock states, which are the eigenstates of occupation numbers, which are either 0 or 1.

This is too sloppy. Occupation numbers are not operators, hence have no eigenvalues. The integrals in quastion have as eigenvalues a sum of occupation numbers, which is a nonnegative integer.

 

[vanhees71]

For me the occupation numbers are represented by the number operators
$$\hat{N}_a(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma) \hat{a}(\vec{p},\sigma).$$
Strictly speaking, you have to first use some regularization (the most simple is to use the "finite-box quantization", i.e., put the system in a cubic box with periodic boundary conditions for the field modes) and afterwards take the infinite-volume limit. For the finite box the $\hat{N}$ are compatible observables and their eigenvectors span a basis of the Fock space. The eigenvalues for the fermion case are 0 and 1 for each occupation number due to the anti-commutation relations of the creation and annihilation operators. That's how we physicists construct the Fock space in our lectures. Maybe it's somehow not rigorous for mathematicians, but I guess that can be made rigorous (in fact the only thing that can be made rigorous in 1+3 dimensions seems to be construction of the free-particle Fock space).

 

[A. Neumaier]

For me the occupation numbers are represented by the number operators
$$\hat{N}_a(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma) \hat{a}(\vec{p},\sigma).$$
Strictly speaking, you have to first use some regularization (the most simple is to use the "finite-box quantization", i.e., put the system in a cubic box with periodic boundary conditions for the field modes) and afterwards take the infinite-volume limit. For the finite box the $\hat{N}$ are compatible observables and their eigenvectors span a basis of the Fock space. The eigenvalues for the fermion case are 0 and 1 for each occupation number due to the anti-commutation relations of the creation and annihilation operators. That's how we physicists construct the Fock space in our lectures. Maybe it's somehow not rigorous for mathematicians, but I guess that can be made rigorous (in fact the only thing that can be made rigorous in 1+3 dimensions seems to be construction of the free-particle Fock space).

But then you need extra explanatory work and an infinite volume limit to ensure that the integral over the operator-valued density with spectrum 0,1 is indeed an integer. (Classically, the integral over a characteristic function of a set has no reason to be an integer!)

Whereas if you take the integral (which is a standard 1-particle operator) as a whole and apply it to an N-particle state you immediately see that it gives N times the same state, revealing the spectrum. There is no need to invoke any particular basis in Fock space, and it works the same way for bosons and for fermions! The number operator is something more basic than the occupation number representation.

 

[strangerep]

$$
\hat{Q}=-e \int \mathrm{d}^3 \vec{x} :\hat{\bar{\psi}}(t,\vec{x}) \gamma^0 \psi(t,\vec{x}): ~=~ -e \sum_{\sigma=\pm 1/2} \int \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} [\hat{a}^{\dagger}(\vec{p},\sigma)\hat{a}(\vec{p},\sigma)-\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma)] $$

In the free case, it is a consequence of the fact that the total charge operator is $e$ times the difference of the number operators for positrons and for electrons. Since the latter have a nonnegative integral spectrum and commute, the statement follows. [...]

Oh,... I was indeed aware of that much. I thought you were referring to something deeper.

This is then just a consequence of the tensor product constructions involved in building a multiparticle Fock space: the integer charge spectrum is "by construction". This is significantly different (imho) from the case of angular momentum, where one merely asks that $SO(3)$ be represented unitarily on Hilbert space, and derives the half-integral spectrum without further input. In the latter case, the half-integers were nowhere inserted by hand.

There is also no group theoretic analysis (afaik) that can simultaneously derive integer charges for leptons, and fractional charges for quarks (without putting it all in by hand at the start).

So I think it is not correct to say:

[...] Thus the analogy with angular momentum is complete

The analogy certainly is not complete since the angular momentum case does not involve inserting half-integers somewhere by hand.

one doesn't need a particle concept

But one does use a (field theoretic version of) a particle concept in that the Fock space is built up by tensoring elementary systems.

 

[cave man]

I am not a physicist, but what happens to the photon when the energy of light is spent and matter cannot be created nor destroyed?

 

[A. Neumaier]

I am not a physicist, but what happens to the photon when the energy of light is spent and matter cannot be created nor destroyed?

Please ask your questions in a new thread and delete them here, where it is completely off-topic!

 

[A. Neumaier]

This is then just a consequence of the tensor product constructions involved in building a multiparticle Fock space: the integer charge spectrum is "by construction". This is significantly different (imho) from the case of angular momentum, where one merely asks that $SO(3)$ be represented unitarily on Hilbert space, and derives the half-integral spectrum without further input. In the latter case, the half-integers were nowhere inserted by hand.

Well, instead of $SO(3)$ you just need to consider a Heisenberg group and proceed in the same way. In the simplest case, where you have just one oscillator its representation theory gives you a unique regular unitary representation, e.g., realized on the dense subspace of Schwartz function of the Hilbert space $L^2(R)$. Inside the algebra of linear operators on this dense subspace you can find ladder operators as linear combinations of $q=x$ and $p=-i\partial_x$ that generate the discrete spectrum of $N=\frac12(p^2+q^2)-$const. It is not built in into the construction of $L^2(R)$, unless you count everything as built-in that can be mathematically deduced! But the the discrete spectrum of $SO(3)$ is also built in!

Thus the analogy is really complete! One can even get the Heisenberg case as a limiting case of either $SO(3)$ or $SO(1,2)$; see Section 22.2 of my online book Classical and Quantum Mechanics via Lie algebras.

For field theory one simply takes a much bigger Heisenberg group with an infinite number of independent oscillators, and pick the simplest of the now uncountably many inequivalent unitary representations. One never encounters particles unless one starts to look more closely at the eigenstates of $N$.

 

[A. Neumaier]

the modern Einselection program [...]
I don't know if this research ever reconsidered the exact same setup as the Mott paper, you are invited to do a literature search as I am unfortunately swamped for time..

Figari and Teta recently wrote a book about Mott's setting and variations, Quantum Dynamics of a Particle in a Tracking Chamber, expanding on an earlier arXiv paper. Although they say that

Mott’s analysis can be considered the original prototype of the
modern approach to the theory of environment-induced decoherence.

their ''environment'' consists (as in Mott's analysis) of 2 electrons only - far from the macroscopic heat bath needed to get irreversible amplification. True decoherence would presumably happen when a liquid drop condensates around the ions produced, but this step is only mentioned, not even superficially discussed. On the level of the formal discussion, one has a 3-particle system that ends up in a pure, entangled state, predicting by Born's rule probabilities consistent only with a straight path. Thus the relation to decoherence is superficial only. (They do, however, in Chapter 3 something with a macroscopic array of spins, which I haven't digested yet.)

But they make a remark that I found interesting:

It is worth emphasizing that a modification of the environment is the only experimental output one can observe. Contrary to what is often stated, one should not “trace out” the environment degrees of freedom, but rather those of the particle.

It fits the theme of this thread in that particles appear to be ghostlike and only macroscopic (hence field-like) things can be observed.

 

[A. Neumaier]

The analogy certainly is not complete since the angular momentum case does not involve inserting half-integers somewhere by hand.

In the case of the Heisenberg group (or equivalently but better, the slightly bigger oscillator group which also contains $N$ as a generator) of a single oscillator, nothing is inserted by hand, except for the choice of the Hermitian operator. This is because there is more freedom than in the $SO(3)$ case, due to the non-compactness of the Heisenberg group. If you pick $q$ or $p$ to determine the spectrum you get all reals; if you pick $N$, you get only the nonnegative integers.

From the interpretation in terms of a harmonic osciilator, the meaning of the eigenvalues of $N$ is the number of excitations of the eigenfunctions. Not the number of particles - after all, we have only a single oscillator, not enough to make up a particle. In a classical analogy, they count overtones - the number of zeros of a standing harmonic wave clamped at both ends.

When you increase the number of oscillators, the eigenvalues of $N$ (now summed over the oscillators) still count the number of excitations. Why should this interpretation suddenly change in the limit of infinitely many oscillators? It doesn't. Therefore the eigenvalues of the number operator in a free quantum field theory count the number of excitations, and nothing else.

In particular, they never count the number of particles, since so far, particles don't even make sense in our construction. To make sense of it we must impose a - somewhat weird and only historically justified - particle interpretation. In this particle interpretation, one says that an elementary excitation of the quantum field (i.e., a state in the $N=1$ eigenspace) constitutes an elementary particle, and defines the meaning of a single particle in this way! It is an arbitrary (only historically sanctioned) name for these states. It just amounts to using the word ''particle'' for ''elementary excitation'', thereby suggesting a sometimes appropriate, sometimes very misleading imagery.

Note that the particle interpretation is possible only when $N$ exists as an operator - i.e., in the free case, or, in the interacting case, asymptotically in the limit of infinite times for bound clusters in a scattering experiment! Therefore, in the real world, where one can never scatter in infinite time, the resulting particle picture is strictly speaking never appropriate - except in an approximate way!

Poincare invariance, Locality, and the uniqueness of the vacuum state now imply that the newly christened single particle space furnishes a causal unitary irreducible representation of the Poincare group, which were classified by Wigner in 1939. This is why particle theorists say that elementary particles are causal unitary irreducible representations of the Poincare group, Thus elementary particles are something exceedingly abstract, not tiny, fuzzy quantum balls!

For spin $\le 1$, these representations happen to roughly match the solution space of certain wave equations for a single relativistic particle in the conventional sense of quantum mechanics, but only if one discards the contributions of all negative energy states of the latter. This already shows that there is something very unnatural about the relativistic particle picture. Problems abound if one tries to push the analogies further, and quantum field theorists in their right mind will never do so.

Thus from a quantum field perspective, particles are ghosts from the past still haunting us as long as we continue to believe in them. It is historical baggage that carries no real weight - except in the terminology, which grew historically and is difficult to change.

To be fair, the particle picture has a very practical use. But only as an approximate, semiclassical concept valid when the fields are concentrated along a single (possibly bent) ray and the resolution is coarse enough. But whenever these conditions apply, one is no longer in the quantum domain and can already describe everything classically, perhaps with small quantum corrections. Thus the particle concept is useful when and only when the semiclassical description is already adequate. Note that this domain of validity excludes experiments with beam-splitters, half-silvered mirrors, double slits, diffraction, long-distance entanglement, and the like. Thus it is no surprise that in the interpretation of experimens involving these, particle imagery leads to mind-boggling features otherwise only knowns from dreams and ghost stories. The latter they are!

 

[Jilang]

A.N, on the subject of buckyballs etc you mentioned once that we shouldn't limit our idea of fields. I am struggling with this a bit. Is there a buckyball field?

 

[A. Neumaier]

A.N, on the subject of buckyballs etc you mentioned once that we shouldn't limit our idea of fields. I am struggling with this a bit. Is there a buckyball field?

Yes. There is an effective field for every molecule. Technically, for every bound state of the fundamental fields. Molecules are such bound states.