# A Is there a quantum uncertainty to the number of atoms in a marble?

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#### jfizzix

Gold Member
Summary
Quantum field Theory gives each Fermion and boson its own field, which can be in different quantum states.. what does that mean for the uncertainty of particle number applied to atoms?
I know precious little about quantum field theory, but want to understand the following.

If each Fermion and Boson has its own field..

..and as an example, the EM field can be in a coherent state, which is a superposition over many photon number states..

.. then can a fermion field, or multiple fermion fields be in a similar superposition?

If the answer to all of these is yes, then can the quantum state of a marble made of quintillions of atoms actually be in a superposition of having different numbers of atoms, or is there some physical reason that forces such objects to have an exact, but very large number of fermions?

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Staff Emeritus
is there some physical reason that forces such objects to have an exact, but very large number of fermions?
Yes, conservation of charge, lepton number and baryon number.

#### A. Neumaier

If the answer to all of these is yes,
It is, when your statements are made suitably precise.
then can the quantum state of a marble made of quintillions of atoms actually be in a superposition of having different numbers of atoms
Yes. This is always so in a grand canonical ensemble, the quantum model in which the equilibrium properties of the marble are usually investigated. In the nonrelativistic quantum field theory employed there, the number operator has a spectrum consisting of nonnegative integers, but there is no superselection rule that would force all states to be eigenstates of this operator (hence forcing a definite number of atoms) .

Of course one may idealize and take the marble to be a perfect crystal with perfectly defined boundary - then the number of atoms is fixed by the geometry together with the idealization involved in this description.
Yes, conservation of charge, lepton number and baryon number.
These conservation laws imply a determined number of atoms in the marble only if this number was already determined at the time of the marble's creation - which is not the case for an ordinary marble.

#### vanhees71

Gold Member
Well, I'd say the PTB is rather close to your "ideal marble", though the silicon sphere of the Avogadro project is adnmittedly a pretty unordinary marble given the care with which it is built over many years of hard work ;-).

Staff Emeritus
These conservation laws imply a determined number of atoms in the marble only if this number was already determined at the time of the marble's creation - which is not the case for an ordinary marble.
Of course it is. I challenge you to show a calculation where the quantum field theoretic effects on the number of atoms show a probability of a number other than N being greater than, say, one in a million.

#### DarMM

Gold Member
In general particle number is undefined in QFT except at asymptotic times. So strictly speaking there isn't a fixed $N$ in any piece of matter, except for how it impinges on asymptotically placed detectors. You can have funny events like so called "dark counts" where a particle detector clicks in response to no matter at all.

However all these effects are suppressed by a term exponential in the size of the device (for massive particles) and most of our devices are placed far enough away so as to count as asymptotic. So ignoring this is one of the most accurate approximations in physics.

#### A. Neumaier

I challenge you to show a calculation where the quantum field theoretic effects on the number of atoms show a probability of a number other than N being greater than, say, one in a million.
Equilibrium statistical mechanics in the grand canonical ensemble gives the thermodynamic laws in terms of a state $\rho=e^{-\beta(H-\mu N+PV)}$, for which variants of the central limit theorem imply that for large systems,
$N$ is approximately Gaussian, with a relative uncertainty of the order of $\langle N\rangle^{-1/2}$, with a material dependent factor. For $\langle N\rangle=10^{23}$, this leaves a probability very close to 1 for the number of atoms to differ from the mean.

Staff Emeritus
That is entirely appropriate for answering the question "what is the thermodynamic variation on the marble production line". It has nothing to do with your claim in #3, and nothing to do with my challenge, which I will repeat: I challenge you to show a calculation where the quantum field theoretic effects on the number of atoms show a probability of a number other than N being greater than, say, one in a million.

#### Demystifier

2018 Award
Equilibrium statistical mechanics in the grand canonical ensemble gives the thermodynamic laws in terms of a state $\rho=e^{-\beta(H-\mu N+PV)}$, for which variants of the central limit theorem imply that for large systems,
$N$ is approximately Gaussian, with a relative uncertainty of the order of $\langle N\rangle^{-1/2}$, with a material dependent factor. For $\langle N\rangle=10^{23}$, this leaves a probability very close to 1 for the number of atoms to differ from the mean.
Why do you think that in this case the relevant ensemble is the grand canonical one and not the canonical one? In other words, what is the physical mechanism by which the number of atoms fluctuates? What, instead of $10^{23}$ atoms in a crystal lattice, we had a C60 molecule https://en.wikipedia.org/wiki/Fullerene in thermal equilibrium with its environment? Are you saying that the number of atoms in C60 would fluctuate by the order of $\sqrt{60}\approx 8$?

#### A. Neumaier

It has nothing to do with your claim in #3, and nothing to do with my challenge, which I will repeat: I challenge you to show a calculation where the quantum field theoretic effects on the number of atoms show a probability of a number other than N being greater than, say, one in a million.
Your challenge (as far as not met by my previous post) has nothing to do with my assertions.

You misunderstood my claim, which was that a quantum system in a superposition of states with distinct particle numbers will - under the usual nonrelativistic Hamiltonians that commute with the number operator - remain in such a state at all times. If the system has initially the probability $p_n$ to have $n$ particles with given probabilities summing to 1 (and nothing in quantum field theory forbids this), this will remain so at all times.

Note that particle number is conserved in any such process. Thus conservation laws do not guarantee that
particle number is definite. In particular, this holds for the number of atoms a marble (an object defined only in thermodynamic terms).

#### A. Neumaier

Why do you think that in this case the relevant ensemble is the grand canonical one and not the canonical one?
Because one can talk about the pressure of a marble, and pressure is defined only in the grand canonical ensemble. In the canonical ensemble, to get a fully well-defined pressure, one would need to take the thermodynamic limit of infinite volume - a marble filling the whole universe!
In other words, what is the physical mechanism by which the number of atoms fluctuates?
This is a completely different question, as meaningless as the quest for the physical mechanism by which the position of a single atom in the ground state of an external potential fluctuates. Quantum mechanical fluctuations are not fluctuations in time but properties of the quantum state at any fixed time!
What if, instead of $10^{23}$ atoms in a crystal lattice, we had a C60 molecule https://en.wikipedia.org/wiki/Fullerene in thermal equilibrium with its environment? Are you saying that the number of atoms in C60 would fluctuate by the order of $\sqrt{60}\approx 8$?
A C60 molecule has by definition exactly 60 atoms, so this number does not fluctuate. The molecule cannot be in thermal equilibrium, since it is not a macroscopic object. Certainly there is no natural way to define its pressure.

But a marble is not a well-defined crystal - it has far too many defects of various kinds.

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#### DennisN

Summary: Quantum field Theory gives each Fermion and boson its own field, which can be in different quantum states.. what does that mean for the uncertainty of particle number applied to atoms?
A great and interesting question!

In general particle number is undefined in QFT except at asymptotic times. So strictly speaking there isn't a fixed NN in any piece of matter, except for how it impinges on asymptotically placed detectors. You can have funny events like so called "dark counts" where a particle detector clicks in response to no matter at all.
Interesting! Are these "dark counts" something only in theory or can it actually happen in an experiment, I wonder?

#### vanhees71

Gold Member
It should be
$$\hat{\rho}=\frac{1}{Z} \exp[-\beta (\hat{H}-\mu \hat{N})].$$
The $PV$ term makes no sense. In the usual treatment $V$ enters as an external parameter (the "quantization volume") via spatial periodic boundary conditions.

Also the grand canonical ensemble applies to an open system, coupled to a heat bath, where energy as well as particle exchange is possible. This does not imply that particle number is not conserved. If you want this stat. op. to be an equilibrium stat. op. it should be a conserved quantity. Thus for a closed system it is indeed conserved. From the relativistic point of view the corresponding conserved quantum numbers are charge like, and there indeed are superselection rules according to which superpositions of states of different charge numbers do not exist in nature.

#### A. Neumaier

It should be
$$\hat{\rho}=\frac{1}{Z} \exp[-\beta (\hat{H}-\mu \hat{N})].$$
The $PV$ term makes no sense. In the usual treatment $V$ enters as an external parameter (the "quantization volume") via spatial periodic boundary conditions.
By taking partial derivatives to get the expectations and comparing with the Euler formula, the Gibbs Duhem equation, and the first law of thermodynamics, one finds that $Z=\exp[\beta PV]$ (or equivalently $P=k_BT V^{-1}\log Z$), so that your formula and mine are the same. (One can also check this formula by considering an ideal gas). For full details of equilibrium thermodynamics in terms of the grand canonical ensemble see my online book.

Also the grand canonical ensemble applies to an open system, coupled to a heat bath, where energy as well as particle exchange is possible. This does not imply that particle number is not conserved.
In the context of the present discussion, total particle number is conserved. This is independent of the consideration of the grand canonical ensemble.

#### vanhees71

Gold Member
Exactly that's my point. The grand canonical ensemble is not applicable to the case considered here, because we just discuss a marble as a closed system, if I understand the OP right. So if anything thermal applies here at all, it's the canonical ensemble with a fixed number of particles.

It's very confusing to write out $\ln Z=\beta P V$, which is correct, but it's derived and put in at the beginning!

#### A. Neumaier

The grand canonical ensemble is not applicable to the case considered here, because we just discuss a marble as a closed system.
The grand canonical ensemble reproduces the full content of equilibrium thermodynamics, hence is applicable independent of the boundary conditions. The boundary conditions just tell which macroscopic extensive or intensive variable is kept fixed. But this does not affect anything in the formalism.
It's very confusing to write out $\ln Z=\beta P V$, which is correct, but it's derived and put in at the beginning!
The chemical potential and the inverse temperature are also put in at the beginning, without producing any confusion. Their meaning becomes apparent later only.

Why then should the pressure term cause confusion? It makes the exponent read $-S/k_B$, where
$TS=H-\mu N+PV$ is the operator version of the well-known Euler identity; everything very natural.
The q-expectation of $S$ is the thermodynamic entropy, just as the q-expectation of $H$ is the thermodynamic internal energy.

One may consider the external parameter $V$ as an operator which has only one eigenvalue - the system volume. My formula gives precisely the right intuition needed for generalizing the density operator to one that represents nonrelativistic local equilibrium, where the intensive variables $\beta,~ \mu,~ P$ become $\beta(x), ~\mu(x), ~p(x)$ and the extensive operators $H,~ N,~ V$ become $h(x)dx, ~n(x)dx$, and $dx$, and the product is interpreted as an integral over a given bounded spatial domain delineating the system.

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#### charters

You can have funny events like so called "dark counts" where a particle detector clicks in response to no matter at all.
But doesn't this, perhaps questionably, assume the detector makes strictly local measurements or is itself inherently strictly localized in a way QFT would not permit? See https://arxiv.org/abs/0805.0806, in particular pages 3-5.

Also, if we treat the detector as a closed system, where does the energy come from for a dark count, in particular sufficient energy to create a pair of massive charged particles?

#### DarMM

Gold Member
But doesn't this, perhaps questionably, assume the detector makes strictly local measurements
It assumes the detector makes measurements in a finite region yes.

is itself inherently strictly localized in a way QFT would not permit?
Why would QFT not permit a detector to be localized?

#### charters

Why would QFT not permit a detector to be localized?
I think the question is how it *would* permit it, given the only localized states in QFT are the so-called Knight states. See https://arxiv.org/abs/math-ph/0607044. These do not seem achievable with fermionic fields, which constitute the main part of any detector/material object in our universe. Or, nonrelativistically, the wavefunction of a detector, or any atom in the detector, has infinite tails. I don't believe these states become strictly localized in QFT, where localization is generally more challenging.

As David Wallace says in https://arxiv.org/abs/quant-ph/0112148:

"From the perspective of this paper, the problem with this approach [in Halvorson and Clifton 2001] is its a priori assumption that what we measure are always exactly localised operators. This is, of course, an interpretive axiom of AQFT as it is often presented, but it effectively assumes the presence of outside observers whose measurements cannot be treated within the ordinary dynamics of the QFT. We shall instead construct an account which treats observers as part of the internal dynamics of the system (although, apart from that difference of emphasis, the solution below will be rather similar in character to that of Halvorson and Clifton). If we wish instead to treat our QFT as a closed system, and the measurement process as part of the internal dynamics of that system, then it is an open question whether or not we must treat our measurements as localised. Furthermore, we have at least some reason to think that the answer to the question is negative — for we believe our measuring devices (including ourselves) to be made out of particles, and we have already noted the fact that particles are never exactly localised."

#### DarMM

Gold Member
I think the question is how it *would* permit it, given the only localized states in QFT are the so-called Knight states
I know Knight's theorem. It says that field quanta cannot be localised. How is that relevant for a detector? Generally in interacting field theory things cannot fundamentally be made out of quanta as the notion is not even defined in general.

#### charters

I know Knight's theorem. It says that field quanta cannot be localised. How is that relevant for a detector? Generally in interacting field theory things cannot fundamentally be made out of quanta as the notion is not even defined in general.
Well an atom as an interacting state (which at least to good approximation is made of N electrons/protons/neutrons) is not exactly localized even non-relativistically. A molecule is not exactly localised either. So it stands to reason a macroscopic object is also not exactly localized. AFAIK, the interacting Hilbert space is not well understood, so there is a degree of conjecture on both sides here. As Wallace says, one can indeed take it as an axiom that the detector is perfectly localized, but it seems the more natural and safe assumption is that it isn't, as localized states in field theories seem generically more rare and fragile.

Put another way, while I'd gladly welcome the fact that interacting theory states naturally have better localization properties, it seems too optimistic.

For context, I will also echo Wallace's point that one's sensitivities on this issue are closely related on one's disposition toward whether open vs closed system dynamics constitute adequate and complete physical theories. I am not at all convinced a closed system version of QFT is tractable, but I am interested in the question of what such a version would require, or where the standard formalism is implicitly committed to an open system approach. My sense is your view is that an open system approach is already satisfying, which is why I like/appreciate getting your take on topics like this, which I see as real pressure points of the open/closed issue. But if this sort of discussion is boring or annoying for you, I can drop it.

#### DarMM

Gold Member
Well an atom as an interacting state (which at least to good approximation is made of N electrons/protons/neutrons) is not exactly localized even non-relativistically. A molecule is not exactly localised either. So it stands to reason a macroscopic object is also not exactly localized
What precisely do you mean by localized here? Non-zero amplitudes being confined to some points in space? (Typo corrected as per charters post below)
QFT does have localized states, those obeying the DHR condition, they just don't have a particle decomposition. Strocchi's brief comments about these states:
In conclusion, the locality property of the field algebra $\mathcal{F}$, with respect to which
the vacuum is cyclic, means that one may have a description of the states of $\mathcal{H}$ in
terms of localized states.
Localized states are common enough in QFT. Localized particle states are a problem.

AFAIK, the interacting Hilbert space is not well understood, so there is a degree of conjecture on both sides here
It's known that particle number is not well defined in an interacting field theory for all times. That's not conjecture. The number operator is not a well defined self-adjoint operator. See Strocchi's book "An Introduction to Non-Perturbative Foundations of Quantum Field Theory" Section 1.3 for a basic account about this.

For context, I will also echo Wallace's point that one's sensitivities on this issue are closely related on one's disposition toward whether open vs closed system dynamics constitute adequate and complete physical theories. I am not at all convinced a closed system version of QFT is tractable, but I am interested in the question of what such a version would require, or where the standard formalism is implicitly committed to an open system approach. My sense is your view is that an open system approach is already satisfying, which is why I like/appreciate getting your take on topics like this, which I see as real pressure points of the open/closed issue. But if this sort of discussion is boring or annoying for you, I can drop it
Could you expand on this more? What's meant by a "closed system version of QFT"?

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#### jfizzix

Gold Member
In general particle number is undefined in QFT except at asymptotic times. So strictly speaking there isn't a fixed $N$ in any piece of matter, except for how it impinges on asymptotically placed detectors. You can have funny events like so called "dark counts" where a particle detector clicks in response to no matter at all.

However all these effects are suppressed by a term exponential in the size of the device (for massive particles) and most of our devices are placed far enough away so as to count as asymptotic. So ignoring this is one of the most accurate approximations in physics.
Under what circumstances can an a fermion field in an isolated region of space be in a superposition of different numbers of fermions? Is this something that can be demonstrated experimentally?

#### DarMM

Gold Member
I think you just made a typo
Could you point it out, just to correct my post.

I'm not claiming the free theory number operator is defined in the interacting theory, I know it isn't. The question is whether the class of interacting states which are sufficiently long-lived to serve as detectors are localized - just because these states are not free number states doesn't entail they are localized. Moreover, these states should describe atoms, molecules, and other larger bound systems, which are countable as such, and have ground states which are energy/Hamiltonian eigenstates. They seem susceptible to the same sort of anti-localization arguments as free theory quanta. In particular Hegerfeldt's theorem.
Particles are global pure free idealizations appropriate for asymptotic detection events. Real localized states in interacting QFT are necessarily mixed and simply don't factorize into multiple copies of simpler states.

In interacting QFTs it is simply not possible to "count" states in the Hilbert space by seeing them as being composed of $N$ copies of some basic set of states. They're not combinations of quanta. Things like Hegerfeldt's theorem use projection operators and pure states, again an idealization inappropriate for localized field theoretic states which must be mixed.

I don't think there's any reason to doubt that detectors, as pieces of matter in an interacting QFT, are localized.

The issue of formulating QFT in the closed manner you mention would require a lot of exposition. I might return to it in another thread.

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#### vanhees71

Gold Member
The grand canonical ensemble reproduces the full content of equilibrium thermodynamics, hence is applicable independent of the boundary conditions. The boundary conditions just tell which macroscopic extensive or intensive variable is kept fixed. But this does not affect anything in the formalism.

The chemical potential and the inverse temperature are also put in at the beginning, without producing any confusion. Their meaning becomes apparent later only.

Why then should the pressure term cause confusion? It makes the exponent read $-S/k_B$, where
$TS=H-\mu N+PV$ is the operator version of the well-known Euler identity; everything very natural.
The q-expectation of $S$ is the thermodynamic entropy, just as the q-expectation of $H$ is the thermodynamic internal energy.

One may consider the external parameter $V$ as an operator which has only one eigenvalue - the system volume. My formula gives precisely the right intuition needed for generalizing the density operator to one that represents nonrelativistic local equilibrium, where the intensive variables $\beta,~ \mu,~ P$ become $\beta(x), ~\mu(x), ~p(x)$ and the extensive operators $H,~ N,~ V$ become $h(x)dx, ~n(x)dx$, and $dx$, and the product is interpreted as an integral over a given bounded spatial domain delineating the system.
In thermodynamics we learn about three standard ensembles with different realms of applicability. Of course, in the thermodynamic limit, $V \rightarrow \infty$, with densities kept constant, all these ensembles are equivalent since the fluctations are small compared to the mean values of the macroscopically relevant quantities.

If you discuss the question whether a marble consists of a determined number (determined value of charges) of particles, you cannot argue with the grand-canonical ensemble since by definition the grand-canonical ensemble deals with the situation, where particles (charges) and energy are exchanged with a larger reservoir.

For a system closed under particle (charge) exchange but exchanging energy with a reservoir you need to use the canonical ensemble, for a completely closed system the microcanonical ensemble.

A marble should in principle be describable with a canonical ensemble as long as the temperature is not that large that particle creation and annihilation processes become relevant, but that's at so high temperatures that this is not relevant here.

You can of course write the $P V$ term, but at least I was confused. There's a difference between $V$ on the one hand and $T$ and $\mu$ on the other: $V$ is a given constraint, while $T$ and $\mu$ or rather $\beta$ and $\mu \beta$ are Lagrange parameters to maximize the entropy under the given constraint of average energy and particle number (net charge).