A Ensembles in quantum field theory

[A. Neumaier]

TL;DR Summary

How does one model an ensemble of particles in quantum field theory?

But a fundamental description would have to come from relativistic quantum field theory, where there are no ensembles. One cannot repeatedly prepare a quantum field extending over all of spacetime.

I don't know of a single paper explaining how the transition in conceptual language from a single quantum field to an ensemble of particles can be justified from the QFT formalism. Maybe you can help me here with a reference?

Within QFT you can as well prepare single Ag atoms as you can in QM. QFT is also used since it's conception to describe scattering cross sections, and that's also due to the standard interpretation of the quantum state in terms of Born's rule. It's explained in any QFT textbook, e.g., Weinberg, QT of Fields vol. 1.

One can model a single silver atom in this way. But how do you model an ensemble of 100 silver atoms moving at well separated times along a beam by quantum field theory? You cannot prepare multiple instances of a field extending over all of spacetime. The only use! of Born's rule in Weinberg's Vol. 1 (namely where he interprets the scattering amplitutes) doesn't address this issue - scattering has nothing to do with this question!

Relativistic QFT has the same probabilistic interpretation as any QT (in fact there is only one overall conceptual framework and a non-relativistic (in both a "1st-quantization formulation" and a field-theoretical one as well as a special relativistic realization in terms of local QFTs).

Of course one can prepare ensembles within all kinds of QTs and, more importantly, in the lab. Relativistic QFT is among the best tested physical theories ever discovered. This would be impossible to achieve if it were not possible to prepare ensembles of the physical systems described by it, and these are particles and nuclei in particle accelerators, with which you can do scattering experiments with high precision. Another application is atomic physics. A specific quantum-field theoretical result is the explanation of the Lamb shift to several significant digits of accuracy, etc.

I don't understand, how one can claim that one cannot build ensembles within relativistic QFT, given all these successes. After all the first goal in all introductions to QFT is to establish the calculations of S-matrix elements, which precisely describe scattering processes, and obviously these can be realized with high accuracy in the lab.

Then please explain how the transition in conceptual language from a single quantum field (extending all over spacetime, or at least over the lab during a day) to an ensemble of particles can be justified from the QFT formalism.

 

[vanhees71]

I don't know, what is there to be explained. One prepares particles with an accelerator, lets them interact and detects the products. QFT predicts the corresponding cross sections. To measure these cross sections you do such scattering experiments and compare the distribution of the scattered particles for given processes.

 

[A. Neumaier]

I don't know, what is there to be explained. One prepares particles with an accelerator, lets them interact and detects the products. QFT predicts the corresponding cross sections. To measure these cross sections you do such scattering experiments and compare the distribution of the scattered particles for given processes.

Well, that's the few particle view. But I am asking for a more fundamental description. In a lab we perform a very simple experiment:

A source produces silver particles at an (average or equidistant) rate of one per second, which move along the z-axis towards a glass surface, adsorbing the silver atoms milliseconds after they were produced, until a faint silver spot is visible under the microscope.

The whole lab is a quantum system that can in principle (though not in practice) be modeled by quantum field theory. Thus one has a time-independent Heisenberg state for the lab, an a number of quantum fields describing the stuff in the lab during the day considered.

The silver field in some region around the beam may be considered to be a free scalar field. No scattering is involved. What is the Heisenberg state of the quantum field in this region of spacetime, given the above description?

 

[Morbert]

Ensembles presumably don't have to be real. A prepared state encompassing the whole lab + quantum fields + silver field can be conceptualised as a fictitious ensemble of macroscopic systems. The extreme of this would be quantum cosmologists conceptualising an ensemble of universes.

 

[vanhees71]

As you write yourself, this you cannot achieve. It's simply impossible to describe the quantum state of a macroscopic object in that detail, but that's for sure not necessary to describe your scenario.

For that it's sufficient to simply describe the silver atoms alone. The initial state is the vacuum (no silver atoms present). The probability for the occurance of the Ag atoms would be something like
$$\langle \Omega \hat{\rho}_1(t_1,\vec{x}_1) \hat{\rho}_2(t_2,\vec{x}_2) \cdots |\Omega \rangle,$$
where $\hat{\rho}(t,\vec{x})$ is the particle-density operator.

 

[A. Neumaier]

As you write yourself, this you cannot achieve. It's simply impossible to describe the quantum state of a macroscopic object in that detail, but that's for sure not necessary to describe your scenario.

Thats why I only ask for the description of the neighborhood of the beam during the whole, very simple experiment. There the quantum field is free (ignoring air), and a detailed description is possible.

For that it's sufficient to simply describe the silver atoms alone. The initial state is the vacuum (no silver atoms present). The probability for the occurance of the Ag atoms would be something like
$$\langle \Omega \hat{\rho}_1(t_1,\vec{x}_1) \hat{\rho}_2(t_2,\vec{x}_2) \cdots |\Omega \rangle,$$
where $\hat{\rho}(t,\vec{x})$ is the particle-density operator.

I don't want a probability, I want the Heisenberg state. If a QFT state can be assigned to systems as big as the sun through millions of years (used, e.g., to derive from it a coarse-grained hydromechanical description), one can surely be assigned to a faint silver beam in the lab through a day. (In fact I know the answer. Not every detail, but detailed enough so that one can discuss the experiment in its terms.)

 

[A. Neumaier]

Ensembles presumably don't have to be real. A prepared state encompassing the whole lab + quantum fields + silver field can be conceptualised as a fictitious ensemble of macroscopic systems. The extreme of this would be quantum cosmologists conceptualising an ensemble of universes.

How can scientific descriptions depend on fictitious stuff? Surely Nature doesn't care about our fictions...

 

[Demystifier]

How can scientific descriptions depend on fictitious stuff? Surely Nature doesn't care about our fictions...

Sure, but it doesn't mean that our fictions are independent on Nature. Our fictions care about Nature, they are chosen with intention to resemble Nature. Any human description of any phenomenon in Nature is necessarily a simplification, but the art of physics consists in finding simplifications that still keep some essential properties of Nature.

 

[Demystifier]

The whole lab is a quantum system that can in principle (though not in practice) be modeled by quantum field theory. Thus one has a time-independent Heisenberg state for the lab, an a number of quantum fields describing the stuff in the lab during the day considered.

The von Neumann theory of measurement models the macroscopic measuring apparatus in a very crude way. I think you want a more detailed model, but this is not only hard in practice, but also unnecessary because it turns out that the crude models capture essential properties very well. It is a general principle in physics that properties of macroscopic systems usually don't depend much on microscopic details.

 

[vanhees71]

Thats why I only ask for the description of the neighborhood of the beam during the whole, very simple experiment. There the quantum field is free (ignoring air), and a detailed description is possible.

I don't want a probability, I want the Heisenberg state. If a QFT state can be assigned to systems as big as the sun through millions of years (used, e.g., to derive from it a coarse-grained hydromechanical description), one can surely be assigned to a faint silver beam in the lab through a day. (In fact I know the answer. Not every detail, but detailed enough so that one can discuss the experiment in its terms.)

I don't understand, what you think is wrong with this standard description. All we have are probabilities. In the case you described, the state in the Heisenberg picture is the vacuum, since the state in the Heisenberg picture is time independent, and the time evolution is entirely in the field operators. The probabilities, which are what's observable in QT, are given by the matrix elements of the statistical operator. For photons that's described in any textbook on quantum optics. See, e.g.,

J. Garrison and R. Chiao, Quantum optics, Oxford University
Press, New York (2008),
https://doi.org/10.1093/acprof:oso/9780198508861.001.0001

which has a very detailed chapter on photon-detection theory. For photons it's of course the energy density, because there are no charge-like quantum numbers for photons.

 

[Morbert]

How can scientific descriptions depend on fictitious stuff? Surely Nature doesn't care about our fictions...

as if is a powerful conjunction in physics. E.g. Fermi liquids are fictitious, but under a broad set of conditions, many-body systems behave as if they are a Fermi liquid. We've made all sorts of technological advancements because physicists are willing to premise their understanding of reality (whatever it may be) in these fictions.

 

[A. Neumaier]

Sure, but it doesn't mean that our fictions are independent on Nature. Our fictions care about Nature, they are chosen with intention to resemble Nature. Any human description of any phenomenon in Nature is necessarily a simplification, but the art of physics consists in finding simplifications that still keep some essential properties of Nature.

The von Neumann theory of measurement models the macroscopic measuring apparatus in a very crude way. I think you want a more detailed model, but this is not only hard in practice, but also unnecessary because it turns out that the crude models capture essential properties very well. It is a general principle in physics that properties of macroscopic systems usually don't depend much on microscopic details.

Crude but sufficient approximation is something quite different from fictitious.

as if is a powerful conjunction in physics. E.g. Fermi liquids are fictitious, but under a broad set of conditions, many-body systems behave as if they are a Fermi liquid. We've made all sorts of technological advancements because physicists are willing to premise their understanding of reality (whatever it may be) in these fictions.

So both of you say that single systems behave as if they were ensembles. But in an ensemble, the term probability $p$ has a meaning for every $p$ between 0 and 1, while for a single system, only probabilities 0 and 1 have meaning. What meaning do you give the other probabilities?

 

[A. Neumaier]

I don't understand, what you think is wrong with this standard description. All we have are probabilities. In the case you described, the state in the Heisenberg picture is the vacuum, since the state in the Heisenberg picture is time independent, and the time evolution is entirely in the field operators.

This is strange since the region under discussion contains a beam, not a vacuum. Since at least sometimes it contains particles, the state of the region cannot be the vacuum state.

The probabilities, which are what's observable in QT, are given by the matrix elements of the statistical operator.

But the statistical operator is not an observable. it transforms under time translations not like observables in quantum theory but with a wrong sign. So you cannot apply Born's rule to its matrix elements.

For photons that's described in any textbook on quantum optics.

I know how textbook detection theory works, but formulas there are derived there in a piecemeal fashion. I asked for a more fundamental description.

The derivation in the quantum optics books does not start from the state of the lab, which is the fundamental description in qunatum field theory. (If the lab were a single block of ice, you'd also start with the state of the block of ice, and then make approximations to get the standard thermodynamic results.)

Though only incompletely known, we know a large amount about the state of the lab, namely (slightly averaged) values of the effective fields at essentially all points of the lab. This is enough to make the relevant deductions.

See, e.g.,

J. Garrison and R. Chiao, Quantum optics, Oxford University
Press, New York (2008),
https://doi.org/10.1093/acprof:oso/9780198508861.001.0001

which has a very detailed chapter on photon-detection theory. For photons it's of course the energy density, because there are no charge-like quantum numbers for photons.

What is ''it''? Do you mean the energy density replaces the statistical operator? How can that be, given their opposite behavior under time translations? Moreover, unlike the statistical operator, the energy density is not normalizable.

Please be more precise!

 

[Demystifier]

But in an ensemble, the term probability $p$ has a meaning for every $p$ between 0 and 1, while for a single system, only probabilities 0 and 1 have meaning. What meaning do you give the other probabilities?

For a frequentist interpretation of probability, only 0 and 1 are possible probabilities for a single system. But for a Bayesian interpretation, a single system can "have" any probability in the interval [0,1]. A standard example is flipping one coin only once and not looking at the outcome, in which case the Bayesian probability of heads for one coin is 1/2.

 

[A. Neumaier]

For a frequentist interpretation of probability, only 0 and 1 are possible probabilities for a single system. But for a Bayesian interpretation, a single system can "have" any probability in the interval [0,1]. A standard example is flipping one coin only once and not looking at the outcome, in which case the Bayesian probability of heads for one coin is 1/2.

But the physics in a lab does not depend on whether someone looks at the outcome.

 

[vanhees71]

This is strange since the region under discussion contains a beam, not a vacuum. Since at least sometimes it contains particles, the state of the region cannot be the vacuum state.

In the Heisenberg picture the statistical operator is time-independent. I describe only the particles, and according to your assumption at $t=0$ there are no particles, i.e., the initial statistical operator is the vacuum.

But the statistical operator is not an observable. it transforms under time translations not like observables in quantum theory but with a wrong sign. So you cannot apply Born's rule to its matrix elements.

Of course, the time evolution of the stat. op. is different from the time evolution of observable operators. It's more or less arbitrary how you put the time-dependence on the stat. op. and the observable operators. That's the choice of the "picture of time evolution". What's observable are the matrix elements of the stat. op. wrt. the eigenvectors of the observed observable.

The states transform with $\hat{C}(t,t_0)$ according to
$$\hat{\rho}(0)=\hat{C}(t,t_0) \hat{\rho}(t_0) \hat{C}^{\dagger}(t,t_0),$$
where $\hat{C}$ obeys the EoM.
$$\mathrm{i} \partial_t \hat{C}(t,t_0)=\hat{H}_1(t) \hat{C}(t,t_0).$$
The (formal) solution is
$$\hat{C}(t,t_0)=\mathcal{T}_c \exp [-\mathrm{i} \int_{t_0}^{t} \mathrm{d} t' \hat{H}(t')].$$
Where $\mathcal{T}_c$ is the "causal time-ordering operator", ordering operator products with increasing time arguments from right to left.

The observable operators transform with $\hat{A}(t,t_0)$,
$$\hat{O}(t)=\hat{A}(t,t_0) \hat{O}(t_0) \hat{A}^{\dagger}(t,t_0),$$
where
$$-\mathrm{i} \partial_t \hat{A}(t,t_0) = \hat{H}_2(t) \hat{A}(t,t_0)$$
with the formal solution
$$\hat{A}(t,t_0) = \mathcal{T}_c \exp[+\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}_2(t')].$$
Finally
$$\hat{H}=\hat{H}_1+\hat{H}_2$$
is the Hamiltonian of the system.

In the Heisenberg picture $\hat{H}_1=0$ and $\hat{H}_2=\hat{H}$, while in the Schrödinger picture it's $\hat{H}_1=\hat{H}$ and $\hat{H}_2=0$. The "interaction picture" is the split $\hat{H}_1=\hat{H}_I$ (the "interaction part") and $\hat{H}_2=\hat{H}_0$ (the "free part", i.e., the part in the Hamilton density with is quadratic in the field operators and their derivatives).

I know how textbook detection theory works, but formulas there are derived there in a piecemeal fashion. I asked for a more fundamental description.

The derivation in the quantum optics books does not start from the state of the lab, which is the fundamental description in qunatum field theory. (If the lab were a single block of ice, you'd also start with the state of the block of ice, and then make approximations to get the standard thermodynamic results.)

You cannot describe the lab quantum mechanically, and fortunately that's not necessary. The description in the quantum optics textbooks is pretty sufficient to explain the high-precision results of quantum-optical experiments.

Though only incompletely known, we know a large amount about the state of the lab, namely (slightly averaged) values of the effective fields at essentially all points of the lab. This is enough to make the relevant deductions.

Indeed. If this were not the case, I doubt that mankind would have ever discovered any "fundamental laws" of nature ;-).

What is ''it''? Do you mean the energy density replaces the statistical operator? How can that be, given their opposite behavior under time translations? Moreover, unlike the statistical operator, the energy density is not normalizable.

Now, the energy density is not the statistical operator but representing an observable. In classical as well as quantum theory the intensity is physically nothing else than the (time-averaged) energy density of the em. field. Of course the corresponding distribution can be normalized to 1 since the total field energy is always finite.

Please be more precise!

I try my best. I'm a bit surprised that we have to discuss again the very basics of QT.

 

[A. Neumaier]

In the Heisenberg picture the statistical operator is time-independent. I describe only the particles, and according to your assumption at $t=0$ there are no particles, i.e., the initial statistical operator is the vacuum.

OK. But then, since we use the Heisenberg picture, the statistical operator is constant, and your claim

The probabilities, which are what's observable in QT, are given by the matrix elements of the statistical operator.

just says that what's observable in QT is given by the vacuum expectation values. Then this is a triviality. (I had misunderstood you, I though you were taking the statistical operator as argument between the vacuum states.) So we can continue. I assume therefore you mean by

For photons it's of course the energy density, because there are no charge-like quantum numbers for photons.

that in the case of photons we are looking for the vacuum expectation value of the energy density at different times. But the query is about a silver beam (for further use in a Stern-Gerlach setting). Is here what is observable the vacuum expectation value of the energy density or that of the mass density?

The observable operators transform with $\hat{A}(t,t_0)$,
$$\hat{O}(t)=\hat{A}(t,t_0) \hat{O}(t_0) \hat{A}^{\dagger}(t,t_0),$$
where
$$-\mathrm{i} \partial_t \hat{A}(t,t_0) = \hat{H}_2(t) \hat{A}(t,t_0)$$
with the formal solution
$$\hat{A}(t,t_0) = \mathcal{T}_c \exp[+\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}_2(t')].$$
Finally
$$\hat{H}=\hat{H}_1+\hat{H}_2$$
is the Hamiltonian of the system.

In the Heisenberg picture $\hat{H}_1=0$ and $\hat{H}_2=\hat{H}$.

This is the textbook treatment for isolated systems. But things are here not so simple since the neighborhood of the beam is not a closed system, and one has to consider the incoming boundary conditions. Otherwise one would never get any signal...

Also some things simplify since we are working exclusively in the Heisenberg picture. To simplify further we may assume that $t_0=0$.

You cannot describe the lab quantum mechanically

This is surely false. Why can astronomers describe the sun quantum mechanically but not the much smaller lab? It is composed of the same stuff as the sun, just in a more complex state. We'd therefore be able to get from whatever knowledge we have of that state, quite as in the case of the sun, using techniques like the maximum entropy principle the simplified description used in the quantum optics books by restricting attention to the region of interest, in the present case the surrounding of the beam.

I'm a bit surprised that we have to discuss again the very basics of QT.

Only to the point where the verbal precision is enough to make clear what you mean.

 

[Demystifier]

But the physics in a lab does not depend on whether someone looks at the outcome.

True, but someone's knowledge about the outcome depends on it. This is what probability means when one uses a fictive ensemble to describe a single system. But that's probably not very important for what really bothers you in this thread. You want an explicit quantum state of the laboratory described by QFT. In practice it's too hard, but in principle I don't see what's exactly the problem with having this state too. If the laboratory is made of $N$ particles, then the state is a product of $N$ creation operators acting on the vacuum, I don't see what's the problem in principle.

 

[A. Neumaier]

it's still not clear to me what's your original motivation to study ensembles, do you want to consider a single system or many systems?

The set-up is a single lab containing for 24 hours a source with a very faint beam along which silver particles travel.

I am asking for its complete quantum field description in terms of the Heisenberg state of the lab (restricted to a small, easily analyzable part), in order to be able to discuss how the single silver quantum field of the lab models what is usually discussed in terms of an ensemble of 86400 silver atoms generated by the source in 24 hours.

(This should give a better understanding of what happens on a fundamental level at the screen of a Stern-Gerlach experiment. But for the moment there are no magnets in the setting; thus we do not need to consider spin.)

 

[Demystifier]

The set-up is a single lab containing for 24 hours a source with a very faint beam along which silver particles travel.

I am asking for its complete quantum field description in terms of the Heisenberg state of the lab

The Heisenberg state does not depend on time, so it does not depend on the fact that it works 24 hours. You can think of Heisenberg state as the initial state Schrodinger state of the lab. Since the lab is made of atoms, which are made of elementary particles described by QFT, the lab is made of N particles, so the Heisenberg state is described by a product of N creation operators acting on the vacuum. In principle, I don't see what's the problem.

 

[A. Neumaier]

The Heisenberg state does not depend on time, so it does not depend on the fact that it works 24 hours. You can think of Heisenberg state as the initial state Schrodinger state of the lab. Since the lab is made of atoms, which are made of elementary particles described by QFT, the lab is made of N particles, so the Heisenberg state is described by a product of N creation operators acting on the vacuum. In principle, I don't see what's the problem.

Yes, there should be no problem; I deliberately kept the setting very simple. But this does not mean that nothing needs to be discussed.

We consider explicitly only the part of the lab containing the beam, assuming the beam goes through a perfect vacuum. Restricting the state to this region, it is a vacuum state before the source is switched on. Your creation operators only affect the state at other positions where there is lab matter, hence have no effect on the beam. (This holds since the Hilbert space of the quantum field theory is the completion of a tensor product of the Hilbert spaces of the regions into which one partitions space.)

Thus as long as we exclude what happens at the source (and later at a magnet or a screen) we only need to work with the vacuum state and a free field Hamiltonian. We may ignore the composition of a silver atom (made from elementary particles). Moreover, since silver is heavy, we may use nonrelativistic quantum field theory, so problems with renormalization are absent. This makes the discussion feasible.

Everything that happens is encoded in the Heisenberg equation for the dynamics of the observables. Since the beam is directed along the $z$ axis, the behavior of the source (and hence the detailed properties of the beam) enters solely through the time-dependent ingoing boundary conditions at $z=0$.

 

[Demystifier]

@A. Neumaier So do you think that the problem, in principle, is solved now?

 

[A. Neumaier]

@A. Neumaier So do you think that the problem, in principle, is solved now?

Not until the discussion has concluded. We have only begun to discuss the details, and there is a lot to be learnt from studying this simple experiment.

 

[Demystifier]

Not until the discussion has concluded. We have only begun to discuss the details, and there is a lot to be learnt from studying this simple experiment.

OK. But as a preliminary conclusion, it would be useful if you could pinpoint one or few things which were not clear (to you) at the beginning of this thread, but are clear now.

 

[vanhees71]

OK. But then, since we use the Heisenberg picture, the statistical operator is constant, and your claim

just says that what's observable in QT is given by the vacuum expectation values. Then this is a triviality. (I had misunderstood you, I though you were taking the statistical operator as argument between the vacuum states.) So we can continue. I assume therefore you mean by

The statistical operator can of course be everything meeting the conditions, not only the vacuum state. In equilibrium you can use the canonical or grand-canonical statistical operator $\propto \exp(-\beta \hat{H}+\alpha \hat{Q})$.

that in the case of photons we are looking for the vacuum expectation value of the energy density at different times. But the query is about a silver beam (for further use in a Stern-Gerlach setting). Is here what is observable the vacuum expectation value of the energy density or that of the mass density?

It depends on the detector. What you have in mind most probably is described by $N$-point auto-correlation functions of the charge density.

This is the textbook treatment for isolated systems. But things are here not so simple since the neighborhood of the beam is not a closed system, and one has to consider the incoming boundary conditions. Otherwise one would never get any signal...

Sure, but that you can mimic by using some external (classical) source and solve for the operator Klein-Gordon equation (or whatever equation you want to use to describe silver atoms) such that you get your (operator-valued) wave packets with at different times around some place.

More realistic is mimicking the original setup by using a beam of thermal Ag atoms emitted from an oven at temperature $T$. Then the initial state is the corresponding stat. op. describing this thermal beam coming out of the oven.

Also some things simplify since we are working exclusively in the Heisenberg picture. To simplify further we may assume that $t_0=0$.This is surely false. Why can astronomers describe the sun quantum mechanically but not the much smaller lab? It is composed of the same stuff as the sun, just in a more complex state. We'd therefore be able to get from whatever knowledge we have of that state, quite as in the case of the sun, using techniques like the maximum entropy principle the simplified description used in the quantum optics books by restricting attention to the region of interest, in the present case the surrounding of the beam.

Sure, you always have to simplify the description sufficiently to be able to describe anything. The Sun is in a sense simple, because you can describe it as a fireball of plasma in thermal equilibrium held together by gravitation.

Only to the point where the verbal precision is enough to make clear what you mean.

 

[A. Neumaier]

OK. But as a preliminary conclusion, it would be useful if you could pinpoint one or few things which were not clear (to you) at the beginning of this thread, but are clear now.

I did just this in my post #21.

 

[A. Neumaier]

The statistical operator can of course be everything meeting the conditions, not only the vacuum state. In equilibrium you can use the canonical or grand-canonical statistical operator $\propto \exp(-\beta \hat{H}+\alpha \hat{Q})$.

In general, yes. But in the current experiment you already agreed that the state was the vacuum state.

It depends on the detector. What you have in mind most probably is described by $N$-point auto-correlation functions of the charge density.

The detector is a glass bottle, on which I find after sufficient time a faint spot of silver. In this simple experiment, correlations cannot be measured, hence we only need 1-point functions. Since silver is uncharged. I guess you regard the mass as a charge.

Thus do you agree that the correct operator to look at is the mass 4-current, and especially its zero component, the mass density?

Sure, but that you can mimic by using some external (classical) source and solve for the operator Klein-Gordon equation (or whatever equation you want to use to describe silver atoms) such that you get your (operator-valued) wave packets with at different times around some place.

Please mimic it for me, so that we can discuss its consequences. Since silver is heavy, I want to use a nonrelativistic QFT, with an antisymmetrized Fock space to account for the spin 1/2 of silver.

More realistic is mimicking the original setup by using a beam of thermal Ag atoms emitted from an oven at temperature $T$. Then the initial state is the corresponding stat. op. describing this thermal beam coming out of the oven.

But we already agreed that the initial state of the beam is the vacuum state, before the oven is switched on. Since we are in the Heisenberg picture, this will be its state throughout the whole day modeled. Thus you somehow need to get the thermal input into the boundary conditions for the observable of interest.

Sure, you always have to simplify the description sufficiently to be able to describe anything. The Sun is in a sense simple, because you can describe it as a fireball of plasma in thermal equilibrium held together by gravitation.

And the lab is in a local equilibrium state described by the properties of the stuff inside. We are only looking at the region where the beam is, where the state is even simpler - just the vacuum state.

 

[vanhees71]

As I said, the switching-on of the oven/any other source of Ag atoms is described by the time evolution of the field operators. The spot on the screen is given by integrating the current $\langle \vec{j}(t,\vec{x}) \cdot \vec{n} \rangle$ ($\vec{n}$ is the normal vector of the screen) over time.

 

[A. Neumaier]

As I said, the switching-on of the oven/any other source of Ag atoms is described by the time evolution of the field operators.

Here more details are required. Which time-dependent Hamiltonian are you using?

Note that the oven itself is outside of the spatial domain of the free field describing the beam, and the Hamiltonian inside this domain is independent of time.

Thus the behavior of the source must be in a time-dependent term of the Hamiltonian at the left border $z=0$ of the domain where the beam is modeled.

The spot on the screen is given by integrating the current $\langle \vec{j}(t,\vec{x}) \cdot \vec{n} \rangle$ ($\vec{n}$ is the normal vector of the screen) over time.

By assumption, $\vec{n}$ is the unit vector along the $z$ axis. Thus the spot on the screen has intensity distribution
$$I(x_1,x_2)=\int_{t=0}^T\Big\langle \vec j_3(t,x_1,x_2, Z) \Big\rangle dt,$$
where $Z$ is the $z$ coordinate of the point where the beam meets the screen?

 

[Demystifier]

I did just this in my post #21.

It was not clear to me what in that post was not clear to you at the beginning. Now I guess - most of it.

 

[A. Neumaier]

It was not clear to me what in that post was not clear to you at the beginning. Now I guess - most of it.

Well, this seems to be nowhere in the literature - at least I haven't seen a discussion; the discussion is always in terms of an ensemble of single particles, without reference to quantum fields. I want to see how things look like a level deeper.

Thus I am thinking about how to model it on the level of quantum fields while the discussion evolves. Most things are evident once one thinks a little about it, but one needs to do it the right way to get something intuitive and manageable, while still be formally precise.

 

[vanhees71]

Yes and for this a single-particle-observable expectation value is not sufficient. For this you need a manyparticle quantity as I said before. That's described in quantum-optics textbooks.

 

[Fra]

Well, this seems to be nowhere in the literature - at least I haven't seen a discussion; the discussion is always in terms of an ensemble of single particles, without reference to quantum fields. I want to see how things look like a level deeper.

Thus I am thinking about how to model it on the level of quantum fields while the discussion evolves. Most things are evident once one thinks a little about it, but one needs to do it the right way to get something intuitive and manageable, while still be formally precise.

It seems to me, unless i totally misinterpret you, is that the problem lies in how making sense of a quantum field? Do we have a field of "local measurements" (indexed by classical spacetime), or do we have local measurement where the illusion of a field with an index is emergent somehow in some holographic sense?

I must admit that while perhaps obsolete, I like the formal idea of nth quantization, as it makes more sense for the latter possibility. It's how i learned to "understand" things, as an higher order of information processing withing the agent. After all, the way qft is conventionally introduced when going from QM, you really presume the existence of the classical spacetime (compliant with special relativity). And if you do, the field of local measurements makes sense. But that whole abstraction gets vary vague, if you do not presume the classical index. So it seems this get very complex without also motivating 4D spacetime (which takes us right into QG and other things??)

If this made no sense between you, ignore this post.

/Fredrik

 

[vanhees71]

Of course, QFT is not restricted to relativistic QFT but can be as well formulated for non-relativistic QM as well. In the cases, where you only deal with systems, where the conserved current has the meaning of a particle-number current, this QFT formulation of non-relativistic QM is equivalent to the first-qantization formulation since then you always stay in the subspace of the Fock space with the particle number that was fixed in the beginning.

The reason, why there is no viable 1st-quantization formulation for relativistic QT and interacting particles is that the conserved currents of relativistic wave equations, which follow from invariance under proper orthochronous Poincare transformations as the symmetry group of Minkowski space, provide no positive definite densities and thus are charge rather than particle-number densities. Further to ensure causality in terms of the microcausality constraint in connection with local realizations of the Poincare group you need both annihilation and creation operators in the mode expansion of the free fields, and in the interacting case not particle number but the charges are conserved, and thus you need the full Fock space to get a consistent description.

 

[A. Neumaier]

It seems to me, unless i totally misinterpret you, is that the problem lies in how making sense of a quantum field? Do we have a field of "local measurements" (indexed by classical spacetime), or do we have local measurement where the illusion of a field with an index is emergent somehow in some holographic sense?

We just have a collection of $N$-point functions for each Heisenberg state.

 

[A. Neumaier]

Yes

To what did you here agree?

and for this a single-particle-observable expectation value is not sufficient. For this you need a many particle quantity as I said before.

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

That's described in quantum-optics textbooks.

I also don't see how quantum optics is relevant for the nonrelativistic QFT modeling of a silver beam.

 

[Demystifier]

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

First, you want to describe the lab. Second, if a late particle does not exist at earlier times, then it's created from something and this something is also made of particles.

 

[A. Neumaier]

First, you want to describe the lab.

No, I only wanted to describe the beam in the lab, represented by a free field; see posts #3, #19, #21. (The Heisenberg state of the lab would not be the vacuum state and its fields would have to be interacting.)

Second, if a late particle does not exist at earlier times, then it's created from something and this something is also made of particles.

That something is the silver oven, which is outside the region modeled. Its influence can therefore be solely through the boundary conditions for the free field operators at the beginning of the beam. Specifying this boundary condition at all times (together with the free field equations) fully captures the dynamics of the beam alone.

 

[vanhees71]

To what did you here agree?

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

I also don't see how quantum optics is relevant for the nonrelativistic QFT modeling of a silver beam.

I agreed with the statement that usually one considers single-particle states and the corresponding probabilities in the standard description of the SGE. I don't know, what you mean by the statement that it's not described within QFT since of course it is described by QFT when it's necessary. For photons you find this in all details in standard textbooks on quantum optics, and not only for single-photon observations but also for two- and multi-photon observations. The latter usually is needed when dealing with entangled states of all kinds or for Hanbury-Brown twiss ("intensity interferometry") etc.

So it depends on which description you want. First you said you want to describe a situation, where every 1 second a silver atom is emitted from some source. This is described by a many-body state, i.e., for $N$ particles (to be "delivered" within $N$ seconds), i.e., an $N$-point correlation function. Which one depends on the detector. I'd say a good detector is measuring the charge/current density.

 

[A. Neumaier]

I don't know, what you mean by the statement that it's not described within QFT

I only meant that in the literature I haven't seen a discussion that treats the process as a long-time nonstationary process involving a sequence of several generates particles. In the present discussion I want to inquire about this nonstationary description in terms of QFT, which indeed exists, I believe.

First you said you want to describe a situation, where every 1 second a silver atom is emitted from some source.

Yes, that's the standing assumption.

This is described by a many-body state, i.e., for $N$ particles (to be "delivered" within $N$ seconds), i.e., an $N$-point correlation function.

Since we don't generate entangled particles and don't perform coincidence tests, the use of multiparticle correlation functions seems not warranted. For photons, the spatially integrated response rate to a single beam is proportional to the 1-point function of the energy density, which can be expressed in terms of two-point functions of the electromagnetic field. For silver, you had suggested instead the integrated 3-component of the silver current (which can be converted to an integral over silver density using the conservation of mass). Neither involves correlation functions.

Moreover, I thought we had agreed that the Heisenberg state is the vacuum state, and the dynamics is solely in the operators. Thus the information about how often particles are emitted must be in the incoming boundary conditions of the operators. How would you accommodate your many-body-state view in the operator boundary conditions?

 

[vanhees71]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second. Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle|\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$. These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

 

[A. Neumaier]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

Since silver is heavy, there is indeed no need for relativistic QFT.

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second.

This seems conceptual progress on our question, since now the dynamics is in the operators. But in the setting under discussion we do not have a magnetic field, just occasional silver particles. So what does have a peak every second?

Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N |\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$.

The typical lab experiment does not resolve the times and positions of each impacting silver particle but only checks the areas where a silver spot forms after some time; thus only an integrated version of this is needed and everything should reduce to single particle matrix elements. But let us consider the probabilities later. At the moment I am primarily interested in getting a correct description of the beam before it hits the screen.

These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

Agreed.

 

[vanhees71]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second. Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle|\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$. These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

Correction: This is of course nonsense, because it's zero ;-). Right is
$$P(t_1,\vec{x}_1;\ldots;t_2,\vec{x}_2)=\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle.$$
This can be written in terms of the $N$-point correlation function of the density $\hat{\phi}^{-}(t,\vec{x}) \hat{\phi}^{(+)}(t,\vec{x})$.

 

[vanhees71]

Since silver is heavy, there is indeed no need for relativistic QFT.This seems conceptual progress on our question, since now the dynamics is in the operators. But in the setting under discussion we do not have a magnetic field, just occasional silver particles. So what does have a peak every second?

The field operators. Without magnetic field you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

The typical lab experiment does not resolve the times and positions of each impacting silver particle but only checks the areas where a silver spot forms after some time; thus only an integrated version of this is needed and everything should reduce to single particle matrix elements. But let us consider the probabilities later. At the moment I am primarily interested in getting a correct description of the beam before it hits the screen.

Sure, that's much simpler. There you only need the current operator [EDIT: technical typo corrected]
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})
$$
changing the notation for the usual Schrödinger field operators, which are pure annihilation operators $\hat{\psi}$ and creation operators $\hat{\psi}^{\dagger}$ and as initial state an appropriate one-particle state. Then you integrate the expectation value over the time of observation to get the accumulated probability distribution on the screen.

 

[A. Neumaier]

The field operators. Without magnetic field you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

Ok, this looks more satisfying.

Sure, that's much simpler. There you only need the current operator
$$\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m} \hat{\psi}^{\dagger}(t,\vec{x}) \leftrightarrow{\nabla} \hat{\psi}(t,\vec{x})$$
changing the notation for the usual Schrödinger field operators, which are pure annihilation operators $\hat{\psi}$ and creation operators $\hat{\psi}^{\dagger}$ and as initial state an appropriate one-particle state. Then you integrate the expectation value over the time of observation to get the accumulated probability distribution on the screen.

Your latex formula doesn't display as it should since $\leftrightoverarrow$ is not recognized and a curly bracket is missing. Is my correction what you had intended? It doesn't quite make sense...

I'll think about your formulas to see what they mean in terms of my intuitive picture of the situation.

 

[Fra]

Of course, QFT is not restricted to relativistic QFT but can be as well formulated for non-relativistic QM as well. In the cases, where you only deal with systems, where the conserved current has the meaning of a particle-number current, this QFT formulation of non-relativistic QM is equivalent to the first-qantization formulation since then you always stay in the subspace of the Fock space with the particle number that was fixed in the beginning.

The reason, why there is no viable 1st-quantization formulation for relativistic QT and interacting particles is that the conserved currents of relativistic wave equations, which follow from invariance under proper orthochronous Poincare transformations as the symmetry group of Minkowski space, provide no positive definite densities and thus are charge rather than particle-number densities. Further to ensure causality in terms of the microcausality constraint in connection with local realizations of the Poincare group you need both annihilation and creation operators in the mode expansion of the free fields, and in the interacting case not particle number but the charges are conserved, and thus you need the full Fock space to get a consistent description.

My perspective was more the foundational one - ie the issues of interpretation and mapping elements of theory to elements of "reality" (which by I mean elements of the the observer/agent) is existing already in normal QM. But the problems become more accentuated when you consider a quantum field theory.

Suppose you have something that a copenhagen interpretation of QM, what is the copenhagen interpretation of QFT?

I think a bit like this: If we think the quantum state, represents the "state of the knowledge" the observer has about a "particle", then the fock space corresponds to the "state of the knowledge" the observer has about the "field", but in "second quantization" the "field" is not a classical field, but reprents just one step in an induction chain - it's a higher level knowledge about the lower level knowlwedge; ie the observer starts (as part of internal processes, which are physical of course) to "reflect" over it's own information, noting that say probabilities are not conserved, and he lower level of logical processing seems to not be viable, so a higher level construction is to consider the "probability" for the "probability following from a simplere inference system" combined with a changed of dependent variables, which can be conceptually thought of as a change of coding/representation, into something more efficient (that is say easier to truncate). Change of variables from KG -> Dirac equation is I think an "example".

Once you get the point of this, there is notthing to stop you also from a third quantisation and thus n'th quantization; each level represents more complex inference system withint the agent, affecting also it's interaction properties. At with n does this stop? - possible when the additional computational complexity of higher orders stalls the agent, more than it helps it?

The conventional method that you get taught isn't like this, it's more

Often, the first step in introducing QFT is "trying" to just plug operators into the klein gordon equation in order to illustrate that, we get the strange negative energy solutions and conservation of probability just goes down the drain - in short, it does not seem to make any sense. So we make say make the ansatz of the linear dirac equation, "relabel" some of the "particle states" as one beeing the anti-particle etc.

But when doing this change of "dependent variable", even mixing it up with the prior spacetime dynamics - what is happening to the "interpretation" of the information of the agent?

I think the conventional method is to just conclude that the 1st quantization doesn't make sense, and neither does the above, so instead one just arbitrarily (by ansatz) introduces more dependent variables, whose "interpretation" to the prior non-viable level is ignored.

One can also just say that the "ansatz" of the original observer was "wrong". ie the observer was convinved it was a one-particle system, but as the predictions based one that ended up non-correctable by changing the intial conditions, he was "wrong" as it was a multi-particle system.

This disturbs me alot as it avoids the problems. I think one can not questions the observers "best knowledge" unless it's of the "ignorance/Bell type" the incompleteness that is due to that is seems physically impossible for an agent to infer with certainty, say initial conditions and the effective laws, will unavoidable I think give predictable consequences.

We just have a collection of $N$-point functions for each Heisenberg state.

If by "we" here mean the "observer/agent", the question I ask is: How did the agent infer the baggage implied here, without adding in "external information" (which at least is what I strive to avoid, but which is of course very difficult it not impossible).

Or if we by "we" instead mean the collection of all "classical" measurementdevices, distributed throughout the universer that can build the records post-aquistion, then that line of thinking deviates from the path I try to stay on

/Fredirk

 

[A. Neumaier]

If by "we" here mean the "observer/agent", the question I ask is: How did the agent infer the baggage implied here, without adding in "external information"

"we" is the user of QFT.

We use a lot of external information from books, journals, discussions, and experiments to create the model from which to calculate the N-point functions.

For example, in this thread we try to figure it out for a simple silver beam on the way from the source to the screen.

 

[vanhees71]

Ok, this looks more satisfying.

Your latex formula doesn't display as it should since $\leftrightoverarrow$ is not recognized and a curly bracket is missing. Is my correction what you had intended? It doesn't quite make sense...

I'll think about your formulas to see what they mean in terms of my intuitive picture of the situation.

I corrected it:
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})$$
It's a usual symbol in QFT it means
$$\hat{\psi}^{\dagger}\overleftrightarrow{\nabla} \hat{\psi} = \hat{\psi}^{\dagger} \nabla \hat{\psi} -(\nabla \hat{\psi}^{\dagger}) \psi.$$
It's the standard probability current of the Schrödinger equation. From the Noether symmetry under multiplication with space-time independent phase factors it follows that
$$\partial_t \rho +\nabla \cdot \vec{j}=0, \quad \rho =\psi^{\dagger} \psi.$$
I find it much more intersting to include the magnet in the analysis. It's one of the few examples, where you can solve the Schrödinger equation quasi analytically, treating the spin-flipping part of the magnetic field (which must exist because of $\vec{\nabla} \cdot \vec{B}=0$) as a perturbation. Then you can put the screen at a place before the magnet, where you have a single beam of unpolarized Ag atoms (provided you initialize the state in this way) or behind it, where you have two beams with the position entangled with the spin component in direction of the magnetic field.

 

[A. Neumaier]

I corrected it:
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})$$
It's a usual symbol in QFT it means
$$\hat{\psi}^{\dagger}\overleftrightarrow{\nabla} \hat{\psi} = \hat{\psi}^{\dagger} \nabla \hat{\psi} -(\nabla \hat{\psi}^{\dagger}) \psi.$$
It's the standard probability current of the Schrödinger equation. From the Noether symmetry under multiplication with space-time independent phase factors it follows that
$$\partial_t \rho +\nabla \cdot \vec{j}=0, \quad \rho =\psi^{\dagger} \psi.$$

OK. That's what I thought; I just didn't know how to write it in latex.

I find it much more interesting to include the magnet in the analysis.

We'll come to that, but first I want to be sure that we agree on the simpler setting.

you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

Please give details. I couldn't find a Gaussian solution of the free Schrödinger equation.

 

[vanhees71]

Here are some details, but not in QFT language but as a simple one-particle treatment with wave functions (in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/quant.pdf

The Gaussian wave packet for free particles is in Sect. 1.2. A simplified treatment of the idealized SGE is in Sect. 6.12. Here the "spin-flipping" is neglected. I guess, one could treat it in time-dependent perturbation theory or refer to the numerical solutions by Potel et al:

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.71.052106