I How Does Quantum Mechanics Relate to Quantum Field Theory in Particle Physics?

[vanhees71]

Only in a very vague sense.

Nonrelativistic QFT is usually taken to be the statistical mechanics of gases, liquids, and solids made of nuclei and electrons, with electromagnetic interaction modeled as external field only. (To handle photons needs at least a partially relativistic setting.) As such it shares the abstract features of relativistic QFT, except that it takes the limit $c\to\infty$ to simplify the dynamics. However, the fields appearing in nonrelativistic QFT (the electron field and one interacting spacetime field for every nuclide appearing in the model - or an external periodic potential if none is modeled) are very different from those appearing in relativistic QFT (one space-time field for every elementary particle). I haven't seen any derivation of the former from the latter. There is a chain of reasoning going from quarks to hadrons to nuclides, considered as asymptotic free fields, but as far as I have seen none that would allow me to say that interacting nonrelativistic QFT is derivable from the relativistic version. It is regarded as an effective theory for the latter, but notebecause of a derivation but based on plausibility reasoning only.

No; I completely disagree!

It only works in the opposite direction, presented in all textbooks on statistical mechanics, by a two-step process of generalization and abstraction during which some features of QM are lost. First one generalizes the setting of QM by turning the number of particles - which in QM is a parameter only - into an operator acting on Fock space whose spectrum are the nonnegative numbers. Then one gets rid of all measurement issues by replacing the Born rule by the definition of ensemble expectations via $\langle X\rangle:=\mbox{tr} ~\rho X$, which no longer refers to observation and measurement. This allows one to consider arbitrarily large systems - which constitutes the second generalization - and the thermodynamic limit of infinite volume (which is needed to make it a QFT proper). Then one is at the level of field expectations and field correlations, which are the subject of QFT. Note that the notions of observation and measurement - the most controversial features of QM - are lost during this abstraction process.

Because of this loss, one can go only part of the way back if one tries to reverse the direction, going from nonrelativistic QFT to QM. One can consider a fixed number of particles and restrict to the eigenspace of the number operator with fixed eigenvalue $N$. This produces (restricting for simplicity to a single scalar field) the Hilbert space of totally symmetrized wave functions in $N$ 3-dimensional position coordinates $x_i$. On this Hilbert space, only those operators (constructed from the field operators in Fock space) have a meaning that commute with the number operator. This is not enough to construct position and momentum operators for the individual particles but only for their center of mass. One sees already here that one needs to make additional assumptions to recover traditional quantum mechanics.

Worse, since in the QFT description both observers and measurements are absent, one has to introduce observers and measurements and their properties by hand! In particular, the Born rule of QM, that tells what happens in a sequence of ideal measurements, must be postulated in addition to what was inherited from QFT! Unless the concept of observers and measurement are fully defined in quantum mechanical terms so that one could deduce their properties. While this seems not impossible, it certainly hasn't been done so far!

In non-relativistic QFT you often have the special case that a particle number is conserved. Then, in the microcanonical ensemble, you can entirely work in the eigenspace of the total-number operator, and then everything can be formulated in the "first-quantization formulation" aka. wave mechanics. In this special case QM and nrel. QFT are fully equivalent. However, even in the non-relativistic many-body theory the QFT formulation is more flexible and you can work with quasiparticles (phonons, plasmons and various other, partially pretty exotic, quasiparticles come to mind) whose number is not conserved, and this is the usual way condensed-matter problems are (approximately) solved with great success.

I'm, however, not aware of any systematic treatment of the non-relativistic limit of relativistic many-body QFT.

 

[meopemuk]

You can calculate a lot at finite time; the problem is the proper interpretation of the results. All this gets indeed worst when massless particles are involved.

Could you please mention some examples of finite-time calculations in a relativistic renormalized QFT, such as QED? Are they comparable with experiments?

 

[Demystifier]

I have heard that it can not have position presentation for photon.But photon is experimental point particle,why there is not a probability notion for photon(there is not wave function for photon)?

Even without a position operator one can introduce a POVM for a photon-position measurement.

 

[A. Neumaier]

to avoid IR and UV divergences of QFT, it is almost unavoidable to break Lorentz invariance in one way or another.

Only during the intermediate calculations. In the final renormalized result expressed in terms of cohrent states, where the continuum limit (for UV) and the infinite volume limit (for IR) are taken, Lorentz invariance of the S-matrix is restored exactly.

if such an idealization leads to a physically unacceptable result, it is theoretical physicists who will first give up of such an idealization.

But the idealization of Lorentz symmetry didn't lead to physically unacceptable results. On the contrary, it is verified to extreme accuracy and assumed to be valid by almost all physicists working on the smallest and the largest scales. Enen when giving up continuous spacetime one doesn't give up Lorentz symmetry! If we did, we wouldn't have any guidance left for restricting the possibilities...

is it justified to claim that relativistic QFT is more fundamental than non-relativistic QM? If yes, then how would you justify it?

it is more fundamental since at the level of unitary evolution (i.e., for isolated systems where external measurements are impossible due to lack of interaction), quantum mechanics is clearly visible to be a low energy approximation of QFT.

The differences only show in the treatment of measurement. QFT is silent about measurement and only talks about mean fields and correlations, whereas QM makes additional assumptions that allow the analysis of particles with macroscopic devices at finite times. These additional assumptions are in conflict with unitary evolution, which is considered acceptable because of the unavoidable interaction wih the measurement device. Again QFT is more fundamental since it is conceptually more parsimonious and does not require (and strictly speaking not even allow) an exernal classical world.

It is very likely that some time in the future people will be able to show in which way the additional assumptions of QM can be fully justified from QFT, by modeling the system of few particles + detector as a pure QFT system. The current trend (decoherence theory) treats it instead as a pure quantum mechanical system, with the unavoidable result that it can only shift the Heisenberg cut between system and observer to a different location. Since the measurement postulate is built in directly into the foundations, it is impossible to resolve the measurement riddle within QM! This is the deepest root of the interpretation problem in quantum mechanics. It cannot go away unless QM is understood as an approximation to a theory whose axioms are independent of measurement. QFT (with the Wightman axioms) is such a theory, and I predict that some time in the future, it will solve the measurement problem in a satisfactory way.

 

[Demystifier]

it is more fundamental since at the level of unitary evolution (i.e., for isolated systems where external measurements are impossible due to lack of interaction), quantum mechanics is clearly visible to be a low energy approximation of QFT.

OK, we can agree on that.

The differences only show in the treatment of measurement. QFT is silent about measurement and only talks about mean fields and correlations, whereas QM makes additional assumptions that allow the analysis of particles with macroscopic devices at finite times. These additional assumptions are in conflict with unitary evolution, which is considered acceptable because of the unavoidable interaction wih the measurement device. Again QFT is more fundamental since it is conceptually more parsimonious and does not require (and strictly speaking not even allow) an exernal classical world.

It is very likely that some time in the future people will be able to show in which way the additional assumptions of QM can be fully justified from QFT, by modeling the system of few particles + detector as a pure QFT system. The current trend (decoherence theory) treats it instead as a pure quantum mechanical system, with the unavoidable result that it can only shift the Heisenberg cut between system and observer to a different location. Since the measurement postulate is built in directly into the foundations, it is impossible to resolve the measurement riddle within QM! This is the deepest root of the interpretation problem in quantum mechanics. It cannot go away unless QM is understood as an approximation to a theory whose axioms are independent of measurement. QFT (with the Wightman axioms) is such a theory, and I predict that some time in the future, it will solve the measurement problem in a satisfactory way.

But we cannot agree on that. It is true that QFT books usually don't talk about measurement axioms, but I think it's only because the writers of these books don't want to repeat what has already been said in books on QM. If QFT could offer some new insight on the measurement problem, writers of QFT books would not miss the opportunity to say something about it.

 

[fxdung]

So what we must account for statistical ensemble for QFT?We only make into acount on the particles(quanta) or also on the quantum states of every particles?I think quantum statistical mechanics account both.

 

[A. Neumaier]

So what we must account for statistical ensemble for QFT? We only make into acount on the particles(quanta) or also on the quantum states of every particles?I think quantum statistical mechanics account both.

In relativistic QFT one talks about the state of a system extending over all spacetime. The (canonical, grand canonical, etc.) ensemble is the label attached in statistical mechanics to particular macroscopic states of the form $e^{-S/\hbar}$ with a nice expression for $S$. Thus in QFT, the notion of an ensemble can be taken as a synonym for the state of the macroscopic system. Nowhere in statistical mechanics (and hence in QFT) is made use of the assumption that an ensemble is interpreted in the sense of a collection of many identically prepared macroscopic objects - it is only an ensemble of many microscopic particles! Therefore the predictions of QFT apply to each single macroscopic object. Already Gibbs, who introduced the notion of an ensemble towards the end of the 19th century, noted (in his 1901 book - still quite readable!) that one must consider a statistical mechanics ensemble as a fictitious collection of copies of which only one is realized!

 

[A. Neumaier]

It is true that QFT books usually don't talk about measurement axioms, but I think it's only because the writers of these books don't want to repeat what has already been said in books on QM. If QFT could offer some new insight on the measurement problem, writers of QFT books would not miss the opportunity to say something about it.

Indeed they say something new [compared to QM foundations, but very old in terms of the physics] on measurement, and they don't miss the opportunity to say it!

In books on nonequilibrium statistical mechanics it is very obvious that whatever they compare with experiment has nothing at all to do with the kind of idealized measurements considered in QM. They talk about field expectations (such as the energy density and mass density) and certain coefficients in the formula for the state of the macroscopic system (such as local temperature and local chemical potential), and relate them to thermodynamic observables, which are measured in the ordinary engineering way. If mentioned at all, Born's rule with its assumption of external measurement is eliminated in the very first few pages of the books in favor of the formula $\langle X\rangle = \mbox{tr}~\rho X$. This formula is a much more general and much more useful axiom for quantum physics! It doesn't have the problematic baggage that the traditional, ill-defined foundations of QM have.

 

[Demystifier]

Indeed they say something new [compared to QM foundations, but very old in terms of the physics] on measurement, and they don't miss the opportunity to say it!

In books on nonequilibrium statistical mechanics it is very obvious that whatever they compare with experiment has nothing at all to do with the kind of idealized measurements considered in QM. They talk about field expectations (such as the energy density and mass density) and certain coefficients in the formula for the state of the macroscopic system (such as local temperature and local chemical potential), and relate them to thermodynamic observables, which are measured in the ordinary engineering way. If mentioned at all, Born's rule with its assumption of external measurement is eliminated in the very first few pages of the books in favor of the formula $\langle X\rangle = \mbox{tr}~\rho X$. This formula is a much more general and much more useful axiom for quantum physics! It doesn't have the problematic baggage that the traditional, ill-defined foundations of QM have.

It seems to imply that quantum statistical mechanics is based on QFT, not on QM.

But one can certainly study quantum statistical mechanics based on QM, without using QFT. (If not for photons, one can certainly do that for non-relativistic electrons.) So can quantum statistical mechanics based on QM tell us something about measurement which "pure" QM can't?

 

[A. Neumaier]

It seems to imply that quantum statistical mechanics is based on QFT, not on QM.

But one can certainly study quantum statistical mechanics based on QM, without using QFT. (If not for photons, one can certainly do that for non-relativistic electrons.) So can quantum statistical mechanics based on QM tell us something about measurement which "pure" QM can't?

Well, quantum statistical mechanics for macroscopic objects is always based on expectations and correlations only, which is the QFT setting. Even though one starts with the QM1 setting - since this is already known, by the way physics education happens everywhere -, one drops the connection to the QM foundations once the QFT fondations are established. Thus the former serve only as a motivation.

However there is a mix of quantum mechanical and field theoretic reasoning in some treatments of the measurement process. They treat the environment as a macroscopic system - typically heavily idealized as an infinite size heat bath -, and then treat system + detector + environment by statistical mechanics. See, e.g., the Lectures on dynamical models for quantum measurements by Nieuwenhuizen, Perarnau-Llobet, and Balian. It is this line of reasoning that ultimately should solve the measurement problem.

 

[Demystifier]

Well, quantum statistical mechanics for macroscopic objects is always based on expectations and correlations only, which is the QFT setting.

It is not only a QFT setting. Even in QM you have correlation functions such as $\langle x(t_1) x(t_2)\rangle$.

 

[fxdung]

What is the difference between the predictions of quantum statistical mechanics and of classical physics?Is classical physics is the result of classical statistical mechanics?

 

[A. Neumaier]

It is not only a QFT setting. Even in QM you have correlation functions such as $\langle x(t_1) x(t_2)\rangle$.

In the Schroedinger picture, which is the basis of the usual axiomatization of QM, this object doesn't exist.

Time correlations only exists after casting QM in the form of a 1+0-dimensional QFT (the Heisenberg picture), where state vectors do not evolve in time and therefore the Born rule no longer applies.

 

[A. Neumaier]

What is the difference between the predictions of quantum statistical mechanics and of classical physics?Is classical physics is the result of classical statistical mechanics?

The quantum predictions difeerr form the classical predictions by corrections of order $\hbar$. Since this is a very small quantity in macroscopic units, the corrections are negligible in macroscopic cases that can both be described classically and quantum mechanically.
In particula, one gets classical hydrodynamics and elasticity theory as macroscopic limits of classical or quantum statistical mechanics applied to the appropriate conditions, without or with the quantum corrections, respectively.

 

[Demystifier]

In the Schroedinger picture, which is the basis of the usual axiomatization of QM, this object doesn't exist.

Time correlations only exists after casting QM in the form of a 1+0-dimensional QFT (the Heisenberg picture), where state vectors do not evolve in time and therefore the Born rule no longer applies.

Then we only disagree on terminology. If by "QFT" you really mean Heisenberg picture and by "QM" you really mean Schrodinger picture, then I can agree with you. But I would prefer to use the standard terminology. Besides, did you know that QFT can be formulated in the Schrodinger picture?

 

[fxdung]

It may be that QM and classical physics applied to macro objects have the same results,but classical and quantum statistical mechanics give different results because the systems is agregate of quantum and classical particles(where two different physics applied to micro particles).So I do not know classical physics is results of what.

 

[A. Neumaier]

[

Then we only disagree on terminology. If by "QFT" you really mean Heisenberg picture and by "QM" you really mean Schrodinger picture, then I can agree with you. But I would prefer to use the standard terminology.

For the purposes of foundations, I call QFT that part of quantum theory where only expectations and correlation functions are asserted to have meaning related to experiment, and QM that part of quantum theory where the Schroedinger equation is used and Born's rule relates it to experiments. This naturally divides quantum physics in two nearly disjoint parts with completely different ontologies.

Besides, did you know that QFT can be formulated in the Schrodinger picture?

Well, there is the so-called functional Schroedinger equation, which is occasionally useful. But it is an amputation rather than a formulation of QFT since one loses in the process not only manifest covariance but also all time-correlation information. But covariance (in the relativistic case) and correlation functions (in general) are the bread and butter of most QFT applications.

 

[Demystifier]

For the purposes of foundations, I call QFT that part of quantum theory where only expectations and correlation functions are asserted to have meaning related to experiment, and QM that part of quantum theory where the Schroedinger equation is used and Born's rule relates it to experiments.

Then you should have said that at the beginning, to avoid all the misunderstandings that this non-standard terminology caused.

 

[vanhees71]

Could you please mention some examples of finite-time calculations in a relativistic renormalized QFT, such as QED? Are they comparable with experiments?

For a pretty academic example, see

http://arxiv.org/abs/1208.6565

There we were modest and came to the conclusion that one has to define quantities carefully using the idea of asymptotic states.

 

[vanhees71]

[
For the purposes of foundations, I call QFT that part of quantum theory where only expectations and correlation functions are asserted to have meaning related to experiment, and QM that part of quantum theory where the Schroedinger equation is used and Born's rule relates it to experiments. This naturally divides quantum physics in two nearly disjoint parts with completely different ontologies.

Well, there is the so-called functional Schroedinger equation, which is occasionally useful. But it is an amputation rather than a formulation of QFT since one loses in the process not only manifest covariance but also all time-correlation information. But covariance (in the relativistic case) and correlation functions (in general) are the bread and butter of most QFT applications.

That's a bit strange a view. Usually you distinguish nonrelativistic QT in the "first-quantization" and the "second-quantization formalism". The former describes systems of a fixed number of particles and can be formulated as wave mechanics, realizing the Hilbert space as $L^2(\mathbb{R}^{3N},\mathbb{C}^{2s+1})$ for $N$ particles of spin $s$ and the latter describes any many-body system of particles and/or quasiparticles be their number conserved or not. The 2nd-quantization formalism is fully equivalent for the 1st-quantization formalism if particle number is conserved and you deal with states of a fixed particle number.

Also there is no difference between the Schrödinger and the Heisenberg picture (at least not as far as I'm aware of, because I've not heard about problems like with the interaction picture in the case of relativistic QFT, where the latter strictly speaking doesn't exist due to Haag's theorem). It's just two equivalent mathematical descriptions of the same theory. They are just related by a unitary time-dependent transformation, and observables (including correlation functions of gauge invariant observables) thus do not depend on which picture you use to evaluate them.

In all cases the Born rule is used to associate formal quantities with real-world observables.

 

[Demystifier]

Also there is no difference between the Schrödinger and the Heisenberg picture (at least not as far as I'm aware of, because I've not heard about problems like with the interaction picture in the case of relativistic QFT, where the latter strictly speaking doesn't exist due to Haag's theorem). It's just two equivalent mathematical descriptions of the same theory. They are just related by a unitary time-dependent transformation, and observables (including correlation functions of gauge invariant observables) thus do not depend on which picture you use to evaluate them.

As long as you only study unitary evolution of matrix elements, there is no difference between the Schrödinger and the Heisenberg picture. But when you attempt to say something more specific about the measurement problem, then, depending on the interpretation you use, some subtle differences between the two pictures may occur.

 

[stevendaryl]

Indeed they say something new [compared to QM foundations, but very old in terms of the physics] on measurement, and they don't miss the opportunity to say it!

In books on nonequilibrium statistical mechanics it is very obvious that whatever they compare with experiment has nothing at all to do with the kind of idealized measurements considered in QM. They talk about field expectations (such as the energy density and mass density) and certain coefficients in the formula for the state of the macroscopic system (such as local temperature and local chemical potential), and relate them to thermodynamic observables, which are measured in the ordinary engineering way. If mentioned at all, Born's rule with its assumption of external measurement is eliminated in the very first few pages of the books in favor of the formula $\langle X\rangle = \mbox{tr}~\rho X$. This formula is a much more general and much more useful axiom for quantum physics! It doesn't have the problematic baggage that the traditional, ill-defined foundations of QM have.

I don't see how the treatment of measurement by QFT is any different, conceptually, than the treatment in QM. In QFT, the field is an operator, and we can get a statistical interpretation by considering expectation values of the field. How is that different (conceptually) from saying, in non-relativistic QM, that position \hat{x} is an operator, and we can get a statistical interpretation by considering expectation values of x? That's fine as far as it goes, but in NRQM, there are other operators, as well, such as \hat{p} and various combinations of \hat{p} and \hat{x}. We can't simultaneously give a statistical interpretation to all such operators (that would violate the Kochen-Specker theorem), so we have to limit our statistical interpretation to those variables that are actually measured in an experiment. That's how measurement comes in.

I don't see how the situation is any better in QFT. We similarly have incompatible field operators (in scalar field theory, \hat{\phi} and the conjugate field momentum \hat{\pi}, for example).

 

[stevendaryl]

In the Schroedinger picture, which is the basis of the usual axiomatization of QM, this object doesn't exist.

Such quantities as \langle A(t_1) A(t_2) \rangle can be computed in the Schrodinger picture: It's just

\langle \psi| e^{i H t_1} A e^{i H (t_2 - t_1)} A e^{-i H t_2}|\psi\rangle

 

[stevendaryl]

As long as you only study unitary evolution of matrix elements, there is no difference between the Schrödinger and the Heisenberg picture. But when you attempt to say something more specific about the measurement problem, then, depending on the interpretation you use, some subtle differences between the two pictures may occur.

There is something especially nice about the Heisenberg picture in relativistic quantum field theory, though. In the Schrodinger picture, the state of the universe is described by a wave function, which is an amplitude function on configuration space (configuration of fields), rather than a function in physical 4-D spacetime. So it's hard to understand what it would even mean for QFT to be "local", since the states don't exist in the physical world. In the Heisenberg picture, however, the equations of motion describe the field operators, which are (or can be, if you choose a position basis) localized operators existing in each point in space. They evolve in a purely local way, affected only by other operators in the neighborhood. So it's clear that the field operators are local. There is still a wave function, or state, in the Heisenberg picture, and it's as nonlocal (or "a-local"--the word "local" doesn't even apply to it) as the wave function in the Schrodinger picture. But the state in the Heisenberg picture is constant. It doesn't evolve. So who cares whether it's local or not?

 

[Demystifier]

Such quantities as \langle A(t_1) A(t_2) \rangle can be computed in the Schrodinger picture: It's just

\langle \psi| e^{i H t_1} A e^{i H (t_2 - t_1)} A e^{-i H t_2}|\psi\rangle

I think he might want to give an ontological status to either $A(t)$ or $|\psi(t)\rangle$, but not to both. From such an ontological point of view, which may be relevant in the context of measurement problem, the two pictures are not equivalent.

 

[atyy]

There is something especially nice about the Heisenberg picture in relativistic quantum field theory, though. In the Schrodinger picture, the state of the universe is described by a wave function, which is an amplitude function on configuration space (configuration of fields), rather than a function in physical 4-D spacetime. So it's hard to understand what it would even mean for QFT to be "local", since the states don't exist in the physical world. In the Heisenberg picture, however, the equations of motion describe the field operators, which are (or can be, if you choose a position basis) localized operators existing in each point in space. They evolve in a purely local way, affected only by other operators in the neighborhood. So it's clear that the field operators are local. There is still a wave function, or state, in the Heisenberg picture, and it's as nonlocal (or "a-local"--the word "local" doesn't even apply to it) as the wave function in the Schrodinger picture. But the state in the Heisenberg picture is constant. It doesn't evolve. So who cares whether it's local or not?

Don't we still have to collapse the operators when a measurement is done?

 

[stevendaryl]

Don't we still have to collapse the operators when a measurement is done?

Not if we go the Many-Worlds route.

 

[atyy]

Not if we go the Many-Worlds route.

I don't think that's possible in the Heisenberg picture.

 

[Demystifier]

I don't think that's possible in the Heisenberg picture.

Good point! More generally, in the realm of interpretations the choice of the picture matters a lot.

 

[stevendaryl]

I don't think that's possible in the Heisenberg picture.

Why not? I haven't actually tried to develop a Many Worlds theory in the Heisenberg picture (I'm not 100% sure I understand it in the Schrodinger picture, either), but it seems to me that you could take the wave function in Many-Worlds and distribute the information to the field operators in a Heisenberg picture, which would give an equivalent description of the same state.

 

[A. Neumaier]

Then you should have said that at the beginning, to avoid all the misunderstandings that this non-standard terminology caused.

I couldn't do this in the beginning, as I found it out only during the discussion. I wouldn't spend so much time in discussing these things here if everything were already crystal clear in my mind. It is being clarified through the attempt to express myself clearly and seeing through the responses how well I succeeded.

 

[A. Neumaier]

In QFT, the field is an operator, and we can get a statistical interpretation by considering expectation values of the field. How is that different (conceptually) from saying, in non-relativistic QM, that position ^xx^\hat{x} is an operator, and we can get a statistical interpretation by considering expectation values of xxx?

It would not be so different if one would consider the expectation of operators in QM as something measurable to a certain accuracy, which is how the field expectations are interpreted in statistical mechanics. But in the QM foundations, measuring is something completely different! There one measures by collapse to an eigenstate (or its statistical version), which is completely foreign to measurement in statistical mechanics.

This is why the formal structure of Hilbert spaces and states is similar in QFT and QM, but the ontology is not.

 

[A. Neumaier]

In all cases the Born rule is used to associate formal quantities with real-world observables.

In all cases?

Then please explain for the following two explicit examples, the first from relativistic QFT, the second from nonrelativistic statistical mechanics:

• (i) How is the Born rule used to associate poles of the renormalized propagators with observable masses?
• (ii) How is the Born rule used in case of a real-world observation of temperature of a bucket of water?

 

[atyy]

Why not? I haven't actually tried to develop a Many Worlds theory in the Heisenberg picture (I'm not 100% sure I understand it in the Schrodinger picture, either), but it seems to me that you could take the wave function in Many-Worlds and distribute the information to the field operators in a Heisenberg picture, which would give an equivalent description of the same state.

I'm not sure MWI in Schroedinger works either. In the Heisenberg picture, one would have all possible observables evolving in time, including the simultaneous evolution of non-commuting observables. In MWI one has to (roughly) pick a preferred basis, and then let those branch. Picking a preferred basis in the Heisenberg picture would be like choosing a set of commuting observables. Since in a number versions of MWI decoherence picks the preferred basis, maybe we can avoid the difficulties of MWI by trying to discuss:

Can decoherence be formulated in the Heisenberg picture?

 

[A. Neumaier]

Such quantities as \langle A(t_1) A(t_2) \rangle can be computed in the Schrodinger picture: It's just

\langle \psi| e^{i H t_1} A e^{i H (t_2 - t_1)} A e^{-i H t_2}|\psi\rangle

This is just the Heisenberg picture.

In terms of the Schroedinger picture this is a meaningless mess, evaluated for the state at time $t=0$. Given only the conventional axioms of QM, one can dream up this expression. But one cannot interpret it as something related to measurements at times $t_1$ and $t_2$ without silently leaving the interpretation framework defined by the axioms.

 

[stevendaryl]

I'm not sure MWI in Schroedinger works either. In the Heisenberg picture, one would have all possible observables evolving in time, including the simultaneous evolution of non-commuting observables. In MWI one has to (roughly) pick a preferred basis, and then let those branch.

I always thought that the description of MWI as different "branches" is just a subjective interpretation. In MWI, there is just the universal wave function, evolving smoothly, and we are free to think of it as a superposition of "possible worlds", but that's not inherent.

 

[A. Neumaier]

Can decoherence be formulated in the Heisenberg picture?

I think you should do this in an independent threat, not to overload this one.

 

[atyy]

I always thought that the description of MWI as different "branches" is just a subjective interpretation. In MWI, there is just the universal wave function, evolving smoothly, and we are free to think of it as a superposition of "possible worlds", but that's not inherent.

Yes, or at least that's my understanding of Wallace's approach. That's why I put in "roughly" in my statements. That was just the motivation for getting to rephrasing the question in a more technical way:

Can decoherence be formulated in the Heisenberg picture?

 

[atyy]

I think you should do this in an independent threat, not to overload this one.

Yes, will do that.

 

[vanhees71]

As long as you only study unitary evolution of matrix elements, there is no difference between the Schrödinger and the Heisenberg picture. But when you attempt to say something more specific about the measurement problem, then, depending on the interpretation you use, some subtle differences between the two pictures may occur.

How can that be? The different pictures are just equivalent mathematical formulations of the QT formalism. How can the physical interpretation be different for the very same theory in different mathematical formulations?

 

[vanhees71]

In all cases?

Then please explain for the following two explicit examples, the first from relativistic QFT, the second from nonrelativistic statistical mechanics:

• (i) How is the Born rule used to associate poles of the renormalized propagators with observable masses?
• (ii) How is the Born rule used in case of a real-world observation of temperature of a bucket of water?

Ad (i). The definition of masses as poles of the propagators is derived from unitarity of the S-matrix. The S-matrix is defined as transition-probability amplitudes from the asymptotic into the asymptotic out states. The probabilities are evaluated via Born's rule.

Ad (ii). Temperature is not an observable in the quantum-theoretical sense. You measure a temperature by putting a thermometer in thermal contact with the heatbath whose temperature you want to measure (more precisely for the relativistic case comoving with the corresponding "fluid cell"). The temperature is a "coarse-grained macroscopic quantity" making sense as an average of some macroscopic quantity (e.g., the average energy density of an ideal gas).

 

[Demystifier]

How can that be? The different pictures are just equivalent mathematical formulations of the QT formalism. How can the physical interpretation be different for the very same theory in different mathematical formulations?

I was not sufficiently precise. What I meant is that in some interpretation only one of the pictures may be appropriate. For example, in the many-world interpretation only the Schrodinger picture is appropriate.

 

[A. Neumaier]

http://arxiv.org/abs/1208.6565

I couldn't find out where in this paper you are using the Born rule to associate your formal quantities with real-world observables (photoproduction). while you had claimed in your post #70 that this is always the case. Instead I noticed that you use a number operator expectation for evaluating photon number in (41), and you used pair correlators in (49), in accordance with what I had claimed is typical for QFT.

 

[A. Neumaier]

Ad (i). The definition of masses as poles of the propagators is derived from unitarity of the S-matrix. The S-matrix is defined as transition-probability amplitudes from the asymptotic into the asymptotic out states. The probabilities are evaluated via Born's rule.

Ad (ii). Temperature is not an observable in the quantum-theoretical sense. You measure a temperature by putting a thermometer in thermal contact with the heatbath whose temperature you want to measure (more precisely for the relativistic case comoving with the corresponding "fluid cell"). The temperature is a "coarse-grained macroscopic quantity" making sense as an average of some macroscopic quantity (e.g., the average energy density of an ideal gas).

ad (i) The unitarity of the S-matrix is independent of the Born rule and suffices for interpreting masses. That one can interpret the S-matrix elements in terms of the Born rule doesn't contribute anything to this interpretation.
ad (ii) Here your explanation uses expectations but not the Born rule.

Thus nothing is left in your explanation that needs the Born rule.

 

[A. Neumaier]

The different pictures are just equivalent mathematical formulations of the QT formalism.

They are not equivalent. The Heisenberg picture is far more general, as it allows to discuss time correlations. The Schroedinger picture addresses only single-time dynamics.

 

[dextercioby]

The S-matrix comes from requiring that time-evolution is unitary, i.e. it conserves probability. Since Born rule is the existence of a probabilistic interpretation of QM mathematics, it follows that the asking the S-matrix to be unitary cannot be independent from the Born rule, it's a consequence of it.

 

[Demystifier]

I couldn't find out where in this paper you are using the Born rule to associate your formal quantities with real-world observables (photoproduction). while you had claimed in your post #70 that this is always the case. Instead I noticed that you use a number operator expectation for evaluating photon number in (41), and you used pair correlators in (49), in accordance with what I had claimed is typical for QFT.

If $A$ is a projector, then the probability is $P(A)={\rm Tr} \rho A$. In standard terminology this is the Born rule. In this form the Born rule does not depend on the picture (Schrodinger, Heisenberg) or type of theory (QM, QFT, quantum gravity, string theory).

 

[Demystifier]

The Schroedinger picture addresses only single-time dynamics.

There is a generalization of the single-time Schrodinger picture to a many-time Schrodinger picture. See e.g.
http://lanl.arxiv.org/pdf/0912.1938
and Refs. [15, 16, 17, 18, 19] therein.

 

[A. Neumaier]

There is a generalization of the single-time Schrodinger picture to a many-time Schrodinger picture. See e.g.
http://lanl.arxiv.org/pdf/0912.1938
and Refs. [15, 16, 17, 18, 19] therein.

One can generalize everything to weaken arguments aimed at the ungeneralized version. The conventional axioms of QM say how the state of a system changes through a perfect measurement. [See, e.g., Messiah I, end of Section 8.1, or Landau & Lifschitz, Vol. III, Chapter I, Par. 7.] This is a context that makes sense only in the ordinary Schroedinger picture.

 

[A. Neumaier]

The S-matrix comes from requiring that time-evolution is unitary, i.e. it conserves probability. Since Born rule is the existence of a probabilistic interpretation of QM mathematics, it follows that the asking the S-matrix to be unitary cannot be independent from the Born rule, it's a consequence of it.

No. The unitarity of the S-matrix is something that follows from asymptotic completeness alone, without reference to the Born rule. Weinberg gives a proof in Vol. I at the end of Section 3.2 (p.115 in the 1995 edition), long before he invokes the Born rule in (3.4.7) to give an experimental meaning to the absolute values of certain S-matrix elements.