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

[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]

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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?