QFT vs. Relativistic quantum mechanics

[Schreiberdk]

Hi PF

How does Quantum Field Theory differ from relativistic quantum mechanics? Ain't Quantum field theory just relativistic quantum mechanics or are there more to the subject? :)

Thanks in advance

\Schreiber

 

[bapowell]

Relativistic quantum mechanics, as a single particle theory, is plagued by various inconsistencies: negative energy states and negative probability amplitudes. This forces one to move to a multiparticle interpretation of quantum theory in which fields replace single particle states. In this framework, negative energy wavefunctions are instead seen as one of two components of a quantum field -- they represent negative frequency operators that destroy positive energy quanta. (The other component, the positive energy wavefunction, is a positive frequency operator that creates positive energy quanta.)

One simple way to understand the necessity of the field viewpoint is to recognize that the relativistic relation $E=mc^2$ implies e+/e- pair creation. This process cannot be understood using single particle QM.

 

[cgk]

Additionally to bapowell's comment, the term "relativistic quantum mechanics" can refer to calculations in which only some relativistic effects are taken into account to some degree. This is commonly done in molecular and solid state calculations.

Some possible options in this domain are:
o Do non-relativistic calculations taking into account some relativistic corrections perturbatively
o Do calculatings including only scalar relativistic effects self-consistently (i.e., the modified kinetic energy operator)
o Do calculations including scalar-relativistic effects and spin-orbit coupling (using the Pauli-Hamiltonian to describe molecules), but neglecting some magnetic couplings or using gauge-dependent approximations etc.
o Do fully relativistic 2-component or 4-component calculations of self-consistent field wave functions of the Dirac equation, again neglecting some small, inconventient terms which, however, might break Lorentz- or gauge-invariance of the results and so on.

In total, there are many sensible ways of doing relativistic QM without referring to QFT. Within their domain of applicability these calculations typically give results indistinguishable from fully relativistic treatments. QFT, however, will be required if you want to model the actual interaction processes of, say, electrons with photons. And it will be required to interpret the sometimes rather ad-hoc approximations and decisions used in the non-QFT relativistic QM (e.g., what to do with the negative energy states and the gauges).

 

[exponent137]

Can we say that Dirac equation is quantum relativistic theory with a little QFT?

 

[dextercioby]

In a rightful manner, I would say that Dirac equation's predictions cannot be fully correct unless the equation is written in terms of quantum fields (for a definition, see any book on axiomatical field theory).

 

[exponent137]

I agree. I only said that Dirac equation is what is "relativistic quantum mechanics" and is yet not full QFT.

 

[element4]

Ain't Quantum field theory just relativistic quantum mechanics or are there more to the subject? :)

Quantum field theory is much more general than just relativity + quantum physics. There also exist non-relativistic quantum field theories, which are essential in many areas of physics.

 

[bapowell]

I agree. I only said that Dirac equation is what is "relativistic quantum mechanics" and is yet not full QFT.

The solution to the Dirac Equation must be interpreted as a quantum field. Why do you say it is not "full QFT"?

 

[A. Neumaier]

How does Quantum Field Theory differ from relativistic quantum mechanics? Ain't Quantum field theory just relativistic quantum mechanics or are there more to the subject? :)

Quantum field theory is the quantum theory of field operators.

Relativistic quantum mechanics is the quantum theory of processes that are (at least approximately) Poincare invariant.

Relativistic quantum field theory is the intersection of these theories. It is the quantum theory of field operators
that are covariant under the Poincare group.

See also the entry ''Is there a multiparticle relativistic quantum mechanics?'' in Chapter B1 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#multi

 

[exponent137]

The solution to the Dirac Equation must be interpreted as a quantum field. Why do you say it is not "full QFT"?

Yes, it should be interpreted as quantum field. But this step can be ignored here, and we can look it only as Poincare invariant.
But Fock space cannot be easily imagined as to be Poincare invariant. So beginners needs more steps, and quantum field and Poincare invariance can be used as two steps.

BTW: Imagination of Dirac equation is easier, if we also can rotate spinor. Can you help me?
https://www.physicsforums.com/showthread.php?t=488277

 

[Demystifier]

Relativistic quantum mechanics, as a single particle theory, is plagued by various inconsistencies: negative energy states and negative probability amplitudes. This forces one to move to a multiparticle interpretation of quantum theory in which fields replace single particle states.

It's true that relativistic QM in its naive form has some internal inconsistencies, and it's true that QFT is a way to avoid them, and it's true that QFT agrees with experiments. Yet, it is not true that QFT is the only logically possible way to avoid these internal inconsistencies. There are other ways as well. The other ways may not be so successfull from a phenomenological point of view, which is why QFT is to be preferred. Yet, QFT cannot be obtained from QM and relativity by a pure demand of internal consistency. Some textbooks try to convince the readers that it can, but they cheat. The true reason why QFT is the right escape from the problems of naive relativistic QM is the fact that it agrees with experiments.

By the way, QFT has some internal inconsistencies as well. For some of them see e.g. the introduction of
http://xxx.lanl.gov/abs/0705.3542 [Europhys. Lett.85:20003, 2009]

 

[Fredrik]

Relativistic quantum mechanics, as a single particle theory, is plagued by various inconsistencies: negative energy states and negative probability amplitudes.

I don't agree with this. What you're saying is true if we define "relativistic quantum mechanics" as the theory of a wavefunction satisfying a relativistic wave equation, but the things you mentioned are good reasons not to. Instead, each single-particle theory (both the non-relativistic ones and the relativistic ones) in the framework of QM should be defined in terms of irreducible representations of groups.

I actually don't know exactly how to do that, but roughly, what you have to do is to replace the axiom that says that state vectors change with time according to the Schrödinger equation, with an axiom that says something about irreducible representations of a group that has the group of translations in time as a subgroup. If the group has the Galilei group as a subgroup, it's a non-relativistic single-particle theory. If it has the Poincaré group as a subgroup, it's a (special) relativistic single-particle theory.

 

[martinbn]

I actually don't know exactly how to do that, but roughly, what you have to do is to replace the axiom that says that state vectors change with time according to the Schrödinger equation, with an axiom that says something about irreducible representations of a group that has the group of translations in time as a subgroup. If the group has the Galilei group as a subgroup, it's a non-relativistic single-particle theory. If it has the Poincaré group as a subgroup, it's a (special) relativistic single-particle theory.

Sorry for the side question, but that is very interesting. Where could I see more detail?

 

[dextercioby]

Sorry for the side question, but that is very interesting. Where could I see more detail?

Weinberg's book (1st volume, 2nd chapter) is a solid reference on relativistic symmetry in the 'orthodox' formulation of QM.

 

[A. Neumaier]

It's true that relativistic QM in its naive form has some internal inconsistencies, and it's true that QFT is a way to avoid them,

Only the ''motivational'' textbook version of relativistc QM is inconsistent.
There is fully consistent non-naive relativistic QM which is not QFT.
See the entry ''Is there a multiparticle relativistic quantum mechanics?'' in Chapter B1 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#multi[/QUOTE]

By the way, QFT has some internal inconsistencies as well. For some of them see e.g. the introduction of
http://xxx.lanl.gov/abs/0705.3542 [Europhys. Lett.85:20003, 2009]

Although this paper appeared in a peer-reviewed journal, this doesn't guarantee that its content is ciorrect,

The leading sentence in the abstract, [Some] ''Practically measurable quantities resulting from quantum field theory are not described by hermitian operators'' is also true of ordinary quantum mechanics. An example is the half-life of an unstable particle (a resonance in a potential with a barrier). and hence would discredit all of QM if it were taken as a serious defect.

The introduction of the paper errs in several respects. At the bottom of p.1 of the arXiv version it says: ''An example is the energy-momentum tensor. Nevertheless, although such operators correspond to quantities that should be measurable in principle, they are not quantities measured in practice.'' - This is far from true. The energy-momentum tensor is one of the basic observables in relativistic therrmodynamics.

On p.2, it claims that QFT has no position operator. But this is incorrect: The Newton-Wigner construction defines one for every massive representation of the Poincare group, which includes QFT.

The remainder of the paper does not address the above topic, and anyway represents a minority view in quantum mechanics.

 

[Demystifier]

The leading sentence in the abstract, [Some] ''Practically measurable quantities resulting from quantum field theory are not described by hermitian operators'' is also true of ordinary quantum mechanics. An example is the half-life of an unstable particle (a resonance in a potential with a barrier). and hence would discredit all of QM if it were taken as a serious defect.

I agree. Yet, there are some ways to incorporate time (and thus half-life as well) as a hermitian operator in QM. See e.g.
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

At the bottom of p.1 of the arXiv version it says: ''An example is the energy-momentum tensor. Nevertheless, although such operators correspond to quantities that should be measurable in principle, they are not quantities measured in practice.'' - This is far from true. The energy-momentum tensor is one of the basic observables in relativistic therrmodynamics.

I guess here you mean CLASSICAL relativistic thermodynamics, while the statement in the quotation marks refers to QUANTUM field theory.

On p.2, it claims that QFT has no position operator. But this is incorrect: The Newton-Wigner construction defines one for every massive representation of the Poincare group, which includes QFT.

A more correct statement in that context would be that "... QFT has no CONSISTENT position operator". Most researches in the field agree that the Newton-Wigner operator is not consistent because it is not relativistic covariant.

Anyway, in an appropriately reformulated form of QFT as in
http://xxx.lanl.gov/abs/0904.2287 [Int. J. Mod. Phys. A25:1477-1505, 2010]
a consistent (relativistic covariant) position operator can be introduced in the same way as in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

 

[A. Neumaier]

I agree. Yet, there are some ways to incorporate time (and thus half-life as well) as a hermitian operator in QM. See e.g.
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

wheras the usual silent agreement is that whatever you can calculate from quantum mechnaics is observable (though not ''an observable'' in the formal sense). Without that agreement, the shut up and calculate interpretation would not be so extremely successful.

Creating artificial Hermitian observables is one of the trickeries that reflect a certain philosophy but not how QM really works.

I guess here you mean CLASSICAL relativistic thermodynamics, while the statement in the quotation marks refers to QUANTUM field theory.

But classical relativistic therrmodynamics is the result of the statistical mechanics of quantum field theories. The only difference between the classical case and the quantum case is in the concrete equation of state. For gases at medium or high temperature, the deviations are negligible.

A more correct statement in that context would be that "... QFT has no CONSISTENT position operator". Most researches in the field agree that the Newton-Wigner operator is not consistent because it is not relativistic covariant.

Nonsense. The N_W position operator is fully consistent with Poincare invariance; indeed, it is derived from that. But it is frame dependent: each observer has its own position operator in the 3-space orthogonal to its trajectory through space-time.

 

[Demystifier]

Nonsense. The N_W position operator is fully consistent with Poincare invariance; indeed, it is derived from that. But it is frame dependent: each observer has its own position operator in the 3-space orthogonal to its trajectory through space-time.

Well, that's one opinion, which I respect. But many researchers in the field do not accept that an observer-dependent operator is a physically acceptable one. Or more generally, many physicists do not accept that quantum mechanics makes physical sense only when observers are present. So I think we can agree that these issues are at least controversial.

 

[A. Neumaier]

Well, that's one opinion, which I respect. But many researchers in the field do not accept that an observer-dependent operator is a physically acceptable one. Or more generally, many physicists do not accept that quantum mechanics makes physical sense only when observers are present. So I think we can agree that these issues are at least controversial.

If this were so then all relativity theory would be controversial. it is well-known that many observables in relativity are observer-dependent, and apparently nobody objects - apart from you.

Nobody needs to be actually present to predict that whenever some observer measures lenths, the results will depend on the observer frame.

Thus we have here no point of agreement.

 

[Demystifier]

If this were so then all relativity theory would be controversial. it is well-known that many observables in relativity are observer-dependent, and apparently nobody objects - apart from you.

Nobody needs to be actually present to predict that whenever some observer measures lenths, the results will depend on the observer frame.

One view of relativity, which I prefer, is that observer-dependent quantities such as length do not have any FUNDAMENTAL meaning. By analogy, one could say that in relativistic QM the NW position operator does not have a fundamental meaning either. Would you agree with the latter statement?

 

[A. Neumaier]

One view of relativity, which I prefer, is that observer-dependent quantities such as length do not have any FUNDAMENTAL meaning. By analogy, one could say that in relativistic QM the NW position operator does not have a fundamental meaning either. Would you agree with the latter statement?

Yes. In fact, it is nowhere needed except when one derives nonrelativistic approximations at some low order in c^{-1}.

Even in nonrelativistic physics, positions are observer-dependent and hence have no fundamental meaning since thery depend on the choice of a coordinate system. (Thus that Bohmian mechanics puts so much emphasis on positions is not justified.)

 

[meopemuk]

Even in nonrelativistic physics, positions are observer-dependent and hence have no fundamental meaning since thery depend on the choice of a coordinate system.

Almost all physical observables (position, momentum, energy, spin projections, etc.) are observer-dependent. Does this mean that all of them "have no fundamental meaning"?


Eugene.

 

[Demystifier]

Almost all physical observables (position, momentum, energy, spin projections, etc.) are observer-dependent. Does this mean that all of them "have no fundamental meaning"?

There are different levels of "observer dependence". I would say that tensor quantities (e.g., the energy-momentum 4-vector) have a "softer" form of observer dependence and that such quantities do have a fundamental meaning.

 

[A. Neumaier]

Almost all physical observables (position, momentum, energy, spin projections, etc.) are observer-dependent. Does this mean that all of them "have no fundamental meaning"? .

I was just echoing demystifier's words, intending to draw into question the very notion of ''fundamental meaning''.

In my opinion, it is quite natural that physical obervables depend on a number of observer's choices. It begins with the choice of units, continues with the choice of coordinates, and with the choice of gauges.
Theory only must tell how the results depend on these choices.