Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Or, why is QFT "better" than QM?

There may be many answers.

For one example of an answer to a parallel question, GR is better than Newtonian gravity (NG) because it gets the perihelion advance of Mercury right.

You could also say that GR predicts a better black hole than NG, but that's a harder sale.

For QFT versus QM, I've heard of the Lamb shift, but what else makes QFT superior?

share|improve this question
I think this is a wrong way of thinking about this. In some sense, QFT is a special case of QM, its just quantum mechanics of fields (object with essentially infinite number of degrees of freedom). –  Heidar Jul 8 '12 at 23:08
More on relation between QFT and QM: physics.stackexchange.com/q/26960/2451 –  Qmechanic Sep 23 '12 at 16:54

3 Answers 3

up vote 12 down vote accepted

The most important thing is that nonrelativistic QM (as it is formulated traditionally) cannot deal with changing particle number, because the position basis Hilbert space changes dimension as you increase the particle number. Quantum field theory allows particle number to change, and this is the main difference.

This is also important in cases where the number of particles is indefinite, like the fixed phase description of a BEC or a superfluid. In this case, the state of fixed phase macroscopic matter wave is a superposition of different numbers of particles. For this reason, nonrelativistic Schrodinger fields are useful in condensed matter physics, even in cases where the particle number is technically conserved, because the states one is interested in are in a classical limit where it is better to assume that the particle number is indefinite. This is exactly analogous to introducing a chemical potential and pretending that the particle number can fluctuate in statistical mechanics, in those cases where the particle number is exactly fixed. Or introducing a temperature in those cases where the energy is exactly fixed. It is still mathematically convenient to do so, and it does no harm in the thermodynamic limit. So it is convenient to use quantum fields to describe nonrelativistic situations where the particle number is fixed, but the behavior is best described by a classical collective wave motion.

share|improve this answer
Stupid question: What precisely holds you from building a non-relativistic formalism which computes an object "$\psi(y,z,t',x,t)$", interpreted as a transition from "$(x,t)$" to "$(y,t')$ as well as $(z,t')$"? –  NikolajK Jul 9 '12 at 8:15
+1 for changing particle number. I guess this allows one to deal with events like collisions at LHC. –  Jim Graber Jul 9 '12 at 11:09
I have heard that one of the main advantages of Feynman diagrams over other techniques is that they make calculations much easiier. This is relevant to the question, if you consider QED to be part of QFT, but not part of QM –  Jim Graber Jul 9 '12 at 11:13
@JimGraber: The major real simplification of Feyman diagrams (when all is said and done, and other than being automatic) is combining particle and antiparticle. This reduces the number of terms by 2 to the number of lines, compared to Old-fasioned relativistic Hamiltonian perturbation theory. But "Dixon diagrams" (this is not a term), i.e. using S-matrix relations to calculate the amplitudes from tree amplitudes, are truly more efficient, by orders and orders of magnitude, compared to Feynman diagrams, especially for gravity, and it was discovered only a few years ago. –  Ron Maimon Jul 9 '12 at 17:00
@NickKidman: that's possible, but $\psi$ is not what you wrote: it's $\psi(y,z,t)$ and a separate $\psi(x,t)$ with transitions between them. This is an equivalent desription of the nonrelativistic multi-particle Hilbert space, and if you introduce operators that add/remove a particle at a point, you are in nonrelativistic QFT. So there is nothing more to nonrelativistc QFT than changing particle number. For relativity, you don't have a simple particle picture, the particle picture is zig-zagging in time, and only makes 100% sense in S-matrix or in complicated formalisms. –  Ron Maimon Jul 9 '12 at 17:12

Any quantum system at relativistic speeds is described in the framework of QFT. Or I should say: quantum systems at high energies or short distances (sub-nuclear) or short-time processes (sub-micro second stuff). For example relativistic Compton scattering is described by the Klein-Nishina formula which particle physics students derive in their QED course.

share|improve this answer
+1 for mentioning relativistic capability. So QFT covers more than QM. But I am still looking for more examples where QM and QFT predict different results. –  Jim Graber Jul 9 '12 at 11:18
@JimGraber: That's like asking "I am looking for a case where classical mechanics and fluid dynamics predict different results". Fluid dynamics is a special case of classical mechanics, QFT is quantum mechanics with commuting local observables. QFT is not a modification of QM, it isn't a different theory, it's an application of QM to fields. The insights that come with QFT are not insights into QM, but insight about continuum limit and renormalization, which are just as applicable to classical statistical mechanics. –  Ron Maimon Jul 9 '12 at 17:04
@Ron: I think you and I might have a slight vocabulary difference re what QM means. I still remember hearing that "the old theory" predicted zero Lamb shift, and that the "new theory", ie QED, "got it right". Mathematically, creation/anhiliation operators are added, and physically, the vacuum becomes "nonempty", resulting in the non-zero Lamb shift. –  Jim Graber Jul 10 '12 at 14:24
@JimGraber: That's not exactly right--- the "old theory" predicted infinite Lamb shift, so people assumed it must be zero. The new theory showed which part of the infinity was the same as the bare electron and should be subtracted off, and the rest is a real residual effect. This was calculated by Bethe without modern methods, and in a nonrelativistic approximation, in the late 40s, and this set the stage for modern renormalization to emerge. The vacuum is nonempty even in the old theory, but you don't know how to deal with the infinities. –  Ron Maimon Jul 10 '12 at 19:54

QFT is just a more powerful version of wave function QM.

In statistical mechanics, it allows one to do many computations that would be awkward in terms of wave functions.

In relativistic theories, it allows one to handle correctly the multiparticle situation which, for more than 2 particles, is extremely awkward to do without fields.

Also, it provides the connection between spin and statistics and the CPT theorem, which have to put in by hand in wave function QM.

Finally, QED and the standard model (with all their predictions) cannot be formulated without QFT.

share|improve this answer
Let's say we're looking at a simple Hydrogen atom, for instance. Is there even really a valid formulation with QM and not QFT? What does that even mean? If the electric interaction was instantaneous and continuous, it seems like that would be completely invalid. You could assume the nucleus to be a stationary point and solve the electron given that field, but such a solution would only be approximately correct, no? –  Alan Rominger Jul 9 '12 at 13:16
Of course, everything in physics is approximate. One first removes the center of mass motion, and may assume an external field problem, as the proton is much heavier than the electron. Thus gives a fairly good spectrum, and QED gives only small but experimentally verifiable corrections to it. –  Arnold Neumaier Jul 9 '12 at 13:59
"external field" is useful wording. I am not familiar with any QM problems that are formulated without an external field. It would then be logical to wonder if particle-particle interactions of any sort are possible without QFT. –  Alan Rominger Jul 9 '12 at 14:18
@AlanSE: The standard Hamiltonian for atomic matter has no external fields but can be (and was, historically) written without QFT. –  Arnold Neumaier Jul 9 '12 at 15:20
@AlanSE: Do you know a method to compute the Hydrogen spectrum (for real, and not perturbatively) in field theory? It's a nasty problem, renormalization is annoying in atoms. One approach (which I don't like, but it works) is to just consider relativity as higher order corrections to nonrelativistic system. The issue is that nonrelativistic systems have unambiguous spatial wavefunctions, and this helps do real calculations. We don't have any real first-principles description of the quark/gluon distribution in a single proton or pion, or even in the QCD vacuum. –  Ron Maimon Jul 9 '12 at 17:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.