Two concepts about quantum electrodynamics and quantum field theory

**guo_xiaobo** · Aug8-04, 05:03 AM

One question has disturbed me long time, I don't know the distinction between quantum electrodynamics and quantum field theory.
By the way, which quantum field thoery or quantum electrodynamics textbook is prefer?

µ³ · Aug8-04, 05:14 AM

I'm not too sure but quantum electrodynamics (along with quantum chromodynamics, and quantum electroweak theory) are branches of quantum field theory. QED deals primarily with the electromagnetic force.

http://en.wikipedia.org/wiki/Quantum_electrodynamics
http://en.wikipedia.org/wiki/Quantum_field_theory

**humanino** · Aug8-04, 07:00 AM

Quantum field theory is the general theory resulting from attempts to put quantum mechanics and special relativity together.

Quantum electrodynamics is the resulting quantum field theory when applied to electromagnetic phenomena. But $LaTeX Code: \\mu^3$ had already said it.

selfAdjoint · Aug8-04, 08:58 AM

Originally Posted by guo_xiaobo

One question has disturbed me long time, I don't know the distinction between quantum electrodynamics and quantum field theory.
By the way, which quantum field thoery or quantum electrodynamics textbook is prefer?

Quantum field theory is the general study of quantized fields. There are a number of different ones, for different fields, some only studied for educational purposes. QED is one such theory, the quantized relativistc electromagnetic field. There were originally three formulations of QED, due to the three discoverers Schwinger, Tomonaga, and Feymann. But Dyson showed they were all equivalent, and the only difference was how you developed the equations.

Other quantum field theories that are studied are the Electroweak theory, which is the unification of electromagnetism with the weak force, and QCD, the field theory of the strong force. The standard model is a unification of these last two theories. There are conjectures of theories beyond the standard model, but none of them has yet emerged as a true competitor for it.

**vanesch** · Aug8-04, 03:00 PM

Originally Posted by guo_xiaobo

By the way, which quantum field thoery or quantum electrodynamics textbook is prefer?

I just finished reading "Quantum Field Theory in a Nutshell" by A. Zee and I have to say it is simply brilliant. The only problem (but also the strong point) is that he jumps over a lot of calculations, to give you the basic ideas, but if you want to sit down with pencil and paper, it is a lot harder. The book that takes the opposite approach (a classic by now) is Peskin and Schroeder, Introduction to quantum field theory. They explain you all the calculations in all details, but sometimes you get a bit lost between all the calculational technicalities. Both books will probably get you going.

cheers,
patrick.

**humanino** · Aug8-04, 06:13 PM

I fully agree with Patrick. I ate Zee's Nutshell in a few hours. One gets a clean overview from the introduction of Fields as matress vibrations, to advanced subject such as supersymmetry unification strings gravity which are usually not even mentinoned in good introductory texts. There are numerous examples and application in condensed matter. This compares well to Peskin and Scrhoeder one which a I was already working at that time. Having those two books together to discover the subject is fascinating. One completes what misses in the other.

**Eye_in_the_Sky** · Aug9-04, 05:01 PM

Consider the following statement:

Originally Posted by humanino

Quantum field theory is the general theory resulting from attempts to put quantum mechanics and special relativity together.

And now consider this one:

Originally Posted by selfAdjoint

Quantum field theory is the general study of quantized fields.

The first description is in terms of "historical events with regard to human efforts". The second is in terms of "the objects of the theory". Each description is given from a different perspective, and each one, relative to the other, enhances our appreciation of "what QFT is".

To my eye, it is no coincidence that the former description is offered by one called "humanino", while the latter is offered by one called "selfAdjoint".

Furthermore, the first description tells us that, in the nonrelativistic domain, we have no particular need for QFT, while the latter suggests to us that, nevertheless, in that domain, QFT can still be applied.

**marlon** · Aug9-04, 05:11 PM

in the nonrelativistic domain QFT has to be valid. It has <to produce the same results as QM. Just as QM has to give the same results as Newtonian mechanics when you look at distances that are not to small that interference-effects come into play... This is the correspondence-principle.

regards
marlon

**vanesch** · Aug9-04, 05:56 PM

Originally Posted by marlon

in the nonrelativistic domain QFT has to be valid. It has <to produce the same results as QM. Just as QM has to give the same results as Newtonian mechanics when you look at distances that are not to small that interference-effects come into play... This is the correspondence-principle.

I suppose that Eye meant the applications of the methods of quantum field theory in condensed matter physics where the fields are not "elementary fields" such as the electron field or so, but constructs that exhibit collective phenomena.

cheers,
patrick.

**Eye_in_the_Sky** · Aug10-04, 11:33 AM

Originally Posted by Eye_in_the_Sky

... the first description tells us that, in the nonrelativistic domain, we have no particular need for QFT, while the latter suggests to us that, nevertheless, in that domain, QFT can still be applied.

The above remark was intended to dispel the not uncommon misconception in which the notion of a "quantized field" is thought to apply strictly to a relativistic domain.

So, what is an example of a nonrelativistic domain in which QFT can be applied, yet for which it may be said that QFT satisfies no particular need?

Well, quite simply ... take the single-particle Schrödinger field and quantize it.

Upon doing that ... what do we get?

(note: x is a spin-space index; i.e. x=(s,x), δ(x-x')=δ³(x-x')δ_ss')

case (1): [Ψ(x,t),Π(x',t)]=ih_barδ(x-x') → Fock space for identical Bosons

case (2): {Ψ(x,t),Π(x',t)}=ih_barδ(x-x') → Fock space for identical Fermions

In QM, on the other hand, the Fock space is constructed by taking a direct sum of a one-dimensional Hilbert space plus a series (n = 1,2, ...) of appropriately symmetrized n-particle Hilbert spaces. Thus, QFT gives us the exact same thing back (except (with the "bonus") that the symmetrization postulates for multiparticle states in QM have been replaced by the commutator-anticommutator rules of QFT (which themselves are on par with (i.e. no stronger than) QM's rule [Q_j,P_j'] = ih_barδ_jj')).

**Haelfix** · Aug10-04, 08:15 PM

For a nice little challenge, try writing down what the infinite square well potential from vanilla quantum mechanics would be in the QFT context.

**Eye_in_the_Sky** · Aug11-04, 06:24 AM

Assuming the particle to be spinless, I get:

Ψ(x,t) = Σ_n b_n exp(-iE_nt/h_bar) φ_n(x)

Ψ^†(x,t) = Σ_n b_n^† exp(iE_nt/h_bar) φ_n^*(x)

where

E_n = k_n² (h_bar²/2m)

k_n = n (π/L)

φ_n(x) = sqrt(2/L) sin k_nx

n = 1, 2, 3, ...

and each b_n^† (b_n) is a creation (annihilation) operator, relative to the single-particle energy eigenstates, for a particle with energy E_n.

-------------------

If a^†(k) and a(k) are creation and annihilation operators with respect to momentum, then from

<x|a^†(k)|0> = [1/sqrt(L)] e^ikx

and

<x|b_n^†|0> = sqrt(2/L) sin k_nx

it follows that

b_n^† = -i/sqrt(2) [a^†(k_n) - a^†(-k_n)]

b_n = i/sqrt(2) [a(k_n) - a(-k_n)] .

-------------------

Pure "vanilla" ... and not a drop of "fudge"!

(I have the sneaking suspicion that somehow I circumvented the "challenge".)

**guo_xiaobo** · Aug11-04, 09:47 AM

Thank you for your answers!
I think the QED is just an application of QFT, in other words, it's an special example,
of cause, the application to electronicmagnetism is very successfully. we called it QED.
and, the QFT contains the weakforce , strongforce and electromagneism force.

**ZapperZ** · Aug11-04, 10:35 AM

Originally Posted by guo_xiaobo

Thank you for your answers!
I think the QED is just an application of QFT, in other words, it's an special example,
of cause, the application to electronicmagnetism is very successfully. we called it QED.
and, the QFT contains the weakforce , strongforce and electromagneism force.

Actually, QFT is even more general than that.

QFT is a treatment or a description of the interaction in quantized fields of anything. In fundamental interactions, this reduces to the basic interactions of strong, weak, and electromagnetic. However, these are NOT the only types of interactions that QFT are used for. In condensed matter physics (where field theoretic methods are used to the full extent), the quantized fields can range from phonons, magnons, polarons, spinons, etc... These are the "collective" fields that cause the relevant interactions in a condensed matter system. In fact, a lot of the the methodologies and techniques used in condensed matter have been adopted by particle physics and field theories (eg. Anderson's broken symmetry principle).

So to put it crudely, QFT is a description of ANY kind of interaction.

Zz.

**vanesch** · Aug11-04, 12:20 PM

Originally Posted by guo_xiaobo

the QFT contains the weakforce , strongforce and electromagneism force.

To add to what Zapper said, QFT is a framework. QED is a theory within that framework about electromagnetism. What you write above is another theory within the same QFT framework, called the standard model.
It is a matter of definition of words. If you want an analogy: QFT is like the laws of mechanics (Newton's law and so on), while QED is like the mechanics of a weight on a spring, and the standard model is like the mechanics of two bodies with gravitational interaction. The frame is mechanics, and the specific applications are weight+spring, or two bodies and gravity.

cheers,
Patrick.

**Haelfix** · Aug11-04, 03:22 PM

Yea the challenge is to do it with the path integral formulation, and I just realized the subtetly is with the finite square well potential =)