# QFT vs QM 101

1. Feb 2, 2012

### waterfall

I'm trying to understand the basics of convensional QFT versus QM. There are too many books about QM in the introductory level for layman but too rare for QFT. But the public needs to be adept about QFT too not just particle-wave duality, entanglement and other attractions in QM.

Let's start by a table or FAQ of some kind distinguishing QFT and QM. Maybe QFT is not so hard after all.

1.
QM uses Hilbert Space.
QFT uses Fock Space.

(Since Hilbert Space is not in physical 3D, then Fock Space is not in physical 3D either, it is in so called abstract configuration space.. therefore automatically quantum fields are not physical in convensional QFT, is this reasoning correct?)

2.
QM has position as observable.
QFT has position as operator (in other words, you can consider these as self-observing, isn't it)
How about momentum and spin? Are these observables or operators in QFT?

3.
QM uses no relativity.
QFT uses relativity in the sense of mass converting to energy and vice versa even if the speed is not near light (so the SR sense is more of E=mc^2 and not speed, correct?)

4.
QED, QCD, and EWT is an application of convensional QFT. In QED. It is natural to quantize the electromagnetic wave or field and produce the harmonic oscillators as photons. What's oscillating are magnetic field and electric field and displacement current via the Maxwell Equations. Steve Weinberg mentioned all particles are actual energy and momentum of the fields. But in electron, what is the equivalent of the electromagnetic field in QED that uses Maxwell Equations? What's oscillating in electron wave/field or the magnetic field/electric field counterpart of it?

(if you can add some basic FAQ of difference between QM and QFT, please add it so we can enable the millions of laymen in QM to understand QFT too in the basic level, thanks.)

Thanks.

2. Feb 2, 2012

### Matterwave

2) QFT has position as simply a parameter. It's QM that has a position operator and observable.

3. Feb 2, 2012

### waterfall

I think this is related to how QFT accomodate changing references frames as it is relativistic versus QM newtonian spacetime.

But QFT having position as parameter? I heard space and time is a parameter in newtonian space but are coordinate in minkowski spacetime.. so how come QFT still have position (or space) as parameter?

4. Feb 2, 2012

### The_Duck

It's worth pointing out that QFT is a subset of quantum mechanics. QFT is specifically the quantum mechanics of fields. So in discussions of "QM vs QFT", QM must be understood to mean "quantum mechanics of nonrelativistic point particles," and QFT must be understood to mean "quantum mechanics of relativistic fields" (one can have non-relavistic QFTs).
Fock space is a Hilbert space. QFT is just the quantum mechanics of fields, and all quantum mechanics uses Hilbert space.
This seems a bit strange; what does it mean for something to be "in physical 3D" and what does it mean for something to be "physical?"
I think you can make a strong case that at least the electromagnetic field is "physical"--it is fairly directly measurable. And the electromagnetic field, properly treated, is a quantum field.
Observables are (represented by) operators in both QM and in QFT. In QM, position is an observable; there is a position operator.

In QFT, people usually say that, in contrast to the case in QM, position is a "label" on operators. A quantum field is really a set of operators, one at each point in spacetime; i.e., an infinite set of operators, each "labelled" by a spacetime position.

I don't know what you mean by "self-observing."
As I mentioned above, observables are operators. There are momentum and angular momentum operators in QFT just as in QM, so in both cases these are observables.
You can include some effects of relativity in regular QM, but to a get a completely consistent accounting for special relativity you need relativistic QFT. QFT includes special relativity in all its aspects. All phenomena of special relativity--time dilation, length contraction, mass-energy equivalence, etc.--appear in QFT, as they must.

In QFT we define an "electron field" whose quantized oscillations are electron particles. The electron field is a bit of a weird thing, though. For instance it is not directly observable. For another its components are "Grassman numbers," as opposed to the electromagnetic field whose components are real numbers.

5. Feb 3, 2012

### waterfall

Can you give an example of non-relativistic QFT?

So Hilbert Space of Fields become Folk space.

Our radio can pick up electromagnetic field.. it is real.. but we can't pick up electron field.. what is the equivalent of EM in electron field. If it is not observable. Why did (convensional) QFT equate the two together?

Position in QM can only collapse upon measurement. In QFT, the position seems to be self-collapsing on its own. It's kinda self-observing or self-measuring.

Time is a parameter in QM, in QFT time is a coordinate. How about space, any idea how QM (of non-relativistic particle) and QFT (of relativisic fields) treat space?

If electron wave is not observable and its components are "Grassman numbers". Why put electron wave in same category as electromagnetic wave which clearly has real numbers as you mentioned?

Last edited: Feb 3, 2012
6. Feb 3, 2012

### Matterwave

What's your definition of parameter vs coordinates? I am using the word parameter to mean coordinates. What I mean is that there is no "position operator" in QFT (as far as I know).

7. Feb 3, 2012

### waterfall

I read that in non-relativistic bohmian mechanics, time is a parameter, not a coordinate, the single time parameter is "shared" by all the particles, i.e., the multi-particle wavefunction evolves in "time" the same way the single-particle one does.

Are you saying these parameter vs coordinate is only words used in Bohmian Mechanics and not a standard usage in QFT? How do you define them anyway?

Parameter = ?
Coordinate = ?

8. Feb 3, 2012

### sheaf

Condensed matter physics uses a lot of NR quantum field theories

Fock space is the Hilbert space for non interacting quantum fields, labelled by occupation numbers giving the numbers of excitations of various particles.

Although the electron field itself is not observable, various Dirac biliears constructed from it do constitute observables

Not sure I get this. You can try to construct position operators in QFT, but it's harder.

I don't understand the parameter/coordinate distinction. You can forumulate QFT Hamiltonian-style and peform time evolution, just like in QM

The Grassmanian-ness comes from the Fermionic nature of the electron.

9. Feb 3, 2012

### sheaf

For me, a parameter is just a variable that a some function depends on. A coordinate is just a label. For x and t, I don't see the distinction. I'm not familiar with the Bohmian terminology - maybe they do make a distinction.

10. Feb 3, 2012

### waterfall

But it is stated that in non-relativistic QM, there is no "spacetime". Time is a parameter, not a coordinate, and there is only one time parameter. It's more appropriate to use that saying there is only one time coordinate in non-relativistic QM. Do you see this now?

11. Feb 3, 2012

### Matterwave

The distinction between coordinate and parameter is very often not made because essentially they are the same thing. I suppose if you view the non-relativistic space-time structure as a fiber bundle structure where the 3-D slices of constant times are the fibers and the 1-D base manifold is time, and you look explicitly at each fiber, then you would call the space coordinates "coordinates" on these fibers, and the time coordinate a "parameter" which specifies which fiber you are on. But of course you can very simply just consider the entire fiber bundle and now you simply have 4 coordinates. This is especially true since this fiber bundle is isomorphic to R^4, so it's a trivial fiber bundle (as far as I know, somebody correct me if I'm wrong here).

I don't see much merit in making this distinction. But, I have not really studied Bohmian mechanics, so I don't know if it's useful there.

12. Feb 3, 2012

### Fredrik

Staff Emeritus
There is. It's called Galilean spacetime. The main difference between that and the special relativistic Minkowski spacetime is the value of the invariant speed, i.e. the speed that all inertial observers agree is the same). For Galilean spacetime, that's ∞. For Minkowski spacetime, it's 1 (at least if we use units such that c=1).

By the way, I like to define "QM" as the framework in which quantum theories are defined. That stuff about wavefunctions and the Schrödinger equation that we all study in our first QM course is just the simplest possible quantum theory, the theory of a single spin-0 particle in Galilean spacetime that's influenced by a potential. I prefer to call that "wave mechanics", at least when it's clear that I'm not talking about classical waves. I guess I'd call it "Schrödinger's theory" or something like that otherwise. (It's perfectly fine to call that theory "QM". There are no standard definitions that everyone uses. I'm just saying that we don't all use the terminology you used in the OP). To me, each QFT is a theory defined in the framework of QM.

To me, relativistic QM and non-relativistic QM are just subsets of the set all quantum theories. A quantum theory is relativistic if it includes operators that represent the symmetries of Minkowski spacetime, and non-relativistic if it includes operators that represent the symmetries of Galilean spacetime.

13. Feb 3, 2012

### waterfall

Superconductivity for instance? I think the BCS uses NR QFT?

If Fock space is the Hilbert space for non interacting quantum fields, then what is the corresponding space for interacting quantum fields? And what is it supposed to mean the quantum field is not interacting?

I read it. But the electron field can't still be measured. Is there a possibility our QFT that uses the concept of fields being more primary to particles being momentum and energy of the field is faulty? What motivated the grandfathers of QFT to equate electromagnetic field with electron field (when this latter is not observable). What's the rationale for this?

In the absence of measurement to determine its position, a particle has no position. But in QFT, the particle has position and vibrating kinda like in harmonic oscillator.

What's the counterpart of magnetic field and electric field in the electron that can travel in free space?

14. Feb 3, 2012

### Demystifier

15. Feb 3, 2012

### Fredrik

Staff Emeritus
This is a question that most physicists don't know the answer to, because the question is only answered in rigorous approaches to QFT. The standard textbooks are very non-rigorous by comparison. I don't know the answer myself, but it's been discussed here before. You could search for it. (You can probably ignore threads that DarMM hasn't posted in. He's the only one here who really seems to know these things).

It's when the Lagrangian doesn't contain any terms where more than two field components or derivatives of field components are being multiplied together. The number of factors determines the number of lines meeting at a point in a Feynman diagram. In a non-interacting theory, particle numbers never change. So they are pretty much useless, but still a good starting point from a pedagogical point of view.

This is wrong. To say that particles in QFTs have positions is even less accurate than to say that the particles in Schrödinger's theory do.

16. Feb 3, 2012

### waterfall

I'm familiar with Feynman diagrams having studied particle physics (in visualization only as all laymen do). In between the interaction vertex or points, virtual particles are being exchanged, and the coupling constants determine how strong are the interaction say between the electron and EM field. So they all interact.. using this context.. please explain what you mean quantum fields never interact using Feynman diagrams.

We may never know the particles exact location but one can imagine quantum fields as like the surface of speaker in full blast where it vibrates very fast and sound waves come in quanta just like the fields having the particles as quanta with creation annihilation going on amidst them.

17. Feb 3, 2012

### mysearch

Hi,
Going back to your opening post, the following site may provide a point of initial reference regarding the main permutations of quantum theory, especially in terms of identifying the significance of the various underlying concepts/parameters:

http://www.quantumfieldtheory.info/
Chapter-1 taken from the site above provides an initial breakdown in Chart 1-2 on page 7/8:
http://www.quantumfieldtheory.info/Chap01.pdf [Broken]
While Chapter-2 provide a further, more extensive comparison on page 20/21:
http://www.quantumfieldtheory.info/Chap02.pdf [Broken]
Other chapters are available that cover ‘free fields’ and ‘interacting fields’ that I haven’t really reviewed in any detail, but didn’t see any obvious description of Fock space. Possibly somebody might be able to comment on their impressions of this site for “the millions of laymen”, in which I include myself, trying to understand the transition from NRQM->RQM->QFT or even offer up alternatives.

Last edited by a moderator: May 5, 2017
18. Feb 3, 2012

### waterfall

Thanks. I'm presently reading on Teller "Interpretative Introduction to QFT". The above will be very useful as we go to the heart of what really is QFT.

Last edited by a moderator: May 5, 2017
19. Feb 3, 2012

### Fredrik

Staff Emeritus
What I'm saying is that you need at least three lines meeting at a point to have an interaction. For example, the diagram representing two electrons exchanging a photon looks like an H. There are two points where three lines meet. If there are no points where three or more lines meet, then all your diagrams look like this: | Such diagrams are present in interacting theories too, but they're ignored because they don't contribute to anything observable, except the energy density of the vacuum.

I don't think that's a good way to think about quantum fields. Neither Schrödinger's theory nor any QFT says that particles have positions, so in my opinion, neither should we.

20. Feb 3, 2012

### The_Duck

In noninteracting theories, there are no vertices in Feynman diagrams. As a result the only Feynman diagrams you can draw consists of a bunch of straight lines that don't touch each other, representing particles that simply travel along without interacting with each other. You can see why this is a simple but boring kind of theory.

21. Feb 3, 2012

### waterfall

Why, doesn't Fock space involve this 3 lines meeting at a point or standard Feynmann Diagram with interaction? Are you (and The_Duck) saying Fock space just involves noninteracting vertical lines? Is this related to perturbation theory which is Fock space pretending to have interaction when it doesn't really?

Oh. So quantum fields don't have particles before measurement. If this is so. Then quantum field is just like the classical electromagnetic field with only photons appearing during measurement? Then what's the advantage of QFT? I thought it involves particles being created and annihilated as part of the quantum field.

22. Feb 3, 2012

### Fredrik

Staff Emeritus
A Fock space is constructed from the Hilbert space associated with the single-particle theory. You use the single-particle space to construct a space of 2-particle states, a space of 3-particle states, and so on, and then you combine them all into a Hilbert space that contains all the 1-particle states, all the 2-particle states, and so on. This Hilbert space is called a Fock space. So it's just an algebraic construction. You need nothing more than the Hilbert space from the single-particle theory to define it, and the single-particle theory can be defined using a Lagrangian with no products of more than two field components or derivatives of field components.

However, in non-rigorous QFT, I think the idea is just to ignore that the interacting Hilbert space is really a different Hilbert space, and just introduce operators that can take n-particle states to (n+1)-particle states for example. In this context, Fock space is, as you put it, "pretending to have interaction when it doesn't really". I really suck at QFT beyond the most basic stuff, so I can't explain it better, and I might even be wrong (about the stuff in this paragraph).

I didn't say that. I said that the theory doesn't give us any reason to think that particles have positions. A position operator can be defined for massive particles, but it's kind of weird. I suppose we could use it to define "approximately localized" states, in a way that's similar to how its done in Schrödinger's theory. But there's no position operator for massless particles.

Anyway, "particle" doesn't mean "classical particle", so you can't assume that something has properties like position just because a quantum theory calls it a "particle".

Last edited: Feb 3, 2012
23. Feb 3, 2012

### atyy

What do you think of this comment in Collins's notes? He does acknowledge that a point of view different from his is more common.
http://www.phys.psu.edu/~collins/563/LSZ.pdf [Broken]: "Note that in both formulae, the vacuum state |0> is very definitely and strictly the true vacuum. This is just the same as in the definition of the coefficient c, (3), where the vacuum and one-particle states are definitely the true vacuum and one-particle states, i.e., the true physical states. In contrast, many textbook treatments appear to suggest that the state |0> should be the free-field unperturbed vacuum; if that approach is tried, very delicate limits involving adiabatic switching of the interaction are called for."

Edit: Here's another presentation by Srednicki that starts off with the more common point of view, but he goes on to discuss that it's wrong, and says that renormalization computes the corrections to having started the calculation with the wrong ground state. http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf, p51: "However, our derivation of the LSZ formula relied on the supposition that the creation operators of free field theory would work comparably in the interacting theory. This is a rather suspect assumption, and so we must review it."

Last edited by a moderator: May 5, 2017
24. Feb 3, 2012

### Fredrik

Staff Emeritus
I don't know QFT well enough to answer that, so I'll leave it for someone who does.

25. Feb 3, 2012

### waterfall

Are you saying not all physicists with Ph.D. are experts in QFT? I thought they all wer. But using Fock space noninteracting terms, how could they make the Large Hadron Collider function and successfully predict those scattering angles and interactions of the numerous particles. Is Fock space enough to analyze them including predicting the mass of the Higgs? Or do Large Hadron Collider physicists use purely rigorous QFT that normal physicists don't tackle?

So how does one imagine a quantum field? I thought it should have particles vibrating like harmonic oscillator.. but now saying particles don't have position.. then how does one picture it? Or is it possible space and time only occur during interaction with the quantum field, and without interaction, space and time doesn't really exist as we know it in the quantum field? And it is just a blob of untime and unspace?