# Very basic questions about QFT

1. Sep 22, 2007

### nrqed

A few months ago I started a thread asking some basic questions about QFT and it led to a large number of posts that were extremely interesting and that I am still going through. But the thread started to go in several directions (all very interesting!) away from my initial basic questions which I haven't answered to my satisfaction yet. So I will ask some of them again, hoping that nobody will be annoyed by this.

One thing that has always frustrated me with QFT is the lack of clarity on two points (at least they are unclear to me).

First, the connection between the formalism and quantities that can be measured in a lab. In QM, there is a clear formalism. Observables are associated to operators, the eigenvalues of these operators give the possibel outcomes of a measurement, the coefficients of the wavefunction expanded over the eigenstates give the probabilities, etc.

Now, in QFT, what is the equivalent? As far as I can tell, only one thing is ever actually calculated in QFT: the transition probability for a system to go from a certain number on non-interacting particles (with certain momenta and quantum numbers) in the far past to a certain number of free particles in the far future. This is the quantity that is used to calculate scattering cross sections or decay rate. It seems as if all a QFT book typically does is to develop techniques to do this unique calculation for more and more complicated system! (One exception that comes to mind as I am typing this is the use of path integrals to calculate the ground state energy of a meson, say, using lattice gauge theory...I have to think about it a bit more)

However, as soon as I try to think of any other question that one solves in QM I can't think of any way to do the equivalent using QFT. And this brings me to my second point: How does one start from QFT and show in a certain limit that the results of QM are recovered? This lack of connection is frustrating. It's like if we would learn GR and the books would refuse to show that in some limit one recovers Newtonian gravity. Or learning about SR and never be shown how it reduces to Galilean and Newtonian physics when c is taken to infinity!

The two questions are clearly related...If I knew how to recover the results of QM from QFT I would obviously know how to measure energy, momentum, etc in the QFT context.

As one example, consider the infinite square well. That's the first thing one does in QM so it seems that it should be the very first thing one should recover from a QFT approach. And yet I have never seen that done. The worse is that it's not even clear to me how I would set up the problem to start with. Let's say we ask the very simple question: if a particle has that energy, what is the probability of observing it between x=0 and x=1/3, or "if the particle has this probability distribution at t=0, what is the probability that it has an energy equal to the ground state energy". That sort of question. Questions which are very basic in QM. But how would one go about answering them in QFT??

Consider even the simple case of a free particle. IN QM, one can build a wavepacket for a single particle. Now if you start with QFT, it seems that in order to build a wavepacket, one necessarily mixes diffrent states in teh Foxk space, thereby getting a state involving a particle number which is not well defined. No problem. But if the particle has an energy much less than its rest mass (I consider a massive particle), one should be able to show that if the momentum of the particle (say) is measured, using a QFT approach, one recovers the answer of QM.

Those are questions which I would expect to be done at the very beginning of a QFT book. But the question of measuring observables besides cross sections (like energy, momentum, etc) is not discussed in QFT books as far as I can tell. The same thing is true for the connection to QM.

So my questions are: does that bother anyone else or is it that my questions do not make any sense to others? And: any books/papers discussing these points (for example the connection with QM)?

There are several extremely knowledgeable people on these boards so I am really looking forward to getting some feedback/comments/criticisms.

Patrick

2. Sep 22, 2007

### meopemuk

Hi Patrick,

yes, these are exactly the questions that bothered me very much when I first learned QFT. You are absolutely right that QFT is designed only to calculate the scattering (S-) matrix and doesn't concern itself with proper definitions of the Hamiltonian and particle wave functions, which can be compared with usual QM definitions in the low-energy limit.

For myself I found the solution of this problem (i.e., the QFT <-> QM correspondence) in the "dressed particle" approach, which I tried to advertize in many posts on this Forum. In the "dressed particle" formulation QFT becomes, basically, indistinguishable from QM of interacting particles. The only significant difference is that traditional QM doesn't consider interactions that change the number of particles (decays, annihilation, bremsstrahlung, etc.). All these interactions find a proper place in the "dressed particle" formulation of QFT. Moreover, the "dressed particle" approach avoids such unphysical features of QFT as particle self-energies, vacuum polarization, and divergences.

The "dressed particle" approach is not covered in textbooks, except a few brief mentions in chapter 12 of

S. S. Schweber, "An introduction to relativistic quantum field theory", 1961

You can learn more details and find further references in journal articles:

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim. 8 (1958), 378.

A. V. Shebeko, M. I. Shirokov, "Unitary transformations in quantum field theory and bound
states", Phys. Part. Nucl. 32 (2001), 15; http://www.arxiv.org/abs/nucl-th/0102037

E. V. Stefanovich, "Relativistic quantum dynamics" http://www.arxiv.org/abs/physics/0504062

Eugene.

3. Sep 23, 2007

### BoTemp

Good references, I think these'll keep me busy for a while. Thanks Eugene

4. Sep 24, 2007

### Demystifier

5. Sep 24, 2007

### Avodyne

There is an exact map between nonrelativistic quantum mechanics and nonrelativistic quantum field theory; this is treated in the books by Srednicki and Brown. So, then you have to get the nonrelativistic field theory as a limit of the relativistic one; Brown sketches how this works.

In general, the rules of measurement in QFT are the same as in QM: observables are represented by hermitian operators, etc. (QFT is just a particular kind of QM.) When you measure the strength of an electric or magnetic field in a lab, you are making a measurement of a hermitian operator (the field strength) in quantum electrodynamics.

6. Sep 24, 2007

### genneth

Do you mean 2nd quantization to be well defined? Or to put a finer point on it, functorial? It's just that as I've got the book by Srednicki, I wonder where it actually says that. I do not doubt the truth, but it seems my reading comprehension has been amiss...

7. Sep 24, 2007

### Avodyne

Near the end of chapter 1.

8. Sep 24, 2007

### meopemuk

Hi Avodyne,
I think you are talking about http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

This is a good place to discuss similarities and differences between QFT and quantum mechanics. The crucial point is in introduction of interaction operators of the type (1.36). Any relativistic quantum field theory (QED, QCD, electroweak theory) has these kinds of interaction operators, and I dare to say that they are "the source of all evil".

Srednicki presented (1.36) as an example of interaction that changes the number of particles. However, processes described by these operators are impossible in nature. For example, the term a*aa describes "merging" of two free particles into one. The hermitean conjugated term a*a*a describes decay of one free particle into two products. Such processes violate the momentum and energy conservation laws, and they have never beed observed experimentally. Terms like these (plus a*a*a* and aaa) are responsible for such unphysical features of quantum field theories as "particle self-energies" and "vacuum polarization". This means that in QFT we don't have a well-defined notion of vacuum and 1-particle states. So, it becomes almost impossible to make a connection to the ordinary quantum mechanics.

The idea of the "dressed particle" approach is that "bad" terms of the types aaa, a*aa, a*a*a, and a*a*a* should never be present in the interaction Hamiltonian. If you wish to describe particle emission you are welcome to use interaction terms like a*a*a*aa which correspond to the perfectly physical process in which an extra particle is emitted in the collision of two particles.

Eugene.

9. Sep 29, 2007

### nrqed

thanks, I will have a look.
To make my point, consider the following question: Let's say we have a free scalar AFT. The eigenstates of H are the multiparticle states of the Fock space.

Consider a very very simple situation...the system is described by some multi- particle state. What is the probability of observing one particle in a certain volume element dV? Whatis the probability of observing two particles in a certain colume element dV?

There are clear answers to that in QM. But I don't see a clear answer from QFT.

I simply don't see clearly a conenction between QFT and QM, in the sense of some limit that one may take to recover all of QM from QFT.

One problem is that there are two key differences between QM and QFT(here I am talking about QFT as applied in partcile physics, not in condensed matter, say).:

a) QFT involves Lorentz invariant or covariant quantities

b) QFT involves a Fock space, so it accomodates for non-fixed number of particles.

Now, it's obvious how to take a nonrelativistic limit of relativistic quantities (part a). It's basically the same as recovering classical quantities from SR.

However, I don't understand how to recover the fixed number theory of QM from QFT in some limit. One should be able to show that as the energies become ninrelativistic, QFT necessarily becomes a theory of a fixed number of particles!! But I have never seen tbhis done!

I would like to be proven otherwise, but to me it feels as if QFT is {bf NOT} a true relativistic generalization of QM. I see QFT more as an {\bf extra layer} on top of QM to allow for a varying number of particles. But I don't see any way to recover QM from QFT in some limit.

So I see QFT not as a full fledged formalism. QFT seems to me to be an extra sets of rules put on top of QM. In other words, QFT is not a more general theory which encompasses QM in some limit (the way GR encompasses Newtonian gravity, say). I would say that QFT is really QM with an extra structure: the possibility of having the number of particles changing.

Because of that, within QFT itself some questions can not be answered. If I have a bunch of intereacting particles in a box and I decrease the energy, I know that a will have a fixed number of real (as opposed to virtual) particles. But how is that shown in QFT? How does one show that the number of particles becomes conserved in the low energy limit?

this is what bothers me. How is the varying number approach of QFT connected in some limit with the fixed number view of QM. And how are all the results of QM recovered from QFT (including position measurements)

10. Sep 29, 2007

### meopemuk

Traditional QFT does not answer this question, because it doesn't pay attention to multiparticle wave functions. Why? Because it doesn't need them. The goal of QFT is to calculate the S-matrix, and this can be done without considering particle wave functions in the position space. However, it is not difficult to define wave functions of many-particle states in the Fock space and study their properties.

Eugene.

Last edited: Sep 29, 2007
11. Sep 29, 2007

### meopemuk

In my opinion, you are not going to establish the connection QFT <-> QM if QFT is formulated in the traditional bare particle representation. As I said earlier, the reason is in the structure of interaction, which contains terms like a*a*a*, a*a*a, a*aa, aaa. Due to these terms, particles get created and annihilated in infinite numbers even in the 0-particle and 1-particle states. There are permanent "self-interaction" effects that never go away. So, even a single non-relativistic bare particle cannot be described within the fixed-number-of-particles QM formalism.

Probably, I sound like a broken record, but I don't see any other way to establish the connection QFT <-> QM except formulating QFT in the "dressed particle" formalism, where the simplest interaction terms that can change the number of particles are a*a*a*aa and a*a*aaa, which means that particles cannot be created spontaneously. They are created only in collisions of two or more other particles. If the energy of colliding particles is less than the rest energy of the particle a, then the new particle cannot be created. Therefore, for particles moving with low velocities, the processes of creation and annihilation can be ignored, and non-relativistic quantum mechanics appears as a natural limit of the "dressed particle" QFT.

Eugene.

12. Oct 2, 2007

### reilly

Many articles over the past 35 years, and books, Weinberg's in particular, dissertations, mine in particular, and seminars and private conversations have dealt with multi-particle states in QFT, some with considerable fluency.

Given an almost universal recognition of connections between multiparticle states and
QFT, how can you apparently say that QFT, anthropomorphically speaking, need not deal with such states?

And how could you state that QFT is only for computing S-Matrices? Nothing could be further from the truth. Check out quantum optics, and bound state theory, for example. I believe that some have looked at the vacuum of QFT in terms of superconductivity, which is definitely a many body problem.

By the way, based on real experience, I can assure you that computing S-Matrices is generally very difficult, often very tedious. As I'm sure you know, the spin stuff can drive you to madness -- the worst I ever had to do was exchange of a spin 2 particle between two spin one particles. Took about three weeks to get the spin stuff right. Never again.

And by the way your take on QFT interactions is not completely on solid ground. For the moment I suggest you look at Franz Gross's Relativistic QM and Fields theory, both with respect to three point interactions, as an example, with respect to wave functions, and, for that matter S-Matrices.
Regards,
Reilly Atkinson

Last edited: Oct 3, 2007
13. Oct 3, 2007

### meopemuk

I would definitely want to learn more about quantum optics, but my impression was that this field doesn't need to use the full renormalized interaction Hamiltonian of QED. The multi-photon Fock space can be explored without reference to electron-photon interactions. So, the difficult questions of renormalization, self-energies, vacuum polarization, etc. can be avoided. Please correct me if I am wrong.

As far as I know, the bound state theory in QFT (e.g., the Bethe-Salpeter approach) is limited to calculations of energies of bound states. This information is contained already in the S-matrix (the S-matrix has poles at negative energies corresponding to bound states). So, the success of these calculations does not contradict my statement that renormalized QFT is good only for S-matrices.

However, I haven't seen any reliable QFT calculation of wave functions of bound states from first principles. (The knowledge of bound state energies and their wave functions is equivalent to the knowledge of the full Hamiltonian, the time evolution, i.e., it goes beyond the limited scope of the S-matrix). In most cases, e.g., in Lamb's shift calculations in atoms, people simply reuse non-relativistic or Dirac's wave functions. I don't claim that I know everything done in this field. If you aware of works in which both energies and wave functions of bound states were calculated directly from the Hamiltonian of renormalized QED, I would appreciate you giving me the reference.

Are you talking about the "spontaneous symmetry breaking" of the vacuum? I don't think this speculative idea has any experimental proof. Yes, it is a part of a generally accepted (Standard) model. But without direct experimental proof, you can't forbid me to look for other models in which the vacuum is nothing but empty space without particles.

Thank you for the reference.

Eugene.