About bare and physical mass, Juan R

In summary: In Weinberg, the lagrangian of QED appears in volume 1 equation 8.6.1. It contain m.Yes, but that is the bare mass, not the physical mass.The problem with infinites arises when calculating higher order processes using the Lagrangian with the physical mass m. This can be solved by using the bare mass m0 instead in the Lagrangian, and then renormalizing to obtain the correct physical results.The physical mass of the electron is not the "nude" mass, as it is affected by the presence of virtual particles. This also applies to the physical charge.Some argue that this issue of changing mass through renormalization is a flaw in current theoretical physics
  • #1
EL
Science Advisor
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Since my discussion with Juan R in the thread "photon's mass is zero?" under Special and General relativity went off topic I will try to continue it here:

Juan R said:
One always begins with real physical mass m. After one may apply renormalization for eliminate infinites, but real electron mass is m, what is the mass that appears in tables of universal constants.

...

On any case i see no signifcant error in my initial claim one begins with mass m before renormalization. You said "NO, one begins with bare mass", but in at least three standard books one begins with the same m i used in my definition.

As I wrote before (#88) books often start with assuming that m in the Lagrangian is the ordinary mass (i.e. the one you can find in tables) just to later find out that this leads to infinities when calculating higher order processes.
Then this problem can be solved by noticing that if we in the Lagrangian substitute m with the bare mass m0 instead, the amplitudes turn out to be finite when we express them in terms of the physical mass. (Of course we also have to do a charge renormalization, but let's just stick to the mass for simplicity.)

Hence the correct Lagrangian density should include the bare mass (as well as the bare charge), and not the physical (since that leads to infinities). However, all results will of course be expressed in terms of the physical mass (i.e. the one we find in tables).

Please could any mentor or advisor verify or crank down on what I am saying, so we can get an end to this...
 
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  • #2
EL,

I haven't been following the discussion in the other thread, but I think I can answer your question. The Lagrangian is written in terms of the bare mass, bare field, etc. (i.e. the nice pretty looking Lagrangians in your book always refers to the bare quantities). Let us focus on the behavior of the bare mass. The physical mass of a particle is usually defined in terms of the pole of the propagator, and at tree level the pole is located at the bare mass. Therefore, at tree level the bare mass is the physical mass. However, if the propagator is evaluated beyond tree level then one finds that the pole shifts. In other words, the pole of the propagator is not given by the simple parameter m^2 that appears in the Lagrangian. Mass renormalization is then the procedure whereby we correct the pole structure of the propagator so that the propagator maintains its pole at the physical mass.

The really interesting thing to me is that renormalization is not inherently associated with removing infinities. Even in a theory where all momentum integrals converged, mass renormalization would still be necessary because you still have to correct the pole.

If I may, let me recommend Weinberg's magnificent text on quantum field theory to the interested people out there. Everything is wonderfully clear. In particular, an excellent discussion of this very subject (including a reference to the importance of renormalization apart from infinities) can be found around p. 438 in volume 1.
 
  • #3
Juan R, have a look at this:

Physics Monkey said:
The Lagrangian is written in terms of the bare mass, bare field, etc. (i.e. the nice pretty looking Lagrangians in your book always refers to the bare quantities).

Thanks Physics Monkey. Your answer was the one I wanted!
I will try Weinberg as soon as I get some time over.
 
  • #4
Physics Monkey said:
The really interesting thing to me is that renormalization is not inherently associated with removing infinities. Even in a theory where all momentum integrals converged, mass renormalization would still be necessary because you still have to correct the pole.

So in that case the relation beween bare and physical mass won't include any "infinities"?
 
  • #5
EL said:
Since my discussion with Juan R in the thread "photon's mass is zero?" under Special and General relativity went off topic I will try to continue it here:



As I wrote before (#88) books often start with assuming that m in the Lagrangian is the ordinary mass (i.e. the one you can find in tables) just to later find out that this leads to infinities when calculating higher order processes.
Then this problem can be solved by noticing that if we in the Lagrangian substitute m with the bare mass m0 instead, the amplitudes turn out to be finite when we express them in terms of the physical mass. (Of course we also have to do a charge renormalization, but let's just stick to the mass for simplicity.)

Hence the correct Lagrangian density should include the bare mass (as well as the bare charge), and not the physical (since that leads to infinities). However, all results will of course be expressed in terms of the physical mass (i.e. the one we find in tables).

Please could any mentor or advisor verify or crank down on what I am saying, so we can get an end to this...

Thanks by continue this interesting discussion!

My point is as follow. The mass of an electron is m, its rest mass, which appears in handbooks or tables of universal constants.

The Lagrangian contains that m. I already cited to you three textbooks. In Weinberg, the lagrangian of QED appears in volume 1 equation 8.6.1. It contain m.

Then, there are problems with infinites in the computation of interactions and that mass may be changed via renormalization procedure. However "physical" mass after procedure and measured exp is not the mass of the electron "nude". It is really the mass of the electron nude more cloud of virtual particles surrounded the electron. The same about the physical charge.

I also cited a book on quantum physics where this problem of changing of mass -mass before renormalization is different from mass after it- is claimed to be one of main flaws of current theoretical physics. The infinites are artificial.
 
  • #6
Juan R. said:
The mass of an electron is m, its rest mass, which appears in handbooks or tables of universal constants.
Yes but that is the physical mass.

The Lagrangian contains that m.
Yes but that is the bare mass.


However, as Physics Monkey pointed out, at tree level they are the same, and that's why it's possible to calulate first order processes without encountering infinities. But for the theory to be consistent at higher order corrections you need to start from a Lagrangian where m is the bare mass. Hence the correct Lagrangian is written in terms of the bare mass (and charge).


I also cited a book...
You can cite as many books as you want, the problem is that it seems you don't understand them.
 
  • #7
Juan R,

Unfortunately, Weinberg is being a bit careless here. He does not specify whether this is the physical or bare mass. As I said, it is true that the the paramters of the Lagrangian correspond to the physical particle parameters at tree level. Notice that he does not calculate anything beyond tree level in Ch. 8, so his statement is harmless at that point: physical and bare mass are the same. However, please note eq. 11.1.1 in Weinberg where he explictly indicates that the mass, etc. in the Lagrangian you are talking about is the bare mass, etc. He proceeds to break the Lagrangian into a term which looks like the free Lagrangian (but now written with physical paramters), a renormalized interaction term, and the renormalization counterterms.

EL,

You right about the corrections being finite if the momentum integrals are all finite. All this means is that in any interacting field theory, the bare charge and mass get 'dressed' by the interaction.
 
  • #8
Physics Monkey said:
You right about the corrections being finite if the momentum integrals are all finite. All this means is that in any interacting field theory, the bare charge and mass get 'dressed' by the interaction.

Great. Could you give an example of a theory where this happens? I would guess this could occur when using quantum field theory in solid state physics?
 
  • #9
EL said:
Great. Could you give an example of a theory where this happens? I would guess this could occur when using quantum field theory in solid state physics?

Just a guess: any finite field theory would be good enough, no ? Like phi^4 in less than 4 spacetime dimensions ?
 
  • #10
EL said:
Yes but that is the bare mass.

No, you are not fixing my point of the change of mass!

I already cited several books on the topic saying the same i said, including the "Fisica quantica" where clearly states that re-definition of mass and charge is one of flaws of QFT that nobody has solved.

I began again now from Weinberg

Equation (8.6.2) for Lagrangian (i ignore the field "FF" contribution by comodity)

[tex]
\mathcal{L} = - \overline{\Psi} (\gamma^{\mu}[\partial_{\mu} + ieA_{\mu}] + m) \Psi
[/tex]

[tex]m[/tex] is the rest mass that appears in special relativity, in Maxwell electromagnetism, and in the Dirac equation.

Now, this does not work well for the computation of higher orders in scattering, and then one may ad hoc change it a posteriori.

Then the (11.1.1) is (again I omit "FF" terms)

[tex]
\mathcal{L} = - \overline{\Psi_{B}} (\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi_{B}
[/tex]

where the new variables are related to previous one via (11.1.2), (11.1.3), etc. For example the mass is (11.1.5)

[tex]
m \equiv m_{B} +\delta m
[/tex]

But i maintain again my initial points: the bare mass [tex]m_{B}[/tex] is not identical to initial [tex]m[/tex] mass that appears in the lagrangian. Note also that [tex]m[/tex] does not disappear of Lagrangian after of the use of bare constants even if equation (11.1.1) suggests that at the first look. In the (11.1.7) the free Lagrangian is defined in terms of [tex]m[/tex]. instead of [tex]m_{B}[/tex].

Also the E.g. equation 12.1.1 of Michel le Bellac. Quantum and statistical field theory. Oxford university Press, 1991. i cited begin with [tex]m[/tex] and change after to [tex]m_{B}[/tex].

Also the J. Sanchez Guillen and M. A. Braun. Física cuántica. Alianza Editorial S.A., 1993. does and add in its pag 362.

Con ello las constantes m and e que caracterizan al electrón 'desnudo' y entran en el Hamiltoniano en realidad están mál definidas, lo que constituye una cierta debilidad teórica de la física cuantica relativista que no se ha sabido superar todavía.

It is saying that [my translation]

the constants m and e that initially characterize to the "nude" electron and enter on the Hamiltonian [or, of course, the Lagrangian] are really incorrectly defined, which is a theoretical flaw of relativistic quantum physics that nobody has corrected still.

In fact, that Weinberg is doing in chapter 11 that Physics Monkey cited is adding ad hoc counter terms via [tex]\mathcal{L}_{2}[/tex] related to new renormalized constants Z, etc to the initial lagrangian (8.6.2).

It was my point in the other thread, that one may begin with physical rest mass and after changes it ad hoc if one want compute higher orders, if one want compute lower orders, one may use the Lagrangian (8.6.2). In fact, if one want obtain the Dirac equation one may use the (8.6.2) with [tex]m[/tex], and i do not see because i would am wrong. Please any comment!
 
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  • #11
Juan R. said:
It was my point in the other thread, that one may begin with physical rest mass and after changes it ad hoc if one want compute higher orders, if one want compute lower orders, one may use the Lagrangian (8.6.2).

Yes. But why not start from the same Lagrangian in both lowest order and higher order calculations? Since both the Lagrangian with the physical mass and the Lagrangian with the bare mass gives the same result for tree diagrams, which one would you say is the "correct" one? The one which is correct to all orders, or the one which only works in a special case?
 
  • #12
vanesch said:
Just a guess: any finite field theory would be good enough, no ? Like phi^4 in less than 4 spacetime dimensions ?

That's fine to me. :smile:
 
  • #13
EL,

Most field theories in condensed matter still have divergent integrals, but the cutoff is a very physical thing. The lattice spacing is the natural limit to the high energy behavior of the system. As vanesch said, you usually need to look at field theories in low space-time dimensions to see examples of finite theories.

Juan R,

You are correct that bare mass and physical mass are not the same. What is also true is that the Lagrangian,
[tex]\mathcal{L} = - \overline{\Psi} (\gamma^{\mu}[\partial_{\mu} + ieA_{\mu}] + m) \Psi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} [/tex] does not describe particles of mass m. One finds that at tree level approximation, the theory does contain a particle of mass m. However, as the interaction energy is included in higher order loop calculations, the mass of the particle predicted by the theory is altered. Now we want the mass of the particle predicted by theory to be the mass we observe, so we renormalize. We adjust the parameters in the Lagrangian in such a way that the pole of the propagator is always located at the physical mass no matter what order of perturbation theory we are doing. The confusion stems in part from the fact that the electron we observe is not really described by a free Lagrangian but rather is described by the interacting Lagrangian "summed to all orders". The electron we see is already dressed by electromagnetic interactions.
 
  • #14
EL said:
Yes. But why not start from the same Lagrangian in both lowest order and higher order calculations? Since both the Lagrangian with the physical mass and the Lagrangian with the bare mass gives the same result for tree diagrams, which one would you say is the "correct" one? The one which is correct to all orders, or the one which only works in a special case?

I'm sorry but I do not understand to you.

If you want compute self-reaction effects and radiative corrections correctly you may work with

[tex]
\mathcal{L} = - \overline{\Psi_{B}} (\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi_{B}
[/tex]

but if want compute energy spectra of H atom to the Dirac level of precision (without self-reaction), the Dirac-like equation for the electron field [tex]\Psi(x)[/tex] is

[tex]
(\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m) \Psi(x) = 0
[/tex] (1)

not

[tex]
(\gamma_{\mu}[\partial^{\mu} + ieA^{\mu}] + m_{B}) \Psi(x) = 0
[/tex] (2)

or similar

unless you assumed a priori that [tex]\delta m = 0[/tex] and [tex]Z_{2} = 1[/tex] and then counterterms from (11.1.1) are cancelled. But then you are assuming that [tex]m_{B}[/tex] and [tex]m[/tex] are the same and by using (2) you are really using (1) which is the correct.

Renormalization may be done order by order. There is not general prescription. Moreover the correction of mass cannot be computed, only obtained via experiments. I always thought that correct mass was [tex]m[/tex] and introduction of other mass was a "mathematical trick" for accounting effects like interaction of electron with itself and polarization of vacuum. I think that is the reason that all textbooks I know begin with Lagrangian defined in terms of rest mass of the electron [tex]m[/tex] and only in precision computation change the mass, charge used.
 
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  • #15
Physics Monkey said:
However, as the interaction energy is included in higher order loop calculations, the mass of the particle predicted by the theory is altered. Now we want the mass of the particle predicted by theory to be the mass we observe, so we renormalize. We adjust the parameters in the Lagrangian in such a way that the pole of the propagator is always located at the physical mass no matter what order of perturbation theory we are doing. The confusion stems in part from the fact that the electron we observe is not really described by a free Lagrangian but rather is described by the interacting Lagrangian "summed to all orders". The electron we see is already dressed by electromagnetic interactions.

Yes i agree. This was precisely my point in the other thread. The mass and charges of electron nude are altered due polarization of vacuum. E.g. the induced virtual cloud around the electron modifies his initial mass. Electron when moves may also move the cloud surrounding it and his inertial properties vary.

However, standard QFT states that only physical electron is the observed electron, which is the "dressed electron". I claim that

dressed electron = nude electron + virtual cloud

In fact, standard QFT claims that [tex]\delta m[/tex] has no physical sense, since virtual cloud form part of that QFT calls the "observed electron". I think that above decomposition between electron and polarization of vacuum would offer full physical value to [tex]\delta m[/tex]

What is your opinion?
 
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  • #16
Juan R. said:
However, standard QFT states that only physical electron is the observed electron, which is the "dressed electron". I claim that

dressed electron = nude electron + virtual cloud

In fact, standard QFT claims that [tex]\delta m[/tex] has no physical sense, since virtual cloud form part of that QFT calls the "observed electron". I think that above decomposition between electron and polarization of vacuum would offer full physical value to [tex]\delta m[/tex]

What is your opinion?

I agree that this there is a temptation to do so, especially in theories which have small coupling constants (such as QED) and where successive orders seem to have physical meaning (so that the lines in a Feynman diagram seem to have some physical meaning). However, I think it is fundamentally misleading. After all, what do we have ?

We have a theory that is supposed to crank out quantum amplitudes for different measurements (usually scattering experiments) as a function of a few parameters, here mB and eB. It turns out that certain behaviours of those quantum amplitudes are very similar to those of a free field theory, or have in other ways behaviours which make us think of classical particle theories. So we identify certain approximate properties of these quantum amplitudes (in certain limiting conditions) as defining something we call "the physical mass" mP or "the physical charge" eP of the particle. This comes down in setting up an experiment (satisfying the said limiting conditions) to extract these quantities from the experimental results (which are predicted by the quantum amplitudes). Let us for the moment assume that our theory is finite.

This means that the theory gives us a function f1(mB,eB) which gives us mP and f2(mB,eB) which gives us eP:

mP = f1(mB,eB)
eP = f2(mB,eB)

We could then use our experimental knowledge of mP and eP to fix the parameters mB and eB.

However, in the perturbative approach, we introduce an extra perturbation parameter lambda in our theory, and do a series development wrt lambda. So we've now introduced a new function f1(mB,eB,lambda) such that:

f1(mB,eB) = f1(mB,eB,lambda=1), and we write the second term out in a series in lambda:

f1(mB,eB,lambda) = f1_0(mB,eB) + lambda f1_1(mB,eB) + lambda^2 f1_2(mB,eB) + ...

It now turns out that f1_0(mB,eB) = mB

So we have that to zeroth order, mP is equal to mB.

In the same way, we can inverse the relation, and use mP as an input. mB will now be a function of mP and eP:

mB = g1(mP,eP) (the inverse of f1 and f2)
eB = g2(mP,eP)

In a similar series devellopment, we now have:

mB = g1_0(mP,eP) + lambda g1_1(mP,eP) + ...

and it turns out that g1_0 = mP.

All other quantum amplitudes, for other experiments, are of course just a function of eB and mB: A(mB,eB) = A( g1(mP,eP), g2(mP,eP) ) = A{mP,eP}


It is a priori not clear to me why the mathematical trick of writing f1 as a series in lambda, should give physical meanings to the different orders in lambda.

In the case of infinite (but renormalizable) theories, the function f1(mB,eB) is ill-defined, so we introduce an extra parameter C (cutoff). With C, the theory becomes finite, and we are again in the same situation as above, only now with f1(mB,eB,C).

This means that we can have a g1(mP,eP,C) ...

It also means that OTHER quantum amplitudes, A(mB,eB) now become a function of C too: A(mB,eB,C), and the trick of renormalization is that:

A( g1(mP,eP,C), g2(mP,eP,C), C) is, in the limit of large C, asymptotically not dependent anymore on C. If that's true for all A, we say that the theory is renormalizable, because it means that in the end, A is only a function of mP and eP.

The point I'm trying to make is that it is somehow a coincidence that the 0-th order terms have mP and mB coincide. It is not really a coincidence, because these parameters where choosen so that they corresponded to the corresponding parameters in free field theories, which themselves were of course designed to describe free particles with mass mB for instance, so we shouldn't be surprised in fact that this comes out of it again when we look at zeroth order. But there's nothing in fact wrong or magical that the approximate property we understand as "physical mass" is a complicated function of the parameters of our theory.
 
  • #17
Ok, now I'm lost again. Question to you who know this subject well:
Who is right, me or Juan R? Or are we partly right both of us? Or maybe none of us?
 
  • #18
EL said:
Ok, now I'm lost again. Question to you who know this subject well:
Who is right, me or Juan R? Or are we partly right both of us? Or maybe none of us?

My opinion is that you are right, but Juan is also right of course if you stick to lowest-order interactions because both are the same. On top of that, there's an extra confusion if you ask different people, because of the technique of counter terms in the Lagrangian. In that case, you keep the physical mass and charge in the lagrangian, but you add correction terms to the lagrangian. So people doing that would say that you have the physical mass in the lagrangian, not the bare one.
 
  • #19
vanesch said:
My opinion is that you are right, but Juan is also right of course if you stick to lowest-order interactions because both are the same.
Nice to here that! (And of course Juan is right to lowest order, I never doubted that.)

On top of that, there's an extra confusion if you ask different people, because of the technique of counter terms in the Lagrangian. In that case, you keep the physical mass and charge in the lagrangian, but you add correction terms to the lagrangian. So people doing that would say that you have the physical mass in the lagrangian, not the bare one.
Yes sure it's possible to express the Lagrangian in terms of the physical mass if we add correction terms instead, but I don't think that is the case we have been discussing, right Juan R?
 
  • #20
vanesch said:
I agree that this there is a temptation to do so, especially in theories which have small coupling constants (such as QED) and where successive orders seem to have physical meaning (so that the lines in a Feynman diagram seem to have some physical meaning). However, I think it is fundamentally misleading. After all, what do we have ?

We have a theory that is supposed to crank out quantum amplitudes for different measurements (usually scattering experiments) as a function of a few parameters, here mB and eB. It turns out that certain behaviours of those quantum amplitudes are very similar to those of a free field theory, or have in other ways behaviours which make us think of classical particle theories. So we identify certain approximate properties of these quantum amplitudes (in certain limiting conditions) as defining something we call "the physical mass" mP or "the physical charge" eP of the particle. This comes down in setting up an experiment (satisfying the said limiting conditions) to extract these quantities from the experimental results (which are predicted by the quantum amplitudes). Let us for the moment assume that our theory is finite.

This means that the theory gives us a function f1(mB,eB) which gives us mP and f2(mB,eB) which gives us eP:

mP = f1(mB,eB)
eP = f2(mB,eB)

We could then use our experimental knowledge of mP and eP to fix the parameters mB and eB.

However, in the perturbative approach, we introduce an extra perturbation parameter lambda in our theory, and do a series development wrt lambda. So we've now introduced a new function f1(mB,eB,lambda) such that:

f1(mB,eB) = f1(mB,eB,lambda=1), and we write the second term out in a series in lambda:

f1(mB,eB,lambda) = f1_0(mB,eB) + lambda f1_1(mB,eB) + lambda^2 f1_2(mB,eB) + ...

It now turns out that f1_0(mB,eB) = mB

So we have that to zeroth order, mP is equal to mB.

In the same way, we can inverse the relation, and use mP as an input. mB will now be a function of mP and eP:

mB = g1(mP,eP) (the inverse of f1 and f2)
eB = g2(mP,eP)

In a similar series devellopment, we now have:

mB = g1_0(mP,eP) + lambda g1_1(mP,eP) + ...

and it turns out that g1_0 = mP.

All other quantum amplitudes, for other experiments, are of course just a function of eB and mB: A(mB,eB) = A( g1(mP,eP), g2(mP,eP) ) = A{mP,eP}


It is a priori not clear to me why the mathematical trick of writing f1 as a series in lambda, should give physical meanings to the different orders in lambda.

In the case of infinite (but renormalizable) theories, the function f1(mB,eB) is ill-defined, so we introduce an extra parameter C (cutoff). With C, the theory becomes finite, and we are again in the same situation as above, only now with f1(mB,eB,C).

This means that we can have a g1(mP,eP,C) ...

It also means that OTHER quantum amplitudes, A(mB,eB) now become a function of C too: A(mB,eB,C), and the trick of renormalization is that:

A( g1(mP,eP,C), g2(mP,eP,C), C) is, in the limit of large C, asymptotically not dependent anymore on C. If that's true for all A, we say that the theory is renormalizable, because it means that in the end, A is only a function of mP and eP.

The point I'm trying to make is that it is somehow a coincidence that the 0-th order terms have mP and mB coincide. It is not really a coincidence, because these parameters where choosen so that they corresponded to the corresponding parameters in free field theories, which themselves were of course designed to describe free particles with mass mB for instance, so we shouldn't be surprised in fact that this comes out of it again when we look at zeroth order. But there's nothing in fact wrong or magical that the approximate property we understand as "physical mass" is a complicated function of the parameters of our theory.


Are you claiming that, in rigor, there are not particles?
 
  • #21
vanesch said:
My opinion is that you are right, but Juan is also right of course if you stick to lowest-order interactions because both are the same. On top of that, there's an extra confusion if you ask different people, because of the technique of counter terms in the Lagrangian. In that case, you keep the physical mass and charge in the lagrangian, but you add correction terms to the lagrangian. So people doing that would say that you have the physical mass in the lagrangian, not the bare one.

Personally, i prefer to think that real mass is [tex]m[/tex] because

i) at lower orders one works with [tex]m[/tex]

ii) In rest of physics: EM, Dirac, KG, Schrödinger, etc. one works with [tex]m[/tex] and i like unification schemes.

iii) I believe that counterterms add to the standard Lagrangian containing [tex]m[/tex] are not really physical and wait for a future theory without infinites where renormalization was not needed.
 
  • #22
EL said:
Yes sure it's possible to express the Lagrangian in terms of the physical mass if we add correction terms instead, but I don't think that is the case we have been discussing, right Juan R?

Well i always said that one begins with [tex]m[/tex] and after one may change, ad hoc, to [tex]m_{B}[/tex] by consistency (e.g. infinites that may be cancelled).

In the post #10 here, i already expressed that in the full regime one has the Lagrangian defined via [tex]m[/tex] more counterterms added ad hoc.

In fact, that Weinberg is doing in chapter 11 that Physics Monkey cited is adding ad hoc counter terms via [tex]\mathcal{L}_{2}[/tex] related to new renormalized constants Z, etc to the initial Lagrangian (8.6.2).
 
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  • #23
Juan R. said:
Are you claiming that, in rigor, there are not particles?

You mean, in QFT ? In free field theory, there is an equivalence between both viewpoints: a quantum field as a quantized classical field is equivalent to the bookkeeping device of relativistic particles (at least, that's what I understand from Weinberg). But interacting field theories are much stranger. If you could calculate the spectrum of the set of commuting observables (H, Px, Py, Pz) you could answer the question unambiguously whether there are particles or not. Of course, perturbative approaches ALWAYS assume that we can take particles to be there, because we use free field theory propagators between the vertices ; but I don't know if it is strictly required that any QFT has particles as such in all rigor. Clearly, the low-energy limit of QED has well-defined particles (electrons and positrons). For QCD this is already much less evident. But I'm not sure that this must always be the case, so I don't take it a priori for granted.
 
  • #24
vanesch said:
You mean, in QFT ? In free field theory, there is an equivalence between both viewpoints: a quantum field as a quantized classical field is equivalent to the bookkeeping device of relativistic particles (at least, that's what I understand from Weinberg). But interacting field theories are much stranger. If you could calculate the spectrum of the set of commuting observables (H, Px, Py, Pz) you could answer the question unambiguously whether there are particles or not. Of course, perturbative approaches ALWAYS assume that we can take particles to be there, because we use free field theory propagators between the vertices ; but I don't know if it is strictly required that any QFT has particles as such in all rigor. Clearly, the low-energy limit of QED has well-defined particles (electrons and positrons). For QCD this is already much less evident. But I'm not sure that this must always be the case, so I don't take it a priori for granted.

Therefore, the concept of particle is well defined only for free fields (free particles) but when interaction is introduced one can sure that.

Due to that coupling constant on QED is weak one can do

dressed electron = nude electron + virtual cloud.

and asume that concept of particle (free particle) is valid when *modified* (perturbation theory) by interactions with others field of self-field.

That is one asummes that dressed electron is aproximatedly equal to nude electron and the concept of particle remains valid.

But what about the total sum (infinite order) on the QED perturbation series? Can one talk of particle/field still?

Even in the small orders of QED perturbation series. It is not the concept of particle just an approximation, since really there is no such one thins like a field/particle due to interaction. That is factorization of quantum state on Hilbert products of electronstate * EMstate is an approximation, correct?
 
  • #25
It is a fact of experience that we observe experimentally something we call an electron that has a pretty well defined mass and charge. Any theory that is useful for predicting phenomenon involving our observed electron must be able to be viewed as 'containing' the electron we see. As vanesch said, for non-interacting field theories and for interacting field theories treated perturbatively, we can make sense of the theory in terms of particles we observe experimentally (using renormalization). In free field theories, the electron is defined as an eigenstate of the free Hamiltonian, but no one knows any exact eigenstates of the interacting Hamiltonian. It isn't totally clear what the exact energy eigenstates of QED look like. Note for instance that the number operator (defined in terms of the old 'electron' and 'photon' operators) doesn't commute with the interacting Hamiltonian, so it would seem that the exact energy eigenstates of the theory don't contain definite numbers of the quanta described the 'electron' and 'photon' operators.

The question of the equivalence between fields and particles (and what we even mean by particles) in QFT has been attacked in a rigorous manner using the tools of so called general or axiomatic quantum field theory. I would recommend the book by Rudulf Haag, "Local Quantum Physics: Fields, Particles, Algebras" as good place to start if you are interested in this approach.
 
  • #26
Physics Monkey said:
It is a fact of experience that we observe experimentally something we call an electron that has a pretty well defined mass and charge. Any theory that is useful for predicting phenomenon involving our observed electron must be able to be viewed as 'containing' the electron we see. As vanesch said, for non-interacting field theories and for interacting field theories treated perturbatively, we can make sense of the theory in terms of particles we observe experimentally (using renormalization). In free field theories, the electron is defined as an eigenstate of the free Hamiltonian, but no one knows any exact eigenstates of the interacting Hamiltonian. It isn't totally clear what the exact energy eigenstates of QED look like. Note for instance that the number operator (defined in terms of the old 'electron' and 'photon' operators) doesn't commute with the interacting Hamiltonian, so it would seem that the exact energy eigenstates of the theory don't contain definite numbers of the quanta described the 'electron' and 'photon' operators.

The question of the equivalence between fields and particles (and what we even mean by particles) in QFT has been attacked in a rigorous manner using the tools of so called general or axiomatic quantum field theory. I would recommend the book by Rudulf Haag, "Local Quantum Physics: Fields, Particles, Algebras" as good place to start if you are interested in this approach.


Thanks! I am now talking of memory but i think that Lubos Motl wrote an Amazon review on that book, and if i remember correctly Motl said (in his well know style) that Haag was "nonsense".

Effectively, in Motl words

Axiomatic Field Theory has given no physical predictions and it has led to no conceptual developments. Today, Axiomatic Field Theory is not an active field of physics anymore. Moreover, most of its conclusions are believed to be incorrect.

The discovery of the Renormalization Group (RG) showed that many exact - and seemingly rigorous - ideas about the operator algebras were too naive to be true... There exists almost no useful quantum field theory that would satisfy the axioms of Axiomatic Field Theory, and therefore the "theorems" derived within the framework of Axiomatic Field Theory have almost no physical impact. Although there are many correct and useful statements in the book, the number of incorrect and misleading sections is too large and it makes the book useless.
 
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  • #27
Motl did indeed review the book and and gave it a poor score. However, Motl can be rather arrogant and degrading sometimes. Many of the other reviewers have very positive things to say. Somebody like John Baez, who has worked in the field, might have a somewhat different characterization of the relevance of "local quantum physics".
 
  • #28
but I don't know if it is strictly required that any QFT has particles as such in all rigor.
It isn't necessary, I think. An action for a free quantum field can be reformulated to be an action for an infinite number of decoupled harmonic oscillators. One then takes the tensor product of all the Hilbert spaces and obtains a Fock space. That can be separated so that each particle gets a Hilbert space. But in curved spacetime this isn't always possible. This is due to the fact that in Minkowski spacetime we have the Poincare group to pick out a preferred vacuum state which we can't use in curved spacetime.
 
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  • #29
Berislav said:
It isn't necessary, I think. An action for a free quantum field can be reformulated to be an action for an infinite number of decoupled harmonic oscillators. One then takes the tensor product of all the Hilbert spaces and obtains a Fock space. That can be separated so that each particle gets a Hilbert space. But in curved spacetime this isn't always possible. This is due to the fact that in Minkowski spacetime we have the Poincare group to pick out a preferred vacuum state which we can't use in curved spacetime.

But our experimental knowledge is about particles, like Weinberg states. One newer detects any field. Even in classical theory one newer detect fields, one always works with particles. The field is a theoretical interpretation and, by definition, unobservable.
 
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  • #30
Berislav said:
It isn't necessary, I think. An action for a free quantum field can be reformulated to be an action for an infinite number of decoupled harmonic oscillators. One then takes the tensor product of all the Hilbert spaces and obtains a Fock space. That can be separated so that each particle gets a Hilbert space. But in curved spacetime this isn't always possible. This is due to the fact that in Minkowski spacetime we have the Poincare group to pick out a preferred vacuum state which we can't use in curved spacetime.
I think this is always possible. However the definition is not unique for curved spacetimes. But I agree in that the notion of particles may not make always sense. A great example is cosmology on a de-Sitter background where superhorizon modes exists with a radius greater than the Hubble length.
 
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What is the difference between bare mass and physical mass?

Bare mass refers to the intrinsic property of an object that determines its resistance to acceleration, while physical mass refers to the total mass of an object including its energy and momentum. In other words, bare mass is the mass that an object would have at rest, while physical mass takes into account the object's motion and energy.

How is bare mass measured?

Bare mass is typically measured using a balance scale, which compares the mass of an object to a known standard. It can also be measured using other methods such as inertia balance or mass spectrometry.

What factors can affect an object's bare mass?

The bare mass of an object is a fundamental property and is not affected by external factors. However, the physical mass of an object can change due to factors such as velocity, energy, and mass-energy equivalence (E=mc^2).

What is the importance of understanding bare and physical mass in physics?

Understanding bare and physical mass is crucial in physics as it helps us accurately describe and predict the behavior of objects in motion. It also plays a significant role in theories such as relativity and quantum mechanics.

Is there a relationship between bare mass and physical mass?

Yes, there is a relationship between bare mass and physical mass. In the theory of relativity, it is stated that an object's physical mass increases as its velocity increases, approaching its bare mass as it approaches the speed of light. This relationship is also described by the famous equation E=mc^2, where the physical mass of an object is equal to its energy divided by the speed of light squared.

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