Is renormalization the ideal solution?

  • #36
meopemuk
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The non-relativistic Hamiltonian is well defined and self-adjoint. In QED there isn't really a Hamiltonian for hydrogen, just the QED Hamiltonian.

I am OK with that. In the dressed theory there is also a full Hamiltonian ##H= H_0+V## in the entire Fock space. This Hamiltonian is obtained by a unitary transformation from the "QED Hamiltonian". ##H## is a normally-ordered polynomial in particle creation and annihilation operators, like in eq. (4.4.1) of Weinberg's vol. 1. This polynomial is Poincare invariant, cluster separable and has huge (infinite) number of different terms. However, if we are interested only in the hydrogen atom (which I define as a bound state of two particles -- an electron and a proton), we don't need the full interaction ##V##, we can choose only few terms that can be relevant to this physical system.

Let me introduce the following creation/annihilation operators: ##a^{\dagger}, a## for electrons, ##d^{\dagger}, d## for protons, ##c^{\dagger}, c## for photons. Then in a pretty decent approximation, the part of the polynomial ##V## that is relevant to the description of the hydrogen atom is

##V = d^{\dagger}a^{\dagger}da + d^{\dagger}a^{\dagger}dac + d^{\dagger}a^{\dagger}c^{\dagger}da + \ldots##

For brevity, I left only operator symbols and omitted coefficient functions, integral signs and other paraphernalia. If necessary, I can supply the missing info. These three terms have clear physical meanings:

## d^{\dagger}a^{\dagger}da## is the direct electron-proton interaction potential. Its leading part is the usual Coulomb potential, plus there are relativistic (contact, spin-orbit, spin-spin) corrections, plus -- starting from the 4th perturbation order -- there are corrections responsible for the Lamb shift. If we diagonalize ##H_0 + d^{\dagger}a^{\dagger}da## in the electron+proton sector of the Fock space, we will get a decent energy spectrum of hydrogen.

## d^{\dagger}a^{\dagger}dac## is interaction by which the atom can absorb a free photon and jump to a higher energy level

## d^{\dagger}a^{\dagger}c^{\dagger}da## is interaction by which photons are emitted from excited atomic levels. So, the energy widths of the excited levels can be properly described.

In brief, the dressed approach provides a clear physical picture of the hydrogen atom with all important interactions. The relevant part of the Hamiltonian is resembling the familiar non-relativistic hydrogen Hamiltonian. However, there are also relativistic and radiative corrections, which ensure that the whole theory is Poincare invariant and cluster separable.

Now, you are saying that the "proper" field-based QED does not use the ideas of particles, their creation/annihilation operators and their Fock space, when interacting systems (such as the hydrogen atom) are involved. On the other hand, you claim that this theory possesses a finite well-defined cutoff-independent Hamiltonian. I am really interested in learning how this Hamiltonian looks like and how one can extract from it useful info, like the energy spectrum, lifetimes, wave functions, etc?

Eugene.
 
  • #37
DarMM
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Ten years ago I went through all of this with you, giving you the relevant papers, links etc that mathematically prove these facts and how Fock space and creation and annihilation operators will not work and explaining every point in detail. I'm not repeating it.
 
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  • #38
A. Neumaier
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@A. Neumaier so to be clear, it sounds like you're saying that we should abandon the idea of a Fock space
No. A Fock space is alright for approximations, but not for the limit of these approximations.
 
  • #39
A. Neumaier
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I deliberately avoided infrared aspects.
You had to, because you only work perturbatively. As a result, your 2-point function for the electron has a pole at the electron mass while the correct QED 2-point function has there a branch point (as found by a renormalization group analysis). This conclusively proves that your dressed particle QED is not equivalent to QED proper.
In Appendix E.2 of volume 2, I copied Weinberg's proof that 10 Poincare generators of QED ##(\boldsymbol{P}_0, \boldsymbol{J}_0, \boldsymbol{K}_0 + \boldsymbol{Z}, H_0+V)## satisfy the required commutation relations. This proof is non-perturbative. Ten generators of the dressed theory are obtained from QED generators by means of a unitary "dressing" transformation. This means that all commutators are preserved and the dressed theory is rigorously relativistic.
Weinberg has no dressing transform, hence his derivation is alright. But your dressing transforms are only perturbatively defined, and do not have a limit when the order goes to infinity. Thus they are no true unitary transformations in the limit relevant for true QED. This is indeed impossible by Haag's theorem.
In practice, this unitary transformation can be performed only at low perturbation orders, so -- you are right -- some violations of the Poincare invariance are to be expected.
This is not different from the Poincare-non-invariance of QED in finite orders.
It is very different from standard renormalized QED, which is Poincare invariant at every loop order.
 
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  • #40
meopemuk
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...you only work perturbatively. As a result, your 2-point function for the electron has a pole at the electron mass while the correct QED 2-point function has there a branch point (as found by a renormalization group analysis). This conclusively proves that your dressed particle QED is not equivalent to QED proper.

The 2-point function is a field concept, but the dressed particle theory is about particles, not fields. It deals with particle positions, momenta, spins, wave functions, etc. But n-point functions is a foreign concept for the dressed particle theory.

As for the non-equivalence to QED, I can argue that one can make the dressed particle theory as close to QED as one wishes. For example one quick and dirty way to get the dressed Hamiltonian is simply to fit its interactions to the S-matrix calculated at the specific loop order in QED.

But your dressing transforms are only perturbatively defined, and do not have a limit when the order goes to infinity. Thus they are no true unitary transformations in the limit relevant for true QED.

You claimed that the renormalized QED has a finite cutoff-independent Hamiltonian. However this is not what textbooks say. You can find the Lagrangian of QED in eqs. (11.1.6) - (11.1.9) of Weinberg's vol. 1 and in eq. (10.38) of Peskin & Schroeder. (Obviously, the QED Hamiltonian ##H_{QED}## has a similar structure.) These expressions have counterterms that depend on divergent (=cutoff-dependent) constants ##Z_1, Z_2, Z_3, \delta m##. The counterterms are divergent, but (in Weinberg's words) "these terms just suffice to cancel the ultraviolet divergences that arise from loop integrals." That's the whole idea of renormalization, as far as I understand it.

Now, the dressing approach says: I would like to make a unitary transformation of the QED Hamiltonian

##H_d = e^{i\Phi} H_{QED} e^{-i \Phi}##

where the operator ##\Phi## is chosen to satisfy 3 conditions:

1. It is Hermitian in each perturbation order. This means that the dressing transformation ##e^{i\Phi}## is unitary in each perturbation order, thus your objection about the non-unitarity is not valid.
2. It is sufficiently smooth. Then one can prove that the S-matrix computed with ##H_d## is exactly the same as the S-matrix computed with ##H_{QED}##
3. It cancels order-by-order certain "bad" terms in ##H_{QED}##. The "bad" terms are those which act non-trivially on the vacuum and 1-particle states. They are the ones responsible for self-interactions and ultimately for the divergences in the traditional QED.

At the end of this procedure one obtains the dressed Hamiltonian ##H_d## without self-interactions. This Hamiltonian makes exactly the same predictions as ##H_{QED}## about scattering and bound state energies. Not surprisingly, ##H_d## resembles the familiar Hamiltonian of the non-relativistic theory. For example, the leading interaction term is the Coulomb potential between charged particles. As a little bonus, ##H_d## is finite, i.e., all the divergent renormalization constants cancel out in all orders. More precisely, they get absorbed into the transformation operator ##\Phi##, which becomes divergent now, but we shouldn't be bothered, because this operator has no physical meaning.

Now we can use ##H_d## to make all kinds of electrodynamics calculations without ever thinking about cutoffs, counterterms and renormalization. All integrals are ultraviolet-convergent.

Eugene.
 
  • #41
A. Neumaier
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The 2-point function is a field concept, but the dressed particle theory is about particles, not fields. It deals with particle positions, momenta, spins, wave functions, etc. But n-point functions is a foreign concept for the dressed particle theory.
The 2-point function is the integral kernel of the inner product in the renormalized 1-particle Hilbert space (which exists since a single electron does not scatter), hence can be compared with the inner product in the dressed particle theory. They do not agree.
As for the non-equivalence to QED, I can argue that one can make the dressed particle theory as close to QED as one wishes. For example one quick and dirty way to get the dressed Hamiltonian is simply to fit its interactions to the S-matrix calculated at the specific loop order in QED.
You claimed that the renormalized QED has a finite cutoff-independent Hamiltonian.
This is assuming that QED exists as a Wightman field theory. (You need to assume as well that QED exists since your symmetry arguments depend on it...)
However this is not what textbooks say. You can find the Lagrangian of QED in eqs. (11.1.6) - (11.1.9) of Weinberg's vol. 1 and in eq. (10.38) of Peskin & Schroeder. (Obviously, the QED Hamiltonian ##H_{QED}## has a similar structure.)
This is only the approximate, cutoff-dependent unrenormalized Hamiltonian in Fock space describing a noncovariant approximation of QED; without cutoff the Hamiltonian and the other generators satisfy formally the Poincare commutation relations, but mathematically these expressions are meaningless. The limiting Hamiltonian of true QED is an operator on the renormalized Hilbert space, and cannot given in terms of Fock space but in terms of the Wightman reconstruction theorem, as the infinitesimal generator of the translations on the constructed states.
Now, the dressing approach says: I would like to make a unitary transformation of the QED Hamiltonian
##H_d = e^{i\Phi} H_{QED} e^{-i \Phi}##
which is a cutoff-dependent approximation to QED only. No Poincare invariance there; the generators don't satisfy exact Poincare relations.
where the operator ##\Phi## is chosen to satisfy 3 conditions:

1. It is Hermitian in each perturbation order. This means that the dressing transformation ##e^{i\Phi}## is unitary in each perturbation order, thus your objection about the non-unitarity is not valid.
2. It is sufficiently smooth. Then one can prove that the S-matrix computed with ##H_d## is exactly the same as the S-matrix computed with ##H_{QED}##
3. It cancels order-by-order certain "bad" terms in ##H_{QED}##. The "bad" terms are those which act non-trivially on the vacuum and 1-particle states. They are the ones responsible for self-interactions and ultimately for the divergences in the traditional QED.

At the end of this procedure one obtains the dressed Hamiltonian ##H_d## without self-interactions.
Well, you proceed formally without any consideration of limits, thus you only prove formal relations, not true ones. Note that in general, limits are not interchangeable, which is one of the reasons why subtle issues like the existence of QED are technically so difficult.
This Hamiltonian makes exactly the same predictions as ##H_{QED}## about scattering and bound state energies.
No. It makes exactly the same noncovariant predictions as the noncovariant approximations used at the start of the procedure. You don't even try to analyze the limit. Indeed, the limit does not exist, by Haag's theorem.
 
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  • #42
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Abdus Salam was writing before the more rigorous understanding of QFTs we possesses today. Renormalization does result in well-defined Hamiltonians that is a mathematically established fact.

Indeed. Interestingly as was shown by Hawking Zeta Function renormalization is equivalent, and in some cases maybe better than dimensional regularization. What is more, no counter-terms are required and it's basis is a mathematically rigorous and a very well understood area of math called analytic continuation:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.570.4579&rep=rep1&type=pdf
Evidently something similar was tried by Schwinger but he couldn't make it work.

Thanks
Bill
 
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  • #43
A. Neumaier
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There are many works in dressed/clothed particle theory. They usually pay tribute to the seminal paper

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

You can use Google Scholar to find all citations of this work. Currently this search shows 113 results.
Why don't you say that the best exposition is in your [meomepuk = Eugene Stefanovich] 3 volume treatise on quantum electrodynamics ? I think this is a valid statement. Your treatise shows both the potential and the limitations of the dressed particle approach.
Thank you for the generous introduction of my humble book.
The interesting part about your books is the extent to which the dressed particle approach can be made to work, as a (though somewhat pedestrian) application of what nowadays is usually called similarity renormalization. (The linked Google scholar search for this turns up over 1400 results.)

The problematic part of your books is that they lack a self-critical attitude and do not discuss the (to experts obvious) limitations of the approach, but sell them as nonexistent or even progress:
The superluminal effects are not bugs, but unavoidable features of the dressed particles approach.
4. This Hamiltonian is relativistically covariant: there exists a corresponding interacting boost operator such that all commutators of the Poincare Lie algebra are satisfied.
At the end of this procedure one obtains the dressed Hamiltonian ##H_d## without self-interactions. This Hamiltonian makes exactly the same predictions as ##H_{QED}## about scattering and bound state energies.
No. It makes exactly the same noncovariant predictions as the noncovariant approximations used at the start of the procedure. You don't even try to analyze the limit. Indeed, the limit does not exist, by Haag's theorem.
And, most sadly, you seem to be de facto uninterested to learn what is outside your horizon:
Now, you are saying that the "proper" field-based QED does not use the ideas of particles, their creation/annihilation operators and their Fock space, when interacting systems (such as the hydrogen atom) are involved. On the other hand, you claim that this theory possesses a finite well-defined cutoff-independent Hamiltonian. I am really interested in learning how this Hamiltonian looks like and how one can extract from it useful info, like the energy spectrum, lifetimes, wave functions, etc?
Ten years ago I went through all of this with you, giving you the relevant papers, links etc that mathematically prove these facts and how Fock space and creation and annihilation operators will not work and explaining every point in detail. I'm not repeating it.
 
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  • #44
@bhobba that's interesting, but I lost hope once I read the passage "This form of re-calculate a series has been widely used in Physics, one of the most famous examples being the calculation of the force due to “Casimir Effect” using the Zeta regularized value zeta (-3)=-1/120, or as another example, the value zeta(-1)=-1/12 for the divergent sum 1+2+3+4+5+6+7+…. That appears in theoretical physics."

I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect. It seems this paper does something similar at first glance, and if this paper uses that same "form of re-calculate a series" then it should contain the same error. With that in mind I stopped reading it, but if this isn't the case then please let me know.
 
  • #45
A. Neumaier
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I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect.
This caricature is of course incorrect, but the sum of ##kx^k## for all positive integers defines an analytic function with the value of -1/12 at x=1 after removal of the singularity, and its Taylor expansion around zero at x=1 agrees with the series you mention. Something similar happens in all cases where the series of interest is a special case of a power series expansion of an analytic function, and then one can use the method to find the intended sum, though the literal sum diverges. (Note that the singular terms in individual infinite contributions to Feynman diagrams cancel by design of any renormalization procedure, so dropping them in each term of the sum is justified!)
 
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  • #46
@A. Neumaier I wish I was as educated as you, I'm not sure what you mean by "intended sum" and "literal sum" (Edit: Oh, OK). More importantly, are you saying that you think that paper provides a legitimate technique?
 
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  • #47
@A. Neumaier regarding everything else, I only have a Bachelor's in physics at the moment, but I understood enough to find your arguments to be very persuasive. I admire your willingness to go to the lengths that you have here to contribute to sites like this one for the sake of spreading the truth.

The fact that I didn't understand everything makes me want to learn more. You have me asking myself the question: What are all of the ways that we can test the limitations of a quantum theory? I know of many, but clearly, I'm not aware of all of them. Which is very worrisome, not knowing all of these methods can determine whether or not someone spends years confused about an important topic. Unfortunately, it might be impossible to adequately discuss such a broad question on Physics Forums, but maybe it would be appropriate to make a new post regarding that question to avoid going too far off topic. Do you think this question is too broad for Physics Forums?
 
  • #48
Fyi, if you don't ignore me at some point then I will never shut up. I have a divergent number of questions about physics (I also have subtle jokes), and a desire to avoid making myself seem uneducated in public in order to find the answers to them is nothing in comparison to my desire to find those answers.
 
  • #49
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@bhobba I've seen people use the zeta function to claim that the sum of all positive integers is equal to -1/12, but it's my understanding that this is actually incorrect. It seems this paper does something similar at first glance, and if this paper uses that same "form of re-calculate a series" then it should contain the same error. With that in mind I stopped reading it, but if this isn't the case then please let me know./

Dr Neuaimer explained it correctly. If its true or not is often argued about. Its simply how you define infinite sums. If you replace the terms as the special case of a series where the terms involve a variable, and interpret the variable as a complex number, you can often sum it for some values of that variable - let's call it x. Then you can use analytic continuation to define the sum for a much larger range of values of x. Then you substitute the values you started with and viola - you get an answer. There are a number of ways of doing it - my favorite is Borel Summation because explaining it is so simple (it will not work on the Zeta Function though). There other ways to sum it - the general one I like best is Ramanujan Summation - despite Hardy waning against it - he wrote a textbook on summing divergent series.

As an example let's look at Borel Summation. ∑an = ∑an*n!/n! = ∑an*∫(t^n)*e^-t/n!
= ∑∫(an*t^n)*e^-t/n! where we have used n! = ∫(t^n)*e^-t.

The above is rigorously correct but sometimes in physics and applied math we get sloppy and reverse integrals and sums without showing you can do it. This gives ∫∑(an*t^n)*e^-t/n! and is called the Borel Sum. It turns out for series that converge normally you can reverse the sum and integral. So we have a more general summation method allowing the summation of things like 1-1+1-1... A deeper analysis than I will give shows its really a form of analytic continuation. The great mathematician, Borel, when an unknown young man, discovered that his summation method gave the 'right' answer for many classical divergent series. He decided to make a pilgrimage to Stockholm to see Mittag-Leffler, who was the recognized lord of complex analysis. Mittag-Leffler listened politely to what Borel had to say and then, placing his hand upon the complete works by Weierstrass, his teacher, he said in Latin, 'The Master forbids it'. So even experts get fooled by this. Mittag-Leffler eventually came around and cam up with his own method, Mittag-Leffler summation.

We are very lucky having Dr Neumaier and others like him post here - you will find you will look back after a while and be surprised at what you have learned.

Thanks
Bill
 
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  • #50
The fact that we're talking about debatable math is crazy to me. Math is supposed to be the one thing that is black and white, but I guess we've reached the realm of theoretical math where that's no longer the case.

What is more, no counter-terms are required

This is very interesting, honestly, I expected someone to shoot this down. Now I'm extremely curious about this, and I'm reading the paper very closely. If this is legit then what happens to renormalization conditions? I've never been a fan of their presence in QFT, and if this means that we no longer need them then that would be interesting. That's currently what's going through my mind... I'm going to have to think about this some more. Thank you for your input Bill.
 
  • #51
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The fact that we're talking about debatable math is crazy to me. Math is supposed to be the one thing that is black and white, but I guess we've reached the realm of theoretical math where that's no longer the case.

What was it the great polymath (and of course mathematician both pure ad applied) John Von-Neumann said: “Young man, in mathematics you don't understand things. You just get used to them.”

A bit sad really - but the truth. I did a math degree undergrad and was always asking questions with Hilbert's view in mind that was put on his gravestone - "We must know, we will know". Thing is, after more experience and lecturers that said - I knew you were going to ask that, I just knew it - forget it or you are letting yourself in for a whole heap of hurt - the answer at the rigor you are after is contained in tomes you simply would not read. Sometimes I have read such tomes eg Geflands 6 volume set on Generalized Functions and I do not recommend it even though in hindsight I did come out of it with many questions resolved.

That's one of the good things about a site like this - those that have been through the 'wars' can steer others away from pitfalls.

There is much more that can be said about what you wrote - but you have a math based bachelors, best if you research it yourself. If you can't find a satisfactory answer, then post here and we will help.

Thanks
Bill
 
  • #52
ftr
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No. it just means that when one renormalizes it one has much more freedom in choosing the details in the theory. it is somewhat analogous to the freedom in choosing functions analytic at zero. The renormalizable case corresponds to functions that are representable by quartics, while the nonrenormalizable case corresponds to functions that are representable by an arbitrary power series. In the latter case, many more parameters are to be chosen. For an effective theory, only the first few matter. The high order terms only affect the theory at energies too high to be deemed relevant.

Thus the standard model (being renormalizable) is completely fixed by fixing slightly over 30 constants, all of which are known to some meaningful accuracy. Whereas canonical gravity needs infinitely many constants to single out the unique true theory, and only the lowest order constant (the gravitational constant) is known. To determine the next constant we already need to observe quantum gravity effects, which is still beyond the capability of experimenters.

Two questions.
1. Are you saying(or might) that gravity needs the high energy components in the interaction which interact at huge distances, yet in scattering theory where particles collide and they are at "zero" distance can be ignored.
2. AFAIK the constants origins are not known , hence they are non calculable, So how are they energy dependent?(OK alpha is energy dependent from experiment, but still)
 
  • #53
A. Neumaier
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What are all of the ways that we can test the limitations of a quantum theory?
[...]
maybe it would be appropriate to make a new post regarding that question to avoid going too far off topic. Do you think this question is too broad for Physics Forums?
Fyi, if you don't ignore me at some point then I will never shut up. I have a divergent number of questions about physics (I also have subtle jokes), and a desire to avoid making myself seem uneducated in public in order to find the answers to them is nothing in comparison to my desire to find those answers.
Generally, you should stick within one thread to one questions and minor variations of it, and ask sgnificantly different questions in a new thread.
 
  • #54
A. Neumaier
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Two questions.
1. Are you saying(or might) that gravity needs the high energy components in the interaction which interact at huge distances, yet in scattering theory where particles collide and they are at "zero" distance can be ignored.
2. AFAIK the constants origins are not known , hence they are non calculable, So how are they energy dependent?(OK alpha is energy dependent from experiment, but still)
Every QFT needs a suitable high energy formulation to be consistent, but different effective versions of the same theory differ in the high energy details. Energy dependence can be formulated even when the values of the constants are unknown. It is one of the strength of mathematics that one can formulate relations even when they involve unknown quantities.
 
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  • #55
@bhobba I see what you're saying. I realize that there are many lines of thought that we can follow that won't lead to an answer without coming to understand ridiculously complex math, and even then, a definite answer likely won't be found. I'm not a mathematician and so I won't try to pursue that type of problem, but it is likely that my curiosity will lead me to learn about more complex theoretical math in order to find certain deeper understandings about physics. You seem to be advising me against diving in too deep, and I am trying to avoid that since I am not a talented mathematician. My only advantages as an aspiring physicist has been my extreme tenacity and curiosity. I've spent over a decade now (I started at 13) trying to come up with an idea for a research project and I finally found something that got me excited a year before I graduated with a B.S. I've spent the last 3 years trying to develop it into something useful. It has been rough without an advisor as I decided against going to grad school for various reasons. Despite that, I'm just about finished developing it, but there are still a few details left for me to work out. I am obviously an amateur in comparison to tenured professors with PhDs, but nevertheless, I might have something interesting to show you superior physicists. We'll see, I'm sure I'll end up talking to you gentlemen about the details some time very soon. I can't express how eager I am to do that as I have not discussed it with anyone yet... Hopefully it can withstand the scrutiny, my heart tells me that it will as it appears to be elegant, but if not then I'll shed a tear and move on... After all, humility is important if you actually care about finding the truth.

My goal has been to learn only the things that I need to know in order to progress, but the problem with that is that there might be something important that I'm missing. Fortunately, after talking to you gentlemen, I'm no longer extremely suspicious of renormalization... I'm only mildly suspicious, but that's all I need to sleep at night.

@A. Neumaier thank you for taking the time to give me your opinion. I was worried that the question might be too broad. If I do post about it then I'll be sure to write my own answer to the question first and just leave the blanks for other members to fill in.
 
  • #56
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You seem to be advising me against diving in too deep,

I am advising you to dive in as deep as you want - or do not want. Do not get too worried about things that are unclear, but have known answers, if they are not really germane to what you are studying. If answering those kind of questions interests you by all means study them - but you do not have to. An extreme example is Wittgenstein. He was a good aeronautical scientist and to advance his knowledge decided to do a PhD in math at I think Manchester University. Here he came into contact with Bertrand Russell who interested him in the foundations of mathematics and he switched from studying math relevant to aeronautical science to the foundations of math and then onto philosophy. This also happened to Russell himself - he was 7th Wrangler at Cambridge and a career in math seemed inevitable. But he, like Wittgenstein, became interested in its foundations and the rest is history. Still it was well known that the one thing that would be sure to engage Russell was math - he particularly liked talking to Ramanujan, but Hardy and Littlewood were the main mathematicians working with Ramanujan - it was just an interest for Russell.

Thanks
Bill
 
  • #57
Do not get too worried about things that are unclear, but have known answers, if they are not really germane to what you are studying.

I see, thank you for the reassurance/advice.

I've often asked myself the question, especially when I was a kid - how do people shape themselves into good scientists? There's some obvious answers to the question such as constantly studying, but I always felt that the full answer was more complicated and extremely important to think about. I think your advice is definitely part of the right answer if you wish to use your time studying as effectively as possible.

Thanks again for the input @everyone.
 
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  • #58
Elias1960
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Is renormalizability considered a physical quality of a theory or merely a mathematical property? If it's merely mathematical why was it such a crucial guideline in forming the SM?
I would say it is, in the concept of effective field theory, a natural property of the large distance approximation. The non-renormalizable terms decrease faster with increasing distance. So what remains are the renormalizable terms. As long as there are any - for gravity, there are none, so the surviving term is also non-renormalizable. And, therefore, for the same reason, very small in comparison with the other forces. So, this has some explanatory power too.
 

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