"For every complex problem, there is a solution that is simple, neat and wrong"

H.L. Mencken

The Quote Of The Week - A Solution To Every Problem

"For every complex problem, there is a solution that is simple, neat and wrong"

H.L. Mencken

Oy vey. Vladimir, please read Chapter 7 of Peskin and Schroeder.

Heather Logan (not verified) | 09/13/13 | 06:59 AM

Yes, I am reading: "*The quantity ***m** is the exact mass of a single particle - the exact energy eigenvalue at rest." It means, in **this** theory, which is, of course, **implied **to be a correct theory.

Then: "*This quantity will in general differ from the value of the mass parameter that appears in the*

Lagrangian. We will refer to the parameter in the Lagrangian as m_0, the bare mass and refer to**m** as the physical mass of the boson. Only the physical mass m is directly observable." It means, if we leave **m** in our equations, our results will be wrong. So we subtract the wrong corrections to **m**, but we present it as "absorbing" them by m_0 to obtain **m**.

There was no m_0 in nature, nor in our theory project. We just advanced a wrong interaction (self-action), we obtained wrong corrections (original coefficients of equations changed), and in order to make ends meet, we blame the original**m** to be m_0 (!?) and at the same time we say that our corrections to it are right! This is exactly cheating described in my previous answer.

Interaction in QFT should not change the equation coefficients, it should change the occupation numbers. If one fails to obtain the occupation number dynamics, it is not a nature property, but a one's fault.

Then: "

Lagrangian. We will refer to the parameter in the Lagrangian as m_0, the bare mass and refer to

There was no m_0 in nature, nor in our theory project. We just advanced a wrong interaction (self-action), we obtained wrong corrections (original coefficients of equations changed), and in order to make ends meet, we blame the original

Interaction in QFT should not change the equation coefficients, it should change the occupation numbers. If one fails to obtain the occupation number dynamics, it is not a nature property, but a one's fault.

Vladimir Kalitvia... | 09/13/13 | 11:49 AM

That's simply because a theory needing (infinite) renormalization is an effective theory...

So, in a way, it's the thery's fault. Anyone who is in a minimal field know agrees on this: nothing fancy or conspirational.

anbar (not verified) | 09/13/13 | 13:09 PM

Dear anbar,

Indeed, nothing fancy or conspiratorial. You have just to believe in bare particles and you have to say: "As long as we do not know physics of short distances, our theory is obliged to be stuffed with wrong results, but it is OK, - we know how to obtain good results from bad ones, see the universal renormalization prescription". Simple, neat, right, and unique!

Anyone agrees but me. Highly energetic excitations, right or wrong, are not excited, sorry for pun. Only in wrongly made theories they cause problems in**any** calculation, even for a free motion.

Indeed, nothing fancy or conspiratorial. You have just to believe in bare particles and you have to say: "As long as we do not know physics of short distances, our theory is obliged to be stuffed with wrong results, but it is OK, - we know how to obtain good results from bad ones, see the universal renormalization prescription". Simple, neat, right, and unique!

Anyone agrees but me. Highly energetic excitations, right or wrong, are not excited, sorry for pun. Only in wrongly made theories they cause problems in

Vladimir Kalitvia... | 09/14/13 | 12:13 PM

Vladimir, you don't seem to be getting the point: the theory being effective is the very reason why you don't have to believe in bare particles... They are just a computational tool

anbar (not verified) | 09/15/13 | 14:38 PM

In QFT they (their constants) are a computational tool serving to "absorb" perturbative corrections, right? So these perturbative corrections are wrong if one gets rid of them with help of inventing bare parameters in course of our calculations. Read my opus here.

Vladimir Kalitvia... | 09/15/13 | 15:00 PM

So you're saying that loop corrections are wrong, right? Therefore there is no Higgs decay in two photons, right?

Please, can you explain me the observation of the H -> γγ made by ATLAS and CMS?

Claudio Corti (not verified) | 09/21/13 | 14:48 PM

Vladimir Kalitvia... | 09/22/13 | 03:40 AM

Higgs decay in two photons is an impossible process without loop corrections, still require renormalization to come up with a finite results. What's wrong with this example?

The Standard Model is a coherent theory, if renormalization is wrong for the photon or electron self-energy, than it's wrong also for Higgs decay.

If you prefere QED, then let's talk about the anomalous magnetic moment of the muon, verified experimentally with more than 10 significant digits: it's only a coincidence that renormalization in this case works?

Or we can talk about the Lamb shift: another incredible coincidence in which a wrong theory, renormalization, give us the correct result.

Claudio Corti (not verified) | 09/22/13 | 05:06 AM

You touched the very main question: if renormalization is wrong, then why does it work? Can it be another incredible coincidence or it is the true feature of interactions?

My answers are simple ant right:

1) We write a*wrong *interaction because we use analogy rather than physics to write it in case of completely coupled systems,

2) We*remove the wrong part* of our interaction from results by hand, because otherwise it just does not work. This is called renormalization. We force our wrong results to be right. It is not a calculation, but a modification of calculation results. Pretext? Our theory MUST describe and DESCRIBES reality. Another pretext - it luckily works!

3) Such a prescription only "works" in very rare cases (all the others are non renormalizable), and its working is a fluke.

I showed all this on**a simple example** of coupling two equations: our initially wrong ansatz, our wrong results, and our way of obtaining good results from bad ones. My model is exactly solvable, so you can see that success of renormalization is a fluke and there is nothing physical behind its success.

My answers are simple ant right:

1) We write a

2) We

3) Such a prescription only "works" in very rare cases (all the others are non renormalizable), and its working is a fluke.

I showed all this on

Vladimir Kalitvia... | 09/22/13 | 08:00 AM

must describe the reality, thus f(x) is in fact g(x). You can present it as a physical constraint imposed on the theory. Hence, you obtain g(x) in your theory which is, of course, a great success of principles. Simple, neat, and wrong.