Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Renormalization and running coupling constant

  1. Dec 5, 2011 #1
    Hi everyone,
    even if I continue reading books about renormalization, I have the same basic doubts.
    So we have radiative corrections that give infinites etc.. So what we do is regularize and renormalize the theory.
    What is left at the end? Is it true that loops are no longer there since the parameters have eaten them? I mean, the effect of the loops are inside the new parameters?

    As a consequence of renormalization, couplings now depend on a scale μ, and physical observables must not depend on it (but perturbation approximation does).
    When we say that after renormalization the couplings depend on the energy of the process, is it because they depend on μ? In some formulas I see they depend on q (energy of the process), and on μ (scale), what's the difference?

    Last question, often i find mentioned the word "quantum fluctuations". What exactly are these? I mean, it's everywhere written that in the vacuum pair of particles pop up continuously, is there a way to calculate for example how many of them averagely pop up in a certain region of space and interval of time?

    Thank you
     
  2. jcsd
  3. Dec 5, 2011 #2

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    http://www.mat.univie.ac.at/~neum/ms/ren.pdf contains an account of renormalization with examples at a more elementary level than quantum field theory, where one can see in detail how the infinities arise due to poor modeling and are eliminated without the slightest mathematical blemish. Armed with the insight from these examples, you'll have a much easier time to understand renormalization in QFT.
    Quantum fluctuations are a popular buzzword for the statistical triviality that the variance <A^2> of a random variable A with zero mean is typically not zero - except that A is now an operator. Some people therefore think that this deserves a much more mysterious name.

    Nothing more pops in and out of existence than in the case where you repeat a measurement multiple times and get a nonzero mean square deviation.
     
  4. Dec 6, 2011 #3
    Ok, thank you!!!
     
  5. Dec 6, 2011 #4
    Let me just add that renormalization absolutely does not get rid of loops. Probably the easiest way to assure yourself of this is to recognize that there are processes that only first arise at loop orders in the first place.
     
  6. Dec 6, 2011 #5

    tom.stoer

    User Avatar
    Science Advisor

    You are right, the way renormalization is introduced in most QFT textbooks is like a waste product of a dirty trick.

    You should study e.g. Kadanoff's block spin renormalization which was developed in the context of condensed matter physics w/o need to hide infinities; it shows that renomalization is nothing else but "re-definition of coupling constants" when "probing the interaction at different scales".

    http://en.wikipedia.org/wiki/Renormalization_group
     
  7. Dec 7, 2011 #6
    Thank you all, I'm getting a better idea about renormalization.

    I will tell you something about the fields fluctuation, please say if I am correct.

    Suppose our universe is governed by the free real scalar Klein-Gordon field and suppose it is in the vacuum state. Thus there are no real particles, but there are neither virtual particles popping up or loops, those "vacuum fluctuations" that we have only if we put an interaction term as Phi^4. So there is in spacetime actually nothing, right? The field though fluctuates, which means that if we could measure the value of the field at a point, we would have an average given by the field's VEV, which is 0 on the ground state, but with an infinite statistical deviation (the vev of the field squared is infinite). So there is a dynamical object in spacetime, which is the field, but any kind of particles.

    In the case of electomagnetism (with no interaction) the vacuum space implies the Casimir effect. But since we are in the vacuum and there is no interaction, no virtual particles are there, and the Casimir effect is due just to the field fluctuations (different modes). Why then on wikipedia it is said that the effect is due to virtual particles??

    Be free to comment and add references about the nature of fields.
    Thanks
     
  8. Dec 7, 2011 #7

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    Yes. The vacuum state is the unique state in which the universe is empty.
    The free field is a distribution, hence has no value at a point. Measurable values would be means over a small space-time regions. And their mean square deviations would be finite. Thus, indeed, there are
    ''vacuum fluctuations''. But it is better to avoid the word, since it suggests to the imagination weird things without connection to what the formalism really means.
    Only in the view of some who don't understand the nature of the Casimir force.
    Parrotts don't need to understand in order to repeat what someone has said.
     
  9. Dec 7, 2011 #8

    tom.stoer

    User Avatar
    Science Advisor

    Common misconception; there's a nice paper (R. Jackiw) which shows that the Casimir effect is a one-loop quantum correction.
     
  10. Dec 7, 2011 #9
    But how could it be a one-loop effect if in most textbooks it is derived just after quantizing the free electromagnetic field, and so when no interaction is included? (I couldn't find the Jackiw paper).
     
  11. Dec 7, 2011 #10

    qsa

    User Avatar

    Is there any standard answer. And direct. To the OP question apart from everybodys favourite answer. It has been sixty years. I mean what does the majorty say. Exactly.
     
  12. Dec 8, 2011 #11

    tom.stoer

    User Avatar
    Science Advisor

    http://arxiv.org/abs/hep-th/0503158
    The Casimir Effect and the Quantum Vacuum
    R. L. Jaffe
    (Submitted on 21 Mar 2005)
    Abstract: In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero point energies of quantum fields are "real''. On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero point energies. They are relativistic, quantum forces between charges and currents. The Casimir force (per unit area) between parallel plates vanishes as \alpha, the fine structure constant, goes to zero, and the standard result, which appears to be independent of \alpha, corresponds to the \alpha\to\infty limit.
     
  13. Dec 8, 2011 #12

    qsa

    User Avatar

    What is meant by alpha going to zero and infinity. AFAIK alpha is a constant and it runs at high energy with a very limited range.
     
  14. Dec 8, 2011 #13

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    In QED, and in all formulas derived from it, alpha is a free parameter., and one can study what happens to the formulas when alpha goes to zero or infinite, independent of the fact that the value of alpha realized in Nature is fixed.

    It is very common in physics that one studies a particular system with a small parameter by looking at the system for all values of alpha and considering the limiting cases first. For example, The Planck constant is a small constant; if you take the limit where it is zero, you get classical mechanics (which is very useful in many engineering applications) although in Nature it is not zero.
     
  15. Dec 9, 2011 #14

    qsa

    User Avatar

    ok thanks. of course I know about h, but I have never come across alpha variations, maybe because I thought of it had a different nature than h. so is alpha variation is used for other things that I have missed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Renormalization and running coupling constant
  1. Running couplings (Replies: 1)

  2. Running coupling (Replies: 1)

Loading...