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How much is this renormalization business a problem in QFT?

  1. Jul 25, 2005 #1
    How much is this renormalization business a problem in QFT? Always read it’s complete ‘hand-waving’ and arbitrary, but also that QFT is the most precise theory ever.

    Also found this quote:

    "[Renormalization is] just a stop-gap procedure. There must be some fundamental change in our ideas, probably a change just as fundamental as the passage from Bohr's orbit theory to quantum mechanics. When you get a number turning out to be infinite which ought to be finite, you should admit that there is something wrong with your equations, and not hope that you can get a good theory just by doctoring up that number."
    - Paul Adrien Maurice Dirac (in a 1970's radio interview)
     
  2. jcsd
  3. Jul 25, 2005 #2
    In an easy language you can say that renormalization is a way to get rid of infinities in QFT. For example, you know that an harmonic oscillator(HO) has a non zero lowest energy value. A quantumfield can be seen as a mattress consisted out of a gazzillion harmonic oscillators. So basically, a quantumfield has a lowest energy value that is the sum of all the lowest energy values of the HO's, hence an infinite value. Physically this infinity is useless because we only talk about energy differences relative to the vaccuum level. I mean you talk about E = hbar *w above the vaccuum energy level. So the actual value of the vaccuum level does not matter at all. You can say that if this levels is a positive infinity, there has to be a second class of HO (ie anti-matter) that will yield a negative infinite vaccuum energy level. When you add these two infinites, you get zero. This is not really correct mathematically but the physical idea is clear: If you get an infinity, just add another infinity with opposite sign. Physically, the net effect is that you get rid of this infinity.

    Remember that this is just an attempt to explain the general idea but it is not entirely accurate and complete. I did not mention the fact that renormalization is dependent of parameters like energy scale and so on...

    regards
    marlon
     
  4. Jul 25, 2005 #3

    Meir Achuz

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    Renormalization is required when QFT is solved in perturbation theory.
    The theory can be finite, while each order of perturbation is infiinite.
    Renormalization is showing that these infinities can be made to cancel, order by order.
    There are approaches that are not perturbative. Some are related to "Axiomatic Field Theory" and to "S-Matrix Theory". These approaches can be written down in terms of finite physical couplings and masses, with no need to renormalize.
    The perturbation expansion in e^2 in QED was shown by Dyson to be an asymptotic expansion. This means it can be quite accurate up to a certain order, but would eventually get worse. We are not near that order as yet.
    Dirac may be the smartest physicist ever, and was always careful in his statements.
     
  5. Jul 26, 2005 #4

    Haelfix

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    In this case I think Dirac was wrong. Renormalization seemed somewhat mysterious at the time he said that, but then the Renormalization group ideas came around and it more or less made every understand that it wasn't that big a deal and perfectly natural.

    In fact in this day and age a nonrenormalizable theory isn't viewed as catastrophic anymore, it just means we can't make good predictions with field theory and have to be a little careful.
     
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