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Why can we not have non-integer powers of fields in a QFT Lagrangian, eg. $\phi^{5/2}$? Or if we wanted a charged $\phi^3$ theory, could we not have a $(\phi^*\phi)^{3/2}$ term?

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This is a property of renormalization group fixed points---- in usual circumstances, with local Lagrangians, noninteger powers get smoothed out to integer powers. To understand qualitatively why, consider the simplest example of renormalization--- the central limit theorem.

One dimensional fields--- Central Limit Theorem

Suppose I have a quantity X(t) which is determined by the sum of many random quantities over time. The quantities have some mean, which leads X to drift up or down, but subtract out a linear quantity $X'(t) = X(t) - mt$ and fix m by requiring that X(t) have the same mean now as at a later time (in finance/probability language, make X a Martingale). Then all that is left are the fluctuations.

It is well known that these fluctuations average out by the central limit theorem, so that X at a later time $t+\epsilon$ is a gaussian centered around the value of X(t), as long as $\epsilon$ is large enough to include many independent quantities which add up to a gaussian. So the probability density of $X(t+\epsilon)$ is:

$$ \rho(X(t+\epsilon)|X(t)) = e^{-a(x(t+\epsilon)-x(t))^2}$$

Further, the increments are independent at different times, so that the total probability of a path is the product of these independent increments

$$ P(X(t)) = e^{-\int \dot{X}^2}$$

Where you need to choose the scaling of the fluctuations of X appropriately with $\epsilon$ to get to the continuum limit (a trivial case of renormalization), and this P(X) is the weight function in a path integral. Notice that you get an exactly quadratic Lagrangian without any fine-tuning. This is the generic situation. Further, the analytic continuation of this is the particle action for nonrelativistic quantum mechanics, and if you reinterpret time as proper time, this quadratic action is the relativistic propagator as well.

If you make a Lagrangian which has the wrong power, not two, then it will renormalize to the quadratic Lagrangian under very general conditions. For instance, suppose

$$ P(X(t+\epsilon)|X(t) ) = e^{ -|X(t+\epsilon)-X(t)|^{16.7} }$$

This leads to a boxy short-scale motion, but if you average over a few $\epsilon$'s, you recover a quadratic Lagrangian again.

To see why, there is a useful heuristic. Suppose I add a term to the action of the form

$$ \dot X^2 + |(\dot X)|^n $$

If I define the scale of fluctuations in X by setting the $\dot X$ term to 1 (meaning that I normalize the scale of X by the random-walk fluctuations left over when the mean drift is zero), then X has scale dimension -1/2. The new term, integrated in time, has scale dimension which is determined by dimensional analysis to be -n/2 + n-1, so that if n>2, the scale dimension is negative. This is obvious, because the n=2 term sets the scaling, so any higher power must be more negative.

The scale dimension tells you how important the perturbation is at long scales, because it determines how fast the correlations in the quantity fall off. So for integer n, all the correlations disappear at long distances, and you are free to conclude that the theory is perfectly Brownian at long distances.

This is true for integer n. For noninteger n, there are subtleties. If you have exponentials of powers, the resulting distribution for $X(t+\epsilon)$ always has a finite variance, and you always recover the central limit theorem at long distances. But if you make the distribution Levy, you can get new central limit fixed points. These are not exponentials of fractional powers, but the distributions themselves have fractional power tails (the Lagrangian has logarithms of fractional powers).

Since there is no rigorous mathematical theory of quantum fields, these heuristics have to be the guide. You can see them work with lattice models, so they are certainly correct, but proving them is just out of reach of current mathematical methods.

Free field Lagrangians are (free) Central Limit Fixed Points

The central limit points are given by free field Lagrangians. These are quadratic, so they are central-limit stable. If you average a free field at short distances to get a longer distance description, you get the same free field, except that you can recover certain symmetries, like rotational or Lorentz invariance, which might not be there originally. But ignore this. The fundamental renormalization criterion is that you start from a renormalization fixed point, and if you want interactions, you perturb away from this free fixed point by adding nonlinearities.

These fixed points are not all stable to nonlinear interactions, they get perturbed by generic polynomial interactions. If you start with a scalar field

$$\int (\partial\phi)^2 + P(\phi) d^dx$$

The scale dimension of $\phi$ is (d-2)/2 (zero in 2d, 1/2 in 3d, 1 in 4d), which is found by dimensional analysis. This dimensional analysis is assuming that the coefficient of the gradient term is 1, which is exactly for the same reason as in 1d, the coefficient of the gradient term is analogous to the coefficient of the time-derivative term in Brownian motion--- setting it to one normalizes the fluctuations of the field to be those of the free field at long distances. You should be confident that any free field action which is reasonably close to this one will converge to this when you look at it at long distances. This perhaps requires a few discrete symmetries to ensure that you recover rotational invariance, but the intuition is that it is a central limit theorem again).

With this scaling, higher derivative terms have negative scale dimension, so they should disappear when you look at long distances--- their contribution only alters the long-distance central-limit fluctuations, and this just changes the normalization of the field.

Each polynomial interaction also has a scale dimension, which gives you a first order view of how important it is. You can check that in 4d, the naive scale dimension of a term of the form $\phi^n$ is positive only for n=1,2,3, it is zero for n=4, and it is negative for higher n.

This means that there are three coefficients which alter the fluctuations at long distances, which are the linear, quadratic, cubic and quartic terms. The linear term is analogous to the drift in the Brownian motion, and it can be absorbed into a field redefinition. The quadratic term is the mass, and must be tuned to close to zero to approach a continuum limit. The cubic term is a relevant interaction, and the quartic term is a marginal interaction. The space of scalar theories is defined by the quadratic/cubic/quartic couplings.

This is not a theorem, but it should be, and it will be one day. For now, it suffices to say that if you consider such theories, and you look at them at longer distance scales, these coefficients are the only ones that blow up or stay stable. For any other polynomial, the coefficients fall away to zero (I am neglecting polynomial terms of the form $\phi|\nabla\phi|$, or which break rotational invariance, but have positive scale dimension)

If you impose $\phi$ to $-\phi$ symmetry, cubic terms are forbidden, and you only get quadratic/quartic terms. I will assume your model treats positive and negative values symmetrically from now on, although the general case is just slightly more complicated.

The expectation is that adding a term of the form $(|\phi|)^{3/2}$ should renormalize at long distances to a quadratic/quartic coupling of some sort. The reason is that the short distance fluctuations of the field are wild, the $|\phi|^{3/2}$ will get averaged over many neighbors, and if you just look at the correlations of fields at relatively long distances, you will see a RG fixed point, and we know the polynomial ones, and we conjecture that they exhaust the space.

If central limit analogies are not persuasive, one can justify this conjecture using the particle path formulation of field theory. Each polynomial term is a particle point-interaction, of k-particles with each other, and we assume that at long distances, the particles are far apart and free. Whatever the short distance interaction, even if it is a complicated thing, at long distances it should look like the scattering of asymptotic particles, 2->2, 2->3 1->2, and these are all polynomial terms. If there is a complicated superposition of many-particle states which looks like a 2->3.5 interaction, at long distances, you expect the particles to separate, and you recover a polynomial interaction.

None of this is rigorous, and it is very likely that there are counterexamples. In two dimensions, this is completely false, because the scale dimension of phi is zero, so every function of $\phi$ is as good as any other. The condition of being a RG fixed point can be fruitfully analyzed by assuming conformal invariance, and then there is a whole world of examples which differ from the higher dimensional intuition. But in 2d (one space one time dimension) particles don't separate with time, they move together forever, so the idea that you should recover a decomposed scattering picture of interactions at long-distance fails.

Fine Tunings

The scalar example above required fine-tuning of the quadratic term to a special value to reach a nontrivially flucutating limit, because there were terms with positive mass dimension. This is a problem for physics, because to tune things precisely is not physically plausible.

The systems which do not require fine tuning with no space-time symmetry beyond Lorentz invariance exclude scalars, and only have chiral fermions and spin-1 gauge fields, where the fixed points are self-tuning--- there are no terms which have positive scale dimensions for these guys. The standard model is build entirely out of these, with one scalar Higgs field which has a fine tuned.

If you allow supersymmetry, you get a whole world of fluctuating scalar limits, which are protected. These theories are studied widely.

Fractional powers

It is possible to make Levy fields in higher dimensions, by having a fractional power propagator. Such Levy fields have a Lagrangian which has fractional powers of momentum, but not fractional powers of the fields. As far as I know, these nonlocal fields are not studied in the literature, but they appear in a formal way (no physics) in the program of "analytic regularization" of the 1960s, which became dimensional regularization in the 1970s, and in Banks-Zaks fixed points or "Unparticle" theories. There is a recent paper of Haba which analyzes these theories.

There is no full classification of renormalization group fixed points, and it is likely that there are many more that we have missed. But these are not likely to be Lorentz invariant theories.

Once you leave the world of unitary quantum theories, you can get statistical field theories with all sorts of potentials. In particular, the Lifschitz statistical field theory

$$ \int |\nabla^2\phi|^2 + Z|\nabla\phi|^2 + V(\phi) d^4x$$

Gives a dimensionless $\phi$, and should be as rich a statistical theory in 4d for choices of V as conformal theories in 2d. But this is not a unitary theory, and it has been only studied in limited ways. Near eight dimensions, it has an $\epsilon$ expansion which was worked out by Mukhamel in the late 1970s, and which is very close to the normal scalar $\epsilon$ expansion. But in 4d, where it is most interesting, nothing is known.

Even studying this numerically is a challenge, because tuning Z to zero requires a lot of work. These type of Lifschitz points are in vogue right now, because Horava has speculatively connected theories of this sort to quantum gravity. But they are interesting in their own right, even for the nonunitary purely statistical case, because every symmetric renormalization group fixed point is mathematically special and should be understood exhaustively.

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So, if there is no requirement of renormalizability, the interaction Lagrangian can be any combination of fields. S. Weinberg proposed once a non-linear Lagrangian to match the current algebra results. Also, we cannot exclude a possibility to write interaction Lagrangians without invoking filed products. –  Vladimir Kalitvianski Oct 20 '11 at 13:34
    
Ron, can you say something about my toy model of renormalizations in the chat room, please? chat.stackexchange.com/rooms/1606/… –  Vladimir Kalitvianski Oct 20 '11 at 13:49
    
@Vladimir: The nonlinear models which can match current algebra results renormalizes to a free field theory at long distances. They are not a counterexample to the general rule of renormalizability, but explain how it arises. The requirement of renormalizability is assuming that there is no structure at short distances--- that you are in the continuum limit. The point of renormalization is that only certain special polynomial interactions show up as continuum limits, and these are the ones observed. –  Ron Maimon Oct 20 '11 at 16:24
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Ron, I think I love you :-). Seriously, thanks for taking the time to go into this in depth. I have to say, though, I'm still left wondering if I could have a $\phi^3$ term if I wanted charged scalars (i.e. presumably it would have to be $(\phi^*\phi)^{3/2}$)? –  James Oct 22 '11 at 12:37
    
@James: Such a term renormalizes to a mass correction $\phi^*\phi$ plus 2->2 scattering $(\phi*\phi)^2$. There are no other renormalizable terms. You can understand this in particle terms--- charged particles can't go 1->3 because it doesn't conserve charge. You can make a term like this with different charged fields of charge 1 and 2, but it will be polynomial. –  Ron Maimon Oct 22 '11 at 13:53
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