21 Renormalization group
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1 Renormalization group. Renormalization and interpolation Probably because they describe point particles, quantum field theories are divergent. Unknown physics at very short distance scales, removes these infinities. Since these infinities really are absent, we can cancel them consistently in renormalizable theories by a procedure called renormalization. One starts with an action that contains infinite, unknown charges and masses, such as e = e e and m = m m, inwhich e is the charge of the electron and m is its mass. One then uses e and m to cancel unwanted infinite terms as they appear in perturbation theory. Because the underlying theory is finite, the value of a divergent scattering amplitude may change by a finite amount when we compute it at two different sets of initial and final momenta. This happens, for example, in the theory of a scalar field with action density L i m g 4. (.) 4! The amplitude for the elastic scattering of two bosons of initial four-momenta p and p into two of final momenta p and p is A = g g 6 k 3 dk +[k + m tx( x)] n dx [k + m sx( x)] +[k + m ux( x)] o (.) to one-loop order (Weinberg, 995, section.). In this formula, s, t, and u are the Mandelstam variables s = (p + p ), t = (p p ), and u = (p p ), and after a Wick rotation k = k + k + k + k 3.
2 756 Renormalization group The amplitude A(s, t, u) diverges logarithmically, but the di erence between it and its value A = A(s,t,u ) at some point (s,t,u )isfinite g A(s, t, u) A = 6 k 3 dk dx x( x)(s s )[k +m (s + s )x( x)] [k + m sx( x)] [k + m s x( x)] +(s, s! t, t )+ (s, s! u, u ). (.3) The second and third terms within the big curly brackets are the same as the first term but with s and s replaced by t and t in the second term and by u and u in the third. The k integral is g apple m s x( x) A(s, t, u) A = dx ln +ln 3 apple m t x( x) m tx( x) +ln m sx( x) apple m u x( x) m ux( x). (.4) If we choose as the renormalization point s = t = u = 4µ /3, then we get the usual result (Weinberg, 995, 996, sections., 8. ).. Renormalization group in quantum field theory We can use the scattering amplitude (.4) to define a running coupling constant g µ at energy scale µ as the experimentally measured, finite, physical amplitude A at s = t = u = µ.thenwithg µ A(s,t,u ), the scattering amplitude (.4) is finite to order g g ( apple +(µ/m) x( x) A(s, t, u) =g µ 3 dx ln s/m (.5) x( x) apple +(µ/m) apple ) x( x) +(µ/m) x( x) +ln t/m +ln x( x) u/m. x( x) Callan (Callan, 97) and Symanzik (Symanzik, 97) noticed that this scattering amplitude, like any physical quantity, is independent of the sliding scale µ. Thus its derivative with respect to µ vanishes t, 3g 3 dx ln +(µ/m) x( dx (µ/m )x( x) +(µ/m) x( x). x) (.6)
3 . Renormalization group in quantum field theory 757 For µ m, the integral is /µ. So at high energies, the running coupling constant obeys the di erential equation (g µ)= 3g 6 = 3g µ 6 (.7) in which the last equality holds to second order in g µ. Integrating the beta function (g µ ), we get ln E E M = ge M µ = dg µ g M (g µ ) = 6 ge dg µ 3 g M gµ = 6. 3 g M g E (.8) So the running coupling constant g µ at energy µ = E is g M g E = 3 g M ln(e/m)/6. (.9) As the energy E = p s rises above M, while staying below the singular value E = M exp(6 /3g M ), the running coupling constant g E slowly increases, as does the scattering amplitude, A g E. Example. (Quantum electrodynamics) Vacuum polarization makes the one-loop amplitude for the scattering of two electrons proportional to A(q )=e + (q ) rather than to e (Gell-Mann and Low, 954), (Weinberg, 995, section.). Here e is the renormalized charge, q = p p is the four-momentum transferred to the first electron, and apple (q )= e x( x)ln + q x( x) m dx (.) represents the polarization of the vacuum. One defines the square of the running coupling constant e µ to be the amplitude A(q ) at q = µ e µ = A(µ )=e + (µ ). (.) For q = µ m, the vacuum-polarization term (µ ) is (exercise.) The amplitude then is apple (µ ) e 6 ln µ m A(q )=e µ and since it must be independent of µ, wehave = d A(q ) + (q ) = d 5 6. (.) + (q ) + (µ ), (.3) e µ + (µ ) d e µ (µ ). (.4)
4 758 Renormalization group So by di erentiating e µ and the vacuum-polarization term (.), we find deµ =e µ (µ ) e d (µ ) deµ µ =e µ (µ ) e e µ 6 µ. (.5) But by (.) the vacuum-polarization term (µ ) is of order e,whichis the same as e µ to lowest order in e µ. Thus we arrive at the Callan-Symanzik equation which we can integrate to ln E M = E M µ = ee µ de µ (e µ)= e3 µ (.6) e M de µ (e µ ) = ee e M de µ e 3 µ =6 e M e E e E = e M e M ln(e/m)/6. (.7) The fine-structure constant e µ/4 slowly rises from =/37.36 at m e to e (45.5GeV) 4 = ln(45.5/.5)/3 = 34.6 (.8) at p s = 9 GeV. When all light charged particles are included, one finds that the fine-structure constant rises to =/8.87 at E = 9 GeV. Example. (Quantum chromodynamics) The beta functions of scalar field theories and of quantum electrodynamics are positive, and so interactions in these theories become stronger at higher energy scales. But Yang- Mills theories have beta functions that can be negative because of the cubic interactions of the gauge fields and the ghost fields (.79). If the gauge group is SU(3), then the beta function is µ dg µ (g µ)= to lowest order in g µ. Integrating, we find ln E E M = ge M µ = dg µ g M (g µ ) = 6 ge dg µ g M gµ 3 g 3 6 = g 3 µ 6 (.9) = 8 g M g E (.)
5 .. Renormalization group in quantum field theory 759 Asymptotic freedom Figure. The strong-structure constant s (E) as given by the one-loop formula (.5) (thin curve) and by a three-loop formula (thick curve) with = 3 MeV and n f = 5 is plotted for m b E m t. This chapter s Matlab scripts are in Renormalization group at github.com/kevinecahill. and g E = g M apple + g M 8 ln E M (.) which shows that as the energy E of a scattering process increases, the running coupling slowly decreases, going to zero at infinite energy, an e ect called asymptotic freedom (Gross and Wilczek, 973; Politzer, 973). If the gauge group is SU(N), and the theory has n f flavors of quarks with masses below µ, then the beta function is (g µ )= g3 µ N 4 n f 6 (.) which is negative as long as n f < N/. Using this beta-function with N = 3 and again integrating, we get instead of (.) g E = g M apple + ( n f /3)g M 6 ln E M. (.3)
6 76 Renormalization group So with we find (exercise.) M 6 exp ( n f /3)gM s (E) g (E) 4 = (33 n f )ln(e / ) (.4) (.5) which expresses the dimensionless QCD coupling constant s (E) appropriate to energy E in terms of a parameter that has the dimension of energy. Sidney Coleman called this dimensional transmutation. For = 3 MeV and n f = 5, Fig.. displays s (E) in the range 4.9 = m b E m t = 7 GeV. The thin curve is the one-loop formula (.5), and the thick curve is a three-loop formula (Weinberg, 996, p. 56)..3 Renormalization group in lattice field theory Let us consider a quantum field theory on a lattice (Gattringer and Lang,, chap. 3) in which the strength of the nonlinear interactions depends upon a single dimensionless coupling constant g. The spacing a of the lattice regulates the infinities, which return as a!. The value of an observable P computed on this lattice will depend upon the lattice spacing a and on the coupling constant g, and so will be a function P (a, g) of these two parameters. The right value of the coupling constant is the value that makes the result of the computation be as close as possible to the physical value P.So the correct coupling constant is not a constant at all, but rather a function g(a) that varies with the lattice spacing or cuto a. Thus as we vary the lattice spacing and go to the continuum limit a!, we must adjust the coupling function g(a) so that what we compute, P (a, g(a)), is equal to the physical value P. That is, g(a) must vary with a so as to keep P (a, g(a)) constant at P (a, g(a)) = P dp (a, g(a)) da Writing this condition as a dimensionless derivative a dp (a, g(a)) da =. (.6) = da dp (a, g(a)) dp (a, g(a)) = = (.7) d ln a da d ln a we arrive at the Callan-Symanzik equation dp (a, = = d ln ln a + P (a, g(a)). (.8) d ln
7 .3 Renormalization group in lattice field theory 76 The coe cient of the second partial derivative with a minus sign is the lattice -function L(g) dg d ln a. (.9) Since the lattice spacing a and the energy scale µ are inversely related, the lattice -function di ers from the continuum beta-function by a minus sign. In SU(N) gauge theory, the first two terms of the lattice -function for small g are L(g) = g 3 g 5 where for n f flavors of light quarks = (4 ) 3 N 3 n f = (4 ) N 3 Nn f N (.3) N n f and N = 3 in quantum chromodynamics. Combining the definition (.9) of the -function with its expansion L(g) = g 3 g 5 for small g, one gets the di erential equation which one may integrate d ln a =lna ln c = dg d ln a = g 3 + g 5 (.3) dg g 3 + g 5 = g + ln to find that the lattice spacing has an essential singularity at g = a(g) =c + g / e /( g ) g + g g (.3) in which c is a constant of integration. The term g is of higher order in g, and if one drops it and absorbs a power of into a new constant of integration, then one finds a(g) = g / e /( g ). (.33) As g!, the lattice spacing a(g) goes to zero very fast (as long as n f < 7 for N = 3). The inverse of this relation (.33) g(a) ln(a )+( / )ln ln(a ) / (.34) shows that the coupling constant slowly goes to zero with a, whichisa lattice version of asymptotic freedom.
8 76 Renormalization group.4 Renormalization group in condensed-matter physics In classical statistical mechanics, the partition function ( ) isasumover all configurations c of exp( H(c)) in which H(c) is the energy of the configuration c. A configuration might be a d-dimensional array of spins or a field in d-dimensional space (not spacetime) (x) = e ik x (k) d d k (.35) in which the integral is over Fourier modes k with a cuto k< at an inverse lattice spacing. The hamiltonian might be H[ ]= (r ) + g + g g d d x. (.36) The partition function ( ) is an integral over all such configurations ( )= exp [ H[ ]] D. (.37) We change variables from (x) toastretchedfield (x/`) with`> `(x) =a` (x/`) =a` e ik x/` (k) d d k (.38) in which a` is a factor that keeps the partition function ( ) invariant (Fisher, 974, 998; Kosterlitz et al., 976; Kadano, 9; Wilson, 97, 975). To have H[ `] =H[ ], we need each derivative term H i to remain (x/`) H i [ `] = d d x = d d x = `da (x/`) `@x i /` a i d d (x/`) = `d a (x ) d d (x )=H i [ ]. That is, we need a` = `( d)/. We also need the potential terms to remain invariant. They will if the coe cients g n vary suitably with the stretch ` H n [ `] = g n ǹ (x) d d x = g n a ǹ n (x/`) d d x = g n a ǹ n (x/`) `d d d (x/`) = g n `d a ǹ n (x ) d d x = H n [ ], that is, if g n (`) =`d+n( d)/ g n. Similar reasoning shows that a coupling constant g n,p of a term with n factors of the field and p spatial derivative of should vary as g n,p (`) =`d+n( d)/ p g n,p. In quantum statistical mechanics, the partition function (.37) i
9 Exercises 763 a path integral over fields (,x) that are periodic in the inverse temperature = /kt, and the hamiltonian is replaced by a euclidian action. If one stretches space leaving inverse temperature invariant, then the scaling formulas g n (`) =`d+n( d)/ g n and g n,p (`) =`d+n( d)/ p g n,p still apply. Coupling constants with positive exponents d + n(d )/ p > become more important at greater spatial scales and are said to be relevant. Those with negative exponents become less important at greater spatial scales and are said to be irrelevant. Those with vanishing exponents are marginal. Further reading Quantum Field Theory in a Nutshell (ee,, chapters III & VI), An Introduction to Quantum Field Theory (Peskin and Schroeder, 995, chapter ), and The Quantum Theory of Fields (Weinberg, 995, 996, sections. & 8. ). Exercises. Show that for µ m, the vacuum polarization term (.) reduces to (.). Hint: Use ln ab=lna +lnb when integrating.. Show that by choosing the energy scale according to (.4), one can derive (.5) from (.3).
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