Volume Effects in Discrete beta functions
VVolume Effects in Discrete β functions Yuzhi Liu ∗ , Y. Meurice, and Haiyuan Zou Department of Physics and AstronomyThe University of IowaE-mail: [email protected]
We calculate discrete beta functions corresponding to the two-lattice matching for the 2
D O ( N ) models and Dyson’s hierarchical model. We describe and explain finite-size effects such as theappearance of a nontrivial infrared fixed point that goes to infinity at infinite volume or the merg-ing of an infrared and an ultraviolet fixed point. We present extensions of the RG flows to thecomplex coupling plane. We discuss the possibility of constructing a continuous beta functionfrom the discrete one by using functional conjugation methods. We briefly discuss the relevanceof these findings for the search of nontrivial fixed points in multiflavor lattice gauge theory mod-els. The XXIX International Symposium on Lattice Field Theory - Lattice 2011July 10-16, 2011Squaw Valley, Lake Tahoe, California ∗ Speaker. c (cid:13) Copyright owned by the author(s) under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike Licence. http://pos.sissa.it/ a r X i v : . [ h e p - l a t ] D ec olume Effects in Discrete β functions Yuzhi Liu
1. Introduction and Background
With the upcoming experimental effort at the Large Hadron Collider (LHC), there has been alot of activity in the lattice community regarding the existence of non-trivial infrared fixed points(IRFP) [1]. Different methods have been used to look for or exclude IRFP and some results arecontroversial. Calculating the discrete β function is one of the methods used. In the following, wediscuss finite volume effects in discrete β functions for spin models that can be solved by numericalmethods. β functions and Two Lattice Matching In the continuum, the Callan-Symanzik β function, β CS , can be calculated using standardperturbation theory β CS ( α ) = − β α − β α − β α + · · · (2.1)where α = g π . The first two coefficients of Eq. 2.1 are regularization scheme independent. Theyare functions of the number of colors N c and the number of quark flavors N f [2, 3]: β = π ( N c − N f ) (2.2) β = π ( N c − N c N f − N c − N c N f ) (2.3)The theory is asymptotically free if β > β > β < α region where perturbation theory breaks down. Therefore, itis essential to locate IRFP nonperturbatively.One method to calculate a discrete β function nonperturbatively is called the two lattice match-ing [5, 6]. It is a way of measuring the running of bare couplings based on the fact that all the ob-servables will have the same value if the models have the same effective action. One can block-spinthe system of volume V , where V is the volume in lattice units, n times and calculate an observablethat will be denoted R ( g , V ) . One can also block-spin the system of smaller volume V / b D n − R ( g (cid:48) , V / b D ) , and tune the new coupling g (cid:48) until the two observablesmatch. If all the observables give the same value for these two coupling configurations g and g (cid:48) ,then one the two blocked system have the same large distance physics and the same physical cor-relation length. In the models discussed below, we only consider one parameter in the couplingspace. In general, many more couplings are generated during the block-spin transformation. Ifonly one relevant parameter exists, all the other irrelevant couplings will eventually die out as oneblock-spins many times near the critical point. Therefore, the examples showed below should applyto a larger class of system. In Ref. [4], we used the following observable: R ( β , V / a D ) ≡ (cid:68) ( ∑ x ∈ B (cid:126) φ x )( ∑ y ∈ B (cid:126) φ y ) (cid:69) β (cid:68) ( ∑ x ∈ B (cid:126) φ x )( ∑ y ∈ B (cid:126) φ y )) (cid:69) β (2.4)2 olume Effects in Discrete β functions Yuzhi Liu where V is the physical volume of the system, a is the lattice spacing, B and B denote neighborblocks and D is the dimensionality of the system. The reason we chose this particular rationalform is that we do not need to calculate the partition function explicitly since it cancels out inboth denominator and numerator. The matching condition reads R ( β , L D ) = R ( β (cid:48) , ( L / b ) D ) , where b is the scaling factor. The discrete β function is defined as ∆ β ( β , L D → ( L / b ) D ) = β − β (cid:48) . Thisdefinition is consistent with the one used in [5, 6]. β denotes either a quantity proportional to g − or the inverse temperature and should not be confused with β CS . In order to calculate ∆ β ( β ) , westart out from large volume L D at β and tune the coupling at smaller volume ( L / b ) D to β so that R ( β , L D ) = R ( β , ( L / b ) D ) . We then match R ( β , L D ) = R ( β , ( L / b ) D ) and so on. By repeatingthis procedure, we get a sequence of β ’s from which we can calculate the discrete β functiondefined above. In the following, we will work with Dyson’s hierarchical Ising model and the D = O ( N ) sigma model in the large N limit.
3. Dyson’s Hierarchical Ising Model
The Hamiltonian of the Dyson’s hierarchical model with 2 n max lattice sites is defined as H = − n max ∑ n = ( c ) n f ( n ) ∑ B ( n ) ( ∑ x ∈ B ( n ) φ x ) (3.1)where c controls the interaction strength for different block sizes. We can include the dimension-ality D through the relation c / = b − − D [7], where b = / D is the scaling factor. The model hasmany nice properties and is an ideal laboratory to test various ideas before one applies them to themore complicated full QCD case. Here is a list of properties we used [8, 9, 10] • For D > f ( n ) =
1, the model has a second order phase transition, which is similar tothe D = • For D ≤ f ( n ) =
1, the model has no phase transition at finite temperature, which isdifferent from the D = • For D = f ( n ) = log ( n ) , the model is equivalent to Anderson model and has a Thoulesseffect (discontinuity in the magnetization).The cases D = 3, 2, 1.7 with f ( n ) = f ( n ) = log ( n + ) ( models with f ( n ) = log ( n + ) have sameproperties as f ( n ) = log ( n ) in the infinite volume limit).
4. How does an IRFP appear/disappear?
Fixed points correspond to zeros of the β function. It has been pointed out that zeros of the β function can disappear in three ways as one or several parameters change [11]. They are schemat-ically described in Fig. 1. The first two ways are interchangeable depending on what parametersare used ( α ↔ / α ). For the third way, the fixed points that disappeared can be recovered in thecomplex parameter plane [12]. This third way can be easily seen for the hierarchical model bytuning the dimensionality D . Figure 2 shows discrete β functions for f ( n ) = log ( n + ) , D = olume Effects in Discrete β functions Yuzhi Liu β α (a) β α (b) β α (c) Figure 1:
Three mechanisms for the loss of the fixed point(s) described in [11]. -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0 2 4 6 8 10 12 14 ∆ β β Hierarchical Ising D=1.9D=1.994D=2 0 UV IR
Figure 2:
Discrete β functions for f ( n ) = log ( n + ) and D =1.9, 1.994, and 2 hierarchical models (dimen-sion increases from top to bottom). -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.9,n=4 vs n=5
RG flowsZeros n=4Zeros n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=1.994,n=4 vs n=5
RG flowsZeros n=4Zeros n=5 -4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IR-4-2 0 2 4 0 2 4 6 8 10 I m β Re β D=2,n=4 vs n=5
UV IRRG flowsZeros n=4Zeros n=5
Figure 3:
Complex RG flows for f ( n ) = log ( n + ) and D =1.9, 1.994, and 2 hierarchical models. The flowsare constructed with the two lattice matching method described in the text. olume Effects in Discrete β functions Yuzhi Liu ∆ β β D=1.9, f(n)=log(n+1)2 and 2 and 2 and 2 and 2 -0.4-0.3-0.2-0.1 0 0.1 0.2 0.3 0.4 0 2 4 6 8 10 12 14 ∆ β β D=1.994, f(n)=log(n+1)2 and 2 and 2 and 2 and 2 -0.4-0.3-0.2-0.1 0 0.1 0.2 0.3 0.4 0 2 4 6 8 10 12 14 ∆ β β D=2, f(n)=log(n+1)2 and 2 and 2 and 2 and 2 Figure 4:
Discrete β functions for f ( n ) = log ( n + ) and D =1.9, 1.994, and 2 hierarchical models. Pseudofixed points may appear for small volumes. Volume dependence is clearly shown in the figure (volumeincreases from bottom to top).
5. Volume Effects
In [4], we systematically analyzed complex RG flows for the f ( n ) = β plane).Fisher’s zeros in lattice gauge models are discussed in [13]. Here we would like to illustrate volumeeffects in ∆ β . For this purpose, we consider the f ( n ) = log ( n + ) hierarchical model. Figure4 shows how discrete β functions change with the volume. When the volume is small and thedimension parameter D is large enough, pseudo fixed points may appear (Fig. 4 middle and right).However as the volume increases, all the zeros of discrete β functions disappear. This is expectedsince there is no second order phase transition for D = f ( n ) = log ( n + ) case. Fixed pointsappearing in smaller volumes are finite volume artifacts.
6. From Discrete to Continuous Flows
Up to now, we have discussed properties of discrete β functions. If one wants to see howthe corresponding continuous Callan-Symanzik β functions behave, we need to do some interpo-lations. In the following, we will construct continuous β functions from discrete ones for O ( N ) models. In [14], complex RG flows and discrete β functions of O ( N ) models have been constructedfrom both two lattice matching and a rescaling method (where the UV cutoff appearing in the di-mensionless expression of the bare mass is rescaled). The relation between β and g is β ∝ g − forboth gauge models and the O ( N ) model. By changing the energy scale, β → β (cid:48) and β CS → β (cid:48) CS ,we get β (cid:48) CS = Λ ∂∂ Λ g (cid:48) = ∂ g (cid:48) ∂ g ( Λ ∂∂ Λ g ) = g (cid:48) g ∂ β (cid:48) ∂ β β CS (6.1)A more general discussion on how to generate continuous flow from step-scaling function has beenprovided in [15]. ∂ β (cid:48) / ∂ β can be obtained from the discrete β function ∂ β (cid:48) ∂ β = − ∂ ∆ β∂ β (6.2)5 olume Effects in Discrete β functions Yuzhi Liu
From Eqs. 6.1 and 6.2, one can easily get β CS from ∆ β by iteration. The only thing one needsto fix is the initial starting point for the interaction, i. e. β CS . Continuous choice can be obtainedby requiring that the asymptotic behavior agrees with expansions at small or large coupling. Figure5 shows the corresponding continuous β functions from both matching and rescaling methods dis-cussed in [14]. Colored solid lines are obtained from matching discrete β functions with different β CS . A pseudo fixed point appeared in the small g region and is a finite volume effect. It will goaway when we further increase the volume. The black dotted line is obtained from the rescalingdiscrete β function. -1-0.8-0.6-0.4-0.2 0 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 β (g) gContinuous β functions for 2D O(N) modelMatchingRescaling Figure 5:
Continuous β functions constructed from both matching and rescaling discrete β functions.
7. Conclusion
We have successfully constructed discrete β functions for f ( n ) = log ( n + ) hierarchical Isingmodels for different volumes. It has been clearly demonstrated that volume effects may generatepseudo fixed points for systems without phase transition. In order to verify whether the observedfixed point(s) is intrinsic or purely finite volume artifact, one needs to do simulations for largervolumes. An alternative way is to calculate Fisher’s zeros and apply finite size scaling technique.The technique of finding Fisher’s zeros from the constructed density of state has been developedfor SU ( ) [16] and tested for U ( ) [17]. The case of SU ( ) with various N f is in progress. Acknowledgments
This research was supported in part by the Department of Energy under Contract No. FG02-91ER40664. 6 olume Effects in Discrete β functions Yuzhi Liu
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