1090/33
INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS
ON A PHASE STRUCTURE OF A THREE-DIMENSIONAL cp4 FIELD THEORY
G.V. Efimov
INTERNATIONAL ATOMIC ENERGY AGENCY
UNITED NATIONS EDUCATIONAL, SCIENTIFIC AND CULTURAL ORGANIZATION 1990 MIRAMARE-TRIESTE
T T r r
IC/90/33 INTRODUCTION International Atomic Energy Agency In this paper we will investigate the phase structure of the quantum field models and United Nations Educational Scientific and Cultural Organization 1 (M) INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS - g (1.2)
in a space-time of three dimensions R3. The Lagrangians (1.1 and 1.2) describe a one-component scalar field
coupling constant, and tp0 in (1.2) is a real constant. If the coupling constant is small enough, the Lagrangian (I.I) describes the symmetric interaction, but the Lagrangian (1.2) describes the so-called situation of spontaneous symmetry breaking (SSB) (see, for example. Reft. 1 and 2, containing many earlier references). G.V. Efimov ** Our aim is to investigate the behaviour of the systems (1.1 and 1.2) in the strong cou- International Centre for Theoretical Physics, Trieste, Italy. pling limit, i.e., for 0 —> oo. We will investigate these Lagrangians by the method of canonical transformations. We have used this method for the investigation of the same models in R 2 [2].
By comparison with the Lagrangians in R2 we have the additional divergent diagrams ABSTRACT in R3. Nevertheless, the Lagrangians (1.1 and 1.2) remain to be super-renormalizable and we can write down all necessary counter-terms in the explicit form. We construct the Hamiltonians for Two models of a scalar field
l)z Lagrangians (1.1 and 1.2) and introduce operators with counter-terms. We formulate our problem 3 are considered in R . There are phase transitions in these models for a certain g = gc. In the as follows: first model gtp* the symmetry
gc. The effective perturbation coupling constant is small in both cases.
(1.3)
and what physical picture corresponds to this representation?
2 M1RAMARE - TRIESTE It turned out that the picture is quite different from the situation in R . Namely, the system (1.1) keeps the symmetry tp —> -tp (or all values of the coupling constant although a kind
January 1990 of a phase transition takes place. The symmetry
\c where \c is a critical coupling constant At last we want to stress especially that the effective perturbation coupling constant is small in both cases, i.e., the Lagrangians (I.I and 1.2) describe the systems with weak coupling for any values of g. In other words, the strong coupling regime in the sense that the perturbation expansion has a large coupling constant does not exist in the system s described by the Lagrangians To be submitted for publication. Permanent address: Joint Institute for Nuclear Research, Dubna, USSR. (1.1 and 1.2),
T 3 to remove these divergences we should introduce in the interaction Hamiltonian (2.2) an operator 2. HAMILTONIAN v* IN R Rm containing counter-terms which cancel these divergences in perturbation calculations. This
In this section we consider the theory of a scalar field
<*) (2.6) iv L2 C(x) = -
= m Ea = j 4! 9\ Wm - j Vm + ~ 3! (2.7) We shall consider this theory in the Hamiltonian representation. The second-quantized These divergent diagrams can be computed, for example, in the Euclidean metrics (xo Hamiltonian for the Lagrangian (2.1) is written as iky) where the propagator of the field
Here*2 - *2+Jt|+Jt|.(Jta) + x\ + s|,andfc » y^Jt2 + A22 + fcf We have (23)
Here
(2.9) Here an appropriate regularization should be introduced. (2.4) We can now say that the theory (2.2 and 2.3) is defined, i.e., the S-matrix The fields p( 1) and ir( z) are canonical variables and obey the canonical commutation relations S-Tcxpi-if* dtH?(t)\ (2.10) [pUKipCr')] = [ir(z),ir(5')]=0 is finite in each order of the perturbation expansion.
[?<*),«<« ')]=i6(i~i'). (2.5) 3. CANONICAL TRANSFORMATION The integration in (2.3) is performed over a large "volume" V, here it is a square: Our aim is to realize what is this system (2.1) for different values of the coupling constants j3 and 04. In order to do it we shall use the method of canonical transformations. Let us perform and V = (2 L)2. All operators in (2.3) are taken in the form of normal products. the canonical transformation of our fields This theory contains ultraviolet divergences but it is super-renormalizable, i.e., it has only a finite number of the divergent Feynman diagrams. These diagrams are shown in Fig.l. In order
mass is equal to Hz= f dx
M - m( 1 + f) . (32) where
B is a constant field. The explicit form of this transformation (3.1) can be written but we will not S(m,M) =m2 -Af 4! do it here. 2 2 P(m, Af) = tn B + - 3AB) - A) + 4g4B) We now perform this canonical transformation (3.1) for the Hamiltonian (2.2 and 2.3) (3.8) and go over to the normal ordering in the new fields * (a;) and El ( x). We will use the well-known formulas B(m,M) ,Af) + S4(B*-6AB2 + 3A2)+ ^ ) (3.3) In Here: * : denotes the normal ordering of the fields
( x) correspondingly, * is a constant 4! Kim, M) = - and
A = A(m,M) = £>m(0) - D«(0) = . 3 Using the formulas (2.9) one can get
The canonical transformation of the free Hamiltonian looks as 1 1 , M 1 l 64 ' = Ho1 + I dS: \^-=— + TO2B(SC) + = ©2(i) + L(m, j (3.5) Jv I 2 2 —l(4ir)_3 where (3.9) - M2-m21
Let us put the coefficients of the operator :
Using these formulas we are able to do the canonical transformation of the total Hamil- P(m,AO=0 . (310) tonian (2.2). One can obtain This requirement means that the total Hamiltonian cannot contain the linear term (?}, jj —» GA from the S-matrix. We want to stress that it is the main reason so we cannot do the variational 5 i estimations for the Hamiltonian (2.2 and 2.3). At the same time the canonical transformations do Let us divide F into two subregions not introduce new divergences in the Hamiltonian. All lenns in the representation (3.6) are finite and the operator R contains all necessary counter-terms (see alto [4]). In other words, we can u where say that Eqs.(3.10) define the ground-state of our system. Introducing the dimensionless variables if if (3.16) ft . b^l^B (311) , x3 Then we shall consider that our system (2.1) is described by the representation (2.2 and 2.3) if (Xj,X«) £ F+ and by the representation (3.14) if e(X3,X«) e F_, So that the conditions (3.16) and using the formulas (3.9) one can get for Eq.(3.10) defines the phase structure of the system under consideration. 1 In addition we should remark that Eqs.(3.12) can have several solutions. Then we have -/ (l + |) + 3X4 (b - f) + 3X36 + 12X? In(l + /) = 0 to choose the solution which leads to the lowest vacuum energy. 6+ 2A (63 - 3/6) (3.12) 4 Thus our direct problem is to solve Eqs.(3.12) and to investigate the behaviour of the function e(X3)X4) (3.13). The function £( m, M) in (3.6) and (3.8) has sense of the vacuum energy density. It can be written using the definitions (3.11) THE SYMMETRIC INTERACTION (313) In this section we consider the symmetric initial Lagrangian (1.1), i.e., we have to put in 1 2 2 2 2 (2.3) e(X3pX,) = b + / (l + 0 + >4(6* - 6/ft + 3/ ) + 2Xj6(i - 3/)- (4 -1 nj+(1 + /) In(1 + or turn Eqs.(3.12) are in this case where the parameters / and 6 arc defined by Eqs.(3.12). Prom the physical point of view, the parameter 6 is connected with a classical constant field which defines the minimum of the classical potential in the Hamiltonian (2.2). The parameter (4.2) / is connected with the quantum contributions into self-interaction arising in the quantum system (2.3). Using the second equation in (4.2) we can get for the firston e Thus our Hamiltonian (2.2), after the canonical transformation (3.1) becomes the form (4.3) (3.14) Thus we have two solutions: The nonsymmelrical one Our main point consists in the following. In the initial representation (2.2) or (3.14) the vacuum energy equals to zero but after the canonical transformation this energy equals E(m, M). Let us (4.4) 4X consider the function e(X3,X*). The parameters X3 and \< are changed in (he region and the symmetrical one (X3,X4 : -oo< oo, 0) . (3.15) 6 = 0 . (4.5) 8 Let us consider the nonsymmefrical solution (4.4). Excluding b1 according to (4.4) from and the second equation in (4.2) one can get /1 i /\2 (4.6) The effective coupling constant in this second phase is The solution of this equation exists for -1 < / < 0 only. Then we should substitute (4.4) and (4.13) / = /(X) according to (4.6) into the vacuum energy \(b* -6/b1 + 3/)- We have obtained a very amusing picture. In the symmetrical system we have a kind of the phase transition but the symmetry is not breaking. The only result is that the effective pertur- bation coupling constant in both phases is less than the critical one 24X262in(l (4.7) Xt// Now let us consider the symmetrical solution 6 - 0. The second equation (4.2) can be (4.15) written in the form l + 3X + |-m* tw(lV> . (4.8) One can say that the main remit of strong interaction in the system (1.1) is the contribution to the The vacuum energy is mass of our scalar particle. f (l + fj + 3X/* - 18ir2 in(l + /)- 5. THE TWO-WELL POTENTIAL (4.9) Let us consider the Lagrangian (1.2) with two-well potential which leads to the well- The behaviour of e( X) is shown in Fig.2. known situation of the spontaneously broken symmetry. In this Lagrangian we should go to the one of two minima One can see that for / = 0 Then we obtain i.e., there exists the critical value of the coupling constant Xe which equals to X l + =0 437 (5.1) °=8-[ -\/?] - - (4.10) where m2 = ! as it follows from (4.8) for / = 0. One can see that We have the Lagrangian (2.1) and the Hamiltonian (2.2-3) with e(X)>0 if X < XC, E(X)<0 if X>Xc. (4.11) 9* - 9, S3 = rny/2g . (5.2) When X —> oo the behaviour of the vacuum energy is We can now use all the results of Sections 2 and 3. The dimensionless coupling constants are (4.12) X4=x = -2-, X3 = \/X (5 3) 2um 10 and The second equation (S.5) can be written in the form G* =s, 93 U+ —T= (54) (512) Eqs.(3.12) in this case look as One can check that the real solution of this equation exists for X £ \c = 0.7043 ... (5.13) 0- Solving this equation and putting the solution f(X) into the vacuum energy e(X) in (3.13) =0. (5.5) we obtain the picture shown in Fig.3. Far X ~* oo one can get The first equation has two solutions: The nonsymmetrical one /-XV5T5T [1 + 0(5^ (514) b2 + -7= - 3/ + 12X In(1 + f) • 0 (5.6) Thus we can say that for the coupling constant X > Xe the symmetry p —> —tp restores. and the symmetrical one The mass of panicle grows as *-- (5.7) mXV24inX |l + Of TA-11 • (515) The last solution leads to Gi = 0. The effictive coupling constant decreases Let us consider the nonsymmetrical solution (5.6). The second equation in (5.5) becomes X of the form (516 '" 1 + /(X) >-«o Vr24fnX > | /) (5.8) One can see that this equation has the unique solution THE LAGRANGIANS IN R2 AND R1 /-0. We can now compare the behaviour of our Lagrangians (1.1-2) in the space-time R3 and R 2 (see [3]). The situation is presented in Table 1 and looks completely different. OUT system is Then Eq.(5.6) looks as 2 always nonsymmetrical for g > gc in the space-lime R and it is always symmetrical for g > gc in (5.9) the space-time R 3 irrespective of the initial Lagrangian (1.1-2) for the small coupling constant g. Thus we can say that the different ultraviolet divergences in R 2 and R 3 lead to different physical The vacuum energy e( X) is pictures. (5.10) for the parameter b satisfying (5.9). The coupling constant Oi equals 1 IE Jl-i (5il) Thus the nonsymmetircal solution leads to the initial system (1.2) with the spontaneously broken Acknowledgments symmetry. The author is grateful to Professor Abdus Salam for fruitful discussions, the International Let us now consider the symmetric solution (5.7) for which Atomic Energy Agency and UNESCX) for hospitality at the International Centre for Theoretical Physics, Trieste. G3 =0 . 11 12 REFERENCES Table 1 [ 1J Ta-Pei Cheng and Ling-Fong Li, Gauge Theory of Elementary Particle Physics (Claren- don Press, Oxford, 1984). [2] A.A. Grib, Problem of Noninvariance Vacuum in Quantum Field Theory (in Russian) (Atomizdat, Moscow, 1978). 2 4 3 [3] G.V. Efimov, Int. J. Mod. Phy*. 4 (1989) 4977. R J£** + ff* + 4ffB(s) R* ^i (K1 + S*4 + 4ffB(j)*J SSB R3 SSB if ** + g** 13 14 "T T" .: FIGURE CAPTIONS Fig. 1 Divergent diagrams for the Hamikonian (2.3). Fig.2 The ground-state energy e(\) for the symmetrical Lagrangian (1.1). Fig.3 The ground-state energy e{\) for the Lagrangian (1.2). m m m m Fig.l 15 16 0.7043 ,. 0 O5 1 15 2 - 33 -5O 100 I 15O I 2OO 3OO 350 I 4OO 45O 1 Fig.3 Fig.2 17 T TT Stampato in proprio nella tipografia del Centro Internazionale di Fisica Teorica