Path Integrals in Polar Field Variables in QFT
E.N. Argyres, C.G. Papadopoulos, R.H.P. Kleiss, M.T.M. van Kessel
aa r X i v : . [ h e p - t h ] J a n Path Integrals in Polar Field Variables in QFT
E.N. Argyres C.G. Papadopoulos Institute of Nuclear Physics, NCSR ”Demokritos”, Athens, GreeceM.T.M. van Kessel R.H.P. Kleiss IMAPP, FNWI, Radboud Universiteit Nijmegen, Nijmegen, The NetherlandsOctober 21, 2008
We show how to transform a d -dimensional Euclidean path integral in terms of two (Cartesian) fields to a pathintegral in terms of polar field variables. First we present a conjecture that states how this transformationshould be done. Then we show that this conjecture is correct in the case of two toy models. Finally theconjecture will be proven for a general QFT model with two fields. [email protected] [email protected] [email protected] Introduction
Since Feynman presented his path-integral formulation of quantum mechanics [1] there have been a lot ofapplications for this formalism, both in quantum mechanics and in quantum field theory. In the case ofquantum mechanics it is known how to transform a path integral in terms of the normal (Cartesian) fields toa path integral in terms of curvilinear fields. This is very convenient in calculations for models that possessa rotational symmetry. Such transformations are worked in the case of quantum mechanics by Edwards etal. [2], Peak et al. [3], Lee [4], B¨ohm et al. [5], Gerry et al. [6], Grosche et al. [7] and Kleinert [8].Quantum mechanics corresponds mathematically to one-dimensional quantum field theory. This meansthat the path integral in terms of curvilinear fields is known for quantum field theories in one dimension.For dimensions greater than one it is, up to now, not known how the path integral can be transformed intoa path integral in terms of curvilinear field variables.In this paper we will show how this transformation can be performed, in the case of a d -dimensionalEuclidean quantum field theory with two scalar fields.The outline of this paper is as follows. First we shall discuss the difficulties that one encounters whentrying to transform to polar field variables in a d -dimensional path integral. Then we present a conjectureon how this transformation should be done. We will verify this conjecture for two d -dimensional toy models.After this we will actually prove the conjecture. Finally we will specify our calculations to the one-dimensionalcase, to make contact with the literature on one-dimensional path integrals in terms of polar fields. The generic form of a d -dimensional Euclidean path integral P in two scalar fields ϕ and ϕ is: P = Z D ϕ D ϕ ϕ ( x (1) ) · · · ϕ ( x ( m ) ) ϕ ( y (1) ) · · · ϕ ( y ( n ) ) · exp (cid:18) − h Z d d x (cid:18)
12 ( ∇ ϕ ) + 12 ( ∇ ϕ ) + V ( ϕ , ϕ ) (cid:19)(cid:19) (1)Here x ( i ) and y ( i ) are d -dimensional vectors, x ≡ x ... x d , (2)and ∇ is the d -dimensional vector ∇ ≡ ∂/∂x ... ∂/∂x d . (3)In (1) we have written the path integral P in continuum form. We have to realize that this is just ashorthand notation, and the path integral is only defined on the lattice. Therefore we should actually read(1) as: P = M (∆ , N ) Z ∞−∞ N − Y i ,...,i d =0 dϕ i ,...,i d Z ∞−∞ N − Y i ,...,i d =0 dϕ i ,...,i d · ϕ j (1)1 ,...,j (1) d · · · ϕ j ( m )1 ,...,j ( m ) d ϕ k (1)1 ,...,k (1) d · · · ϕ k ( n )1 ,...,k ( n ) d · exp (cid:18) − h S (cid:19) (4)Here we have introduced a d -dimensional lattice with N points in each direction. We denote the latticespacing by ∆. The length of the lattice is ( N − ≡ L in each direction. The discrete coordinates i are2elated to the continuum coordinates x as: x ≡ − L/ i ... − L/ i d , i , . . . , i d = 0 , . . . , N − V contains no products of derivatives of the fields and the fields themselves. This is what we will assume inthe rest of this paper. The fields on the discrete lattice are denoted by: ϕ i ,...,i d ≡ ϕ ( x ) ϕ i ,...,i d ≡ ϕ ( x ) (6)For these fields we shall assume periodic boundary conditions in all directions. For the 1-direction thismeans: ϕ N,i ,...,i d = ϕ ,i ,...,i d ϕ N,i ,...,i d = ϕ ,i ,...,i d (7)The discrete coordinates j (1) up to j ( m ) and k (1) up to k ( n ) correspond respectively to the continuumcoordinates x (1) up to x ( m ) and y (1) up to y ( n ) . The action S in (4) is given by: S = ∆ d N − X i ,...,i d =0
12 1∆ ( ϕ i +1 ,i ,...,i d − ϕ i ,i ,...,i d ) + . . . +12 1∆ (cid:0) ϕ i ,...,i d − ,i d +1 − ϕ i ,...,i d − ,i d (cid:1) +12 1∆ ( ϕ i +1 ,i ,...,i d − ϕ i ,i ,...,i d ) + . . . +12 1∆ (cid:0) ϕ i ,...,i d − ,i d +1 − ϕ i ,...,i d − ,i d (cid:1) + V ( ϕ i ,...,i d , ϕ i ,...,i d ) ! (8)The function M in (4) is a measure, which is quite unimportant for our purposes. In QFT we are neverinterested in the path integral itself, but rather in the ratio of two path integrals. In this ratio the measure M always drops out. Details about this measure can be found in [4, 9]. When dealing with a normal integral a common technique to solve it is to transform to a different integrationvariable. Since the path integral (4) is merely a 2 n d -dimensional integral, one should also be able to transformto different integration variables in this case. In particular one can transform to polar field variables, r and θ , defined by: ϕ ( x ) = r ( x ) cos θ ( x ) ϕ ( x ) = r ( x ) sin θ ( x ) . (9)It is this transformation to polar field variables that we shall study in this paper. We wish to emphasizehere that not the space-time coordinates, but the quantum fields ϕ and ϕ are transformed to polar fields r and θ .The discreteness of the path integral makes such a variable transformation very complicated. Severaldifficult questions immediately pop up: 3. In principle the whole path integral is only defined on a lattice, so the transformation should also bedone with the path integral in discrete form. This means one cannot simply let the transformationwork on the continuum action. Instead one has to write out the action in discrete form, as in (8),and only then let the transformation work. After this one gets a complicated action, with also termsproportional to the lattice spacing ∆. These terms cannot be discarded, since one has to perform thepath integration first and then take the continuum limit ∆ → ∞ . It is not obvious that the termsproportional to ∆ will not give a finite contribution in this continuum limit. In fact, we will see thatthey do give finite contributions to Green’s functions.2. After the transformation the domain of integration is not h−∞ , ∞i . For the r -variables it is [0 , ∞i ,whereas for the θ -variables it is [ − π, π ]. How does one evaluate such a path integral, especially becausewe can only compute path integrals with perturbation theory? To do perturbation theory we have tobe able to identify a Gaussian part, and the fields in such a Gaussian part are always integrated from −∞ to ∞ .3. After the transformation one gets a Jacobian, how does one deal with this? Since we can only doperturbation theory we also have to identify the Gaussian part and the perturbative part of thisJacobian.From these questions it is clear that transforming to polar variables in a d -dimensional path integralis very complicated. For one-dimensional systems, i.e. quantum-mechanical systems, there is quite someliterature on the transformation to polar variables.In his textbook [4] Lee derives the quantum-mechanical path integral in curvilinear coordinates in chapter19. The result (19.49) is a path integral with a new action L eff , which is not equal to the action one wouldfind by transforming to polar coordinates in the continuum action (in Cartesian coordinates).Edwards et al. [2] and Peak et al. [3] also transform to polar coordinates in the discrete quantum-mechanical path integral. They find that terms of order ∆ or higher, which arise when transforming to polarcoordinates in the discrete action (in Cartesian coordinates), cannot all be neglected. From the above it should be clear that transforming a path integral in terms of the normal (i.e. Cartesian)fields ϕ and ϕ to a path integral in terms of polar fields is far from trivial. To be able to do computationsat all we will present a conjecture in this section. In the next sections we will then try to make this conjectureplausible by considering certain toy models where we can see that the conjecture actually works.The generic form of a d -dimensional Euclidean path integral P in two fields is given in (1). Very naively,one could hope that the transformation to polar variables works as: Z ∞−∞ D ϕ Z ∞−∞ D ϕ → Z ∞−∞ D r Y x r ( x ) Z ∞−∞ D θϕ ( x ) → r ( x ) cos θ ( x ) ϕ ( x ) → r ( x ) sin θ ( x ) Z d d x (cid:18)
12 ( ∇ ϕ ) + 12 ( ∇ ϕ ) (cid:19) → Z d d x (cid:18)
12 ( ∇ r ) + 12 r ( ∇ θ ) (cid:19)Z d d x V ( ϕ , ϕ ) → Z d d x V ( r cos θ, r sin θ ) (10)Here one has just extended the integration domains for r and θ to h−∞ , ∞i . In the fourth line one has justtransformed the continuum action to an action in terms of polar fields, disregarding the fact that one shoulddo this on the lattice, where the path integral is defined.To bring all these expressions in continuum form we still have to do something about the Jacobian factor Y x r ( x ) , (11)4ince this is still a lattice expression. We shall write this Jacobian as: Y x r ( x ) = Y x exp (cid:18) − h ( − ¯ h ln r ( x )) (cid:19) = exp − h X x ( − ¯ h ln r ( x )) ! → exp (cid:18) − h d Z d d x ( − ¯ h ln r ( x )) (cid:19) → exp (cid:18) − h (cid:20) π ) d Z d d k (cid:21) Z d d x ( − ¯ h ln r ( x )) (cid:19) (12)In a discrete space (i.e. finite ∆) it can easily be seen that the last step is correct. Now we have also writtenthe Jacobian as a continuum expression, which looks a lot like the exponential of an action.So, we might hope that the continuum form of a path integral in polar field variables is given by makingthe substitutions Z ∞−∞ D ϕ Z ∞−∞ D ϕ → Z ∞−∞ D r Z ∞−∞ D θ · exp (cid:18) − h (cid:20) π ) d Z d d k (cid:21) Z d d x ( − ¯ h ln r ( x )) (cid:19) ϕ ( x ) → r ( x ) cos θ ( x ) ϕ ( x ) → r ( x ) sin θ ( x ) Z d d x (cid:18)
12 ( ∇ ϕ ) + 12 ( ∇ ϕ ) (cid:19) → Z d d x (cid:18)
12 ( ∇ r ) + 12 r ( ∇ θ ) (cid:19)Z d d x V ( ϕ , ϕ ) → Z d d x V ( r cos θ, r sin θ ) (13)in the path integral in Cartesian form.Now the conjecture we are going to make is: Conjecture
It is correct to transform to polar variables naively, as in (13), provided one does thecalculation in a d -dimensional way. This seems a very strange conjecture, and with all the remarks we made it is hard to imagine how it canwork. However in the next two sections we shall demonstrate that indeed this conjecture is true for two toymodels. Also there we will demonstrate what is meant exactly by ‘calculating in a d -dimensional way’. In this section we shall calculate several Green’s functions in the so-called shifted toy model. This modelhas an action S = Z d d x (cid:18)
12 ( ∇ ϕ ) + 12 ( ∇ ϕ ) + 12 m ( ϕ − v ) + 12 m ϕ (cid:19) . (14)This is just the action of a free model with the ϕ -field shifted, hence the name. We shifted this field suchthat the minimum of the action is at ϕ = v, ϕ = 0, and not at r = 0, because the transformation to polarfields becomes singular at this point.To prove that the conjecture indeed works in the case of this model we shall now calculate several Green’sfunctions through the normal, Cartesian, path integral and through the path integral in terms of polar fields.Then we can compare results. 5 .1 Cartesian Results Because the shifted toy model is just a free theory with one field shifted it is very easy to obtain the exactfull
Green’s functions. They are: h ϕ ( x ) i = v h ϕ ( x ) i = 0 h ϕ ( x ) ϕ ( y ) i = v + ¯ h (2 π ) d Z d d k e ik · ( x − y ) k + m = v + ¯ h A m ( x − y ) h ϕ ( x ) ϕ ( y ) i = ¯ h (2 π ) d Z d d k e ik · ( x − y ) k + m = ¯ h A m ( x − y ) h ϕ ( x ) ϕ ( y ) i = 0 (15)Here we wrote the results in terms of standard d -dimensional integrals, which are defined in appendix A. Now we perform the transformation to polar fields in the path integral: ϕ ( x ) = r ( x ) cos (cid:18) w ( x ) v (cid:19) ϕ ( x ) = r ( x ) sin (cid:18) w ( x ) v (cid:19) (16)Here we have used w ( x ) /v instead of θ ( x ) to have a new angular field-variable with also the dimensions ofa field. This is purely a matter of convenience.To calculate the Green’s functions through the path integral in polar fields now we will use the conjecture.According to this conjecture, the new action we have to work with is: S = Z d d x (cid:18)
12 ( ∇ r ( x )) + 12 v r ( x ) ( ∇ w ( x )) + 12 m r ( x ) − m vr ( x ) cos( w ( x ) /v ) (cid:19) . (17)The minimum of the action is at r ( x ) = v and w ( x ) = 0. Expanding around r ( x ) = v and writing r ( x ) ≡ v + η ( x ) (18)we get S = Z d d x (cid:18)
12 ( ∇ η ( x )) + 12 ( ∇ w ( x )) + 1 v η ( x ) ( ∇ w ( x )) + 12 v η ( x ) ( ∇ w ( x )) + m vη ( x ) + 12 m η ( x ) − m v cos( w ( x ) /v ) − m vη ( x ) cos( w ( x ) /v ) (cid:19) . (19)Expanding also around w ( x ) = 0, i.e. expanding the cosine gives: S = Z d d x (cid:18)
12 ( ∇ η ( x )) + 12 ( ∇ w ( x )) + 1 v η ( x ) ( ∇ w ( x )) + 12 v η ( x ) ( ∇ w ( x )) +12 m η ( x ) + 12 m w ( x ) − m v w ( x ) + m v η ( x ) w ( x ) − m v η ( x ) w ( x ) + O (¯ h ) (cid:19) . (20)6he Jacobian gives, according to the conjecture:exp (cid:18) − h I Z d d x ( − ¯ h ln r ( x )) (cid:19) ∼ exp (cid:18) − h I Z d d x (cid:18) − ¯ h ln (cid:18) η ( x ) v (cid:19)(cid:19)(cid:19) =exp − h I Z d d x − ¯ h ∞ X n =1 ( − n +1 n (cid:18) η ( x ) v (cid:19) n !! (21)The standard integral I is defined in appendix A.Now can read off the Feynman rules from the action and the Jacobian, and the conjecture states that wecan calculate everything in the continuum, provided we do a d -dimensional calculation. The Feynman rulesare (up to order ¯ h / ): ↔ ¯ hk + m ↔ ¯ hk + m k k ↔ hv k · k − m ¯ hvk k ↔ hv k · k ↔ m ¯ hv ↔ m ¯ hv ↔ v I ↔ − v I ↔ v I (22)Here all momenta are counted as going into the vertex. With these rules we can now compute Green’sfunctions up to order ¯ h . 7irst we will demonstrate however what we mean exactly by ‘calculating in a d -dimensional way’. Whatwe mean can most easily be seen in the following ‘ d -dimensional calculation’ of a tadpole diagram.= 12 ¯ hm π ) d Z d d k d d l (cid:18) − hv k − m ¯ hv (cid:19) (cid:18) − hv k · l − m ¯ hv (cid:19) (cid:18) ¯ hk + m (cid:19) ¯ hl + m ¯ h ( k − l ) + m = 12 ¯ h m v π ) d Z d d k d d l ( − k − m )( − k · l − m ) (cid:18) k + m (cid:19) l + m k − l ) + m = − ¯ h m v π ) d Z d d k d d l ( − k · l − m ) · k + m l + m k − l ) + m +12 ¯ h v π ) d Z d d k d d l ( − k · l − m ) · (cid:18) k + m (cid:19) l + m k − l ) + m (23)Now use − k · l − m = (cid:2) ( k − l ) + m (cid:3) − (cid:2) k + m (cid:3) − (cid:2) l + m (cid:3) . (24)Then the tadpole diagram becomes: − ¯ h m v π ) d Z d d k d d l ( − k · l − m ) (cid:18) k + m l + m − k + m k − l ) + m (cid:19) +12 ¯ h v π ) d Z d d k d d l ( − k · l − m ) (cid:18) k + m ) l + m − k + m l + m k − l ) + m + − k + m ) k − l ) + m (cid:19) = 92 ¯ h v I (0 , m ) − h m v I (0 , m ) I − ¯ h m v I (0 , m ) I (0 , m, , m ) (25)In the above steps, and in all d -dimensional calculations, one essentially uses three rules: One writes dot-products from the vertices in terms of the denominators of the propagators, to let them cancel as much aspossible, one can shift all loop momenta and one can set Z d d k k i k + m = 0 . (26)Using these three rules is what we mean by a ‘ d -dimensional calculation’.Notice that, for example in dimension one, where k · l becomes a simple product, we could also havecombined momenta coming from different vertices and let them cancel denominators. This simplifies the8alculation of the tadpole diagram considerably, however this is not what we mean by a ‘ d -dimensionalcalculation’. The result in terms of standard integrals is also different . Even the numerical result is different because the diagram contains a divergence, also in d = 1. (The divergence comes from I .) In order for theconjecture to work we have to perform a d -dimensional calculation.Now we compute h ϕ ( x ) i , h ϕ ( x ) i , h ϕ ( x ) ϕ ( y ) i , h ϕ ( x ) ϕ ( y ) i and h ϕ ( x ) ϕ ( y ) i via the polar fields, upto order ¯ h . To do this we first have to express these Green’s functions in terms of the η - and w -Green’s-functions. h ϕ ( x ) i = h ( v + η ( x )) cos( w ( x ) /v ) i = v + h η ( x ) i − v h w ( x ) i − v h η ( x ) w ( x ) i + 124 v h w ( x ) i + O (¯ h ) h ϕ ( x ) i = h ( v + η ( x )) sin( w ( x ) /v ) i = h w ( x ) i + 1 v h η ( x ) w ( x ) i − v h w ( x ) i − v h η ( x ) w ( x ) i + O (¯ h ) h ϕ ( x ) ϕ ( y ) i = h ( v + η ( x ))( v + η ( y )) cos( w ( x ) /v ) cos( w ( y ) /v ) i = v + 2 v h η ( x ) i + h η ( x ) η ( y ) i − h w ( x ) i − v h η ( x ) w ( x ) i + − v h η ( x ) w ( y ) i − v h η ( y ) w ( x ) i + 112 v h w ( x ) i +14 v h w ( x ) w ( y ) i − v h η ( x ) η ( y ) w ( x ) i − v h η ( x ) η ( y ) w ( y ) i + O (¯ h ) h ϕ ( x ) ϕ ( y ) i = h ( v + η ( x ))( v + η ( y )) sin( w ( x ) /v ) sin( w ( y ) /v ) i = h w ( x ) w ( y ) i + 1 v h η ( x ) w ( x ) w ( y ) i + 1 v h η ( y ) w ( x ) w ( y ) i +1 v h η ( x ) η ( y ) w ( x ) w ( y ) i − v h w ( x ) w ( y ) i − v h w ( x ) w ( y ) i + O (¯ h ) h ϕ ( x ) ϕ ( y ) i = h ( v + η ( x ))( v + η ( y )) cos( w ( x ) /v ) sin( w ( y ) /v ) i = v h w ( y ) i + h η ( x ) w ( y ) i + h η ( y ) w ( y ) i − v h w ( y ) i + − v h w ( x ) w ( y ) i + 1 v h η ( x ) η ( y ) w ( y ) i − v h η ( x ) w ( y ) i + − v h η ( y ) w ( y ) i − v h η ( x ) w ( x ) w ( y ) i − v h η ( y ) w ( x ) w ( y ) i + O (¯ h ) (27)Notice that in these formulas all Green’s-functions are full Green’s functions, i.e. all Green’s functionscontain the connected and disconnected part. We wish to stress here that we cannot replace all averages h . . . i by connected averages h . . . i c on both sides of the equality sign in (27). Connected does not meanthe same in the Cartesian- and the polar-field formalism. In other words taking the connected part andtransforming to polar fields are two operations that do not commute.Now we compute all the η - and w -Green’s-functions that we need, up to order ¯ h . All results will beexpressed in the standard integrals from appendix A. From the Feynman rules we can immediately see thatany Green’s function with an odd number of w ’s is zero. η -Tadpole Below we list all the diagrams contributing to h ˜ η i = h η i up to order ¯ h .= − ¯ hm v I + 12 ¯ hv I (0 , m )9 ¯ hm v I = −
12 ¯ h v I (0 , m ) + 14 ¯ h m v I (0 , m, , m ) I (0 , m )= − h m v I (0 , m ) I + 92 ¯ h v I (0 , m ) + − ¯ h m v I (0 , m, , m ) I (0 , m )= 2 ¯ h m v I (0 , m ) I − h v I (0 , m ) = −
12 ¯ h v I (0 , m ) + 14 ¯ h m v I (0 , m, , m ) I (0 , m )= ¯ h m v I (0 , m ) I −
32 ¯ h v I (0 , m ) +12 ¯ h m v I (0 , m, , m ) I (0 , m )= 18 ¯ h v I (0 , m ) = ¯ h m v I (0 , m ) I (28)The complete result for the η -tadpole is: h ˜ η i = 12 ¯ hv I (0 , m ) + 18 ¯ h v I (0 , m ) . (29)Notice that all the (badly divergent) I -integrals from the Jacobian have nicely cancelled against identicalterms from w -loops. 10 .2.2 The η -Propagator Below we list the diagrams contributing to the connected momentum-space η -propagator h ˜ η ( p )˜ η ( − p ) i c .= ¯ hp + m = 2 ¯ h v p + m ) I − ¯ h v p + 2 m ( p + m ) I (0 , m ) + 12 ¯ h v I (0 , m, p, m )= − ¯ h v p + m ) I + ¯ h m v p + m ) I (0 , m )= − ¯ h v p + m ) I (30)For the connected η -propagator we get: h ˜ η ( p )˜ η ( − p ) i c = ¯ hp + m − ¯ h v p + m I (0 , m ) + 12 ¯ h v I (0 , m, p, m ) (31)Notice that also here the I -integrals cancel. w -Propagator For the connected w -propagator h ˜ w ( p ) ˜ w ( − p ) i c we have the following contributions, up to order ¯ h . h ˜ w ( p ) ˜ w ( − p ) i c = + + ++= ¯ hp + m + ¯ h v I (0 , m, p, m ) (32)Notice that in none of the diagrams in the connected w -propagator a (badly divergent) I -integral occurs.The higher-point η - and w -Green’s-functions up to order ¯ h are easy to calculate, for example h ˜ η ( p ) ˜ w ( p ) ˜ w ( p ) i c is just one tree diagram.Knowing the η - and w -Green’s-functions in momentum space we can easily write down the configurationspace Green’s functions needed in (27), up to order ¯ h . For details on this calculation we refer to one of theauthor’s PhD-thesis [10] (chapter 8). Finally substituting all the results in (27) gives us: h ϕ ( x ) i = v h ϕ ( x ) i = 0 h ϕ ( x ) ϕ ( y ) i = v + ¯ h A m ( x − y ) h ϕ ( x ) ϕ ( y ) i = ¯ h A m ( x − y ) h ϕ ( x ) ϕ ( y ) i = 0 (33)These are indeed the correct results for the Green’s functions. So the conjecture is verified for several Green’sfunctions in the shifted toy model up to order ¯ h . 11 The Arctangent Toy Model
As another illustration of the conjecture we now consider the arctangent toy model. The action of this modelis: S = Z dx (cid:18)
12 ( ∇ ϕ ) + 12 ( ∇ ϕ ) + m v ( ϕ + ϕ − v ) + m v ϕ + ϕ ) arctan (cid:18) ϕ ϕ (cid:19)(cid:19) (34)This action has a single minimum at ϕ = v , ϕ = 0 . (35)We want to stress that this model is not at all a physical model. The action has an infinite numberof vertices (by expanding the arctangent), which means this model is not renormalizable. We just want touse this model as a toy model to test the conjecture. Especially because it is not renormalizable, so no bigcancellations can be expected to occur, the arctangent toy model is a very good test of the conjecture. To find the Cartesian Green’s functions we expand the action around the minimum (35): ϕ ( x ) = v + η ( x ) , ϕ ( x ) = η ( x ) . (36)Notice that also the arctangent in the action has to be expanded, this term will give an infinite number ofvertices. Up to order ¯ h / the Feynman rules are: ↔ ¯ hk + µ ↔ ¯ hk + m ↔ − h m v ↔ h m v ↔ − h m v ↔ − h m v ↔ h m v ↔ h m v − h m v (37)Here the solid lines indicate the η -particle, the dashed lines indicate the η -particle and µ is given by µ = √ m . With these Feynman rules we can now compute some Green’s functions up to order ¯ h . We shallnot present all diagrams here, since this Cartesian calculation is straightforward and quite lengthy. h ϕ ( x ) i = v + h η ( x ) i = v −
32 ¯ hv I (0 , µ ) + 12 ¯ hv I (0 , m ) + −
98 ¯ h v I (0 , µ ) −
858 ¯ h v I (0 , m ) + 994 ¯ h v I (0 , µ ) I (0 , m ) + − h m v I (0 , µ, , µ ) I (0 , µ ) + 4 ¯ h m v I (0 , m, , m ) I (0 , m ) +21 ¯ h m v I (0 , µ, , µ ) I (0 , m ) − h m v I (0 , m, , m ) I (0 , µ ) + −
27 ¯ h m v B µµµ − h m v B µmm + 2 ¯ h m v B mµm +3 ¯ h m v D µµµ −
11 ¯ h m v D mmµ h ϕ ( x ) i = h η ( x ) i = 0 h ϕ ( x ) ϕ ( y ) i = v + 2 v h η ( x ) i + h η ( x ) η ( y ) i = v − h I (0 , µ ) + ¯ h I (0 , m ) + ¯ h A µ ( x − y ) + −
21 ¯ h v I (0 , m ) + 48 ¯ h v I (0 , µ ) I (0 , m ) −
18 ¯ h m v I (0 , µ, , µ ) I (0 , µ ) +8 ¯ h m v I (0 , m, , m ) I (0 , m ) + 42 ¯ h m v I (0 , µ, , µ ) I (0 , m ) + −
14 ¯ h m v I (0 , m, , m ) I (0 , µ ) −
54 ¯ h m v B µµµ − h m v B µmm +4 ¯ h m v B mµm + 6 ¯ h m v D µµµ −
22 ¯ h m v D mmµ +6 ¯ h m v I (0 , µ ) C µµ ( x − y ) −
14 ¯ h m v I (0 , m ) C µµ ( x − y ) +18 ¯ h m v B µµµ ( x − y ) + 2 ¯ h m v B µmm ( x − y ) h ϕ ( x ) ϕ ( y ) i = h η ( x ) η ( y ) i = ¯ h A m ( x − y ) + −
14 ¯ h m v I (0 , µ ) C mm ( x − y ) + 8 ¯ h m v I (0 , m ) C mm ( x − y ) +4 ¯ h m v B mmµ ( x − y ) h ϕ ( x ) ϕ ( y ) i = v h η ( y ) i + h η ( x ) η ( y ) i = 0 (38)13 .2 Polar Results According to the conjecture we can just transform the continuum action (34) to an action in terms of thepolar field variables to obtain the Feynman rules for the polar calculation. So the action becomes: S = Z d d x (cid:18)
12 ( ∇ r ) + 12 r v ( ∇ w ) + m v ( r − v ) + 12 m v w r (cid:19) (39)This can be expanded around r = v again. Defining r ( x ) ≡ v + η ( x ) , (40)we find the following Feynman rules for the η - and w -field (up to order ¯ h / ). The vertices from the Jacobianare exactly the same as in the shifted toy model. ↔ ¯ hk + µ ↔ ¯ hk + m ↔ − h m vk k ↔ hv k · k + 1¯ h m v ↔ − h m v k k ↔ hv k · k − h m v ↔ h m v ↔ v I ↔ − v I ↔ v I (41)Here the solid lines denote the η -field, the dashed lines denote the w -field and all momenta are counted intothe vertex. Also we have defined µ = √ m , as in the Cartesian calculation.14ow we can again compute the η - and w -Green’s-functions up to order ¯ h . η -Tadpole For the η -tadpole we find the following result, up to order ¯ h . For a more detailed calculation we refer to[10]. h ˜ η i = + + ++ + ++ + ++ + += −
32 ¯ hv I (0 , µ ) + ¯ hv I (0 , m ) −
98 ¯ h v I (0 , µ ) −
454 ¯ h v I (0 , m ) + 26 ¯ h v I (0 , µ ) I (0 , m ) − h m v I (0 , µ, , µ ) I (0 , µ ) + 8 ¯ h m v I (0 , m, , m ) I (0 , m )+21 ¯ h m v I (0 , µ, , µ ) I (0 , m ) −
14 ¯ h m v I (0 , m, , m ) I (0 , µ ) −
27 ¯ h m v B µµµ − h m v B µmm + 4 ¯ h m v B mµm +3 ¯ h m v D µµµ −
12 ¯ h m v D mmµ (42)Notice that, as in the shifted toy model, also here the badly divergent I -integrals cancel. η -Propagator The momentum-space η -propagator is: h ˜ η ( p )˜ η ( − p ) i c = + + +15 + += ¯ hp + µ + 6 ¯ h m v p + µ ) I (0 , µ ) −
14 ¯ h m v p + µ ) I (0 , m ) + − ¯ h v p + µ I (0 , m ) + 18 ¯ h m v p + µ ) I (0 , µ, p, µ ) +2 ¯ h m v p + µ ) I (0 , m, p, m ) + 2 ¯ h m v p + µ I (0 , m, p, m ) +12 ¯ h v I (0 , m, p, m ) (43) w -Propagator The w -propagator up to order ¯ h is: h ˜ w ( p ) ˜ w ( − p ) i c = + + += ¯ hp + m + 8 ¯ h m v p + m ) I (0 , m ) − h v p + m I (0 , m ) + −
14 ¯ h m v p + m ) I (0 , µ ) + 5 ¯ h v p + m I (0 , µ ) + ¯ h v I (0 , µ, p, m ) − h m v p + m I (0 , µ, p, m ) + 4 ¯ h m v p + m ) I (0 , µ, p, m ) (44)The higher-point Green’s functions are easily calculated up to order ¯ h , they contain at most tree dia-grams.To compute the Cartesian ϕ - and ϕ -Green’s-functions we can again use the expansions (27), since theseexpansions are model independent. First we have to find the configuration-space η - and w -Green’s functionsfrom the momentum-space Green’s functions above. Then these results can be substituted in (27). Doingthis one finds again the results (38). So also in case of the arctangent toy model the conjecture is verifiedfor several Green’s functions up to order ¯ h . For a more detailed calculation we refer to [10] again. In the last two sections evidence for the truth of the conjecture has accumulated. In this section we shallprove this conjecture for a general model in d space-time dimensions.Our proof will be based on the fact that the transformation to polar field variables actually has to beperformed in the path integral on the lattice, i.e. in (4). After transforming to polar fields one gets a path16ntegral in terms of polar fields formulated on a lattice. This path integral gives a (complicated) set ofFeynman rules, and diagrams actually have to be calculated with space-time still discrete. Only in the endresult for the Green’s function one should then take the continuum limit, i.e. ∆ → d -dimensional continuum calculation is correct if one can see that all the steps one performs thereto calculate a diagram correspond to a similar step in a discrete calculation. In a d -dimensional continuumcalculation one performs the following three steps when calculating any diagram:1. One writes momentum-dependent factors from the vertices in terms of the denominators of propagators,such that one can let them cancel. For example: k · l = 12 (( k + l ) + m ) −
12 ( k + m ) −
12 ( l + m ) + 12 m . (45)2. One shifts momenta, for example: Z d d k k + m ( k + l ) + m = Z d d k ( k − l ) + m k + m . (46)3. When there is momentum dependence left in the numerator, which cannot cancel anything in thedenominator anymore, one uses Z d d k k i k + m = 0 . (47)For example: Z d d k Z d d l ( k − l ) + m ( k + m )( l + m )= Z d d k Z d d l ( k + m ) + ( l + m ) − m ( k + m )( l + m )= Z d d k Z d d l (cid:18) l + m + 1 k + m − m ( k + m )( l + m ) (cid:19) (48)If we can somehow see that these steps are also valid in a discrete calculation, then we have proven theconjecture. For then we know that every operation one performs in the d -dimensional continuum calculationcorresponds to a valid operation in a discrete calculation, even though one writes down these steps in acontinuum formalism.The first step is to transform the discrete action (8) to polar field variables. The transformation goes asfollows: ϕ i ,...,i d = r i ,...,i d cos (cid:16) w i ,...,i d v (cid:17) ϕ i ,...,i d = r i ,...,i d sin (cid:16) w i ,...,i d v (cid:17) (49)To keep things readable we define the shorthand notations: i ≡ i , . . . , i d ∇ r i ,...,i d ≡ ( r i +1 ,i ,...,i d − r i ,i ,...,i d )... (cid:0) r i ,...,i d − ,i d +1 − r i ,...,i d − ,i d (cid:1) (50)The action becomes: S = ∆ d N − X i ,...,i d =0
12 ( ∇ r i ) + 1∆ r i (cid:18)(cid:18) − cos ∆ ∇ w i v (cid:19) + . . . + (cid:18) − cos ∆ ∇ d w i v (cid:19)(cid:19) +1∆ r i ∇ r i (cid:18) − cos ∆ ∇ w i v (cid:19) + . . . + 1∆ r i ∇ d r i (cid:18) − cos ∆ ∇ d w i v (cid:19) + V (cid:16) r i cos w i v , r i sin w i v (cid:17) ! (51)17ow we will substitute the series expansion for the cosines in the first two lines. Also we will assumethat the potential is such that the minimum of the complete action is at r = v , where v is some nonzeroconstant (the same v that divides w in the cosine). This assumption is necessary to avoid difficulties withthe singularity at r = 0 in the transformation (49). Because the minimum of the action is at r = v we alsoexpress the action in terms of η , which denotes the deviation from v : r i = v + η i . (52)The final form of the discrete action is: S = ∆ d N − X i ,...,i d =0
12 ( ∇ η i ) + 12 ( ∇ w i ) − ∞ X n =2 ( − n (2 n )! ∆ n − v n − (cid:16) ( ∇ w i ) n + . . . + ( ∇ d w i ) n (cid:17) − ∞ X n =1 ( − n (2 n )! ∆ n − v n − (cid:16) η i ( ∇ w i ) n + . . . + η i ( ∇ d w i ) n (cid:17) − ∞ X n =1 ( − n (2 n )! ∆ n − v n − (cid:16) ( ∇ η i ) ( ∇ w i ) n + . . . + ( ∇ d η i ) ( ∇ d w i ) n (cid:17) − ∞ X n =1 ( − n (2 n )! ∆ n − v n − (cid:16) η i ( ∇ w i ) n + . . . + η i ( ∇ d w i ) n (cid:17) − ∞ X n =1 ( − n (2 n )! ∆ n − v n − (cid:16) η i ( ∇ η i ) ( ∇ w i ) n + . . . + η i ( ∇ d η i ) ( ∇ d w i ) n (cid:17) + V (cid:16) ( v + η i ) cos w i v , ( v + η i ) sin w i v (cid:17) ! (53)Note that this expression is still exact, as long as we keep all the terms in the sums coming from theexpansion of the cosines. Our complete discrete path integral P (defined in (1)), formulated in terms of thepolar fields, now looks like: P = Z ∞− v N − Y i =0 ( v + η i ) dη i ! Z vπ − vπ N − Y i =0 dw i ! O exp (cid:18) − h S (cid:19) , (54)where the action S is given by (53) and O is the product of the ϕ - and ϕ -fields: O ≡ (cid:0) v + η j (1) (cid:1) cos w j (1) v · · · (cid:0) v + η j ( m ) (cid:1) cos w j ( m ) v · ( v + η k (1) ) sin w k (1) v · · · ( v + η k ( n ) ) sin w k ( n ) v . (55)The product N − Y i =0 ( v + η i ) = N − Y i =0 r i (56)is the Jacobian from the transformation to polar fields. This Jacobian factor can be recast in the followingform: N − Y i =0 ( v + η i ) ! = exp N − X i =0 ln( v + η i ) ! (57)Also the domain of integration for the w -fields can be extended from [ − vπ, vπ ] to h−∞ , ∞i , because thewhole integrand is periodic in the w -fields. Finally we can also extend the lower integration boundary forthe η -fields from − v to −∞ , this shift will only have non-perturbative effects.18o we have brought our path integral to the form P = Z ∞−∞ N − Y i =0 dη i ! Z ∞−∞ N − Y i =0 dw i ! O exp − h S + N − X i =0 ln( v + η i ) ! . (58)This is a normal path integral on a lattice, in the sense that it has the same form of a path integral in termsof Cartesian fields. Such a path integral we can calculate in the ordinary way, with perturbation theory. Theaction of this path integral does have an infinite number of vertices however, because the discrete action (53)has an infinite number of interaction terms and also because the expansion of the logarithm coming fromthe Jacobian has an infinite number of terms. But it is still an exact expression, because we keep all terms.Now we have to write down the (discrete) Feynman rules for the action we have found. To write downthe momentum-space Feynman rules we must first transform the configuration-space fields η and w tomomentum-space fields ˜ η and ˜ w . In the continuum such a transformation is given by:˜ η ( k ) = Z d d x η ( x ) e ik · x , (59)and similar for the ˜ w -field. The discrete analogue of this formula is:˜ η k ,...,k d = ∆ d N − X i ,...,i d =0 η i ,...,i d exp (cid:18) πiL ( k (∆ i − L/
2) + . . . + k d (∆ i d − L/ (cid:19) , (60)where we have used that the continuous momentum is related to the discrete momentum as k cont = 2 πk discr L . (61)The inverse transformation of (60) is η i ,...,i d = 1 L d N/ − X k ,...,k d = − N/ ˜ η k ,...,k d exp (cid:18) − πiL ( k (∆ i − L/
2) + . . . + k d (∆ i d − L/ (cid:19) , (62)and similar for the w -field. To see that this is indeed the inverse transformation of (60) one can use theidentity 1 N d N − X i ,...,i d =0 exp (cid:18) πiN ( k i + . . . + k d i d ) (cid:19) = δ k , N · · · δ k d , N . (63)From the relation (61) we can see that when we take L → ∞ the momenta become a continuous set.Their domain is still finite however. Because the discrete momenta are between − N/ N/ −
1, thecontinuous momenta are in the domain k cont ∈ D − π ∆ , π ∆ E . (64)The finiteness of this domain reflects the discreteness of space-time. From now on we shall understand thatwe have taken the limit L → ∞ , such that all sums over momenta become integrals. But of course ∆ is stillfinite.By using (62), in the limit L → ∞ , we can now express the discrete action (53) in terms of the momentum-space fields ˜ η and ˜ w . From this action one can then read of the discrete, momentum-space Feynman rules.Notice that we did not specify the potential V , so we will not include the Feynman rules coming from thispart of the action. This potential V will also determine the masses for the η - and w -field. We shall keepthese masses general, the upcoming proof for the conjecture will not depend on the explicit form of thepotential V and the masses. In the Feynman rules below we will neither include the Feynman rules from theJacobian, the proof of the conjecture will also not depend on the exact form of these vertices.The discrete, momentum-space Feynman rules are then: ↔ ¯ h d ∆ − cos ∆ k − . . . − cos ∆ k d + m η ¯ h d ∆ − cos ∆ k − . . . − cos ∆ k d + m w p k (1) k (2) ↔ − hv (cid:20) (cid:0) e − i ∆ p + 1 (cid:1) (cid:16) e − i ∆ k (1)1 − (cid:17) (cid:16) e − i ∆ k (2)1 − (cid:17) + . . . + (cid:0) e − i ∆ p d + 1 (cid:1) (cid:16) e − i ∆ k (1) d − (cid:17) (cid:16) e − i ∆ k (2) d − (cid:17) (cid:21) ... p k (1) k (2 n ) ↔ ( − n ¯ hv n − (cid:20) (cid:0) e − i ∆ p + 1 (cid:1) (cid:16) e − i ∆ k (1)1 − (cid:17) · · · (cid:16) e − i ∆ k (2 n )1 − (cid:17) + . . . + (cid:0) e − i ∆ p d + 1 (cid:1) (cid:16) e − i ∆ k (1) d − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) d − (cid:17) (cid:21) ... pq k (1) k (2) ↔ − hv (cid:20) (cid:0) e − i ∆ p + e − i ∆ q (cid:1) (cid:16) e − i ∆ k (1)1 − (cid:17) (cid:16) e − i ∆ k (2)1 − (cid:17) + . . . + (cid:0) e − i ∆ p d + e − i ∆ q d (cid:1) (cid:16) e − i ∆ k (1) d − (cid:17) (cid:16) e − i ∆ k (2) d − (cid:17) (cid:21) ... pq k (1) k (2 n ) ↔ ( − n ¯ hv n (cid:20) (cid:0) e − i ∆ p + e − i ∆ q (cid:1) (cid:16) e − i ∆ k (1)1 − (cid:17) · · · (cid:16) e − i ∆ k (2 n )1 − (cid:17) + . . . + (cid:0) e − i ∆ p d + e − i ∆ q d (cid:1) (cid:16) e − i ∆ k (1) d − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) d − (cid:17) (cid:21) ... k (1) k (2) k (3) k (4) ↔ hv (cid:20) (cid:16) e − i ∆ k (1)1 − (cid:17) · · · (cid:16) e − i ∆ k (4)1 − (cid:17) + . . . + (cid:16) e − i ∆ k (1) d − (cid:17) · · · (cid:16) e − i ∆ k (4) d − (cid:17) (cid:21) k (1) k (2 n ) ↔ ( − n ¯ hv n − (cid:20) (cid:16) e − i ∆ k (1)1 − (cid:17) · · · (cid:16) e − i ∆ k (2 n )1 − (cid:17) + . . . + (cid:16) e − i ∆ k (1) d − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) d − (cid:17) (cid:21) ... (65)Here all the (continuum) momenta are counted incoming. Together with these Feynman rules for thepropagator and the vertices we have the rule that every internal momentum should be integrated over from − π/ ∆ to π/ ∆.Now all these vertices can be written in a more convenient form. By combining all the complex exponen-tials (i.e. writing out all products) and using momentum conservation at the vertex one will notice that foreach exponential also its complex conjugate occurs. They can be combined into a cosine, the same cosinethat occurs in the discrete propagator. In this way we can write the vertex expressions above in terms ofthe denominator of the propagator. As a shorthand notation we denote the denominators by Π k and ¯Π k :Π k ≡ d ∆ − cos 2 πk N − . . . − cos 2 πk d N + m w ¯Π k ≡ d ∆ − cos 2 πk N − . . . − cos 2 πk d N + m η (66)To write the vertices into this more convenient form we also have to define an operator P . We denote theset consisting of the j th components of the momenta k (1) , . . . , k (2 n ) by { k j } : { k j } ≡ { k (1) j , k (2) j , . . . , k (2 n ) j } . (67)Then the operator P i working on { k j } returns the sum of i momenta chosen from the set { k j } . There are (cid:0) ni (cid:1) ways to choose i momenta from a set of 2 n momenta, so there are also (cid:0) ni (cid:1) different operators P i . Forexample: P { k j } = k (1) j + k (2) j . (68)With these notations we can write the vertex expressions as follows. p k (1) k (2) ↔ − hv (cid:0) Π k (1) + Π k (2) − ¯Π p − m w + m η (cid:1) ... p k (1) k (2 n ) ↔ ( − n ¯ hv n − − ¯Π p + X P Π P { k } + X P Π p + P { k } + − X P ¯Π P { k } − X P ¯Π p + P { k } + . . . +( − n X P n − Π P n − { k } + ( − n X P n − Π p + P n − { k } +212 ( − n +1 X P n ¯Π P n { k } + 12 ( − n +1 X P n ¯Π p + P n { k } + − n − m w − (cid:0) − n − (cid:1) m η ! ... pq k (1) k (2) ↔ − hv (cid:18) − ¯Π p − ¯Π q +12 Π p + k (1) + 12 Π p + k (2) + 12 Π q + k (1) + 12 Π q + k (2) + − m w + 2 m η (cid:19) ... pq k (1) k (2 n ) ↔ ( − n ¯ hv n − ¯Π p − ¯Π q + X P Π p + P { k } + X P Π q + P { k } + − X P ¯Π p + P { k } − X P ¯Π q + P { k } + . . . +( − n X P n − Π p + P n − { k } + ( − n X P n − Π q + P n − { k } +12 ( − n +1 X P n ¯Π p + P n { k } + 12 ( − n +1 X P n ¯Π q + P n { k } + − n − m w + 2 n − m η ! ... k (1) k (2) k (3) k (4) ↔ hv (cid:18) Π k (1) + Π k (2) + Π k (3) + Π k (4) + −
12 ¯Π k (1) + k (2) −
12 ¯Π k (1) + k (3) −
12 ¯Π k (1) + k (4) + −
12 ¯Π k (2) + k (3) −
12 ¯Π k (2) + k (4) −
12 ¯Π k (3) + k (4) − m w + 3 m η (cid:19) ... k (1) k (2 n ) ↔ ( − n ¯ hv n − X P Π P { k } − X P ¯Π P { k } + . . . +( − n X P n − Π P n − { k } + 12 ( − n +1 X P n ¯Π P n { k } n − m w − (cid:0) − n − (cid:1) m η ! ... (69)Notice that the bars are always placed on the terms containing an operator P i with i even. For the sakeof the argument it is convenient to place these bars in this way. Whether or not a bar is placed on the lastterm, with P n , thus depends on n , whether n is even or odd. Above the bars are placed as if n were even,but it should be clear how they should be placed when n is odd.Now notice that these vertex rules look identical to the rules one would use when doing a d -dimensionalcontinuum calculation. For example, in the continuum the 3-vertex would be: p k (1) k (2) ↔ hv k (1) · k (2) (70)To simplify the dot-product in a d -dimensional calculation one would write it as2¯ hv k (1) · k (2) = − hv (cid:18) (cid:18)(cid:16) k (1) (cid:17) + m w (cid:19) + (cid:18)(cid:16) k (2) (cid:17) + m w (cid:19) + − (cid:18)(cid:16) k (1) + k (2) (cid:17) + m η (cid:19) − m w + m η (cid:19) , (71)which corresponds exactly to the discrete vertex expression given in (69) for the 3-vertex. So when one usesthe continuum rules to rewrite dot-products of momenta, as in the 3-vertex example above, one is actuallydoing a correct calculation, although one is doing a continuum calculation.Another rule that one uses in a continuum calculation is that it is allowed to shift the loop momenta.Also in a discrete calculation this is allowed, because of the periodicity of the discrete propagators and vertexexpressions.What then goes wrong in a continuum calculation? There is one more rule that one uses in a continuumcalculation that we have not mentioned up to now. This rule is:12 π Z d d k k i k + m = 0 . (72)This rule, however, is not correct in a discrete calculation. For example, in dimension 1, we have to realizethat by k and k + m we actually mean: k ↔ i ∆ (cid:0) e − i ∆ k − (cid:1) k + m ↔ − cos(∆ k ) + m (73)So by the integral above we actually mean:12 π Z dk kk + m ↔ π Z π/ ∆ − π/ ∆ dk i ∆ (cid:0) e − i ∆ k − (cid:1) − cos(∆ k ) + m = − i + O (∆) (74)This shows that the rule (72) is not the correct one to use. The only instances that one would use the rule(72) is when a Π (or ¯Π) is left in the numerator, and cannot cancel anything in the denominator anymore.Below we shall show that all such terms, where a Π (or ¯Π) remains in the numerator, cancel when oneadds all diagrams for a certain Green’s function. To this end it is convenient to split up the vertices in (69)as follows: p k (1) k (2 n ) ↔ ( − n ¯ hv n − (cid:0) − ¯Π p (cid:1) k (1) k (2 n ) ↔ ( − n ¯ hv n − Π k (1) ... p k (1) k (2 n ) ↔ ( − n ¯ hv n − Π k (2 n ) p k (1) k (2 n ) ↔ ( − n ¯ hv n − X P Π p + P { k } − X P ¯Π P { k } − X P ¯Π p + P { k } + . . . +( − n X P n − Π P n − { k } + ( − n X P n − Π p + P n − { k } +12 ( − n +1 X P n ¯Π P n { k } + 12 ( − n +1 X P n ¯Π p + P n { k } ! p k (1) k (2 n ) ↔ ( − n ¯ hv n − (cid:0) − n − m w − (cid:0) − n − (cid:1) m η (cid:1) ... pq k (1) k (2 n ) ↔ ( − n ¯ hv n (cid:0) − ¯Π p (cid:1) pq k (1) k (2 n ) ↔ ( − n ¯ hv n (cid:0) − ¯Π q (cid:1) pq k (1) k (2 n ) ↔ ( − n ¯ hv n X P Π p + P { k } + X P Π q + P { k } + − X P ¯Π p + P { k } − X P ¯Π q + P { k } + . . . +( − n X P n − Π p + P n − { k } + ( − n X P n − Π q + P n − { k } +242 ( − n +1 X P n ¯Π p + P n { k } + 12 ( − n +1 X P n ¯Π q + P n { k } ! pq k (1) k (2 n ) ↔ ( − n ¯ hv n (cid:0) − n − m w + 2 n − m η (cid:1) ... k (1) k (2 n ) ↔ ( − n ¯ hv n − Π k (1) k (1) k (2 n ) ↔ ( − n ¯ hv n − Π k (2 n ) k (1) k (2 n ) ↔ ( − n ¯ hv n − − X P ¯Π P { k } + . . . +( − n X P n − Π P n − { k } + 12 ( − n +1 X P n ¯Π P n { k } ! k (1) k (2 n ) ↔ ( − n ¯ hv n − (cid:0) − n − m w − (cid:0) − n − (cid:1) m η (cid:1) ... (75)Having written the vertices in this form it is clear where the problem terms in a certain diagram comefrom. They come from a vertex with a dot in the center or two vertices connected by a line with two dots.That a dotted vertex is a source can immediately be seen from the vertex expressions above. A line withtwo dots and momentum k flowing through it gets two Π k ’s in the numerator, from the vertices, and onlyone Π k in the denominator, from the propagator.It is now easy to derive the following recursion relation, valid for n ≥ n − n − . . . + e ( n ) 2 n − e ( n ) +25 n − n − . . . + e ( n ) + 12 n − e ( n ) − n = 0 (76)In these diagrams it is understood that the outgoing legs should be connected in all possible ways. e ( n ) Isdefined as: e ( n ) ≡ (cid:26) n if n is even n − n is odd (77)Notice that the third and fourth line in the recursion relation above are not there when n = 2, these diagramssimply do not exist.We also have the following two recursion relations:2 n − . . . + e ( n ) 2 n − e ( n ) +2 n − . . . + e ( n ) 2 n − e ( n ) +2 n − . . . + e ( n ) + 12 n − e ( n ) − n − . . . + e ( n ) + 12 n − e ( n ) − n = 0 (78)In this recursion relation n ≥
2. 26 n − . . . + e ( n ) 2 n − e ( n ) +2 n − . . . + e ( n ) 2 n − e ( n ) +2 n − . . . + e ( n ) + 12 n − e ( n ) − n − . . . + e ( n ) + 12 n − e ( n ) − n = 0 (79)In this recursion relation n ≥
1. The first two lines are not there when n = 1.With these recursion relations it is easy to see that the problem terms always cancel in the complete setof diagrams for a certain Green’s function. If somewhere in a diagram an internal line with two dots occurs,then at this same point in the diagram also the dotted vertex can occur. These diagrams then sum up tozero.So finally we have proven that the problem terms cancel in the complete set of diagrams for a certainGreen’s function. If they cancel out anyway it is also correct to treat these problem terms like one would inthe continuum. Of course one makes a mistake for each problem term, but these mistakes cancel out againin the complete set of diagrams. So there is nothing wrong with taking the continuum limit right from thestart and doing a d -dimensional continuum calculation.Notice that it is no problem to add the vertices coming from the potential V and the Jacobian to thisargument. These vertices give no problem terms themselves. They can be combined with the dotted vertices,but the problem terms always come from a clear , separated part of the diagram, either two vertices connectedby a line with two dots, or a dotted vertex.Now we know it is correct to take the continuum limit ∆ → ηww - and ηηww -vertex do not vanish. All the other verticesgo to zero, as can be seen from their expressions in (69), but also, and much quicker, from (65), becausethey all have one or more factors of ∆ in front when the exponentials are expanded.27o, finally we are left with the Feynman rules: ↔ ¯ hk + m η ↔ ¯ hk + m w k k ↔ hv k · k k k ↔ hv k · k (80)plus the Feynman rules coming from the potential V and the Jacobian. These are exactly the Feynman rulesthat one would have read off from the continuum action: S = Z d d x (cid:18)
12 ( ∇ r ) + 12 ( ∇ w ) + 1 v η ( ∇ w ) + 12 v η ( ∇ w ) + V (cid:16) r cos wv , r sin wv (cid:17)(cid:19) = Z d d x (cid:18)
12 ( ∇ r ) + 12 r v ( ∇ w ) + V (cid:16) r cos wv , r sin wv (cid:17)(cid:19) (81)Recapitulating the proof, we have done the following. The basis of our proof is the discrete path integralin terms of the Cartesian fields ϕ and ϕ . In this path integral on the lattice it is completely legitimateto transform to polar fields. After this transformation we get a very big, complicated action. Looking atthe discrete vertex expressions we find how we can simplify these expressions, and how we can let themcancel against propagators in a certain diagram. We notice that these rules are exactly the same as in a d -dimensional continuum calculation. All the rules that we would use in a continuum calculation appear tobe valid in a calculation on the lattice as well, except for one: rule (72). The terms where we would needto use this rule can then be shown to cancel in the complete set of diagrams, by using the three recursionrelations. So by using the incorrect rule (72) we actually make a mistake, but all these mistakes cancel inthe complete set of diagrams. Thus we know that all the rules that we use in d -dimensional continuumcalculation are also valid in a correct, discrete calculation. This means we might as well take the continuumlimit directly in the discrete Feynman rules (65). Then these Feynman rules simplify to (80), and we haveproven that a d -dimensional continuum calculation with the action (81) is correct . To see explicitly how the mechanism described in the previous section works we consider an example. Con-sider the 1-loop η -propagator. There are two types of diagrams (We do not include vertices from the potential V and the Jacobian, because such vertices will never give problem terms.):andHere, dots should still be put on the lines or in the vertices. There are a lot of diagrams, but it is easy tosee that there are only two diagrams that contain problem terms. These are:28 1(2 π ) d Z π/ ∆ − π/ ∆ d d k (cid:18) − hv (cid:19) Π k (cid:18) − hv (cid:19) Π k ¯ h Π k ¯ h Π p − k = 1 v π ) d Z π/ ∆ − π/ ∆ d d k Π p − k Π k = 12 1(2 π ) d Z π/ ∆ − π/ ∆ d d k (cid:18) − hv (cid:19)(cid:18)
12 (Π p − k + Π p + k ) + 12 (Π p + k + Π p − k ) (cid:19) ¯ h Π k = − v π ) d Z π/ ∆ − π/ ∆ d d k Π p − k Π k (82)Indeed these diagrams cancel, as is guaranteed by the recursion relations derived in the previous section. Theother diagrams, contributing to this η -propagator at 1-loop order, have their dots in other places, or havevertices without dots, that can also come from the potential V . These diagrams can never have a problemterm. And thus the whole 1-loop propagator is free of problem terms, and the continuum limit could havebeen taken right from the start. w -Loops In the previous sections it has become clear that it is allowed to work with the continuum Feynman rules(80), as the conjecture states. Together with these Feynman rules we have of course the rules from thearbitrary potential V , and the rules from the Jacobian. From the discrete calculation it is easy to see thatthe Jacobian can indeed be rewritten as: Y x r ( x ) = Y x exp (cid:18) − h ( − ¯ h ln r ( x )) (cid:19) = exp − h X x ( − ¯ h ln r ( x )) ! → exp (cid:18) − h d Z d d x ( − ¯ h ln r ( x )) (cid:19) → exp (cid:18) − h (cid:20) π ) d Z d d k (cid:21) Z d d x ( − ¯ h ln r ( x )) (cid:19) = exp − h (cid:20) π ) d Z d d k (cid:21) Z d d x − ¯ h ∞ X n =1 ( − n +1 n (cid:18) η ( x ) v (cid:19) n !! , (83)as the conjecture states. The Feynman rules coming from this Jacobian are:29 v I ↔ − v I ...n ↔ ( − n +1 ( n − v n I (84)We see that all these vertices give strange integrals I . In the case of the shifted toy model and thearctangent toy model we already saw that these integrals I always cancelled against identical terms comingfrom w -loops. We shall now prove this in general.Consider a w -loop with only ηww - and ηηww -vertices (as given in (80)) on it. So the diagrams we arecalculating can only have external η -legs. Such a diagram would look like:Now these diagrams have a part which is going to cancel the I -integrals from the Jacobian. This part isexactly the worst divergent part of the diagrams above. To calculate this worst divergent part the massesand incoming momenta can be neglected.In this case it is easy to write down the generating functional for such diagrams. This generating functionalis defined as: Z ( x ) = ∞ X n =1 x n n ! , (85)where the diagram symbolizes all n outgoing η -lines.We denote the number of ηww -vertices in the w -loop by n and the number of ηηww -vertices by n .Then the generating functional Z ( x ) for diagrams of this type is given by: Z ( x ) = 1(2 π ) d Z d d k ∞ X n ,n =0 n + n > (cid:18) (cid:19) n (cid:18) (cid:19) n n ! 1 n ! (cid:18) hv ( − k ) (cid:19) n (cid:18) hv ( − k ) (cid:19) n (cid:18) ¯ hk (cid:19) n + n (2 n + 2 n )!!( n + 2 n )! x n +2 n ( n + 2 n )!= − I ln (cid:16) xv (cid:17) = I ∞ X n =1 ( − n ( n − v n x n n ! (86)30o we can read off that a w -loop with n outgoing η -lines has a worst divergent part given by:( − n ( n − v n I (87)This exactly cancels the vertices from the Jacobian. In any diagram, wherever a dotted vertex from theJacobian with n legs occurs, also a w -loop with n outgoing legs can occur, and their part that contains thestandard integral I cancels! In the case that one uses the dimensional-regularization scheme one has that:1(2 π ) d Z d d k k ) m = 0 ∀ m , (88)which means that also our standard integral I becomes zero: I = 0 . (89)This means that in the dimensional-regularization scheme it becomes even easier to work with a path integralin polar field variables. In this case one can also completely forget about the Jacobian one gets from thetransformation. Also one can ignore the integrals I that are generated by w -loops.In this paper we will keep everything general however, and not specify a regularization scheme. In section 8 we have proven the conjecture for a general model with two fields in d space-time dimensions.This conjecture, which is promoted to a theorem by now, enables us to actually calculate things via the pathintegral in terms polar fields for any d -dimensional model. In a d -dimensional model the only analyticalcomputations we can do in practice are continuum calculations. Analytical discrete calculations, i.e. calcu-lations on the lattice, are in practice much too hard to do. That is why we have not bothered to simplifythe discrete d -dimensional path integral, hoping to do a discrete calculation with this simplified form.In one dimension analytical discrete calculations are sometimes possible (as we will see in the nextsection). Therefore it is convenient to have a reasonably simple, discrete path integral in terms of polarfields for the case d = 1. It is this path integral that we shall derive in this section.By deriving this path integral we shall also make contact with the literature on quantum mechanical (i.e.1-dimensional) path integrals in terms of polar fields, e.g. [4, 2, 3].Our starting point will again be the discrete Feynman rules (65), now specified to d = 1 however. Also,for the sake of the argument, we will split up the vertices as given below. The one-dimensional Feynmanrules are then: ↔ ¯ h − cos ∆ k + m η ↔ ¯ h − cos ∆ k + m w p k (1) k (2 n ) ↔ − n ¯ hv n − h(cid:16) e − i ∆ k (1) − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) − (cid:17)i p k (1) k (2 n ) ↔ ( − n ¯ hv n − h(cid:0) e − i ∆ p − (cid:1) (cid:16) e − i ∆ k (1) − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) (cid:17)i q k (1) k (2 n ) ↔ − n ¯ hv n h(cid:16) e − i ∆ k (1) − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) − (cid:17)i pq k (1) k (2 n ) ↔ ( − n ¯ hv n h(cid:0) e − i ∆ p − (cid:1) (cid:16) e − i ∆ k (1) − (cid:17) · · · (cid:16) e − i ∆ k (2 n )1 − (cid:17)i k (1) k (2 n ) ↔ ( − n ¯ hv n − h(cid:16) e − i ∆ k (1) − (cid:17) · · · (cid:16) e − i ∆ k (2 n ) − (cid:17)i (90)Looking at these vertex expressions we notice that only the verticesand (91)have a finite continuum limit, all the other vertices go to zero when ∆ is sent to zero in the Feynman rules.First we are going to consider all diagrams which have at least one of these vertices that vanish in thecontinuum limit. The only way these vertices can survive a continuum limit in a complete diagram is whenthere occur loops that give a 1 / ∆.First consider 1-loop diagrams. All 1-loop diagrams can be built from the vacuum diagram (92)By attaching legs we can build any 1-loop diagram from these. Having an η -line in the loop will never give a1 / ∆, no matter which vertices we use. If the whole loop is a w -line this loop can give 1 / ∆’s. If we constructa diagram from this vacuum graph with at least one of the vertices that go to zero in the continuum limit,one can verify easily that either the whole diagram goes to zero in the continuum limit or diagrams cancelamong each other in the complete set of graphs for a certain process, such that the whole process is zero.The same thing can now be done on 2-loop level. Here we can construct all diagrams from the vacuumgraphs , and (93)One can see that the only diagrams surviving the continuum limit and containing at least one of the verticesthat vanish in the continuum limit can be constructed from the following vacuum graphs by only attachinglines with the vertices (91), because these vertices do not give additional powers of ∆. , , and (94)For the vacuum graphs we have the following expressions, excluding vertex constants and symmetry factors.Only the discrete loop integration is done and the worst behavior in ∆ is kept.32 ¯ h ∆ = −
12 ¯ h ∆= −
12 ¯ h ∆= ¯ h ∆ (95)We now construct all 1PI diagrams from these vacuum graphs by attaching lines with the vertices (91).Because we can only attach lines with these vertices we can only get external η -lines. We shall now calculatethe generating functional of all the diagrams that can be constructed in this way: Z ( x ) = ∞ X n =0 n 1 n ! x n (96)Now this Z , for the first vacuum graph, is given by: Z = 18 ¯ h ∆ ∞ X m ,m ,n ,n =0 (cid:18) ∆ ¯ hv + 2∆ ¯ hv x + 2∆ ¯ hv x (cid:19) ¯ h m + m + n + n (cid:18) − hv (cid:19) m + n (cid:18) − hv (cid:19) m + n (cid:18) (cid:19) m + n (cid:18) (cid:19) m + n m ! 1 n ! 1 m ! 1 n ! (2 m + 2 m )!!(2 n + 2 n )!! x m + n +2 n +2 m = 18 ¯ hv (cid:0) xv (cid:1) = ¯ h ∞ X n =0 ( − n ( n + 1)! v n +2 n ! x n (97)Here 1 / m and m denote respectively the number of ηww -and ηηww -vertices in the left loop and n and n denote the number of ηww - and ηηww -vertices in the rightloop. Now we can read off n = ¯ h − n ( n + 1)! v n +2 (98)for the first vacuum graph.Also the contributions from the three other vacuum blobs can be constructed in the same way. Theirgenerating functions appear to cancel each other. The reason for this shall become clear below. For now the only n -leg diagrams that we find in (98) are exactly the vertices one would get from a term − ¯ h r i (99)in the action. This can easily be seen by substituting r = v + η in this term and expanding it in η . Thismeans, up to 2-loop level, one can discard the vertices that go to zero in the continuum limit and replacethem by the vertices from (99). In the action this means one is left with S = ∆ N − X i =0 (cid:18)
12 ( r i +1 − r i ) ∆ + 12 r i v ( w i +1 − w i ) ∆ − ¯ h r i (cid:19) . (100)Disregarding three- and higher-loop level we have now proven that our one-dimensional discrete pathintegral P is equal to: P = Z ∞−∞ dr . . . dr N − r . . . r N − Z ∞−∞ dw . . . dw N − O exp − h ∆ N − X i =0 (cid:18)
12 ( r i +1 − r i ) ∆ + 12 r i v ( w i +1 − w i ) ∆ − ¯ h r i + V (cid:16) r i cos w i v , r i sin w i v (cid:17) (cid:19)! . (101)This is a form that one can also find in the literature. This same path integral is derived by Lee [4] inchapter 19, formula (19.49). Also Edwards et al. [2] and Peak et al. [3] find a term (99). However they startwith the discrete path integral in terms of Cartesian fields, transform to polar fields and actually performthe angular integration. Only then they find the term (99). We have presented a more general proof of thisterm here, like Lee [4].Up to now we have not proven that at three- and higher-loop level there are diagrams, containing at leaston of the vertices that vanish in the continuum limit, that can not be built also from vertices from the term(99). We shall not prove this in this paper. In this paper we are mostly interested in the transformation topolar fields in d -dimensional models, and the conjecture needed to compute via polar fields in these modelshas already been fully proven. It should be clear however that, to have agreement with the literature, thepath integral (101) is correct, up to all orders. So, although we cannot prove it at this point, there are no diagrams at three- and higher-loop order that cannot also be constructed with only the term (99). Another way to derive the discrete one-dimensional path integral (101) is by first using the conjecture. Soone computes all diagrams in a d -dimensional way in the continuum, with the simple Feynman rules fromthe continuum action, then one lets d →
1. To obtain the discrete version of these diagrams one has to knowwhat the difference is between calculating in the continuum and calculating in the discrete. This difference,we know, comes from the problem terms. If we formulate the (continuum) ηww - and ηηww -vertex with thedots, as we did in section 8, then we have a clear source of problem terms. Because in this case we only havethe ηww - and ηηww -vertex we also only have the recursion relation:+ + = 0 (102)This recursion relation ensures that all problem terms from a w -line with two dots cancel against the dottedpart of the ηηww -vertex. So the problem terms from w -lines are never going to give a difference between adiscrete and continuum calculation. All we have to do is find the problem terms coming from η -lines withtwo dots. 34ow, as in the previous (partial) derivation of (101) we can build all diagrams from the vacuum graphs.At 1-loop order there is no difference between a continuum and discrete calculation. At 2-loop order theonly problem terms come from the vacuum graph (103)Now we can understand why, in the previous derivation of (101), the generating functionals from the lastthree vacuum graphs in (94) cancelled. Only the first graph in (94) corresponds to the vacuum graph above.This correspondence can be seen by pinching the dotted η -line in the vacuum graph above. The last threevacuum graphs in (94) correspond to problem terms from dotted w -lines or a dotted ηηww -vertex. Thesecancel among each other because of the recursion relation (102).Now the difference between a continuum and discrete calculation of the graph (103) can be calculated.Also the generating functional of diagrams where we connect any number of η -legs via the ηww - and ηηww -vertices can be calculated. The result of this generating functional is given by (97). In this way we findthat, to compensate for the differences that we get by doing a discrete instead of a continuum calculation,we have to introduce the term (99) in the action again.Also in this way of deriving (101) we do not know how to show that three- and higher-loop diagrams giveno new differences, but this is not important for the main result of this paper.For a nice illustration of the strictly one-dimensional path integral in terms of polar fields (101) we referto [10].
10 Conclusion
We have presented a way to rewrite a d -dimensional Euclidean path integral, in terms of two scalar fields ϕ and ϕ , in terms of the corresponding polar fields r and w . Our first step was to introduce a conjecturethat stated how to perform this transformation. This conjecture states that (13) is the correct way totransform. After this we showed, by doing explicit calculations for two toy models, that the conjecture isindeed correct for these toy models up to some order in perturbation theory. Finally we gave a general proofof the conjecture, based on a completely discrete (i.e. lattice) calculation. To make contact with the literatureon the transformation to polar fields in the case of quantum mechanics, we also specify our calculation tothis case ( d = 1), and find agreement.In a forthcoming paper we will use the conjecture in actual calculations on the Euclidean N = 2 linearsigma model. A Standard Integrals