Introduction to Lattice QCD: The Basics Ava Khamseh Supervisor: Professor Luigi Del Debbio University of Edinburgh Last Updated: June 28, 2014 i CONTENTS iii Contents 1 Introduction 1 2 Derivatives on the Lattice 1 3 Fourier Transfrom on the Lattice and Different Boundary Conditions 2 4 Free Field Scalar Propagator on the Lattice 4 5 Restoration of Rotational Invariance in 2 Dimensions 6 6 Comparison of Lattice and Continuum Propagators Using Mathematica 7 7 Free Field Naive Fermion Propagator on the Lattice 9 8 Free Field Wilson Fermion Propagator 12 1 1 Introduction This report focuses mostly on the theoretical background required to construct QFT on the lattice. The derivations presented here are done for a free theory, i.e. not involving gauge fields. An important derivation is that of the fermion propagator on the lattice. A naive discretisation of the Dirac operator leads to extra poles that have no continuum analogue. This implies the need for constructing other types of lattice fermions, such as Wilson, Domain Wall and Overlap fermions to describe the physics correctly. 2 Derivatives on the Lattice In this section we will introduce basic mathematical tools required for the formulation of Quantum Field Theory on the lattice. Given lattice spacing a, the lattice coordinates can be written as: xµ = nµ a n = 0, 1, ... µ = 1, 2, ..., D in D dimensions (1) for finite volume box of size L = N a, n takes the values n = 0, 1, ..., N − 1. For a smooth function f (x) in D dimensions one has, in the limit a → 0: f (x) = aD x → dD x f (x) (2) n again, for a finite volume lattice of side L, the corresponding integral would range from 0 to L. We define the forward and backward derivatives on the lattice as follows: φ(x + aˆ µ) − φ(x) (3) ∇µ φ(x) = a φ(x) − φ(x − aˆ µ) ∇∗µ φ(x) = (4) a so that ∇µ = ∂µ + O(a). In principle, one can increase the accuracy of the calculations involved by defining derivatives in different ways. The Central Difference formulation takes the form φ(x + a) − φ(x − a) + O(a2 ) (5) 2a which means that the central difference derivative will have smaller discretisation errors (since it is O(a2 )) compared to the forward derivative (O(a)). Therefore, one can see that there is an ambiguity in the way derivatives can be defined on the lattice. Having derivative formulations which contain higher orders of a will naturally increase accuracy. The drawback however, is that it will become more computationally expensive. ∇mid φ(x) = For the second derivative we have φ(x + a) − 2φ(x) + φ(x − a) a2 which can be easily proved by taylor expanding both sides up to order a4 . φ + O(a2 ) = (6) Note: The latter definition of the first derivative i.e. of order O(a2 ) as in equation 5, is the one that will be used in section 7 when constructing the fermion propagator. As for the scalar field, again the definition of the second derivative which is of O(a2 ) will be used. 23 3 FOURIER TRANSFROM ON THE LATTICE AND DIFFERENT BOUNDARY CONDITIONS Fourier Transfrom on the Lattice and Different Boundary Conditions Recall the integral representation of the Kronecker delta function (i.e. discrete), δnn = zn 1 2πi (7) z n +1 which is simply due to the residue theorem. Making a change of variables z = eik where −π < k < π results in π 1 dkeik(n−n ) (8) δnn = 2π −π Note that the limits make sense because if we set n = n on the LHS we get 1 from the delta and on the right hand side 2π 2π = 1. Making another change of variable k → ak yields δnn = π/a a 2π eika(n−n ) dk (9) −π/a Now consider the discretised fourier transform, f˜(k) = a f (na)e−ikna (10) n Multiplying both sides of the equation by eikn and integrating with respect to k from −π/a to π/a gives π/a π/a f˜(k)eikna dk = a −π/a e−ika(n−n ) f (na)dk = 2π n −π/a δnn f (na) = 2πf (n a) (11) n So the discretised inverse fourier transform is π/a f (na) = −π/a dk ˜ f (k)eikan 2π (12) where we are in fact integrating over the first Brillouin zone. Note that increasing the range does not give any new information. In all the above, the volume was considered to be infinite. If however, we wish to work in a finite volume we would need to impose periodic boundary conditions: f (an + aN ) = f (an) (13) Where there are N points in total. Then, defining n = N + n, we must have f˜(k) = a f ((N + n)a)e−ikna = a n f (n a)e−ik(n −N )a = a n f (n a)e−ikn a eikN a (14) n meaning eikaN = e2πm where m is an integer and so k = 2πm aN . Going back to equation 12 and −N observing the limits, we conclude that m must run from 2 + 1 to N2 since we cannot have more 3 than N points. Replacing dk by its discretised equivalent i.e. the “fundamental” unit k which is k = 2π/aN we get f (na) = 1 L N/2 m=−N/2+1 2πm 2πimn f˜( )e N L (15) where we have used L = aN . The above can be easily generalised to n dimensions. Before we proceed, let us discuss a different boundary condition for fermions on a finite lattice which will turn out to be very useful [1]. The fact that momentum is quantised on the lattice, leads to limitations in different phenomenological applications. Take the 2-body hadron decay as an example. The energies of the decay products is related to the masses of the particles involved by 4-momentum conservation. However, they cannot take their physical value unless the masses are consistent with the momentum quantisation rule. This issue can be resolved by choosing different boundary conditions. The momentum quantisation rule for periodic boundary conditions is well-known: ψ(x + ei L) = ψ(x) , i = 1, 2, 3 (16) for spacial directions. Taking the Fourier transform ˜ d4 p e−ip(x+ei L) ψ(p) = d4 p eipx ψ(p) , i = 1, 2, 3 (17) which implies eipi L = 1 =⇒ pi = 2πni L , i = 1, 2, 3 (18) Now, define the θ-boundary conditions as ψ(x + ei L) = eiθi ψ(x) (19) similar to above, taking the Fourier transform gives ei(pi −θi /L) = 1 =⇒ pi = 2πni θi + L L , i = 1, 2, 3 (20) In other words, the spacial momenta are still quantised for periodic boundary condition but also shifted by an arbitrary amount θi /L which is continuous. It is shown [1] that θ/L does indeed act as a true physical momentum, in particular the physical energy of the mesonic state can be written as Eij (θ, a) = 2 + Mij θ L 2 (21) where Mij is the mass of the pseudoscalar meson made of an i and a j quark anti-quark pair. It is also shown that the continuum extrapolation gives the correct relativistic dispersion relations. 4 4 4 FREE FIELD SCALAR PROPAGATOR ON THE LATTICE Free Field Scalar Propagator on the Lattice The continuum Euclidean action in 4 dimensions for the free scalar field reads as follows d4 x SE [φ] = 1 1 ∂µ φ(x)∂ µ φ(x) + m2 φ2 (x) 2 2 = 1 2 d4 x −φ(x)∂ 2 φ(x) + m2 φ2 (x) (22) Discretising the space-time and using 6 for the definition of the derivative we get SE [φ] = a4 2 −φ(x) x µ 4 = a 2 − x kk φ(x + aˆ µ) − 2φ(x) + φ(x − aˆ µ) + m2 φ2 (x) 2 a eikx ˜ φ(k) a2 4 = = = = a 2 1 2 ei(k+k )x x kk δ(k + k ) kk 1 2 µ −1 ˜ φ(k)2 a2 −1 ˜ φ(k)2 a2 dk φ(k) m2 + 1 2a2 ˜ ) + m2 φ(k) ˜ φ(k ˜ )ei(k+k )x eik .(x+aˆµ) − 2eik .x + eik .(x−aˆµ) φ(k 1 a2 cos(kµ a) + m2 + µ cos(kµ a) + m2 + µ 8 a2 8 a2 ˜ φ(k ˜ ) φ(k) ˜ φ(k ˜ ) φ(k) (2 − 2 cos(kµ a)) φ(k ) µ dk φ(k) a2 m2 + (2 − 2 cos(kµ a)) φ(k ) µ (23) 4 4 d kd k where k.ˆ µ = kµ and kk denotes (2π) 4 (2π)4 . In the last line we have changed variables from k → −k and the integral is over the first Brillouin zone. Therefore, the scalar field propagator is π/a DS (x) = φ(x)φ(0) = −π/a d4 k eik.x (2π)4 m2 a2 + µ 4 sin2 ( kµ a ) 2 = d3 k (2π)3 dk 4 (2π) m2 a2 + eik.x µ (2 − 2 cos(kµ a)) = d3 k (2π)3 dk 4 (2π) m2 a2 + 8 − = d3 k ik.x e (2π)3 (24) eik.x i 2 cos(ki a) − 2 cos(k4 a) dk 4 eik4 x4 (2π) 2M 2 − 2 cos(k4 a) Where we have defined 2M 2 = m2 a2 + 8 − i 2 cos(ki a) noting that m2 a2 + 8 − i 2 cos(ki a) is a positive quantity. The poles will be at the points where the denominator vanishes: 2M 2 − 2 cos(k4 a) = 2M 2 − eik4 a + e−ik4 a = 2M 2 − z − z −1 = 0 (25) dz Note however, the change of variable z = eik4 a ⇒ dk4 = iaz . The z in the denominator of the last latter expression will multiply equation 25. Multiplying top and bottom by -1, ones needs to simply solve z 2 − 2M 2 z − 1 = 0 (26) 5 which gives z = M2 ± M4 − 1 (27) Which is real since M ≥ 1. Figure 1: z = eik4 a contour √ But the contour is the unit circle so only one of the poles i.e. z = M 2 − M 4 − 1 contributes since the other one lies outside the unit circle. Because z is real and positive, we can write it as z = e−ωa where ω is a positive quantity associated with the energy (see the next paragraph). Note that the option z = eωa is omitted since it makes the integral divergent for large k4 and positive t. Indeed, the non-contributing pole can be written as, √ 1 M2 + M4 − 1 √ √ × = M 2 + M 4 − 1 = eωa (28) M2 − M4 − 1 M2 + M4 − 1 Hence the integral 24 is evaluated d3 k ik.x 2πi 1 1 e = (2π)3 −ia2π e−ωa − eωa a d3 k ik.x e−ωx4 e (2π)3 2 sinh(ωa) (29) Also, cosh(ωa) = 1 ωa (e + e−ωa ) = M 2 2 (30) Expanding the both sides up to order O(a2 ) yields 1+ ω 2 a2 1 = m2 a2 + 8 − 2 2 2(1 − i (ki a)2 1 ) = m2 a2 + 2 + k2 a2 2 2 (31) Therefore, for a → 0 ω(k) → m2 + k2 and the expression for the energy in the continuum is recovered. (32) 6 5 5 RESTORATION OF ROTATIONAL INVARIANCE IN 2 DIMENSIONS Restoration of Rotational Invariance in 2 Dimensions Let us now consider the free propagator of scalar field in 2 dimensions which can be written as follows in lattice units [2], DF (x) = φ(x)φ(0) = eipx d2 p 2 2 (2π) m + 4 − 2 cos(p1 ) − 2 cos(p2 ) (33) Consider 2 cases: x → ∞ along √ a lattice direction where x = nt, t → ∞ with n = (1, 0) or along the diagonal with n = (1, 1)/ 2. 1. For n = (1, 0) and denoting 2b = m2 + 4 − 2 cos(p2 ), equation 33 reads π DF = −π π = −π π dp2 π dp1 eip1 t dp2 e−ωt = 2π −π 2π 2b − 2b cos(p1 ) −π 2π 2 sinh(ω) √ 1 dp2 −t ln(b(p2 )+ b(p2 )2 −1) e 2π 2 sinh(ω) (34) where the first integral is done by finding the relevant √ pole in exactly the same way as the previous section with cosh(ω) = b and ω = ln(b + b2 − 1). Now, for t large, one can use the saddle point method to evaluate the major contribution to the integral. To this end, we need to find the point(s) at which the first derivative of ω(b(p2 )) = ln(b(p2 ) + b(p2 )2 − 1) with respect to p2 vanishes. Note that in order to keep the algebra simpler, there is no need to derive an explicit form for the derivative. It merely suffices to identify the point at which the derivative vanishes. So taking 1 + √2b(p2 )2 dω ∂ω ∂b 2 b(p2 ) −1 = = sin(p2 ) = 0 db ∂b ∂p2 b(p2 ) + b(p2 )2 − 1 (35) It is clear from the definition of b that b > 1 meaning the fraction term in the above expression is always greater that zero. Hence, within the Brillouin zone, the solution to the equation corresponds to p2 = 0. This implies b0 = m2 +1 2 ⇒ cosh(ω) = m2 +1 2 (36) Then, equation 34 will be equal to DF 1 2 sinh(ω(b)) b0 =m2 /2+1 e−tω(b0 ) dp2 − 1 tω e 2 2π (b0 )(b(p2 )−b0 )2 ∝ e−tω(b0 ) (37) The correlation length ξ(n) in direction n is identified by DF ∝ e−t/ξ(n) so in this case, ξ −1 (n) = ω. 7 √ 2. For n = (1, 1)/ 2, changing variables to Σ = π π p1 +p2 2 and ∆ = p1 −p2 2 , equation 33 becomes √ dp1 dp2 eit(p1 +p2 )/ 2 p1 +p2 2 (2π)2 m2 + 4 − 4 cos( p1 −p 2 ) cos( 2 ) √ ∝ ei 2Σt dΣd∆ 2 m + 4 − 4 cos ∆ cos Σ ∝ dΣd∆ 2 cos ∆ ∝ dΣd∆ e 2 cos ∆ 2b − 2 cos Σ ∝ d∆ e−t 2ω 2 cos ∆ 2 sinh ω ∝ d∆g(∆)e−t ∝ d∆g(∆)e−t ei √ 2Σt m2 +4 2 cos ∆ − 2 cos Σ √ i 2Σt (38) √ √ √ 2ω 2 ln(b(∆)+ √ b2 (∆)−1) where m2 + 4 (39) 4 cos ∆ Using the same saddle point method as the previous part, the integral is dominated by ∆0 such that sin ∆0 = 0 i.e. cos ∆0 = 1 implying ω = ln(b + b2 − 1) DF ∝ e−t , √ b= 2ω (40) with m2 cosh(ω ) = b0 = 1 + 4 √ −1 In this case, the correlation length, ξ (n) = 2ω . (41) Taking the ratio cosh(ω) ξ m2 √ = =1− + O(m4 ) ξ 48 cosh( 2ω ) In non-lattice units m → ma and so invariance. ξ ξ (42) → 1 as ma → 0 which implies the recovery of the rotational Therefore, in order to restore rotational invariance, one needs to work in a regime where x a, that is the distance between the particle sitting at the origin and the one at distance x from the origin, is large in units of lattice spacing a. This implies x/a 1 as a → 0 which was denoted by nt , t → ∞. 6 Comparison of Lattice and Continuum Propagators Using Mathematica We will now try to compare the lattice formulation of the scalar propagator to the continuum one in infinite volume. 86 COMPARISON OF LATTICE AND CONTINUUM PROPAGATORS USING MATHEMATICA It is known that the continuum scalar field propagator in 4 dimensions can be written in terms of the modified Bessel function K1 [3]. Here will merely quote the results for the continuum case, ∞ DFC (x, n) = −∞ dn p eipx (x2 )1/2 = (2π)−n/2 n 2 2 (2π) m + p m 1−n/2 K1−n/2 m(x2 )1/2 (43) and the lattice case π/a DFL (x, n) = −π/a dn p eipx = n 2 (2π) m + pˆ2 where pˆ2 = 4 a2 n ∞ 2 dα e−m 0 n sin2 µ=1 α 2α 1 e− a2 ( )I xµ a a µ=1 pµ a 2 2α a2 (44) (45) and I xµ is the modified Bessel function of the first kind. a The plot below is the Mathematica [4] implementation of the above propagators in one dimension for particular values of x. Figure 2: 1-D Lattice and Continuum Scalar Propagators As it can observed, the lattice points agree with the continuum fit. What is interesting however, is to examine the deviation of the lattice points from the continuum to be able to analyse the discretisation errors properly and recover the full continuum limit. However, note that the continuum limit has be taken in a meaningful way i.e. in lattice units. In other words, in going from continuum to the lattice, we induce a UV cut-off on the momentum (of order 1/a). In order to discuss the physics properly, one has to work in a region where the physics is insensitive to the imposed cut off (i.e. of order 1/a). This is better illustrated with an example: Let us take mx to be fixed at a value say, α = mx, change the ratio β = xa which takes different values over a range, in which case am = α β and then compare the difference between lattice and 9 continuum results. This is presented in the following graph by a particular value of α. As tt can be seen in figure 3, taking the distance between to source and the sink to be larger lattice spacing results in the error, i.e. the difference between the two propagators, to reach a constant value. Figure 3: Difference between Lattice and Continuum Scalar Propagators Due to time constraints, at this stage, we will not go into more details. The above could also be analysed for the finite volume case. 7 Free Field Naive Fermion Propagator on the Lattice The continuum Euclidean action in 4 dimensions for the free fermion field reads as follows ¯ = SE [ψ, ψ] ¯ ¯ d4 x ψ(x)γ µ ∂µ ψ(x) + mψ(x)ψ(x) (46) The continuum action can be discretised using the “Finite Difference Approximation” for the derivatives. So we get ¯ =a4 SE [ψ, ψ] x a4 = (2π)2 = a4 (2π)2 ψβ (x + aˆ µ) − ψβ (x − aˆ µ) ψ¯α (x) γµαβ + mψ¯α (x)δαβ ψβ (x) 2a x x 4 a (2π) a4 = (2π) kk ¯ (k)γµαβ 1 e−ikx ψ˜ α 2a kk ¯ (k)e−ikx γµαβ 1 eik x eikµ a − e−ikµ a ψ(k )β + mδαβ e−ikx+ikx ψ˜¯α (k)ψ˜β (k ) ψ˜ α 2a eik (x+aˆµ) − eik (x−aˆµ) ˜ )β + mδαβ eikx ψ¯˜α (k)ψ˜β (k ) ψ(k i γµαβ sin(kµ a)ψ(k )β + mδαβ ψ˜¯α (k)ψ˜β (k ) a sin(kµ a) ˜ + γµαβ ψβ (k) a ¯ (k)δ(k − k ) ψ˜ α = kk ¯ (k) mδαβ ψ˜ α k (47) 10 7 FREE FIELD NAIVE FERMION PROPAGATOR ON THE LATTICE where x = na and index µ is summed over. Therefore the lattice fermion propagator in momentum space is 1 ˜ F (k) = D (48) sin(k a) m1 + iγµ a µ In the continuum limit, Minkowski space, one must recover the usual fermion propagator. Therefore in the limit as a → 0 ˜ F (k) → D 1 1 1 Minkowski −−−−−−−→ =− / m + iγµ kµ −(γ0 k0 − γ.k − m) k−m (49) Where we have used the fact that γ0E = γ0M , γ E = −iγ M and tE = itM . We now need to find the inverse fourier transform. Note that the integral is over the first Brillouin zone and for now, we will work in infinite volume and will make it finite at the end of the computation. d4 k (2π)4 m + ¯ ˜ F = ψ(x)ψ(0) D = i a m− eikx × sin(kµ a)γµ m− i a i a sin(kµ a)γµ sin(kµ a)γµ dk 4 d3 k ma2 − ia sin(kµ a)γµ ik4 x4 ik.x e e 2π (2π)3 m2 a2 + µ sin2 (kµ a) = (50) where we have contractions over µ, used γµ γµ = 1 and multiplied top and bottom by a2 . Now we have to find the relevant poles in the denominator. But note that we are integrating with respect to dk 4 , so we need to consider the contour in the k 4 space. Because the integral is over the first Brillouin zone, the real part of k 4 must satisfy −π/a < k 4 < π/a. Defining z = eik4 a the denominator will become 3 sin2 (ki a) + m2 a2 + i=1 −1 2 (z + z −2 − 2) = 0 4 (51) M2 We multiply the numerator and the denominator by −4z 2 . So need to find the solutions to z 4 − 2(1 + 2M 2 )z 2 + 1 = 0 (52) solving it, yields z 2 = (1 + 2M 2 ) ± 2M M 2 + 1 = (−M ± 1 + M 2 )2 (53) Hence the four solutions for z are √ ωa 2 M + √ 1 + M = e M − 1 + M 2 = −e−ωa √ z= −M + 1 + M 2 = e−ωa √ −M − 1 + M 2 = −eωa (54) √ where ω is defined such that M + 1 + M 2 = eωa . The main task now is to identify the relevant poles. Since we are parameterising by z it would be easier to find the poles using the complex z plane. Complex k plane is commented out 11 The limits of the k integral run from −π/a to π/a so z will run from e−iπ to eiπ i.e. the unit circle. As well as that, 1 dk = (55) iaz Therefore, using 52 and the equation above we have an overall factor of z in the numerator. Figure 4: z = eik4 a contour Clearly, the only poles that contribute are the ones the lie within the unit circle 0 ≤ |z| ≤ 1, i.e. z = ±e−ωa . Now, we need to use the residue theorem. Finding the coefficients for the pole at: 1. z =ik4 a = e−wa , the numerator becomes 3 − 4z ma2 − ia sin(ki a)γi − ia i=1 1 −ωa e − eωa γ4 2i (56) 3 = − 4e −ωa 2 ma − ia sin(ki a)γi + a sinh(ωa)γ4 i=1 and the denominator, (z − eωa )(z + e−ωa )(z + eωa ) = 2e−ωa (e−ωa − eωa )(e−ωa + eωa ) = −4e−ωa sinh(2ωa) (57) 2. z =ik4 a = −e−wa the numerator becomes 3 (−)x/a 4e−ωa ma2 − ia sin(ki a)γi − a sinh(ωa)γ4 (58) i=1 Note that since eika = −e−ωa , raising both sees to the power of x/a yields eikx = (−)x/a e−ωx . As for the denominator − 2e−ωa (e−ωa + eωa )(−e−ωa + eωa ) = 4e−ωa sinh(2ωa) (59) 12 8 FREE FIELD WILSON FERMION PROPAGATOR Finally, summing the residues results in ˜F = D d3 k eik.x−ωx4 (2π)3 sinh(2ωa) 3 3 x4 /a ma − i sin(ki a)γi + sinh(ωa)γ4 + (−) ma − i i=1 sin(ki a)γi − sinh(ωa)γ4 i=1 (60) √ 3 For e−ωa = −M + 1 + M 2 where M = m2 a2 + i=1 sin2 (ki a), if we expand both sides for lattice spacing a → 0 up to order a2 , in the continuum limit: 1 − ωa = −a m2 + k2 + 1 + a2 (m2 + k2 ) = −a m2 + k2 + 1 + O(a2 ) (61) i.e. ω(k) → m2 + k2 (62) Note that because of of the term e−ωx4 , where both ω, x4 > 0, the integral above is dominated by e−ωx4 begin close to 1 (largest possible value). However, given the form of M on the RHS we know that it is not zero before taking any limits. This function is largest when M itself is at its smallest 3 possible value (plot if unsure) i.e. when i=1 sin2 (ki a) = 0. This occurs when ki = ni π/a. Since the integration is over the first Brillouin zone where k ∈ (−π/a, π/a], ni can only be either 0 or 1. Therefore, we have 23 minima in total. Expanding for small a, √ 1 1 − ωa = − m2 a2 + (1 + m2 a2 )1/2 = −ma + 1 + m2 a2 + O(a3 ) 2 (63) Therefore, in the limit where a → 0: ω(k) = m (64) So we see that at the energy ω is equal to the mass at more than one point i.e. the rest frame (0, 0, 0). This the source of the fermion double problem. 8 Free Field Wilson Fermion Propagator Consider the following lattice action for massless fermions in momentum space, ¯ ψ(x)D(x − y)ψ(y) S= (65) x,y According the Nielsen-Ninomiya theorem, the properties below cannot hold simultaneously [5]: ˜ 1. D(x) is local i.e. D(p) is a periodic, analytic function of pµ . ˜ 2. D(p) ∝ γµ pµ for a|pµ | 1 ˜ 3. D(p) is invertible for pµ = 0 ˜ 4. {γ5 , D(p)} =0 13 Note that violating locality will result in discontinuities in the derivatives of the propagator and hence the analyticity. The second and third cases are related to having a single flavor of Dirac fermion in the continuum limit. We have come across the third case in our discussion of the doubling problem. Any attempt to solve this issue will involve violation of one of the other conditions. The last point is a statement about Chiral symmetry which will be discussed later on. We will now introduce a simple method by which one can resolve the doubling problem. This method involves adding a term to the mass. It is clear that this will then break the chiral symmetry and hence the Nielsen-Ninomiya theorem will still hold. Let 4 1 1 (1 − cos(kµ a)) = m + M (k) = m + a µ=1 a 3 (1 − cos(ki a)) + i=1 1 1 − cos(k4 a) a a (66) Ω where we have labeled the k4 independent part by Ω. Using the above relabelling, one can observe that the form of the propagator is the same as that in 50 i.e. ¯ ˜ F = ψ(x)ψ(0) D = dk 4 d3 k M a2 − ia sin(kµ a)γµ ik4 x4 ik.x e e 2π (2π)3 M 2 a2 + µ sin2 (kµ a) (67) The denominator can be written as follows: M 2 a2 + µ 1 2 sin2 (kµ a) =a2 Ω2 − Ω cos(k4 a) + 2 cos2 (k4 a) + a a sin2 (ki a) + 1 − cos2 (k4 a) i 3 sin2 (ki a) + 1 =a2 Ω2 − 2aΩ cos(k4 a) + i=1 (68) which vanishes when a2 Ω2 − 2aΩ cos(k4 a) + cos(k4 a) = a2 Ω2 + i [ma + sin2 (ki a) + 1 = 2Ωa i sin2 (ki a) + 1 = 0. Therefore 2 − cos(ki a)) + 1] + 1 + i sin2 (ki a) 2 [ma + i (1 − cos(ki a)) + 1] i (1 (69) ≡ cosh(ωa) One can check that with the above identification the energy ω makes sense by taking the continuum limit of the above expression (also see 62). Note however that the exact relation between ω and k4 is more subtle and is addressed in detail when the poles are computed below. Expanding for small a gives, 1 1 + (k2 + m2 )a2 + O(a4 ) 2 On the other hand the expansion of cosh(ωa) = 1 + ω(k) → as expected. ω 2 a2 2 . m2 + k2 (70) Comparing implies (71) 14 8 FREE FIELD WILSON FERMION PROPAGATOR Now, as before, we start by changing variables to z = eik4 a in 68. Note that the relevant indz . The latter z will multiply tegral will now be over a unit circle in complex z plane and dk 4 = iaz the term in 68 to give z a2 Ω2 − aΩ z + z −1 + sin2 (ki a) + 1 =0 i ⇒−2 a2 Ω2 + sin2 (ki a) + 1 2aΩ i z + z2 + 1 = 0 (72) λ Meaning, z =λ± Note: We need to keep track of the factor λ2 − 1 (73) −1 ia2 Ω . However, at this stage a few checks need to be made: 1. Because Ω > 0, it immediately implies that λ > 0. For z to be real, we must have λ ≥ 1. This condition is also satisfied for aΩ > 0 which can be easily verified by plotting λ as a function of aΩ. See the left graph in figure 5. (Note that to plot the graph we have merely treated λ as a mathematical function without considering its dependence on any physical 2 +1 ). parameter such as m, i.e. we have really plotted y = x 2x Figure 5: Plot of λ vs aΩ and z vs λ 2. Only one pole i.e. z = λ − 5 (b). √ λ2 − 1 lies within the unit circle for λ > 1, right graph in figure 3. Now, we need to consider how to write z as a function of ω correctly. z = ±eωa will result in a divergent integral for large time due to the term eik4 x4 = eωx4 . Therefore, z = ±eωa . 4. The only other possibilities are z = ±e−ωa corresponding to k = iω and k = iω + π. Since √ 2 the solution 0 < λ − λ − 1 < 1 for λ > 1 (from the plot), z = −e−ωa leaving only one possibility: z = e−ωa i.e. k = iω. √ Hence, there is only one contributing pole corresponding to z = λ − λ2 − 1 ≡ e−ωa . Substituting 15 back into 67 and using the residue theorem, 3 d3 k Ωa2 − a cosh(ωa) − ia i=1 sin(ki a)γi + aγ4 sinh(ωa) −ωx4 ik.x e e (2π)3 Ω sinh(ωa) (74) The major contribution to the integral (as time increases), corresponds to the lowest possible value of ω which in turn, according to the plot, corresponds to the lowest possible value of λ. So we need to search for a point as close as possible to aΩ = 1 implying cos(ki a) = 1. Hence ki = 2ni π/a. Within the first Brillouin zone where k ∈ (−π/a, π/a], ni can only take the value 0 ⇒ k = (0, 0, 0). Expanding for small a gives, 1 ¯ ˜ F = ψ(x)ψ(0) D = 2 2a 1 − ωa = 1 − ma + O(a2 ) =⇒ ω(k = 0) = m (75) It is clear that this corresponding to only one point i.e. where k = (0, 0, 0) and so the doubling problem has been resolved. 16 REFERENCES References [1] GM De Divitiis, R Petronzio, and N Tantalo. On the discretization of physical momenta in lattice QCD. Physics Letters B, pages 1–10, 2004. [2] Jan Smit. Introduction to Quantum Fields on a Lattice. Cambridge University Press, 2002. [3] Beatrice Paladini and JC Sexton. Asymptotic expansion of the lattice scalar propagator in coordinate space. Physics Letters B, (February 2008):1–13, 1999. [4] Wolfram. Mathematica 9.0, 2012. [5] David B Kaplan. Chiral Symmetry and Lattice Fermions. December 2009.
© Copyright 2025 ExpyDoc