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.
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