Induced representations of Poincare group on the lattice: spin 1/2 and 1 case

0.16 MB·English

by  M. Lorente

INDUCE REPRESENTATION OF THE POINCARE
GROUP ON THE LATTICE: SPIN 1/2 AND 1 CASE
4
0
Miguel Lorente
0
2 Departamento de F´ısica, Universidad de Oviedo, 33007 Spain

n and Institute fu¨r theoretische Physik. Universit¨at Tu¨bingen, Germany.
a
J
Peter Kramer
5
Institute fu¨r theoretische Physik. Universit¨at Tu¨bingen, Germany.

1
v
8
0
0 Abstract
1
0
Following standard methods we explore the construction of the discrete Poin-
4
0 car´e group, the semidirect product of discrete translations and integral Lorentz
/ transformations, using the Wigner-Mackey construction restricted to the momen-
t
a
tum and position space on the lattice. The orbit condition, irreducibility and
l
-
assimptotic limit are discussed.
p
e
h
: 1 Introduction
v
i
X
r InapreviouspaperofthisSymposium[14]wehavediscussedtheinducedrepresentations
a
of the euclidean group on the lattice via duality and Fourier transform. In this paper
we apply the same method to the induced representation of Poincar´e group on the
lattice. We explore the problem of irreducibility connected with the orbit condition, the
assymptotic limit of the difference equation for the Klein-Gordon and Dirac field.
In section 2 we present the realization of the integral Lorentz transformation, that
can be factorized completely with the help of Kac generators, both for the fundamental
and spin representation. In section 3 we introduce several types of Fourier transform
that will be used to go from momentum to position space. In section 4 we reproduce
the properties of covariant bispinors in discrete momentum space in similar fashion to
the continuous one. In section 5, we elaborate the induced representation of Poincar´e
group on the lattice following standard procedure and we compare the results derived
from the different types of Fourier transform.
1
2 Integral Lorentz transformations
A Lorentz transformation is integral if leaves invariant the cubic lattice and at the same
time keeps invariant the bilinear form
x2 x2 x2 x2 (1)
0 − 1 − 2 − 3
According to Coxeter [1] all integral Lorentz transformations (including reflections)
are obtained by combining the operations of permuting the spatial coordinates x ,x ,x
1 2 3
andchanging thesignsofanyofthecoordinatesx ,x ,x ,x togetherwiththeoperation
0 1 2 3
of adding the quantity x x x x to each of the four coordinates of a point.
0 1 2 3
− − −
These operations can be described geometrically by the reflections on the plans
perpendicular to the vectors
α = e e , α = e e , α = e , α = (e +e +e +e )
1 1 2 2 2 3 3 3 4 0 1 2 3
− − −
where e ,e ,e ,e , is an orthonormal basis.
0 1 2 3
{ }
In matritial form these reflections are
1 0 0 0 1 0 0 0
0 0 1 0 0 1 0 0
S =  , S =  
1 0 1 0 0 2 0 0 0 1
 0 0 0 1   0 0 1 0 
   
 1 0 0 0  2 1 1 1
0 1 0 0 1 0 1 1
S =  , S =  − − − 
3 0 0 1 0 4 1 1 0 1
− − −
 0 0 0 1   1 1 1 0 
 −   − − − 
   
These reflections constitue a Coxeter group, the Dynkin diagram of which is the
following
4 4
a a a a
1 2 3 4
Kac has proved [2]that S ,S ,S ,S generate all the integral Lorentz tranformations
1 2 3 4
that keep invariant the upper half of the light cone.
This result can be used to factorized any integral Lorentz transformationthat belong
to the proper orthochronous group. Let
a e f g
b
L   (2)
≡ c ∗
 d 
 
 
be an integral matrix of determinant +1, satisfying
LgLt = g , g = diag(1, 1, 1, 1) (3)
− − −
and also a 1. From
≥
a2 b2 c2 d2 = 1 (4)
− − −
2
it follows that only one of b,c,d can be zero. Suppose a > 1. Then we apply S ,S ,S
1 2 3
to L from the left until b,c,d become non-positive integers. To the resulting matrix we
apply S . We get
4
a′ e′ f′ g′
b′
L′ =  
c′ ∗
 d′ 
 
 
with a′ = a+b+c+d. Obviously
(a+b+c+d)(a b c d) = 1 2bc 2bd 2cd < 0
− − − − − −
therefore a+b+c+d < 0 or
2a+b+c+d = a′ < a (5)
By iteration of the same algorithm we get
a > a′ > a′′ > ... > a(k) 1 (6)
≥
The last inequality is a consequence of the fact that L and S belong to the complete
4
Lorentz group. Following this process we get an integral matrix with a(k) = 1 which is
a combination of S ,S ,S , giving all the 24 elements of the cubic group on the lattice.
1 2 3
Therefore a general integral Lorentz transformation of the proper orthochronos type
L can be decomposed as
L = PηPθPiS ...S PδPεPζS SαSβSγ (7)
1 2 3 4 4 1 2 3 4 1 2 3
all permutations
n o
where P = S S S S S , P = S S S , P = S are the matrix which change sign of
1 1 2 3 2 1 2 2 3 2 3 3
b,c,d and α,β,γ,δ,ε,ζ,η,θ,ι...= ,
A particular case of integral Lorentz transformations are the boost or integral trans-
formations of two inertial systems with relative velocity. The general expresion for these
transformations can be obtained with the help of Cayley parameters [3]. Let us take
n = p = q = 0 and m,r,s,t, integer numbers. We have two cases, corresponding to two
diophantine equations:
i) m2 r2 s2 t2 = 1
− − −
m2+r2+s2+t2 2mr 2ms 2mt
2mr m2+r2 s2 t2 2rs 2rt
L= 2ms 2r−s − m2 r2+s2 t2 2st  (8)
 2mt 2rt − 2st − m2 r2 s2+t2
 − − 
 
ii) m2 r2 s2 t2 = 2
− − −
m2 1 mr ms mt
m−r r2+1 rs rt
L= ms rs s2+1 st  (9)
 mt rt st t2+1 
 
 
The solutions of the diophantine equation i) and ii) are obtained by applying all the
Coxeter reflections to the vector (1,0,0,0) in case i) and to the vector (2,1,1,0) in case
ii)
From the inspection of (7) if we take the quotient of L with respect the subgroup
of all integral rotations or cubic group we are left with the coset representatives which
3
are not exhausted by the pure Lorentz transformations (8) or (9), because these are
always symmetric matrices. Therefor we have to add all the integral Lorentz matrix
that applied to the vector (1,0,0,0) gives all the integral vectors of the type (m,r,s,t)
with m2 r2 s2 t2 = 1. This is equivalent to say that we apply to (1,0,0,0) not
− − −
only the matrix L given by (8) but also its square root matrix, namely,
m r s t
r 1+ r2 rs rt
√L =  s rsm+1 1+m+1s2 ms+t1  (10)
m+1 m+1 m+1
 t rt st 1+ t2 
 m+1 m+1 m+1 
 
In position space the space-time coordinates of the lattice x are integer numbers.
µ
They transform under integral Lorentz transformations into integral coordinates. The
same is true for the increments ∆x .
µ
In momentum space the components of the four-momentum are not integer numbers
but they can be constructed with the help of integral coordinates, namely,
c∆t ∆~x
p m c ,
µ ≡  ((c∆t) (∆~x))/ ((c∆t) (∆~x))/!
− −
If ∆x transform under integral Lorentz transformations the new p′ will be given
µ µ
in terms of integral ∆x′ .
µ
Using the homomorphism between the groups SO(3,1) and SL(2,C/) we obtain the
representation of integral Lorentz transformations in 2-dimensional complex matrices.
From the knowledge of the Cayley parameters [5] we read off the matrix elements of
α SL(2,C/)
∈
 m+t+i(n λ), p+r +i(q +s)
α = − − (11)
√∆ p+r +i(q s), m t i(n+λ)
(cid:18) − − − (cid:19)
∆ = detα = m r s t +m +p +q λ (12)
− − − −
For instance, we calculate the 2-dimensional representation of the Coxeter reflection
S multiplied by the parity operator P in order to get an element of the proper Lorentz
i
group) identifying its matrix elements with Lorentz matrix written in terms of Cayley
parameters: Easy calculations give the unique solutions:
1 0 1 i
D(PS ) = − − (13)
1 ±√2 1 i 0
(cid:18) − (cid:19)
1 i 1
D(PS ) = (14)
2 ±√2 1 i
(cid:18) − − (cid:19)
i
D(PS ) = (15)
3 ± i
(cid:18) − (cid:19)
1 0 1 i
D(PS ) = − (16)
4 ±√2 1 i 2i
(cid:18) − − (cid:19)
The integral Lorentz transformations without rotations as given in (8) and (9) have
a 2-dimensional representation making n = p = q = λ = 0 in (11) and the choice
∆ = m2 r2 s2 t2 = 1 or 2 in (12).
− − −
4
In order to complete the picture we have to add the 2-dimensional representation of
the matrix √L given in (10) which turns out to be
 m+1+t r is
α = − (17)
(m+) r +is m+1 t
(cid:18) − (cid:19)
We give also the 2 2 matrix representation of the discrete momentum. Let p a
p µ
×
four momentum which is obtained by applying all the integral Lorentz transformations,
given by (7) divided by the cubic group, to the vector (m c,0,0,0). In 2-dimensional
0
matrix form this is equivalent to apply the matrix α given by (17) to the unit matrix
multiply by m c:
0
m c 0 m+t r is
α  α+ = m c − (18)
0 m c  r +is m t
0
(cid:18) (cid:19) (cid:18) − (cid:19)
which is of the standard form if we identify the components of the 4-momentum as
p = m c(m,r,s,t) (19)
µ 0
with m,r,s,t integral numbers, satisfying m2 r2 s2 t2 = 1.
− − −
3 Fourier transform on the lattice
Inordertogofrompositionspacetomomentumspaceonthelatticewecandefineseveral
restrictions of the continuous variables of Fourier transform to the discrete variables on
the lattice.
I type. Discrete position and momentum variables of finite rank
We construct an orthonormal basis [6]
1+ 1iεp j
f (p ) = 2 m , j = 0,1,...N 1 (20)
j m  εp −
(cid:18) −  m (cid:19)
2 π
p = tg m, m = 0,1,...N 1
m
ε N −
that satisfy periodic boundary conditions:
f (p ) = f (p ), p = p ,
0 m N m m+N m
orthogonality relations:
N−
1
∗
N fj (pm)fj(pm′) = δmm′ (21)
j = 
X
and completness relations:
N−
1 
∗
Nε fj (pm)fj′(pm) = εδjj′ (22)
m = 
X
The finite Fourier transform reads
N−
1
Fˆ = f∗ (p )F (23)
m √N j m j
j = 
X
5
for some periodic function F on the lattice
j
F = F
j+N j
Ifwewritef (p ) exp iπmj thistransformcoincidewithstandardfiniteFourier
j m ≡ N
transform [7].
(cid:0) (cid:1)
II type. Discrete position and continuous momentum
When we restrict the position variables in continuous Fourier transform to discrete
values we obtain the Fourier series.
∞
1
Orthonormal basis : eikjε f (k) (24)
j
√2π ≡
(cid:26) (cid:27)j=−∞
π/ε 1
∗
Orthogonality relations : fj (k)fj′(k)dk = εδjj′ (25)
Z−π/ε
∞

Completness relation : f∗ (k)f (k′) = δ(k k′) (26)
j j ε −
j=−∞
X
Fourier expansion: for a periodic function F (k)
∞
∗
F (k) = f F (27)
j j
j=−∞
X
π/a
F = f (k)F′(k)dk (28)
j j
Z−π/a
Now we make the change of variable
2 1
P = tg kε (29)
ε 2
1 1+ 1iεp j
f (p) = 2 (30)
j √2π 1 1iεp
(cid:18) − 2 (cid:19)
and the orthogonality relations become
∞ dp 1
∗
fj (p)fj′(p) 1+ 1ε2p2 = εδjj′ (31)
Z−∞ 4
and the completness relation
∞
1
f∗(p)f (p′) = 1+ ε2p2 δ(p p′) (32)
j j 4 −
j=−∞ (cid:18) (cid:19)
X
Notice that f (p) may not be a periodic function.
j
III type: discrete position and discrete momentum of infinite rank
We construct an orthonormal basis
1+ 1iεk n
f (j) = 2 , k,n (33)
n 1 1iεp ∈ Z
(cid:18) − 2 (cid:19)
6
satisfying
N
1
lim fn(k)fn(k′) = δkk′ (34)
N→∞ 2N +1
n=−N
X
(Proof: For k = k′ we use the identity
6
1 sin N+ 1 θ
1+cosθ+cos2θ+...+cosNθ = + 2
2 sin θ
(cid:0) 2 (cid:1)
with eiθ f∗ (k)f (k′)). (cid:0) (cid:1)
≡ 1 1
The completness relation is now:
∞
fn∗(k)fn′(k) = Ll→im∞ δL(n−n′) (35)
k=−∞
X
where
L
δ (n n′) = f (k)f∗ (k) (36)
L − n n′
j=−L
X
is a δ sequence satisfying
N
1
lim δ (n n′) = 1 (37)
L
N→∞ 2N +1 −
n=−N
X
(Proof:
N
1
δ (n n′) =
L
2N +1 −
n=−N
X
L N
1
∗
= 1+ f (k)f (k)+c.c. =
2N +1 n n′
k=1 n=−N
X X (cid:0) (cid:1)
L 1 sin N + 1 θ 1
= 1+ 2 f∗ (k)+c.c. −→
k=1 2N +1 s(cid:0)in 21θ (cid:1) n′ ! N→∞
X
for all L, as required)
The Fourier transform becomes:
N
1
Fˆ(k) = lim f (k)F (38)
n n
N→∞ 2N +1
n=−N
X
N
where F 0 when n and 1 F −→0
n → → ∞ 2N+1 n N→∞
n=−N
∞P
F = f (k)Fˆ(k) (39)
n n
k=−∞
X
The Fourier transform of type III was introduced in [8]. When
n,n′ , ε 0 mε x
→ ∞ → →
f (k) eikx
n
→
7
the orthogonality relations converges
1 N π
lim f∗(k)f (k′) = e−ikxeik′xdk
N→∞ 2N +1 n n
n=−N Z−π
X
the completness relations becomes
∞ ∞
fn∗(k)fn′(k) → e−ikxeikx′
j=−∞ k=−∞
X X
and the Fourier transform converges to:
Fˆ(k) = eixkF (x)dx
Z
∞
F (k) = eikxFˆ(k)
k=−∞
X
4 Dirac and vector representation of the Lorentz
group and covariant wave equations
Let L(α) be an element of the proper Lorentz group corresponding to the element
α SL(2,C) and I the parity operator. If the components of four-momentum are
s
∈
written as 2 2 matrix
×
p˜ pµσ = p0σ +piσ (40)
µ 0 i
≡
where σ = 1 and σ are the Pauli matrices.
0 i
The transformations of p˜ under parity and SL(2,C) are
I : p˜ p˜s = p0σ piσ = (det p˜)(p˜)−1
s 0 i
→ −
α : p˜ αp˜α+
→
p˜s α+ −1p˜sα−1 (41)
→
It follows (cid:0) (cid:1)
I L(α)I−1 = L (α+)−1 (42)
s s
Therefore the matrix (a+)−1 gives an oth(cid:0)er 2-dim(cid:1) ensional representation of the
Lorentz group non-equivalent to α SL(2,C). In order to enlarge the proper Lorentz
group with space reflection we take∈both representations α and (α+)−1.
Let π I,I the space reflection group and α SL(2,C), then the semidirect
s
≡ { } ∈
product
SL(2,C) π
⊗
with the multiplication law
(α,π) (α′,π′) = (αα′,ππ′) if π = I (43)
(α,π) (α′,π′) = α(α′+)−1,ππ′ if π = I (44)
s
(cid:16) (cid:17)
8
form a group.
This group has a 4-dimensional representation, a particular elements of which is
α 0 0 σ
D(α,I) = , D(e,I ) = 0 (45)
0 (α+)−1 s σ 0
0
(cid:18) (cid:19) (cid:18) (cid:19)
that satisfy
D(e,I )D(α,I)D e,I−1 =, D α+ −1,I (46)
s s
(cid:16) (cid:17)
With respect to this representation a(cid:0)-4-com(cid:1)ponent s(cid:0)pino(cid:1)r in momentum space ψ(p)
transform as follows
U (α,I)ψ(p) = D(α,I)ψ L−1(α)p (47)
U (e,I )ψ(p) = D(e,I )ψ(I ,p) (48)
s s (cid:0) s (cid:1)
Using a similarity transformation we obtain an equivalent representation
D(α,π) = MD(α,π)M−1
with
1 σ σ
M = 0 0
√2 σ0 σ0
(cid:18) − (cid:19)
In this representation
1 α+(α+)−1, α+(α+)−1
D(α,I) = − (49)
2 α+(α+)−1, α+(α+)−1
(cid:18) − (cid:19)
σ 0
D(e,I ) = 0 (50)
s 0 σ
0
(cid:18) − (cid:19)
The new four-spinor
ψ(p) = Mψ(p)
transform as
U (α,I)ψ(p) = D(α,I)ψ L−1(α)p (51)
U (e,I )ψ(p) = D(e,I )ψ(I ,p) (52)
s s (cid:0) s (cid:1)
The Dirac wave equation can be considered as a consequence of the relativistic
invariance [9]
In the rest system we want a projection operator that selects one irreducible repre-
sentation out of the Dirac representatition. This is achieve in the rest system by
1 σ 0
Q = (I +β), β 0 (53)
2 ≡ 0 σ
0
(cid:18) − (cid:19)
In order to get the projection operator in an arbitrary system we apply D(κ) given
by (49) and (17)
1
Q Q(κ) = D−1(κ)QD(κ) = (I +W (κ)) (54)
→ 2
9
where
1 κ+κ+(κ+κ)−1, κ+κ+(κ+κ)−1
W (κ) = − (55)
2 κ+κ (κ+κ)−1, κ+κ (κ+κ)−1
(cid:18) − − − (cid:19)
Using the identities
3
1
κ+κ −1 = σµp
µ
m c
0
µ=0
(cid:0) (cid:1) X (56)
3
1
κ+κ = σ0p σjp
0 j
m c −
0
j=1
X
we find
3
1
W (κ) = γµp (57)
µ
m c
0
µ=0
X
where γµ are Dirac matrices with the realization
σ0 0 0 σj
γ0 = , γϕ =
0 σ0 σj 0
(cid:18) − (cid:19) (cid:18) − (cid:19)
γ0 = γ , γϕ = γ
0 − j
Collecting these result we obtain the Dirac equation in momentum space
1
Q(κ)ψ(p) = (I +W (κ)ψ(p)) = ψ(p)
2
or
3
(γµp m cI)ψ(p) = 0 (58)
µ 0
−
µ=0
X
(An equivalent method can be used applying to the projection operator the Foldy-
Wouthuysen transformation [10])
We apply the operator (γµp m c) from the left to (59) and obtain
µ 0
−
P
pµp m2c2 ψ(p) = 0 (59)
µ − 0
The Dirac equation is invar(cid:0)iant under t(cid:1)he group (α,π) defined before. In other
words, if ψ(p) is a solution of the Dirac equation, so is U(α,π)ψ(p). Put π = I. Then
U (α,I)ψ(p) = D(α,I)ψ L−1p
(cid:0) (cid:1)
D−1(α)Q(κ)D(α) = Q(xα)
Q(κα)ψ L−1p = ψ L−1p
Then (cid:0) (cid:1) (cid:0) (cid:1)
Q(κ)U (α,I)ψ(p) = Q(κ)D(α,I)ψ L−1(α)p = D(α,I)Q(κα)ψ L−1(α)p =
= D(α,I)ψ L−1(α)p = U (α,I)ψ(p)
(cid:0) (cid:1) (cid:0) (cid:1)
(cid:0) (cid:1)
10