florian conrady- spin foams with timelike surfaces
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Spin foams with timelike surfaces
Florian Conrady
Perimeter Institute
ILQG seminarApril 6, 2010
FC, Jeff Hnybida, arXiv:1002.1959 [gr-qc]FC, arXiv:1003.5652 [gr-qc]
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Outline
1 Motivation
2 Coherent states
3 Three ways to simplicity
4 Spin foams
5
Summary
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Motivation
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Restriction of triangulations
In the Lorentzian EPRL model, normals U of tetrahedra are timelike.
timelike U
All tetrahedra are Euclidean and triangles can be only spacelike.
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What we did
We extended the EPRL model to also include spacelike normals U.
timelike U
spacelike
U
Lorentzian tetrahedra are allowed and triangles can be spacelike ortimelike.
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Three cases
normal U
timelike U
spacelike
U
spacelike
U
triangle spacelike spacelike timelike
EPRL
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Canonical perspective
Are restricted triangulations preferred froma Hamiltonian point of view?
In the examples I know of the transition from space tospacetime leads to a 4d lattice with timelike (or null) edges. causal dynamical triangulations Ambjorn, Jurkiewicz, Loll evolution schemes for Lorentzian Regge calculus
Barrett, Galassi, Miller, Sorkin, Tuckey, Williams
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Canonical perspective
Are restricted triangulations preferred froma Hamiltonian point of view?
In the examples I know of the transition from space tospacetime leads to a 4d lattice with timelike (or null) edges. causal dynamical triangulations Ambjorn, Jurkiewicz, Loll
evolution schemes for Lorentzian Regge calculusBarrett, Galassi, Miller, Sorkin, Tuckey, Williams
The Hamiltonian approach to Lorentzian spin foams creates a
sequence of 3d spatial lattices. Han, Thiemann cannot (yet) be directly compared with 4d triangulations
discussed here
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Coherent states
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Little groups
normal U
timelike U
spacelike
U
spacelike
U
gaugefix U = (1, 0, 0, 0) U = (0, 0, 0, 1) U = (0, 0, 0, 1)
littlegroup SO(3) SO(1,2) SO(1,2)
triangle spacelike spacelike timelike
EPRL
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Little groups
normal U
timelike U
spacelike
U
spacelike
U
gaugefix U = (1, 0, 0, 0) U = (0, 0, 0, 1) U = (0, 0, 0, 1)
littlegroup SU(2) SU(1,1) SU(1,1)
triangle spacelike spacelike timelike
EPRL
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Representation theory
SL(2,C) SU(2)
generators Ji
, Ki
J1
, J2
, J3
Casimirs C1 = J2 K2
C2 = 4J KJ2
unitaryirreps
H(,n) Dj
R, n Z+ j Z+/2
C1 =12 (n
2 2 4)C2 = n
J2 = j(j + 1)
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Representation theory
SU(2) SU(1,1)
generators J1, J2, J3 J3, K1, K2
Casimirs J2 Q = (J3)2 (K1)2 (K2)2discrete series continuous series
unitaryirreps
Dj Dj Csj Z+/2 j = 12 , 1, 32 . . . j = 12 + is,
0 < s 1/2 D+
j
0
ds
Cs
n/2
j>1/2 D
j
0
ds
Cs
(,n) =
n/2j>1/2
m=j
+j m+j m +n/2
j>1/2
m=j
j mj m
+ =1,2
0
ds (s)
m= ()s m()s m
(see chapter 7 in Ruhls book)
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Nonnormalizability
States
|j m
in the continuous series irrep
Cs are normalizable, but the
corresponding states ()s m in H(,n) are not.
()s m()s m = (s s)(s) mm
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SU(2) coherent states
Definition
|j g Dj(g)|j j , g SU(2)
|jN D
j
(g(N))|j j ,
N S
2
SU(2)/U(1)
Completeness relation
j = (2j + 1) SU(2)
dg |j gj g| = (2j + 1)S2
d2N |jN
At the level of the SL(2,C) irrep H(,n) this becomes
Pj = (2j + 1)
SU(2)
dg |j g j g| = (2j + 1)
S2
d2N
jN
jN
.
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SU(1,1) coherent states discrete series
Definition
|j g Dj(g)|j j , g SU(1, 1)
|j N Dj(g(N))|j j , N H SU(1, 1)/U(1)
Upper/lower hyperboloid
H
=
{N
|N2 = 1 , N0 0
}
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SU(1,1) coherent states discrete series
Definition
|j g Dj(g)|j j , g SU(1, 1)
|jN D
j
(g(N))|j j ,
N H SU(1, 1)/U(1)
Completeness relation
j = (2j 1)
SU(1,1)
dg |j gj g| = (2j 1)H
d2N |j Nj N|
At the level of the SL(2,C) irrep H(,n) this becomes
Pj = (2j 1)
SU(1,1)
dgj g
j g
= (2j 1)H
d2N
jN
jN
.
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Expectation values
So far the coherent states are the ones introduced by Perelomov.
They have the property that
j N
|J
|j N
= j N, N
S2
j N|F|j N = j N, N H
in accordance with the fact thatJ
2
= j(j + 1) 0 andF
2
= j(j 1) 0.
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SU(1,1) coherent states continuous series
Question
What are the appropriate coherent
states for the continuous series?
In this case Q = F2 =
(s2 + 1/4) < 0, so the classical vector N should
be spacelike.
Perelomov uses the state |j m = 0, resulting in a zero classical vector.
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SU(1,1) coherent states continuous series
Question
What are the appropriate coherent
states for the continuous series?
In this case Q = F2 =
(s2 + 1/4) < 0, so the classical vector N should
be spacelike.
F J3
K1
K2 Build coherent states from eigenstates of K1 or K2!
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Eigenstates ofK1
K1
|j
=
|j
,
< /2
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( )
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SU(1,1) coherent states continuous series
Definition
|j gsp
d1
f( s) Dj(g)|j + , g SU(1, 1)
|j N Dj(g(N))|j + , N Hsp SU(1, 1)/G1
Completeness relation
j =
SU(1,1)
dg
|j g
sp
j g
|sp
=
Hsp
d2N
d1
f( s) |j Nj N|
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S i i
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Smearing in s
At the level of the SL(2,C) irrep H(,n), one also needs a smearing in s.
We define a smeared projector onto the irrep with spin j = 1/2 + is:
Ps() =1,2
m=
0
ds (s) f(s s) ()s m()s m
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SU(1 1) h i i
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SU(1,1) coherent states continuous series
Definition
()s g 0
ds (s) f(s s)
d
1
f( s) D(,n)(g)()s +
Completeness relation
P
s
() = =1,2
SU(1,1)
dg ()s g()
s g
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Th i li i
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Three ways to simplicity
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Th t i li it
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Three ways to simplicity
Derivation of extended simplicity constraints by three methods
1. weak imposition of constraints2. master constraint
(as advocated by EPRL)
3. restriction of coherent state basis (inspired by FK model)
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Cl i l t t h d
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Classical tetrahedron
Describe a tetrahedron by four bivectors
J = B +1
B,
where B is constrained to be simple.
Simplicity constraint: unit fourvector U such that
U
B = 0 .
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Classical simplicity constraints
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Classical simplicity constraints
Express B in terms of the total bivector J:
B = 22 + 1
J 1
J
Starting point for quantization:
U
J 1
J
= 0
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First case: normal U timelike
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First case: normal U timelike
timelike U
In the gauge U = (1, 0, 0, 0), the simplicity constraint takes the form
J +1
K = 0
The little group is SU(2), so we use states of the SU(2) decomposition!
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Spacelike U
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Spacelike U
spacelike
U
In the gauge U = (0, 0, 0, 1), the simplicity constraint becomes
F +1
G = 0
The little group is SU(1,1), so we use states of the SU(1,1) decomposition!
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Basic example of secondclass constraints
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Basic example of secondclass constraints
Phase space and constraints
(qi, pi) , i = 1, 2 q1 q2 = 0
{qi, pi} = ij p1 p2 = 0
Change of variables
q = 12 (q1 q2) q = p = 0
p
= 1
2
(p1
p2)
a = 12 (p iq) a = 0
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Weak imposition of constraints
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Weak imposition of constraints
Impose a|
= 0 on physical states, implying
|a| = |a| = 0 |, | Hphys
Physical Hilbert Hphys space is spanned by |n+ |0, n+ N0.
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Restriction of coherent state basis
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Restriction of coherent state basis
Overcomplete basis of coherent states |+ |, a| = |:
H=
1
2 d+ d |++| ||Restrict basis to states whose expectation values satisfy the constraint, i.e.to labels = 0.
Projector on Hphys
Pphys 1 d+ |++| |00|
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Master constraint
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Master constraint
Master constraint operator
M = aa =1
2
p2 + q
2
+1
2
Define Hphys as the subspace of states with minimal eigenvalue w.r.t. M.
Hphys spanned by |n+ |0, n+ N0.
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From classical to quantum simplicity
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From classical to quantum simplicity
J + 1K = 0
F + 1G = 0
4C3 = n
BIJ(B)IJ = 0
(
n) + n = 0
BIJ(B)IJ = 0 is the diagonal simplicity constraint.
C3 is the Casimir of the little group determined by the normal U.
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Weak imposition of constraints
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Weak imposition of constraints
Consider the case U = (1, 0, 0, 0) and suppose that
Hphys =jJ
Dj H(,n) ,
where J is a subset of the total set of spins {j |j n/2}.Require that
|C| = 0 |, | Hphys.
Unless Hphys is trivial, this implies, that for some j n/2,
j m J3 + 1K3j m = j m J + 1Kj m = 0for all admissible m, m.
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Weak imposition of constraints
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pBy using K = [K3, J] and
K3|j m = (. . .)|j+ 1 m mAj |j m + (. . .)|j 1 m , Aj = n4j(j + 1)
,
one obtains
Aj = or 4C3 = 4j(j + 1) = n .
In conjunction with the constraint B B = 0, this gives 4j(j + 1) = n2 if = n and 4j(j + 1) = 2 if n = .
Approximate solution
= n , j = n/2
Hphys = Dn/2 H(n,n)
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Weak imposition of constraints
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p
Next consider U = (0, 0, 0, 1). Suppose first that the weak constraintholds for some irrep Dj of the discrete series:
j mF0 +
1
G0j m = j mF
+1
Gj m = 0 m, m ,
with F F2 iF1 and G G2 iG1. According to Mukunda
K3
|j m = (. . .)|j + 1 m mAj |jm + (. . .)|j 1 m , Aj = n
4j(j 1) ,
and G = [G0, F], which leads to Aj = or 4j(j 1) = n.
Approximate solution
= n , n 2 , j = n/2
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Weak imposition of constraints
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p
For an irrep
Cs of the continuous series, the equations are the same except
that Aj is replaced by
Aj = n
4j(j + 1)= n
4(s2 + 1/4).
A solution exists only when = n/ < 1 and then
s2 + 1/4 =2
4
=n2
42
.
Overall result for U = (0, 0, 0, 1)
= n , n 2 or = n/ < 1Hphys = D+n/2 Dn/2 Hphys = C1
2
n2/21 C
12
n2/21
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Restriction of coherent state basis
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Case U = (1, 0, 0, 0).
Resolve the identity on H(,n) in terms of SU(2) coherent states,
(,n) =
j=n/2
(2j + 1)
S2
d2N
jN
jN
,
and require that
jN
J +1
K
jN
= 0 .
This implies Aj = , that is, 4j(j + 1) = n.
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Restriction of coherent state basis
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The second condition is either obtained from B B = 0 or alternativelyfrom the requirement of minimal uncertainty in K.From = Aj it follows that
J2 = 12
K2 + O(|J|) and J2 = 1J K ,
so that
(K)2 = 1
(1 2)J K 12
C1 + O(|J|)
=
4 1 1
2C2 +2
C
1 + O(|J|)=
4B B + O(|J|) .
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Restriction of coherent state basis
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Projector on Hphys
Pphys = (n + 1)
S2
d2N
n/2N
n/2N
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Master constraint
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For the case U = (0, 0, 0, 1) the Master constraint reads
M = F +1
G2
= 1 +1
2F2
1
22C1
1
2C2 = 0
The diagonal constraint B
B = 0 is equivalent to
1 12
C1 +
2
C2 = 0 .
By combining the two one arrives at the desired second condition
4F2 = n .
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Master constraint
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In the case of the discrete series, the constraints are therefore
n +n
= 04j(j 1) = n
Approximate solution
= n , n
2 j = n/2
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Master constraint
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For states of the continuous series,
n +
n
= 0
4
s2 +1
4
= n
Solution
=
n
constr. on irreps j = n/2 j = n/2 s2
+ 1/4 = 2
/4
area spectrum
j(j + 1)
j(j 1) s2 + 1/4 = n/2 EPRL
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Spin foams
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Spin foam theory
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Complex:
simplicial complex : 4simplex , tetrahedron , triangles t, . . .
dual complex : vertex v, edge e, face f, . . .
Variables (same as in EPRL):
connection ge SL(2,C)irrep label nf Z+
Additional variables:
Ue = (1, 0, 0, 0) or (0, 0, 0, 1): normal of tetrahedron dual to ef = 1: spacelike/timelike triangle dual to f
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Uniform notation
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To cover the different cases we introduce a uniform notation.
Little group
H(, U) SU(2) , if = 1 , U = (1, 0, 0, 0) ,
SU(1, 1) , if = 1 , U = (0, 0, 0, 1) , , if = 1 , U = (1, 0, 0, 0) .
Spin
j = n/2 , if = 1 , U = (1, 0, 0, 0) ,
n/2 , if = 1 , U = (0, 0, 0, 1) ,
12 + i2
n2/2 1 , if = 1 , U = (0, 0, 0, 1) .
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Uniform notation
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Coherent states
|()j h =
|j h
, if = 1 , U = (1, 0, 0, 0) ,
|j h , if = 1 , U = (0, 0, 0, 1) ,|()j h , if = 1 , U = (0, 0, 0, 1) .
Projector
Pj(, U, ) = dj(, U)
H(,U)
dh ()j h ()
j h
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Vertex amplitude
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e e
v
f
gev gve
Av((f, nf); hef, ef, )
=
SL(2,C)
e
dgev
f
(ef)jefhef
D(f,nf)(gevgve) (ef)jefhef
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Compatibility of path integral and operator formalism
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So far we were mainly guided by the path integral pictureand defined the spin foam sum by simply summing overall choices of the normal U.
From the canonical perspective, however, one would alsorequire that the sum over intermediate states on a trian-
gulations boundary is a projector.
For a give choice of U, this is certainly the case.
Is it also true when we sum over U?
Florian Conrady (PI) Spin foams with timelike surfaces ILQGS April 6 2010 55 / 60
Compatibility of path integral and operator formalism
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normal U
timelike U
spacelike
U
spacelike
U
triangle spacelike spacelike timelike
(, n) = n = n, n
2 n =
>
EPRL
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Compatibility of path integral and operator formalism
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,
= 0
,
= 0
,
= 0
,
= 0
The sum over tetrahedral states is a projector if we exclude .
P = + red>0
The boundary Hilbert space is genuinely larger than the EPRL one.
Florian Conrady (PI) Spin foams with timelike surfaces ILQGS April 6 2010 57 / 60
Summary
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Summary
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extension of EPRL model tetrahedra can be Euclidean and Lorentzian triangles can be spacelike and timelike larger boundary Hilbert space
discrete area spectrum of timelike surfaces
definition of associated spin foam model
coherent states for timelike triangles
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Outlook
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regions with timelike boundaries! Rovelli et al., Oeckl
extension of results on EPRL model? asymptotics, graviton propagator . . . ?
canonical LQG on general boundaries?
comparison with previous work on timelike surfacesPerez, Rovelli
Alexandrov, Vassilevich
Alexandrov, Kadar
Florian Conrady (PI) Spin foams with timelike surfaces ILQGS April 6 2010 60 / 60
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