Space from Superstring Bits 1 Charles Thorn University of Florida Miami 2014 1 Much of this work in collaboration with Songge Sun
A single superstring bit: quantum system with finite # of states Superstring bits make their own space as they form into string
Superstring bit models can provide underlying microscopic dynamics of superstring and hence also can form the basis of quantum gravity.
Superstring bit models provide a concrete way to think about emergent space, an idea inspired by black hole firewalls and other difficulties in quantum gravity.
Since space is absent in the underlying dynamics, locality must be a derived property of the emergent space.
String to Bits(1977, with Giles)[3] H = 1 2 P + 0 dσ [ p 2 + T 2 0 x 2] H = 1 2m M [ p 2 n + T0 2 (x n+1 x n ) 2] k=1
Holography(1991)[4] String bits do not sense x. But for a chain of many bits E G αm β Mm + O(M 3 ) β 2P + So long chains do sense x, which is conjugate to P +. M bit chain with M behaves as though it could move in a longitudinal direction.
Superstring Bits Superstring bit annihilation operators (φ [a1 a n ]) β α n = 0,...,s n even : Boson, n odd : Fermion a i : s-valued spinor index No transverse space; No longitudinal space Finite number = 2 s N 2 degrees of freedom.
Chains of String Bits Consider a Hamiltonian with structure H = 1 V ABCD Tr φ A φb φ C φ D N [(φ A ) β α, ( φ B ) δ γ] = δαδ δ γδ β AB, α, β, γ, δ = 1 N (1) Action of H on single trace states: [ ] 1 N V ABCD Tr φ A φb φ C φ D Tr φ F1 φ FM 0 M = V ABF k+1f k Tr φ F1 φ Fk 1 φa φb φ FM 0 + O(N 1 ) k=1 Note how N singles out n.n. interactions on a linear chain
Illustrative Hamiltonian for s = 1. Should give insight into string formation and holography φ β α, (φ 1 ) β α a β α, b β α Superstring bit Hamiltonian becomes: H = 2 N Tr[ (ā 2 i b 2 )a 2 ( b 2 iā 2 )b 2 + (ā b + bā)ba + (ā b bā)ab ] H commutes with the Grassmann odd operator Q = Tr(ābe iπ/4 + bae iπ/4 ) There is no space: this is ordinary QM with 2N 2 d.o.f.
Energy spectrum for N =, general s Superfield: Φ(θ) = n,a k φ[a1 a n ]θ a1 θ a n H d sm θtrφ(θ 1 ) Φ(θ M ) 0 Ψ(θ 1,...,θ M ) = d M θtrφ(θ 1 ) Φ(θ M ) 0 hψ(θ 1,...,θ M ) + O(1/N) h on next page
h = M s [ k=1 a=1 2iθ a kθ a k+1 2i d dθ a k d dθ a k+1 d 2θk+1 a dθk a 2θ a k d dθ a k+1 2 + 4θ a k d dθ a k ] Cyclic constraint: Ψ(θ 1,...,θ M ) = ( ) s(m 1) Ψ(θ 2,...,θ M, θ 1 ) Energy: E G = s T 0 m M 1 n=1 sin nπ M = st 0 m cot π 2M 2T 0Ms mπ + πt 0s 6Mm (2) Comment: Spectrum of h symmetric about 0, but Cyclic constraint breaks this symmetry.
E θ = T 0 m M 1 n=1 sin nπ M = T 0 m cot π 2M 2T 0M mπ + πt 0 6Mm
Message so far: Chain with lowest energy per bit has M : In other words it is a continuous string! Moreover, single string is stable. The string excitation spectrum is that of s pairs of Grassmann odd worldsheet fields θl a(σ, τ), θa R (σ, τ). For s = 8, this is the Grassmann odd sector of the Green-Schwarz lightcone superstring.
Where can we find d transverse coordinates? Possibility 1: s s + 2d and bosonize each pair of extra θ s: Compactified boson coordinate, nonzero R Possibility 2. Add another index to φ: Heisenberg anisotropic chain.
Bosonizing θ s Puzzle: zero point energy of bosonic worldsheet fields is supposed to be πdt 0 /(6Mm), not +2πdT 0 /(6Mm). Resolution of puzzle: Bosonization coordinate is compactified on a finite radius circle, with KK momenta and winding numbers odd. Thus neither KK momentum terms nor winding number terms can vanish. These facts are captured by Jacobi formula: q 1/6 2 (1 + q 2n ) 2 = q 1/12 n=1 m= q (m+1/2)2 n=1 1 1 q 2n which equates the partition function of two integer-moded fermions on left with 1 integer moded boson on the right.
We must do something more sophisticated to get large compactification radius and the possibility of 0 winding number. Possibility 2 Anisotropic Heisenberg spin chain, 1 < < 1. H = C 2 M ( σ 1 k σk+1 1 + σkσ 2 k+1 2 + σkσ 3 k+1) 3 k=1 (3) C can be +ve or ve. For M even these are equivalent, but not for M odd. String bit models admit all M. Energy eigenvalues solved by Bethe ansatz, and explicitly for large M (Yang and Yang, 1966).
2π sinµ δe f C µ δe af C 2π sinµ π µ [ π 6 + π µ Q 2 + P 2 4 = cosµ, δe = E E B ] 1 π µ + 2π(N L + N R ) M ] 1 [ π 6 + µ 4 Q2 + 1 µ P 2 + 2π(N R + N L ) E B : bulk part of ground energy exactly proportional to M. Distinct cases: Let k, l be integers f : C < 0, Q = k, P = lπ af : C > 0, Q = k, P = { lπ Q even (l + 1/2)π Q odd M,
Compare to energy compactified string coordinate: x x + L: P = 1 2 P + 0 dσ [ P 2 + T 2 0 x 2] given by P = T 0 P + [ ] π6 2π2 + 4L 2 (2k) 2 + L2 T 0 T 0 2 l2 + 2π(N L + N R ) Exact match of Q even sector, with L 2 = 2π T 0 π π µ or 2π T 0 π π µ (4) By varying µ, can reach L in range 2π α < L < Comment: For C > 0, decompactification L sends odd Q sector to infinite energy.
What happens when N is finite? Can we still take M? Songge Sun is studying the s = 1, d = 0 model mentioned earlier at fixed bit number M = 3, 4, 5,.... Comments: Multi-trace basis over-complete for fixed N = integer.. Energies can become complex when N is not an integer. For finite integer N there is probably an upper limit to M < 2N + 1 for a stable ground state. s = 1 Hamiltonian. H = 2 N Tr[ (ā 2 i b 2 )a 2 ( b 2 iā 2 )b 2 + (ā b + bā)ba + (ā b bā)ab ]
A Variational Argument Trial state ψ = Tr b M 0, M odd, ψ H ψ = 2M 0 Trb M Tr b M 0 = 2M ψ ψ ψ ψ = MN (M 1)/2 k=1 (N 2 k 2 ) H = H and ψ ψ > 0 Requires N = Integer> (M 1)/2 Then E G < 2M for M odd and all N > (M 1)/2. No information when N (M 1)/2. At N =, E G 8M/π, consistent with the bound. High bit number stringy states may require very large N!
The following graphs were made by Songge Sun: 15 Energy,M=3 10 5 E 0 5 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1/N
12 Norms of Energy Eigenstates,M=3 10 8 6 Norm 4 2 0 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1/N
40 Energy, M=5 30 20 E 10 0 10 20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1/N
: 10 Norms of Energy Eigenstates, M=5 8 6 Norm 4 2 0 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1/N
References [1] S. Sun and C. B. Thorn, Phys. Rev. D 89 (2014) 105002 [arxiv:1402.7362 [hep-th]]. [2] C. B. Thorn, JHEP 1411 (2014) 110 [arxiv:1407.8144 [hepth]]. [3] R. Giles and C. B. Thorn, Phys. Rev. D 16 (1977) 366. [4] C. B. Thorn, In *Moscow 1991, Sakharov memorial lectures in physics, vol. 1* 447-453, and [arxiv: hep-th/9405069]. [5] C. B. Thorn, Phys. Rev. D 20 (1979) 1435. [6] O. Bergman and C. B. Thorn, Phys. Rev. D 52 (1995) 5980 [hep-th/9506125]. [7] G. t Hooft, Nucl. Phys. B72 (1974) 461. [8] C. B. Thorn, Substructure of string, Strings 96: [arxiv: hep-th/9607204].
[9] R. Giles, L. D. McLerran and C. B. Thorn, Phys. Rev. D 17 (1978) 2058. [10] P. Goddard, J. Goldstone, C. Rebbi and C. B. Thorn, Nucl. Phys. B 56 (1973) 109.