Space from Superstring Bits 1. Charles Thorn

Transcription

1 Space from Superstring Bits 1 Charles Thorn University of Florida Miami Much of this work in collaboration with Songge Sun

2 A single superstring bit: quantum system with finite # of states Superstring bits make their own space as they form into string

3 Superstring bit models can provide underlying microscopic dynamics of superstring and hence also can form the basis of quantum gravity.

4 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.

5 Since space is absent in the underlying dynamics, locality must be a derived property of the emergent space.

6 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

7 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.

8 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.

9 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

10 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.

11 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

12 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 θ 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.

13 E θ = T 0 m M 1 n=1 sin nπ M = T 0 m cot π 2M 2T 0M mπ + πt 0 6Mm

14 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.

15 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.

16 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.

17 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 σkσ 2 k 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).

18 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,

19 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.

20 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 ]

21 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!

22 The following graphs were made by Songge Sun: 15 Energy,M= E /N

23 12 Norms of Energy Eigenstates,M= Norm /N

24 40 Energy, M= E /N

25 : 10 Norms of Energy Eigenstates, M=5 8 6 Norm /N

26 References [1] S. Sun and C. B. Thorn, Phys. Rev. D 89 (2014) [arxiv: [hep-th]]. [2] C. B. Thorn, JHEP 1411 (2014) 110 [arxiv: [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* , and [arxiv: hep-th/ ]. [5] C. B. Thorn, Phys. Rev. D 20 (1979) [6] O. Bergman and C. B. Thorn, Phys. Rev. D 52 (1995) 5980 [hep-th/ ]. [7] G. t Hooft, Nucl. Phys. B72 (1974) 461. [8] C. B. Thorn, Substructure of string, Strings 96: [arxiv: hep-th/ ].

27 [9] R. Giles, L. D. McLerran and C. B. Thorn, Phys. Rev. D 17 (1978) [10] P. Goddard, J. Goldstone, C. Rebbi and C. B. Thorn, Nucl. Phys. B 56 (1973) 109.