Time Crystal. Ranjan Modak. Journal Club Centre for Condensed Matter Theory Department of Physics Indian Institute of Science Bangalore

Transcription

1 Time Crystal Ranjan Modak Journal Club Centre for Condensed Matter Theory Department of Physics Indian Institute of Science Bangalore 13 December, 2012

2 What is Time Crystal? Normal Crystal: Space translation and/or rotation groups are spontaneously broken to some discrete subgroups Time Crystal : Unlike the usual crystals, time translation symmetry is spontaneously broken In terms of infinitesimal transformations, a dynamical degree of freedom φ transforms under space and time translations as δφ = x φ and δφ = t φ respectively. So spontaneous breaking of space translation implies x φ 0 and spontaneous breaking of time translation implies t φ 0 in the ground state

3 It is quite easy to construct models of time-independent, conservative dynamical systems with local ground states in which x φ 0. Examples: V 1 (φ) = κ 1 dφ dx + λ 1 2 (dφ dx )2 (1) V 2 (φ) = κ 2 2 (dφ dx )2 + λ 2 4 (dφ dx )4 (2) where all Greek coefficients are positive. The ground states are respectively dφ 1 dx = κ 1 λ 1, dφ 2 dx = ± κ2 λ 2 In either cases the space translation is spontaneously broken

4 Note that these are only potential energies. Kinetic energy terms are omitted because in the ground states these naturally vanish The question is: can we construct models of time-independent, conservative dynamical systems with ground states which spontaneously breaks the time translation symmetry?

5 The first, most naive models which may possibly yield spontaneous breaking of time translation symmetry in the ground state are given by changing the spacial derivatives in V 1 and V 2 given above into time derivatives, i.e. L 1 (φ, φ) = κ 1 φ + λ 1 2 φ 2 (3) L 2 (φ, φ) = κ 2 2 φ 2 + λ 2 4 φ 4 (4)

6 The associated energy functions are respectively E 1 (φ, φ) = λ 1 2 φ 2 (5) E 2 (φ, φ) = κ 2 2 φ 2 + 3λ 2 4 φ 4 (6) E 1 is minimized at φ = 0, hence there is no spontaneous breaking of time translation in the ground state; E 2 is minimized at φ = ± κ2 3λ 2, so it seems that there is indeed spontaneous breaking of time translation in Model 2 The condition φ 0 at the ground state seems to imply that the system undergoes perpetual motion in its lowest energy state

7 Consider a generic Hamiltonian system. The energy H = H(q, p) is a function in phase space. The ground state condition (i.e. the condition for H to take its extremal value) is H q = H p = 0 (7) This, combined with the standard (canonical) Hamiltonian equations of motion ṗ = H q q = H p indicates that the ground state necessarily satisfies q = ṗ = 0, so perpetual motion in the ground state looks impossible (8) (9)

8 In Model 2, Let κ 2 = κ,λ 2 = 1, we have p = L 2 φ = φ 3 κ φ (10) H 2 (p, φ) = p φ L 2 = p φ + κ 2 φ φ 4 (11) The Hamiltonian H 2 is not written in terms of the phase space variable (φ, p) and indeed it cannot be, Energy regarded as multivalued function of p. Conjugate momentum is not a good variable for writing down the corresponding Hamiltonian uniquely

9 Energy is a multivalued function of p with cusps where p = 0 and φ i.e. p = 2κ3/2 corresponding precisely to the energy minima 3 φ 3/2 = ± κ/3 At the cusps the usual condition that the gradient should vanish at a minimum does not apply. There is NO CONTRADICTION

10 To provide a remedy to the above subtle situations, we rewrite the Lagrangian in a different form L = κ ρ2 2 + λ 4 ρ4 + γ(ρ φ) (12) New coordinates ρ,γ are introduced so that γ plays the role of a Lagrangian multiplier and solving ρ in terms of φ will give us the original Lagrangian L 2 The apparent 6D phase space spanned by(ρ, γ, φ, Π ρ, Π γ, Π φ ) Lagrangian is at most linear in velocity of at least one coordinate, Hamiltonian formalism does not work To remedy this, we employ the Dirac s Generalized Hamiltonian Procedure

11 Diracs theory on constrained Hamiltonian systems 3 primary constraints obtained from definition of canonical conjugate momentum H total = H + G 1 = Π φ + γ 0 (13) G 2 = Π γ 0 (14) G 3 = Π ρ 0 (15) 3 µ i G i = κ 2 ρ2 λ 4 ρ4 γρ + i=1 3 µ i G i (16) Analyze the time evolution of the primary constraints to the total Hamiltonian using canonical poisson brackets Only time evolution of G 3 leads to a secondary constraints i=1 G 4 = κρ + λρ 3 + γ 0 (17)

12 Then we evaluate M αβ = {G α, G β } M = κ 3λρ κ + 3λρ 2 0 All constrainsts are Second class. We introduce Dirac poisson brackets {A, B} DB = {A, B} {A, G α }(M 1 ) αβ {G β, B} (18) Where A,B are any two functions on phase space.now, all second class constraints can be set equal to strong zeros.

13 only remaining phase space variables are (φ, ρ) with {φ, ρ} DB = Final Hamiltonian in reduced phase space Hamiltonian equations of motion are: 1 3λρ 2 κ, {ρ, ρ} DB = {φ, φ} DB = 0 (19) H fin = 1 2 κρ λρ4 (20) φ = {φ, H fin } DB = ρ; ρ = {ρ, H fin } DB = 0 (21)

14 Final Hamiltonian is smooth and single valued. using extremal condition H fin φ = H fin ρ = 0 The minima occur at κ φ G = ρ G = ± 3λ (22) Two distinct families of ground states (φ G, ρ G ) = ( κ 3λ t + φ 0, κ 3λ ) and ( κ 3λ t + φ 0, κ 3λ ) Explicit dependence of φ G on t indicates that time translation symmetry is spontaneously broken Actual ground state must be chosen one of the two families. Concrete choise of ground state breaks time reversal symmetry

15 The above prescription can be used for any Lagrangian of following form: L = k=n k=1 1 2k f k(φ) φ 2k f 0 (φ) for (n > 1) where f n (φ) > 0 and at least one of the f k (φ) < 0 for k = 1,...n 1 Our previous Lagrangian is the simplest form this. with n = 2 and f 2 = λ,f 1 = κ, f 0 = 0

16 Analogy with landau second order phase transition H 1 = 1 2 v 2 + U(x) (23) {x, v} 1 = 1 The equations of motion: ẋ = v and v = U x H 2 = ( 1 2 v 2 + U(x)) 2 + E 0 (24) {x, v} 2 = 1 v 2 +2U(x) for this Hamiltonian, will get same equations of motions.

17 Analogy with landau second order phase transition For H 1, conditions for Hamiltonian to have a local minima v 2 = 0, du dx = 0 and d2 U dx 2 > 0 For H 2, conditions for Hamiltonian to have a local minima v 2 = 0, du dx = 0 and d2 U dx 2 > 0 if U(x) 0 v 2 = 2U(x) if U(x) < 0 For 1st case, time translation symmetry is not broken but for second case it is broken because of nontrivial motion in ground state.

18 Landau free energy = F (T, M) = F 0 (T ) + b(t )M 2 + am 4 where, b(t ) = b 0 (T T c ); M is order parameter and T temperature F (M,T ) Conditions for stable minima of free energy : M = 0 ; 2 F (M,T ) > 0 M 2 M = 0 for (T T c ) b M = ± 0 (T T c) 2a for (T < T c ) Like Landau theory in case of H 2 Hamiltonian v plays equivalent role of M. M is controlled by temperature where as the value of v at the ground state controlled by U(x), and which controlled by x hence indirectly by t

19 Conclusions Models describing Time Crystals In the ground state time reversal symmetry is broken with spontaneous time translation symmetry breaking Analogy with Landau 2nd order phase transition

20 Future directions Recently Wilczek has suggested a model which can be interpreted as Quantum time crystal but it turns out that the Ground state of that model as suggested by Wilczek is not an actual one. Proper quantization of these classical models are yet to be done

21 References Classical time crystal by Alfred Shapere, Frank Wilczek arxiv: Quantum time Crystal by Wilczek arxiv: Comment on Quantum Time Crystals : a new paradigm or just another proposal of perpetuum mobile? by Patrick Bruno arxiv: Hamiltonian description of singular Lagrangian systems with spontaneously broken time translation symmetry by L Zhao etal. Landau meets Newton: time translation symmetry breaking in classical mechanics by L Zhao et al

22 Acknowledgement L Zhao Aninda Sinha (CHEP) Ananyo Maitra Subroto Mukerjee