Time Crystal

Ranjan Modak

Journal ClubCentre for Condensed Matter Theory

Department of PhysicsIndian Institute of Science

Bangalore

13 December, 2012

What is Time Crystal?

Normal Crystal: Space translation and/or rotation groups arespontaneously broken to some discrete subgroups

Time Crystal : Unlike the usual crystals, time translationsymmetry is spontaneously broken

In terms of infinitesimal transformations, a dynamical degree offreedom φ transforms under space and time translations asδφ = ∂xφ and δφ = ∂tφ respectively. So spontaneous breaking ofspace translation implies ∂xφ 6= 0 and spontaneous breaking oftime translation implies ∂tφ 6= 0 in the ground state

It is quite easy to construct models of time-independent,conservative dynamical systems with local ground states inwhich ∂xφ 6= 0. Examples:

V1(φ) = −κ1dφ

dx+λ1

2(dφ

dx)2 (1)

V2(φ) = −κ2

2(dφ

dx)2 +

λ2

4(dφ

dx)4 (2)

where all Greek coefficients are positive. The ground states

are respectively dφ1dx = κ1

λ1, dφ2

dx = ±√

κ2λ2

In either cases the space translation is spontaneously broken

Note that these are only potential energies. Kinetic energyterms are omitted because in the ground states these naturallyvanish

The question is: can we construct models oftime-independent, conservative dynamical systems withground states which spontaneously breaks the time translationsymmetry?

The first, most naive models which may possibly yield spontaneousbreaking of time translation symmetry in the ground state aregiven by changing the spacial derivatives in V1 and V2 given aboveinto time derivatives, i.e.

L1(φ, φ) = −κ1φ+λ1

2φ2 (3)

L2(φ, φ) = −κ2

2φ2 +

λ2

4φ4 (4)

The associated energy functions are respectively

E1(φ, φ) =λ1

2φ2 (5)

E2(φ, φ) = −κ2

2φ2 +

3λ24φ4 (6)

E1 is minimized at φ = 0, hence there is no spontaneousbreaking of time translation in the ground state; E2 is

minimized at φ = ±√

κ23λ2

, so it seems that there is indeed

spontaneous breaking of time translation in Model 2

The condition φ 6= 0 at the ground state seems to imply thatthe system undergoes perpetual motion in its lowest energystate

Consider a generic Hamiltonian system. The energy H = H(q, p) isa function in phase space. The ground state condition (i.e. thecondition for H to take its extremal value) is

∂H

∂q=∂H

∂p= 0 (7)

This, combined with the standard (canonical) Hamiltonianequations of motion

p = −∂H

∂q(8)

q =∂H

∂p(9)

indicates that the ground state necessarily satisfies q = p = 0, soperpetual motion in the ground state looks impossible

In Model 2 , Let κ2 = κ,λ2 = 1, we have

p =∂L2

∂φ= φ3 − κφ (10)

H2(p, φ) = pφ− L2 = pφ+κ

2φ2 −

1

4φ4 (11)

The Hamiltonian H2 is not written in terms of the phase spacevariable (φ, p) and indeed it cannot be, Energy regarded asmultivalued function of p .Conjugate momentum is not a good variable for writing down thecorresponding Hamiltonian uniquely

Energy is a multivalued function of p with cusps where ∂p

∂φ= 0 and

i.e. p = ∓2κ3/2

33/2corresponding precisely to the energy minima

φ = ±√κ/3

At the cusps the usual condition that the gradient should vanish ata minimum does not apply. There is NO CONTRADICTION

To provide a remedy to the above subtle situations, we rewritethe Lagrangian in a different form

L = −κρ2

2+λ

4ρ4 + γ(ρ− φ) (12)

New coordinates ρ,γ are introduced so that γ plays the role ofa Lagrangian multiplier and solving ρ in terms of φ will giveus the original Lagrangian L2

The apparent 6D phase space spanned by(ρ, γ,φ,Πρ, Πγ, Πφ)

Lagrangian is at most linear in velocity of at least onecoordinate , Hamiltonian formalism does not work

To remedy this, we employ the Dirac’s GeneralizedHamiltonian Procedure

Diracs theory on constrained Hamiltonian systems

3 primary constraints obtained from definition of canonicalconjugate momentum

G1 = Πφ + γ ' 0 (13)

G2 = Πγ ' 0 (14)

G3 = Πρ ' 0 (15)

Htotal = H +

3∑i=1

µiGi =κ

2ρ2 −

λ

4ρ4 − γρ+

3∑i=1

µiGi (16)

Analyze the time evolution of the primary constraints to thetotal Hamiltonian using canonical poisson brackets

Only time evolution of G3 leads to a secondary constraints

G4 = −κρ+ λρ3 + γ ' 0 (17)

Then we evaluate Mαβ = {Gα,Gβ}

M =

0 1 0 0−1 0 0 00 0 0 κ− 3λρ2

0 −1 −κ+ 3λρ2 0

All constrainsts are Second class. We introduce Dirac poissonbrackets

{A,B}DB = {A,B}− {A,Gα}(M−1)αβ{Gβ,B} (18)

Where A,B are any two functions on phase space.Now, all secondclass constraints can be set equal to strong zeros.

only remaining phase space variables are (φ, ρ) with

{φ, ρ}DB =1

3λρ2 − κ, {ρ, ρ}DB = {φ,φ}DB = 0 (19)

Final Hamiltonian in reduced phase space

Hfin = −1

2κρ2 +

3

4λρ4 (20)

Hamiltonian equations of motion are:

φ = {φ,Hfin}DB = ρ; ρ = {ρ,Hfin}DB = 0 (21)

Final Hamiltonian is smooth and single valued. using extremalcondition ∂Hfin

∂φ = ∂Hfin∂ρ = 0 The minima occur at

φG = ρG = ±√κ

3λ(22)

Two distinct families of ground states(φG , ρG ) = (

√κ3λ t + φ0,

√κ3λ) and (−

√κ3λ t + φ0,−

√κ3λ)

Explicit dependence of φG on t indicates that time translationsymmetry is spontaneously broken

Actual ground state must be chosen one of the two families.Concrete choise of ground state breaks time reversal symmetry

The above prescription can be used for any Lagrangian of followingform:

L =∑k=n

k=112k fk(φ)φ

2k − f0(φ) for (n > 1)

where fn(φ) > 0 and at least one of the fk(φ) < 0 fork = 1, ...n − 1

Our previous Lagrangian is the simplest form this. with n = 2 andf2 = λ ,f1 = −κ, f0 = 0

Analogy with landau second order phase transition

H1 =1

2v2 + U(x) (23)

{x , v }1 = 1

The equations of motion: x = v and v = −∂U∂x

H2 = (1

2v2 + U(x))2 + E0 (24)

{x , v }2 =1

v2+2U(x)

for this Hamiltonian, will get same equations of motions.

Analogy with landau second order phase transition

For H1, conditions for Hamiltonian to have a local minima

v2 = 0, dUdx = 0 and d2U

dx2 > 0

For H2, conditions for Hamiltonian to have a local minima

v2 = 0, dUdx = 0 and d2U

dx2 > 0 if U(x) ≥ 0

v2 = −2U(x) if U(x) < 0

For 1st case, time translation symmetry is not broken but forsecond case it is broken because of nontrivial motion in groundstate.

Landau free energy = F (T ,M) = F0(T ) + b(T )M2 + aM4

where, b(T ) = b0(T − Tc); M is order parameter and Ttemperature

Conditions for stable minima of free energy :∂F (M,T )∂M = 0 ;

∂2F (M,T )∂M2 > 0

M = 0 for (T ≥ Tc)

M = ±√

b0(T−Tc )2a for (T < Tc)

Like Landau theory in case of H2 Hamiltonian v playsequivalent role of M.

M is controlled by temperature where as the value of v at theground state controlled by U(x), and which controlled by xhence indirectly by t

Conclusions

Models describing Time Crystals

In the ground state time reversal symmetry is broken withspontaneous time translation symmetry breaking

Analogy with Landau 2nd order phase transition

Future directions

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

References

“Classical time crystal” by Alfred Shapere, Frank WilczekarXiv:1202.2537

“Quantum time Crystal” by Wilczek arXiv:1202.2539

“Comment on ”Quantum Time Crystals”: a new paradigm orjust another proposal of perpetuum mobile?” by PatrickBruno arXiv:1210.4128

“Hamiltonian description of singular Lagrangian systems withspontaneously broken time translation symmetry” by L Zhaoetal.

“Landau meets Newton: time translation symmetry breakingin classical mechanics” by L Zhao et al

Acknowledgement

L Zhao

Aninda Sinha (CHEP)

Ananyo Maitra

Subroto Mukerjee