Quantum Entanglement in Plebanski-Demianski

Quantum Entanglement in Pleba´nski-Demia´nski Spacetimes

Hugo Garc´ıa-Compe´an Cinvestav FCFM, UANL II Taller de Gravitaci´on,F´ısica de Altas Energ´ıasy Cosmolog´ıa ICF, UNAM, Cuernavaca Morelos Agosto 2014 Work with Felipe Robledo-Padilla Class. Quantum Grav. 30 (2013) 235012

August 3, 2014 I. Introduction

II. Pleba´nski-Demia´nskiSpacetime

III. Frame-dragging

IV. Spin precession

V. EPR correlation

VI. Spin precession angle in Expanding and Twisting Pleba´nski-Demia´nskiBlack Hole I. Introduction

For many years quantum states of matter in a classical gravitational background have been of great interest in physical models. Examples:

I Neutron interferometry in laboratories on the Earth. It captures the effects of the gravitational field into quantum phases associated to the possible trajectories of a beam of neutrons, following paths with different intensity of the gravitational field. (Colella et al 1975)

I Another instance of the description of quantum states of matter in classical gravitational fields is Hawking’s radiation describing the process of black hole evaporation. This process involves relativistic quantum particles and uses quantum field theory in curved spacetimes. II. Plebański-Demiański Spacetime

The metric

" 1 D 2 ρ2 ds2 = − dt − (a sin2 θ + 2l(1 − cos θ))dφ + dr 2 Ω2 ρ2 D # P 2 sin2 θ + adt − (r 2 + (a + l)2)dφ + ρ2 dθ2 , ρ2 P (1) with ρ2 = r 2 + (l + a cos θ)2, α Ω = 1 − (l + a cos θ)r, ω 2 2 P = sin θ(1 − a3 cos θ − a4 cos θ), α α2 Λ D = (κ + e2 + g 2) − 2mr + εr 2 − 2n r 3 − κ + r 4, ω ω2 3 (2) and where α α2 Λ a = 2a m − 4al (κ + e2 + g 2) − 4 al, 3 ω ω2 3 α2 Λ a = −a2 (κ + e2 + g 2) − a2, 4 ω2 3 κ α α2 Λ ε = + 4l m − (a2 + 3l 2) (κ + e2 + g 2) + , a2 − l 2 ω ω2 3 κl α α2 Λ n = − (a2 − l 2) m + (a2 − l 2)l (κ + e2 + g 2) + , a2 − l 2 ω ω2 3 1 + 2l α m − 3l 2 α2 (e2 + g 2) − l 2Λ κ = ω ω2 . 1 2 α2 a2−l2 + 3l ω2 (3) paremeters: α, ω, m, e, g, Λ, `, a. Equation (1) can be represented by the line element

2 2 2 2 2 ds = g00dt + 2g03dtdφ + g11dr + g22dθ + g33dφ , (4) which has a non-diagonal element that represents the axial symmetry and the metric coeﬃcients are:

−D + Pa2 g = , 00 Ω2ρ2 1 D P g dx i = (a sin2 θ + 2l(1 − cos θ)) − a(r 2 + (a + l)2) dφ, 0i Ω2 ρ2 ρ2 ρ2 sin2 θ 1 D g dx i dx j = dr 2 + ρ2 dθ2 + − (a sin2 θ + 2l(1 − cos θ))2 ij Ω2D Ω2P Ω2 ρ2 P + (r 2 + (a + l)2)2 dφ2. ρ2 (5) In order to describe the motion of spinning particles in a curved spacetime, the vierbein chosen is:

µ 1 0 √ g03 e0 (x) = √ (1, 0, 0, 0), e µ = −g00 1, 0, 0, , −g00 g00 µ 1 1 √ e1 (x) = √ (0, 1, 0, 0), e µ = g11(0, 1, 0, 0), g11 µ 1 2 √ e2 (x) = √ (0, 0, 1, 0), e µ = g22(0, 0, 1, 0), g22 s s 2 µ −g00 g03 3 g03 − g00g33 e3 (x) = 2 − , 0, 0, 1 , e µ = (0, 0, 0, 1.), g03 − g00g33 g00 −g00 (6) µ ν ea (x)eb (x)gµν(x) = ηab, a ν ν e µ(x)ea (x) = δµ , (7) a µ a e µ(x)eb (x) = δ b. III. Frame-dragging

As seen by long distance observers, the hovering position has a four-velocity deﬁned by

µ −1/2 uh = (dt/dτ, 0, 0, 0) = ((−g00) , 0, 0, 0). For a free-falling particle the four-velocity due the frame-dragging, as seen by the same distant observers, is described by r µ −g33 g03 ufd = 2 1, 0, 0, − . g00g33 − (g03) g33 In the present work, the frame-dragging velocity has to be measured by the hovering observer as a local inertial frame velocity and it can be obtained by µ projecting out the four-momentum mufd of the particle over the four-vector velocity uhµ of the hovering observer

µ ufduhµ = −E = −γfd, (8) where E is the relativistic energy per unit mass of the particle with respect to a local (hovering) 2 −1/2 observer. Here γfd = (1 − vfd) . Moreover a ab µ uh = η eb uhµ and therefore a ufd = (cosh η, 0, 0, sinh η). (9) IV. Spin precession

Now we consider two observers and one EPR source on the equator plane θ = π/2. The observers are placed at azimuthal angles φ = ±Φ and the EPR source is located at φ = 0. The observers and the EPR source are assumed to be hovering. From the perspective of a zero angular momentum observers (ZAMO), the local velocity of the entangled particles is given by

a uEPR = (cosh ζ, 0, 0, sinh ζ), (10)

where vEPR = tanh ζ is the speed of particles in the local inertial frame of the ZAMO.

After the pair of entangled spin-1/2 particles is generated at the EPR source, they leave it and follow a circular path around a black hole. In spherical coordinates on the equatorial plane θ = π/2, the velocity of particles has two relevant components, the temporal one and the spatial one with φ-coordinate at constant radius r. Thus, for the hovering observer, the motion is measured by the proper-velocity with v = tanh ξ. That is, ua = (cosh ξ, 0, 0, sinh ξ), therefore the general contravariant four-velocity is t t t u = e0 cosh ξ + e3 sinh ξ, φ φ (11) u = e3 sinh ξ, µ such that u uµ = −1. In order the particles describe circular motion, we must apply an external force that compensates both the centrifugal force and the gravity. The acceleration due to this external force is

µ ν µ a (x) = u (x)∇νu (x). (12) On the equatorial plane the acceleration then becomes r t 2 1 2 a = (e0 ) Γ0 0 cosh ξ t 2 1 φ 2 1 t φ 1 2 + (e3 ) Γ0 0 + (e3 ) Γ3 3 + 2e3 e3 Γ0 3 sinh ξ t φ 1 t 1 +2e0 (e3 Γ0 3 + e3 Γ0 0) sinh ξ cosh ξ, (13) µ where Γρ σ are the usual Christoﬀel’s symbols. The change of the local inertial frame consists of a boost along the 1-axis and a rotation about the 2-axis calculated by a ν a χ b(x) = −u ων b(x), (14) where the connection one-forms are deﬁned as a ν a a ν ωµ b(x) = −eb (x)∇µe ν(x) = e ν(x)∇µeb (x). (15) In our particular situation, the connections of interest are given by: 0 r 0 0 r 0 3 ωt 1 = e1 e tΓ0 1 + e1 e φΓ0 1, 1 t 1 1 φ 1 1 ωt 3 = e3 e r Γ0 0 + e3 e r Γ0 3, 0 r 0 0 r 0 3 (16) ωφ 1 = e1 e tΓ1 3 + e1 e φΓ1 3, 1 t 1 1 φ 1 1 ωφ 3 = e3 e r Γ0 3 + e3 e r Γ3 3. The relevant boost is described by

0 t r 0 0 0 3 χ 1 = −e0 e1 (e t Γ0 1 + e φΓ0 1) cosh ξ

r φ 0 0 0 3 t 0 0 0 3 −e1 e3 (e t Γ1 3 + e φΓ1 3) + e3 (e t Γ0 1 + e φΓ0 1) sinh ξ, while the rotation about the 2-axis is given by

1 t 1 t 1 φ 1 χ 3 = −e0 e r (e3 Γ0 0 + e3 Γ0 3) cosh ξ

1 t t 1 φ 1 φ t 1 φ 1 −e r e3 (e3 Γ0 0 + e3 Γ0 3) + e3 (e3 Γ0 3 + e3 Γ3 3) sinh ξ. The inﬁnitesimal Lorentz transformation 1 λa (x) = − [aa(x)p (x) − pa(x)a (x)] + χa (x). (17) b m b b b The boost along the 1-axis and the rotation about the 2-axis are respectively

0 1 t 2 1 2 t 2 1 φ 2 1 t φ 1 2 λ 1 = e r (e0 ) Γ0 0 cosh ξ + ((e3 ) Γ0 0 + (e3 ) Γ3 3 + 2e3 e3 Γ0 3) sinh ξ t φ 1 t 1 + 2e0 (e3 Γ0 3 + e3 Γ0 0) sinh ξ cosh ξ cosh ξ t r 0 0 0 3 −e0 e1 (e t Γ0 1 + e φΓ0 1) cosh ξ r φ 0 0 0 3 t 0 0 0 3 −e1 e3 (e t Γ1 3 + e φΓ1 3) + e3 (e t Γ0 1 + e φΓ0 1) sinh ξ,

1 1 t 2 1 2 t 2 1 φ 2 1 t φ 1 2 λ 3 = −e r (e0 ) Γ0 0 cosh ξ + ((e3 ) Γ0 0 + (e3 ) Γ3 3 + 2e3 e3 Γ0 3) sinh ξ t φ 1 t 1 + 2e0 (e3 Γ0 3 + e3 Γ0 0) sinh ξ cosh ξ sinh ξ t 1 t 1 φ 1 −e0 e r (e3 Γ0 0 + e3 Γ0 3) cosh ξ 1 t t 1 φ 1 φ t 1 φ 1 −e r e3 (e3 Γ0 0 + e3 Γ0 3) + e3 (e3 Γ0 3 + e3 Γ3 3) sinh ξ. (18) The change of the spin is obtained by computing the inﬁnitesimal Wigner rotation

λi (x)p (x) − λ (x)pi (x) ϑi (x) = λi (x) + 0 k k0 . (19) k k p0(x) + m

In particular, the rotation about the 2-axis through a certain angle reads:

1 1 h t 2 1 2 t 2 1 φ 2 1 t φ 1 2 ϑ 3 = −e r (e0 ) Γ0 0 cosh ξ + ((e3 ) Γ0 0 + (e3 ) Γ3 3 + 2e3 e3 Γ0 3) sinh ξ t φ 1 t 1 i + 2e0 (e3 Γ0 3 + e3 Γ0 0) sinh ξ cosh ξ sinh ξ t 1 t 1 φ 1 −e0 e r (e3 Γ0 0 + e3 Γ0 3) cosh ξ 1 h t t 1 φ 1 φ t 1 φ 1 i −e r e3 (e3 Γ0 0 + e3 Γ0 3) + e3 (e3 Γ0 3 + e3 Γ3 3) sinh ξ (20) sinh ξ n 1 h t 2 1 2 t 2 1 φ 2 1 t φ 1 2 + cosh ξ+1 e r (e0 ) Γ0 0 cosh ξ + ((e3 ) Γ0 0 + (e3 ) Γ3 3 + 2e3 e3 Γ0 3) sinh ξ t φ 1 t 1 i + 2e0 (e3 Γ0 3 + e3 Γ0 0) sinh ξ cosh ξ cosh ξ t r 0 0 0 3 −e0 e1 (e t Γ0 1 + e φΓ0 1) cosh ξ r h φ 0 0 0 3 t 0 0 0 3 i o − e1 e3 (e t Γ1 3 + e φΓ1 3) + e3 (e t Γ0 1 + e φΓ0 1) sinh ξ . Finally, from the tetrad given in Eq. (6), it can be shown, after some algebra, that the previous expression (20), can be expressed as 1 cosh(2ξ) ∂g00 ∂g03 ϑ 3 = − g03 − g00 p 2 2g00 g11[(g03) − g00g33] ∂r ∂r sinh(2ξ) − 2 √ 4g00[(g03) − g00g33] g11 h ∂g00 ∂g33 ∂g00 ∂g03 i × g00 g33 ∂r − g00 ∂r + 2g03 g03 ∂r − g00 ∂r . (21) V. EPR correlation

First of all we recall that in the case of the curved spacetime, the one-particle quantum states |pa(x), σ; xi transforms under a local Lorentz transformation as

a X (1/2) a 0 U(Λ(x))|p (x), σ; xi = Dσ0σ (W (x))|Λp (x), σ ; xi, σ0 (22) a a where W b(x) ≡ W b(Λ(x), p(x)) is the so called local Wigner rotation. If the frame-dragging is taken into account on the local inertial frame velocity ua, it will aﬀect the previous local velocity transformation and then the total velocity will be written as a u± = (cosh ξ±, 0, 0, sinh ξ±), where ξ± = ζ ± η. φ After a proper time Φ/u±, each particle reaches the corresponding observer. Thus the ﬁnite Wigner rotation can be written as  1 0 0 0  a  0 cos Θ± 0 ± sin Θ±  W b(±Φ, 0) =   ,  0 0 1 0  0 ∓ sin Θ± 0 cos Θ± (23) 1 Φϑ 3 where Θ± = φ or u± Φ ∂g00 ∂g03 cosh(2ζ ± 2η) Θ± = g03 − g00 p 3 2 −(g00) g11 ∂r ∂r sinh(ζ ± η) ∂g ∂g ∂g ∂g + g g 33 − g 00 + 2g g 00 − g 03 00 00 ∂r 33 ∂r 03 03 ∂r 00 ∂r cosh(ζ ± η) × . p 2 (g03) − g00g33 Then the required Wigner rotation is given in the following form

(1/2) σy D 0 (W (±Φ, 0)) = exp ∓i Θ , (24) σ σ 2 ± where σy is the Pauli matrix. Now we can deﬁne the four-momentum of the particle as seen by each hovering observer. Thus, the spin-singlet state for entangled particles is given by 1 |ψi = √ [|pa , ↑; 0i|pa , ↓; 0i − |pa , ↓; 0i|pa , ↑; 0i], 2 + − + − (25) Therefore after the ﬁnite Wigner rotation, the new total quantum state is given by |ψ0i = W (±Φ)|ψi. Consequently in the local inertial frames at the corresponding positions φ = Φ and −Φ, each particle state can be written as Θ Θ |pa , ↑; ±Φi0 = cos ± |pa , ↑; ±Φi ± sin ± |pa , ↓; ±Φi, (26) ± 2 ± 2 ± Θ Θ |pa , ↓; ±Φi0 = ∓ sin ± |pa , ↑; ±Φi + cos ± |pa , ↓; ±Φi(27). ± 2 ± 2 ± Thus the entangled state is described by the combination

sinh ζ cosh η ∆ = Φ (2A sinh ζ + B cosh ζ) cosh η − A − 1 , cosh2 η − cosh2 ζ (30) where 1 ∂g00 ∂g03 A = g03 − g00 , p 3 2 −(g00) g11 ∂r ∂r 1 B = p 3 2 2 −(g00) g11[(g03) − g00g33] h ∂g33 ∂g00 ∂g00 ∂g03 i × g00 g00 ∂r − g33 ∂r + 2g03 g03 ∂r − g00 ∂r . (31) VI. Spin precession angle in Expanding and Twisting Plebański-DemiańskiBlack Hole Now we study the spin precession angle of the spin-1/2 systems of entangled particles in the spacetime described by the Plebański-Demiańskimetric with frame-dragging. Thus, it is easy to show that the coefficients A and B from the spin precession angle ∆ on the equator (θ = π/2), given by Eq. (30), are written as √ a D A = [(r 2 + l2)D0 − 2r(D − a2)], PD 2(r 2 + l2)(D − a2)3/2 1 B = PD 2(r 2 + l2)(D − a2)3/2 ×[4Dr(D − a2) − (a2r 2 + Dr 2 + Dl2 + a2l2)D0], (32) where D0 is defined by ∂D α2κ Λ 6nαr 2 D0 = = −4 + r 3 − + 2εr − 2m. (33) ∂r ω2 3 ω The frame-dragging local inertial frame velocity is given by s D cosh η = (r 2+l2) . PD (D − a2) [(r 2 + a2 + 2al + l2)2 − (a + 2l)2D] (34) Non-accelerating Kerr-Newman-(anti)de Sitter-NUT black hole The KN(A)dSNUT spacetime represents a non-accelerating (α = 0) black hole with mass m, electric and magnetic charges e and g, a rotation parameter a and a NUT parameter l in a de Sitter or anti-de Sitter background with non-vanishing cosmological constant Λ. This case contains in turn the two limits: |a| > |l| and |a| < |l|, that correspond to the Kerr solution with NUT and the NUT solution with rotation respectively. After setting the acceleration parameter equals zero i.e. α = 0, the parameters in relation (3) become κ = (1 − l2Λ)(a2 − l2), 1 ε = 1 − a2 + 2l2 Λ, 3 (35) 1 n = l + (a2 − 4l2)lΛ. 3 Thus, the metric (1) is reduced to D θ ρ2 ds2 = − [dt − (a sin2 θ + 4l sin2 )dφ]2 + dr 2 ρ2 2 D P ρ2 + [adt − (r 2 + (a + l)2dφ)]2 + sin2 θdθ2, ρ2 P where

ρ2 = r 2 + (l + a cos θ)2, 4 1 P = sin2 θ 1 + Λal cos θ + Λa2 cos2 θ , 3 3 1 1 D = a2 − l 2 + e2 + g 2 − 2mr + r 2 − Λ (a2 − l 2)l 2 + ( a2 + 2l 2)r 2 + r 4 . 3 3 (36) In the case when e = g = Λ = 0 the spin precession angle explicitly written in terms of the physical parameters is given by

∆Kerr−NUT = ( cosh ηKerr−NUT h p 2a r 2 + a2 − l2 − 2mr(mr 2 + 2l2r − ml2) sinh ζ (r 2 + l2)(r 2 − 2mr − l2)3/2 +[r 5 − 5mr 4 + (6m2 − 4l2)r 3 + (10l2m − 2a2m)r 2 (37) +(−4a2l2 + 3l4 − 2m2l2)r + 2a2l2m − l4m] cosh ζ # ) p sinh ζ +a r 2 + a2 − l2 − 2mr(mr 2 + 2l2r − ml2) − 1 Φ, 2 2 cosh ηKerr−NUT − cosh ζ where

cosh ηKerr−NUT = p (r 2 + l2) r 2 − 2mr + a2 − l2 (38) . p(r 2 − 2mr − l2)[r 4 + (a2 − 2l2)r 2 + (8mal + 8l2m + 2a2m)r + 5l4 + 3a2l2 + 8al3] Kerr black hole From equation (30) with coeﬃcients (32) and parameters e = g = l = Λ = 0 the spin precession angle is reduced to

" √ √ # −2 Dam sinh ζ + (H − mD) cosh ζ Dam sinh ζ cosh ηK ∆ = Φ cosh η + − 1 , (39) K 2 3/2 K 2 3/2 2 2 (r − 2mr) (r − 2mr) cosh ηK − cosh ζ where

D = r 2 − 2mr + a2, (40) H = r 3 − 4mr 2 + 4m2r − a2m, (41) √ r D cosh ηK = 6= 1, (42) p(r − 2m)(r 3 + a2r + 2ma2) with the coeﬃcients A and B being √ Dam H − mD A = − , B = . (43) K (r 2 − 2mr)3/2 K (r 2 − 2mr)3/2 Schwarzschild-NUT Black Hole This is more clearly stated when the Eq. (30) is simpliﬁed by setting the parameters e = g = a = Λ = 0

r 3 − 3mr 2 − 3l2r + ml2 ∆NUT = Φ √ cosh ζ cosh ηNUT − 1 , (r 2 + l2) r 2 − 2mr − l2 (44) where r 2 + l2 cosh ηNUT = √ 6= 1. r 4 − 2l2r 2 + 8ml2r + 5l4 The coeﬃcients A and B are given by r 3 − 3mr 2 − 3l2r + ml2 ANUT = 0, BNUT = √ . (45) (r 2 + l2) r 2 − 2mr − l2

The precession angle ∆NUT diverges precisely at r = rNUT, where

p 2 2 rNUT = m ± m + l . (46) The positive root represents the outer Schwarzschild-NUT horizon.

Schwarzschild-(Anti)de Sitter Black Hole The spin precession angle in this case is given by

  r − 3m   ∆(A)dS = Φ q cosh ζ − 1 ,  2 1 4  r − 2mr − 3Λr (47) with A and B of the following form r − 3m A(A)dS = 0, B(A)dS = q 2 1 4 r − 2mr − 3Λr (48) and cosh η(A)dS = 1.

Reissner-NordströmBlack Hole This case corresponds to a Schwarzschild black hole with non-vanishing charges e and g, after setting l, Λ and a to zero. The Reissner-Nordströmspacetime is also a spherically symmetric solution. The spin precession angle is then reduced to ! r 2 − 3mr + 2e2 + 2g 2 ∆RN = Φ cosh ζ − 1 , (49) rpr 2 − 2mr + e2 + g 2 where the functions A and B are r 2 − 3mr + 2e2 + 2g 2 ARN = 0, BRN = (50) rpr 2 − 2mr + e2 + g 2 and cosh ηRN = 1. This result reproduces completely our previous result after adding the magnetic charge g. Schwarzschild Black Hole Finally it is easy to recover the Schwarzschild spin precession by setting a, e, g, l, Λ = 0. The coefficients and frame-dragging√ are reduced to 2 AS = 0, BS = (r − 3m)/ r − 2mr and cosh ηS = 1. Consequently, the expression (30) is given by

r − 3m ∆S = Φ √ cosh ζ − 1 , (51) r 2 − 2mr which is precisely Eq. (51) of Terashima-Ueda paper. Accelerating and rotating black holes In order to simplify our analysis we shall consider in this subsection the case of vanishing parameters Λ = e = g = l = 0. with an arbitrary α and using the remaining scaling freedom to put ω = a, then the Pleba´nski-Demia´nskimetric is reduced to

1 D ρ2 P sin2 θ ! ds2 = − [dt − a sin2 θdφ]2 + dr 2 + (adt − (r 2 + a2)dφ)2 + ρ2 dθ2 , (52) Ω2 ρ2 D ρ2 P where the parameters (2) and (3) are given by

ε = 1 − a2α2, n = −aαm, (53) P = sin2 θ 1 − 2αm cos θ + a2α2 cos2 θ , and ρ2 = r 2 + a2 cos2 θ, Ω = 1 − αr cos θ, (54) D = a2 − 2mr + (1 − a2α2)r 2 + 2α2mr 3 − α2r 4. The previous metric has four singularities when θ = π/2, that is, we can factorize D as

2 2 D = (r − r+)(r − r−)(1 − α r ), (55) where p 2 2 r± = m ± m − a . (56) The acceleration horizon: 1 r = . (57) acc α After some easy manipulations, the coeﬃcients for the spin precession angle are given by √ a D A = [rD0 − 2(D − a2)], AccRot 2 3/2 2r(D − a ) (58) 1 B = [4D(D − a2) − r(a2 + D)D0], AccRot 2r(D − a2)3/2 where ∂D D0 = = −2m + 2(1 − a2α2)r + 6α2mr 2 − 4α2r 3 (59) ∂r and the frame-dragging velocity is s D cosh η = r 2 . (60) AccRot (D − a2) [(r 2 + a2)2 − a2D]

One can show that we can recover the Kerr spacetime results reviewed in previous section, after setting vanishing acceleration (α = 0). Therefore, we shall consider the eﬀect of acceleration over the spin precession angle. C-metric From the pair of accelerated and rotating black holes represented by the metric, we can consider the limit in which a → 0. In this case, the metric has the form of the C-metric and thus the coeﬃcients reduce to α2mr 2 + r − 3m AC−metric = 0, BC−metric = , p(r 2 − 2mr)(1 − α2r 2) (61) and cosh ηC−metric = 1. (62) Then, the spin precession angle for the C-metric is ! α2mr 2 + r − 3m ∆C−metric = Φ cosh ζ − 1 . (63) p(r 2 − 2mr)(1 − α2r 2)

Moreover it is easy to see that this equation reduces to Schwarzschild case when α = 0. In addition, we can see from Eq. (63) that it is divergent at the Schwarzschild radius r = 2m and at −1 the acceleration horizon, that is ∆C−metric → ∞ as r → α .