Quantum, Classical and Symmetry Aspects of Reciprocity

L.M. Tomilchik

B.I. Stepanov Institute of Physics, National Academy of Sciences of Belarus

Minsk, 16–19 May 2017

1 / 36 Topics

1 Reciprocal Symmetry and Maximum Tension Principle 2 Maximum Force and Newton gravity 3 Extended Phase (QTPH) as a Basic Manifold 4 Complex Lorentz group with Real Metric as Group of Reciprocal Symmetry 5 One-Particle Quasi-Newtonian Reciprocal-Invariant Hamiltonian dynamics 6 Canonic Quantization: Dirac Oscillator as Model of Fermion with a Plank mass

2 / 36 Born Reciprocity version

1 Reciprocity Transformations (RT): xµ pµ pµ xµ → , → − . qe pe pe qe 2 Lorentz and Reciprocal Invariant quadratic form: 1 1 S2 = xµx + pµp , η = diag{1, −1, −1, −1} B 2 µ 2 µ µν qe pe 3 xµ, pµ are quantum-mechanical canonic operators:

[xµ, pν ] = i~ηµν 4 Phenomenological parameter according definition pe = Mc so that

qe = ~/Mc = Λc (Compton length). The mass M is a free model parameter. 5 Two dimensional constants qe[length], pe[] connected by corellation:

qepe = ~ (Plank constant). 3 / 36 Relativistic Oscillator Equation (ROE)

Selfreciprocal-Invariant Quantum mechanical Equation:

 ∂2  − + ξµξ Ψ(ξ) = λ Ψ(ξ), µ µ B ∂ξ ∂ξµ

µ µ where ξ = x /Λc. The (ROE) symmetry ≈ U(3, 1). There are solution corresponding linear discrete mass spectrum.

4 / 36 Maximum Tension Principle (MTP) The space-time and momentum variables are to be just the same dimentions of a quantity. It is necessary to have the universal constant with dimention: momentum/length, or equivalently: mass/time, energy/time, momentum/time. In reality: MTP [Gibbons, 2002] Gibbons Limit: dp c4 Maximum force Fmax = = = FG, dt max 4GN dE  c5 Maximum power Pmax = = = PG. dt max 4GN [momentum]/[length] – Universal constant 3 pe c æ0 = = [LMT, 1974, 2003] qe 4GN −1 −2 It is evident: æ0 = c Fmax = c Pmax. All these values are essentially classic (does not contain Plank constant ~). 5 / 36 Born Reciprocity and Gibbons Limit

Let the parameter a to be some fixed (nonzero!) value of action S. We replace initial Born’s relation

qepe = ~ by πqepe = a without beforehand action discretness supposition. We demand the nonzero area Sfix = πqepe of the phase plane only. The second suggestion is:

3 pe c æ0 = = (universal constant). qe 4GN

6 / 36 Now the parameters qe, pe are defined separately r a raæ qe = , pe = πæ π

Evidently: the phenomenological constants Ee = pec [Energy] −1 and te = c qe [time] satisfy the conditions 5 Ee 2 c Eete = πa, = c æ = te 4GN Under Born’s parametrization

2MGN pe = Mc, qe = = rg(M) – gravitational radius. c2 General mass-action correlation ac M 2 = . 4πGN It is valid as in classical as in quantum case. Under supposition when amin = h we receive s 2 hc 1 2 ~c M = = MP l,MP l = – Plank mass. 4πGN 2 GN 7 / 36 Maximum Force and Modified Newton Gravity

4 mM c 2mGN 2MGN 1 (NG): f = G ≡ N 2 2 2 2 r 4GN c c r 2 rgRg r0 p = FG ≡ FG (r 0, r0 = rgRg) r2 r2 > r2 max (ModNG): f mod = F 0 (r r ), f mod = F N G 2 > 0 N G r r=r0 The corresponding gravitational potential Energy 2 mod r0 U = −FG gr r possess a minimum under r = r0 4 2 c 2GN √ c √ U mod(min) = −F r = − mM = − mM gr G 0 2 4GN c 2 We have dealing (instead of point-like masses) with a pair spacely extended simple massive gravitating objects like the elastic balls or drops with a high surface tension. 8 / 36 r0

M m

Rg rg

√ 2 2 1 2 Erest(M, m) = Mc + mc − mMc 2 9 / 36 The whole energy such a “binary” system is as follows: 1√ E(M, m) = Mc2 + mc2 − mMc2 2  1r µ  = (M + m)c2 1 − , 2 M + m where µ = mM/(m + M).

Mc2, mc2, (M + m)c2 – rest energy 1 1√ ∆E = pµ(M + m)c2 = mMc2 – energy corresponding 2 2 1 √ the “mass defect” ∆m = 2 mM

10 / 36 “Gravitational fusion” picture

[eliminated energy] ∆E λ = = [rest energy] (M + m)c2 √ √ mM 1 δ = = , 2(M + m) 2 1 + δ where m Mmin δ = , δ0 6 δ 6 1, δ0 = M Mmax

∆Eeliminated 1 Maximum λmax = = Erest 4 under δ = 1 (m = M) Such a “fusion” of two equal rest mass M lead to mass = 3M/2, the Energy ∆E = Mc2/2 eliminate (3 : 1).

11 / 36 Extended as a Basic Manifold

Space of states (QTP) and Energy (H) (sec [J. Synge, 1962]) in general: 2 + 2N dimensions. The metric is

N z }| { ηAB = diag{1, −1, −1,..., −1}

(X,Y ) – pair of conjugated pseudoeuclidean vectors XA, YA, A, B = 0, 1,...,N (N + 1 dimention). The Poisson brackets for ϕ(X,Y ), f(X,Y ),

N X  ∂ϕ ∂f ∂ϕ ∂f  {ϕ, f} = − A,B ∂Xk ∂Y ∂Y k ∂X k=0 k k so that

{XA,YA}A,B = ηAB

12 / 36 Dynamic System in (QT, P H) is defined, when the Energy Surface( 2N + 1 dimention)

Ω(X,Y ) = 0 is defined. Solving this equation on Y0 we obtain Energy equation

Y0 = F (Xk,Yl,X0) k, l = 1, 2,...,N

The corresponding Hamilton function of the System is by def:

H ≡ Y0 = H (Xk,Yl,Y0)

We shall consider (2 + 2 · 3) – dimensional version (QT, P H)

A, B → µ, ν = 0, 1, 2, 3, ηµν = diag{1, −1, −1, −1},

Xµ, Xµ are real 4-dimensional pseudoeuclidean SO(3, 1) – vectors.

13 / 36 Complexification of (QT, P H)

Let us introduce the complex 4-vector Zµ according definition:

µ µ µ Z = X + iY , ηµν = diag{1, −1, −1, −1} and define the real norm

µ ∗ µ µ∗ Z Zµ = Z ηµν Z

2 2 2 2 = |Z0| − |Z1| − |Z2| − |Z3| = inv

14 / 36 Complex Lorentz Group with a real metric (Barut Group [Barut 1964]) The group of 4 × 4 complex matrices Λ of transformations Z0 = ΛZ,Z = X + iY satisfying the condition ΛηΛ† = η, († – hermitien conjugation)

Notice: the diagonal matrices ΛR and Λ¯ R ∈ (BG)

ΛR = −iI4 = −idiag{1, 1, 1, 1} and Λ¯ R = −iηµν = −idiag{1, −1, −1, −1} Metric invariant (no positive-defined)

µ ∗ 2 2 2 2 Z Zµ = |Z0| − |Z1| − |Z2| − |Z3| = inv 2 2 Three possibilities: |Z| ≷ 0, |Z| = 0 – isotropic case 15 / 36 SU(3, 1) – subgroup structure

(QTPH) SOv 3, 1

QT = {r, ct} = xµ H L Lk PH = p, E/c = pµ

β~ = ~v/c C v C SO 3 U U * SU 3 center 4 kl ‡ lk (PTQH) H L H L H L  µ PT = p,F0t = Π Mk

µ QH = {r, E/F0} = Ξ SO f 3, 1 β~f = f/F~ 0 H L F0 = Fmax k, l = 1, 2, 3

16 / 36 2 Reciprocal Invariant SB as BG metric invariant µ ∗ Z Zµ

By def: Zµ = Xµ + iYµ 1 1 1 1 Xµ = xµ = {x0, xk},Yµ = pµ = {p0, pk}. qe qe pe pe Then:

3 ! 3 ! 1 X 1 X S2 = x2 − x2 + p2 − p2 = B q2 0 k p2 0 k e k=1 e k=1

2 2 3  2 2  3 x0 p0 X xk pk 2 X 2 µ ∗ = + − + = |Z0| − |Zk| = Z Z q2 p2 q2 p2 µ e e k=1 e e k=1

Let xµ, pµ to be 4-vectors under SOv(3, 1)-transformations

17 / 36 Notice! There exist alternative:

Z¯µ = X¯µ + iY¯µ where

def x0 pk  def p0 xk  X¯µ = Ξµ = , , Y¯µ = Πµ = , . qe pe pe qe It is evident:

2 3 2 2 3 2 2 µ µ x0 X pk p0 X xk |Z¯µ| = Ξ Ξµ + Π Πµ = − + − = q2 p2 p2 q2 e k=1 e e k=1 e 1 1 = xµx + pµp = |Z |2. 2 µ 2 µ µ qe pe

The natural supposition: Ξµ and Πµ are 4-vectors under SOf (3, 1)-transformations

18 / 36 2 Selfreciprocal Case: SB = 0 2 µ ∗ Quadratic form SB = Z Zµ is not positive defined µ ∗ µ ∗ Three possibilities Z Zµ = 0, Z Zµ ≷ 0. Especial interesting is: 1 1 S2 = pµp + xµx = 0 B 2 µ 2 µ pe qe Two possibilities

1 µ µ p pµ = x xµ = 0 (light cone, massless particles, standart clock synchronisation)

2 µ 2 µ 2 x xµ = −qe , p pµ = pe,

2meGN qe = rg(me) = , pe = mec. c2

19 / 36 BG: some kinematic outcome (1) Reciprocal-Invariant time TR 3 −1 pe c zµ = xµ + iæ0 pµ, æ0 = = qe 4GN

def −1q µ ∗ dTR = c z zµ q −1 2 2 −2 2 2 = c dx0 − dr + æ0 (dp0 − dp ) q = dt 1 − β2 − β2 + β2β2 cos2 α v f v f r      2 = dt 1 − β2 1 − β2 − β × β . v f f v where r˙ p˙ c4 d β = , β = ,F = , (·) = , v f G c FG 4GN dt t – is the laboratory time (comoving observer time). α – is the angle between r˙ and p˙ . 20 / 36 (2) In the comoving RF : q dT (β = 0) = dt 1 − β2 R v v There must exist an extra time retardation. (3) cos α = ±1, (β × β ) = 0 f v q q dT k = dt 1 − β2 1 − β2 R v f (4) cos α = 0, (β × β )2 = β2 · β2 f v f v q dT ⊥ = dt 1 − β2 − β2 6= dT k R v f R √ (5) cos α = ± 2/2,

k ⊥ dTR = dTR

(6) The upper limit of |r˙| and |p˙ | in the expression for dTR: √ √ |r˙|max = c/ 2, |p˙ |max = FG/ 2

21 / 36 Complex Lorentz Boost Transformations

Let us represent a BG-transformations using the following parametrization:

ΓΓ · β ! ∗ † Λ(β) = ∗ β·β , Λ(β)ηΛ (β) = η Γ · β I3 + |β|2 (Γ − 1) where β = β + iβ = v/c + iF /F , v F G v – is a const velocity 4 F – is a const force, FG = c /4GN – max force

−1 ∗ Γ = detΛ (β),I3 = diag{1, 1, 1}, β·β – 3×3 tensor-diada.

 Q0 + iP 0   Q + iP  Z0 = Λ(β)Z : 0 0 = Λ(β) 0 0 Q0 + iP 0 Q + iP

22 / 36 The simplest case β = {β , 0, 0}, β = {β , 0, 0} v v F F

 1 β  Λ(β) = Γ(β) , β = β + iβ , β∗ 1 v F

1 v F Γ(β) = q , βv = , βF = , 1 − |β|2 c FG

F – is the constant force uniformly accelerated (noninertial) FR.  0 0    Q0 + iP0 1 βv 0 0 = Γ(β) ∗ Q + iP βv 1  0 β   Q + iP  +i F 0 0 −βF 0 Q + iP

2 2 −1/2 Γ(β) = 1 − βv − βF

23 / 36 c4 βF = 0 βv = 0,FG = 4GN

 ct0   t0  0 0 x p /FG         1 βv ct 1 −βF t = γv , = γF , βv 1 x −βF 1 p/FG  E0/c   E0  0 0 p FGx         1 βv E/c 1 −βF E = γv , = γF , βv 1 p −βF 1 FGx

p 2 p 2 γv = 1/ 1 − βv γF = 1/ 1 − βF

x˙ + V p˙ − F x˙ 0 = : p˙0 = : 2 2 1 +xV/c ˙ 1 − pF/F˙ G

0 βv + βV 0 βf − βF βv = βf = , 1 + βvβV 1 − βf βF dp dE p˙ = = f, = −f dt dx 24 / 36 Hamiltonian selfreciprocal one-particle classical model

The selfreciprocal expression of Born interval

2 2 2 2 2 2 SB = x0 − r /qe + p0 − p = 0 is by definition the Energy equation in the (2 + 2 · 3)-dimensional extended Phase Space (QTPH) [J.Synge, Class. Dyn.] and

2 2 2 2 P0 + T − P + Q     q 2 2 q 2 2 = P0 + P + Q P0 − P + Q = 0 is the energy surface equation. According to (QT P H) definition H(Q,P ,T ) ≡ −P0 is the Hamilton function of system. In classical version we are to choose the positive sign.

25 / 36 So that Hamilton function H(Qk,Pk,T ) is as follows q x p 2 2 k k H(Qk,Pk,T ) = H0 − T ,Qk = ,Pk = , qe pe where

T = x0/qe = vet (ve = v(me) = æ0/me) q 2 2 H0 = Pk + Qk – energy (integral of motion) dH0  = H0(P ,Q), H(P ,Q,T ) = 0 dT q,p The Hamiltonian equations

dQk Pk dPk −Qk = , = p 2 2 p 2 2 dT H0 − T dT H0 − T possess under the spherical-symmetric initial conditions the following solution

|Q~| = sin(ϕ + ϕmin), |P~ | = cos(ϕ + ϕmin),

ϕ = arcsin(T/H0) (0 < ϕ < π/2),

ϕmin 6= 0 – minimal angle ∼ minimal value of peqe 26 / 36 The model: massive pulsating sphere Frequency of pulsation (comoving FR time)

3 38 −1 æ0 c 10 s ω0(me) = = ∼ . H0me 4meGN me Pulsation Period

−38 T0(me) = 2π/ω0(me) ∼ me · 10 s.

Spatial peak value

2 −28 R(me) = 4meGN /c ∼ me · 10 sm.

55 For me = mu ∼ 10 gr,

17 −1 27 T (mu) ∼ 10 s ∼ H0 ∼ ∆tu,R(mu) ∼ 10 sm ∼ Ru

.

27 / 36 Elliptic trajectory: r − p-spase

The eccentricity q √ ~ 2 ~ 2 21/4 Q+ − Q− 2 1 − α ε(α) = = ~  √ 1/2 Q+ 1 + 1 − α2

α2 Q~ 2P~ 2 ε(α)| =∼ 1 − = 1 − , bat 6= 1. α1  2 8 2 Q~ 2 + P~ 2

~ 2 2 ∼ 4P 2p p α |P Q = = ,  æ0 Q~ 2 ræ0 r

28 / 36 P (Q˙ ) – dependence

˙ 2 ˙ dQ~ P~ Q~   Q~ = = → P~ 2 = Q~ 2 − T 2 dT q ˙ P~ 2 + Q~ 2 − T 2 1 − Q~ 2

Q~ 2 − T 2 = 1 – hyperbolic motion! ˙ Q~ 2 ~ 2 max ~ Pmax = . But |Pmax| = 1 ~˙ 2 1 − Qmax √ ~˙ 2 Therefore 2Qmax = 1 and r˙max = c/ 2

29 / 36 2 The case SB = ΛB 6= 0 2 µ µ From SB = P Pµ + Q Qµ = ΛB > 0 to ¯2 2 SB = SB − ΛB = 0. The Energy Integral is:

H0(P,Q)|ΛB=0 → H0(P, Q, ΛB) q 2 2 = H0 = P~ + Q~ + ΛB

pe(m) pe(m), qe(m), = æ0 qe(m)

pe(M) → pe(M), qe(M), = æ0 qe(M) √ Scale transformation of the parameter: M = m 1 + ΛB. Consider the case ΛB = 1. Two versions: Classical and Quantum

30 / 36 The Newtonian Limit q 2 2 Let us consider the Energy integral H0 = P + Q + 1 under conditions |Q|  1, |P |  1, |P |/|Q|  1. Use the dimensional values p2 r2 1/2 E = cp H = cp + + 1 e 0 e 2 2 pe qe s ( −1)1/2 r2 p2  r2  = cp 1 + 1 + 1 + e 2 2 2 qe pe qe 1 1 ≈ mc2 + mv2 + mr2ω2(m). 2 2 0 Omitting the “rest energy” mc2 we receive

1 2 1 2 EN = mv + Irω (m) 2 2 0 2 where Ir = mr – moment of inertia ω0 = æ0/m – angular velocity 2 2 Erot = mr ω0/2– is the kinesthetic energy of rotation. 31 / 36 2 ˆ2 Canonical Quantization SB → SB

Accoding the universal condition   2 cae h ~c 1 2 me = → ae = → = MP 4πGN 2 4GN 4 the parameters pe, qe, Ee, te becomes Planken. The basic Born equation can be written in the form (Fock presentation)

3 ! X SˆB|n0, ni = 2 Hˆ 0 − Hˆ k |n0, ni = λB(n0, n)|n0, ni, k=1 where Hˆ µ (µ = 0, 1, 2, 3) are linear QHOsc Hamiltonians.

32 / 36 The discrete spectrum of (mass)2:   2 2 ~c M (n0, n) = {n0 − (n + 1)} MP = N(n0, n) GN P3 where N(n0, n) = n0 − (n + 1), n = k=1 nk. Each of nµ = 0, 1, 2,... (µ = 0, 1, 2, 3) independently. Eigenstates

3 Y 1 |n0, ni = |nµi,N(n) = (n + 1)(n + 2) 2 µ=0 is the degeneracy degree of |ni ∼ |n1i|n2i|n3i states. In ˆ2 reality the operator SB is looking out as the quantum- mechanical action operator having linear spectrum SN in unit of h: SN = N~ (N = 1, 2, 3,...)

33 / 36 Canonical Quantization

H0(P ,Q) → Hˆ 0(Pˆ , Qˆ )

Let as consider the variables Pˆk, Qˆl in the classical Energy integral q 2 2 H0 = P + Q + 1 as canonically conjugated operators [Pˆk, Qˆl] = −iδkl, the constants pe and qe in definition

xˆl pˆk Qˆl = , Pˆk = qe pe are Planken.

 2 2 1/2 Hˆ 0 → Hˆ 0 = Pˆ + Qˆ + 1

34 / 36 The root extraction according Dirac procedure gives (in noncovariant notations):

± for DO Hamiltonian and its Supersymmetry Partner HDO

ˆ ± ˆ ˆ ˆ  ˆ  H0 (P, Q, 1) = αkPk + β I ∓ iαkQk =

ˆ ±  ˆ ˆ  = HDO = αk Pk ∓ iβQk + β,

where αk, ρk (k = 1, 2, 3) – 4 × 4 Dirac matrices, β = ρ3. ˆ ± ˆ ± HDO = HDO(MP l) – well known Dirac Oscillator Hamiltonian which describes the one-half Spin particle with a Plank mass. The exact solutions of this model are well known. But generally accepted physical interpretation of DO-model is absent up to now.

35 / 36 Thank you for your attention

36 / 36