Quantum, Classical and Symmetry Aspects of Max Born 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 Space (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[momentum] 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 Phase Space 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