Greens function, wavefunction and Wigner function of the MIC-Kepler - PowerPoint PPT Presentation

Green’s function, wavefunction and Wigner function of the MIC-Kepler problem Tokyo University of Science Graduate School of Science, Department of Mathematics, The Akira Yoshioka Laboratory Tomoyo Kanazawa 1

Outline 1. Hamiltonian description for the MIC-Kepler problem • The reduced Hamiltonian system by an S 1 action • The 4-dimensional conformal Kepler problem 2. Green’s function of the MIC-Kepler problem • The infinite series of its wavefunction 3. Wigner function of the MIC-Kepler problem 2

1. Hamiltonian description for the MIC-Kepler problem PSfrag repla emen ts z In 1970, McIntosh and Cisneros studied the dynamical system de- magneti for e _ x scribing the motion of a charged particle under the magnetic force due to Dira 's �eld j � j 2 k x _ k � 2 r due to Dirac’s monopole field of strength − µ and the square inverse 3 mr _ x � B � en trifugal for e centrifugal potential force besides the Coulomb’s potential force. k due to Dira 's �eld 2 r Coulom b's for e = − µ B µ r 3 x r = k x k ∥ B µ ∥ = | µ | y O r 2 x µ is quantized µ ∈ ℏ as 2 Z . 3

PSfrag repla emen ts The Hamiltonian description for the MIC-Kepler problem is given by � 4 _ � 3 _ T R T R T. Iwai and Y. Uwano (1986) as follows. 0 0 ( u ; � ) � 1 ( � ) ( x ; p ) ( u ; � ) The MIC-Kepler problem z 1 � is the 2 4 u 1 0 p ( x ) � 2 0 4 u reduced Hamiltonian system ✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿ 1 x = � ( S � u ) 0 u of the 4-dimensional u 1 O S � u y � conformal Kepler problem x 3 _ R 4 _ by an S 1 action, R if the associated momentum mapping ψ ( u , ρ ) ≡ µ ̸ = 0 . ψ ( u , ρ ) is invariant under the S 1 action, then R 4 be a subset s.t. let ψ − 1 ( µ ) ⊂ T ∗ u ˙ � { } � ψ ( u , ρ ) = 1 ψ − 1 ( µ ) = ( u , ρ ) ∈ T ∗ R 4 � u ˙ 2( − u 2 ρ 1 + u 1 ρ 2 − u 4 ρ 3 + u 3 ρ 4 ) = µ . � � 4

The MIC-Kepler problem is a triple ( T ∗ ˙ R 3 , σ µ , H µ ) where σ µ = dp x ∧ dx + dp y ∧ dy + dp z ∧ dz − µ r 3 ( x dy ∧ dz + y dz ∧ dx + z dx ∧ dy ) , µ 2 1 2 mr 2 − k 2 m ( p x 2 + p y 2 + p z 2 ) + H µ ( x , p ) = r . R � � � � � ( u ; � ) � PSfrag repla emen ts �( x ; p ) Its energy hyper surface : H µ = E ⇔ Φ( x , p ) ≡ r ( H µ − E ) = 0 is equal to � 4 _ � 3 _ T R T R ( u ; � ) � � ( x ; p ) ( ) π ∗ � µ Φ ( u , ρ ) = 0 � � 1 ( � ) where π µ → T ∗ ˙ R 3 , ψ − 1 ( µ ) − µ Φ = u 2 ( ) π ∗ H − E , u 2 ≡ u 12 + u 22 + u 32 + u 42 . 5

The conformal Kepler problem is a triple ( T ∗ ˙ R 4 , d ρ ∧ d u , H ) where 4 4 ( ) 1 1 − k ρ j 2 ∑ ∑ d ρ ∧ d u ≡ dρ j ∧ du j , H ( u , ρ ) = u 2 . 4 u 2 2 m j =1 j =1 Since u 2 = r > 0 (invariant under the S 1 action), π ∗ µ Φ = 0 is equal to H − E = 0 . The energy hyper surface H = E is equivalent to K ( u , ρ )= ϵ where K ( u , ρ ) is the Hamiltonian of 4-dimensional harmonic oscillator: 4 4 { 1 ρ j 2 + 1 m > 0 mass of pendulum 2 mω 2 u j 2 ∑ ∑ K ( u , ρ ) = ω > 0 angular frequency 2 m j =1 j =1 considering only the case where the real parameter E < 0 and putting both the constant mω 2 ≡ − 8 E and a real parameter ϵ ≡ 4 k . 6

We solved the harmonic oscillator by means of the Moyal product , which brought the following functions. Moyal propagator ( ∗ -exponential) ♣ ( u i + u f ) i ℏ ( t + iy ′ ) K , ρ ) − 4 cos ωz ′ tan ωz ′ ( [ ( u i + u f ) ] i 2 2 e = exp ℏ ωK , ρ ∗ 2 2 2 R 4 and u f = ( u f 1 , u f 2 , u f 3 , u f R 4 where u i = ( u i 1 , u i 2 , u i 3 , u i 4 ) ∈ ˙ 4 ) ∈ ˙ denote initial point and final point of motion respectively. ♣ Feynman’s propagator K ( u f , u i ; z ′ = t + iy ′ ) − m 2 ω 2 [ }] 1 1 − i mω ( u 2 i + u 2 { f ) cos ( ωz ′ ) − 2 u i · u f = sin 2 ( ωz ′ ) exp 4 π 2 ℏ 2 sin ( ωz ′ ) 2 ℏ ∞ C n e inωt ∑ = n = −∞ ∫ π/ω C n ≡ ω − π/ω K ( u f , u i ; τ + iy ′ ) e − inωτ dτ . where 2 π 7

♣ Green’s function G ( u f , u i ; ϵ ) ∫ ∞   ∞  e − y ′ + iϵ i ( t + iy ′ ) dt C n e inωt ∑ = lim ℏ  y ′ → +0 ℏ 0 n = −∞ ∞ m 2 ω 2 − mω 1 [ ] 2 ℏ ( u 2 i + u 2 ∑ ∑ = π 2 ℏ 2 exp f ) ϵ − ( N + 2) ℏ ω N =0 l 1 + l 2 + l 3 + l 4 = N 1 (√ mω (√ mω (√ mω (√ mω ) ) ) ) ℏ u f ℏ u f ℏ u i ℏ u i 2 N l 1 ! l 2 ! l 3 ! l 4 ! H l 1 H l 1 H l 2 H l 2 1 2 1 2 (√ mω (√ mω (√ mω (√ mω ) ) ) ) ℏ u f ℏ u f ℏ u i ℏ u i H l 3 H l 3 H l 4 H l 4 3 4 3 4 where n − 2 ≡ N = l 1 + l 2 + l 3 + l 4 ( l 1 , l 2 , l 3 , l 4 ∈ N ∪ { 0 } ), H l ( X ) is the Hermite polynomial : H l ( X ) = ( − 1) l e X 2 d l dX l e − X 2 . 8

Moreover, we denote by Ψ N ( u ) the wave function of 4-dimensional R 4 harmonic oscillator on ˙ 1 mω − mω [ ] Ψ N ( u ) = 2 ℏ ( u 2 1 + u 2 2 + u 2 3 + u 2 exp 4 ) ← − √ π ℏ 2 N l 1 ! l 2 ! l 3 ! l 4 ! (√ mω (√ mω (√ mω (√ mω ) ) ) ) H l 1 ℏ u 1 H l 2 ℏ u 2 H l 3 ℏ u 3 H l 4 ℏ u 4 satisfying   4 4 K = − ℏ 2 ∂ 2  + 1 2 mω 2 u 2 ˆ ˆ ∑ ∑ K Ψ N ( u ) = ϵ Ψ N ( u ) where j .  ∂u 2 2 m j =1 j j =1 Then we verify ∞ 1 ∑ G ( u f , u i ; ϵ ) = ϵ − ( N + 2) ℏ ω Ψ N ( u f ) Ψ N ( u i ) . N =0 9

PSfrag repla emen ts 0 0 0 t ! z = t + iy iz it Harmoni K K ~ ~ e e � � K = � 0 ( y 6 = 0) os illator 2 0 � 1 � � 4 k m! � 8 j E j E < 0 y > 0 F H = E � 4 _ onformal T R L 0 m K ( u ; u ; z ) f i G ( u ; u ; 4 k ) f i Kepler [ � 0 � � = 0 y ! +0 � � 1 ( � ) � = 0 m MIC-Kepler G ( x ; x ; E ) or G ( x ; x ; E ) + f i � f i H = E � � 3 _ T R 10

2. Green’s function of the MIC-Kepler problem − 2 mk 2 We suppose E ̸ = ( N = 0 , 1 , 2 , · · · ) , then reduce ℏ 2 ( N + 2) 2 the Green’s function of the conformal Kepler problem G ( u f , u i ; ϵ ≡ 4 k ) assumed mω 2 ≡ − 8 E to the Green’s function of the MIC-Kepler problem G + ( x f , x i ; E ) or G − ( x f , x i ; E ) by an S 1 action. G + ( x f , x i ; E ) and G − ( x f , x i ; E ) denote the Green’s functions in the following local coordinates τ + and τ − respectively. τ + : π − 1 ( U + ) ∋ u �− → ( π ( u ) , φ + ( u )) = ( x ( r, θ, ϕ ) , exp( iν/ 2)) ∈ U + × S 1  u 1 = √ r cos θ 2 cos ν + ϕ , u 2 = √ r cos θ 2 sin ν + ϕ    x = r sin θ cos ϕ   2 2     y = r sin θ sin ϕ , u 3 = √ r sin θ 2 cos ν − ϕ , u 4 = √ r sin θ 2 sin ν − ϕ z = r cos θ        2 2  where U + = R 3 ; r > 0 , 0 ≤ θ < π , 0 ≤ ϕ < 2 π { } x ( r, θ, ϕ ) ∈ ˙ , 0 ≤ ν < 4 π . ← − 11

τ − : π − 1 ( U − ) ∋ u �− ν/ 2)) ∈ U − × S 1 r, ˜ θ, ˜ → ( π ( u ) , φ − ( u )) = ( x (˜ ϕ ) , exp( i ˜ ˜ ν + ˜ ˜ ν + ˜  √ √ 2 cos ˜ 2 sin ˜ θ ϕ θ ϕ  u 1 = − r sin ˜ , u 2 = − r sin ˜ r sin ˜ θ cos ˜   x = ˜ ϕ   2 2     r sin ˜ θ sin ˜ y = − ˜ ϕ , r cos ˜ √ ˜ ν + 3˜ √ ˜ ν + 3˜ z = − ˜ 2 cos ˜ θ ϕ 2 sin ˜ θ ϕ  θ     u 3 = − r cos ˜ , u 4 = − ˜ r cos    2 2 where R 3 ; ˜ U − = { } r, ˜ θ, ˜ ϕ ) ∈ ˙ r > 0 , 0 ≤ ˜ θ < π , 0 < ˜ x (˜ ϕ ≤ 2 π , 0 ≤ ˜ ν < 4 π . ← − z z x x r = ˜ r r = ˜ r θ ˜ ϕ O y O y x ˜ θ x ϕ 12

Iwai and Uwano also investigated the “quantized” system (1988) : The “quantized” conformal Kepler problem is defined as a pair ( L 2 ( R 4 ; 4 u 2 d u ) , ˆ H ) where L 2 ( R 4 ; 4 u 2 d u ) :  The Hilbert space of square integrable   complex-valued functions on R 4 ,        4 H = − ℏ 2 ∂ 2  1  − k ˆ ∑ : The Hamiltonian operator.    ∂u 2 4 u 2 u 2 2 m   j =1 j   • They introduce complex line bundles L l ( l ∈ Z ) on which the linear connection is induced from a connection on the principal R 4 → ˙ R 3 . fibre bundle π : ˙ • By an S 1 action, L 2 ( R 4 ; 4 u 2 d u ) is reduced to the Hilbert space, denoted by Γ l , of square integrable cross sections in L l over R 3 . ˙ 13