Quantum Mechanics on Heisenberg group. Ovidiu Calin Der-Chen Chang Peter Greiner

Transcription

1 Quantum Mechanics on Heisenberg group Ovidiu Calin Der-Chen Chang Peter Greiner

2 I think I can safely say that no one undestands quantum mechanics Richard Feynman

3 A little bit of History Max Plank:the radiation is emitted in packets of energy of size E = hν, where ν is the frequency and h is Plank s constant Albert Einstein:the photoelectric effect 1924 Louis de Broglie:quantization of the momentum p = h/λ (λ = wavelength) 1925 Werner Heisenberg:Matrix mechanics 1926 Erwin Schrödinger:Wave mechanics 1927 W. Heisenberg:uncertainty principle p x > h

4 Classical oscillator 0 m= 1 A m= 1 A x m= 1 mass: m = 1 amplitude: A elastic potential: U(x) = ω2 2 x2 oscillator total energy: E = ω2 2 A2 Newton s equation: ẍ = U (x) = ω 2 x

5 Classical case Momentum: p = mv Kinetic energy: K = 1 2 mv2 = 2 1 mẋ2 = 1 p 2 2 m Potential energy: U(x) = ω2 2 x2 Total energy: E = K + U conserved m = 1 d ds E = d ds ( 1 2 mẋ2 + 1 ) 2 ω2 x 2 = ẋ(ẍ + ωx) = 0 because of Newton s equation.

6 Translating from Classical into Quantum: classical x p E quantum x = x p = ih x ih t Rewrite the conservation of energy law E = 1 2p 2 m + U(x) using the above operators yields a PDE ih t Ψ = ( mh2 2 2 ) x 2 + U(x) Ψ called the Schrödinger equation.

7 Linear Harmonic Oscillator Wave function: Ψ(x, t) = the probability that the particle is at position x at time t. Schrödinger s equation: ih Ψ t + h2 2 2 x2ψ = U(x)Ψ h = Plank constant, i = 1, U(x) = ω2 2 x2 Quantified energy: E n = (n+ 1 2 )hω, n = 0, 1, 2,... U(x) hω hω Ε 1 Ε 2 0 hω 2 Ε ο x

8 The minimum energy the linear harmonic oscillator may have is E 0 = hω 2. This is called the ground-state energy. The same quantification relation will be obtained in the Heisenberg case. The construction will be done by analogy with the elementary Quantum Mechanics construction, which uses the creation and anihilation operators. The linear harmonic oscillator describes vibrations in molecules and their counterparts in solids. The most eminent role of this oscillator is its linkage to the boson. The boson is one of the conceptual building blocks of microscopic physics, describing the models of the electromagnetic field, providing the basis of its quantization. It is named after the Indian physicist S. N. Bose.

9 Schrödinger s equation: ih Ψ t = Ĥ(Ψ), Ĥ = 1 2 ( p 2 + ω 2 x 2), p = h i d dx, x = x Uncertainty relation: [ p, x] = h i I Creation and anihilation operators: â + ω = 2h x i p, 2hω â = ω 2h x + i 2hω p.

10 They are obtained by a formal factorization of the operator hωĥ. 1 The above operators have the following properties [â, â + ] = I, (1) â â + = 1 Ĥ + 1 I, hω 2 (2) â + â = 1 Ĥ 1 I. hω 2 (3) One may find more about the linear harmonic oscillator in Messiah, Quantum Mechanics. In the next section we shall find similar relations to (1) - (3) on the Heisenberg group.

11 Heisenberg case X 1 = x1 + 2hx 2 t, X 2 = x2 2hx 1 t, where h > 0 is Plank s constant. Heisenberg operator: H = 1 2( X X 2 2 ) Non-commutation relation: [X 1, X 2 ] = 4h t The vector fields Z = 1 2 (X 1 ix 2 ), Z = 1 2 (X 1 + ix 2 ), correspond to the operators â + and â.

12 The vector fields Z and Z have similar properties to operators â + and â given above. Proposition: The operators Z and Z verify the following relations 1) [Z, Z] = 2hi t, 2) ZZ = 1 2 H + ih t, 3) ZZ = 1 2 H ih t. Corollary: The Heisenberg operator can be expressed as 1) H = 2ZZ + 2ih t = 2ZZ + [Z, Z], 2) H = 2ZZ 2ih t = 2ZZ [Z, Z].

13 ψ 0 : R 3 C ground state if Zψ 0 = 0 excited states: ψ 1 = Zψ 0, ψ n+1 = Zψ n = Z n ψ 0. Example: ψ 1 = G(h z 2 + it), G C 1 (R). Let λ = 1 2 (X2 1 + X2 2 ) λ 2 i[x 1, X 2 ] = ZZ + ZZ + λ[z, Z]. Proposition: We have λn ψ n = 0 with λ n 2 = n + 1, n = 0, 1, 2,... 2 For the values λ = 2n + 1 the operator λ ceases to be hypoelliptic. A fundamental solution for λ was constructed by Greiner, Beals Gaveau. The fundamental solution is extended meromorphically in λ.

14 The function ψ 0 : R 3 C satisfies the groundstate equation Zψ 0 = 0. (4) Let ψ 0 = u + iv, where u and v are real-valued functions. The following result is an analog of the Cauchy-Riemann equations. Proposition: ψ 0 = u + iv satisfies the groundstate equation (4) if and only if X 1 u = X 2 v, X 1 v = X 2 u. (5) The relationship between the states ψ 0 and ψ 1 is given by the following result. Proposition: Let ψ 1 = Zψ 0 be the first excited state, where ψ 0 = u + iv is the groundstate. Then 1) ψ 1 = X 1 ψ 0 = ix 2 ψ 0, 2) ψ 1 = 2Zu = 2iZv, 3) Zψ 1 = H ψ 0.

15 The operator τ Z Some non-commutativity relations aroused from the operators Z and Z will be studied. The calculus developed is a consequence of the Heisenberg non-commutativity relation [Z, Z] = i 2 [X 1, X 2 ] = 2ih t. (6) Lemma: For any integer n 1, we have [Z, Z n ] = nz n 1 [Z, Z] = n[z, Z]Z n 1. Theorem: Let f be an analytic function on R. Then [Z, f(z)] = f (Z)[Z, Z] = [Z, Z]f (Z).

16 Proposition: Let ψ 0 be a solution of the groundstate equation Zψ 0 = 0 and let f be an analytic function. 1) We have Zf(Z)ψ 0 = f (Z)[Z, Z]ψ 0 = [Z, Z]f (Z)ψ 0. 2) In general, for any n N, we have Z n f(z)ψ 0 = f (n) (Z)[Z, Z] n ψ 0 = [Z, Z] n f (n) (Z).

17 Theorem Let f be an analytic function and let v = f(z)ψ 0. Then the solution of the Cauchy problem τ u = Zu on [0, ) C u τ=0 = v, is given by u = f(z + [Z, Z]τ)ψ 0.

18 Multiplication rules [Z, Z] = 2hi t For any τ, s R we have e τz e sz ψ 0 = e sz+sτ[z,z] ψ 0. For any α, β C we have e αz e βz = e αz+βz 1 2 αβ[z,z]. is an analog of the Campbell-Hausdorff formula from the Lie groups theory. This says that if X, Y are left invariant vector fields and α, β R, then e αx e βy = e αx+βy +1 2 αβ[x,y ]+higher level brackets.

19 Fundamental solutions Ground state and excited states: Zψ 0 = 0, ψ n = Z n ψ 0 Conjugate ground state and excited states: Zψ 0 = 0, ψ n = Zn ψ 0 Proposition: 1) Let ψ 0 = G(h z 2 + it) where G C 1 (R) arbitrary. Then Zψ 0 = 0. 2) Let ψ 0 = F (h z 2 + it) where F C 1 (R) arbitrary. Then Zψ 0 = 0.

20 We shall look for a fundamental solution of H = 1 2 (X2 1 + X2 2 ) = ZZ + ZZ in the form K = Mψ, ψ = n,k 0 M n ψ n ψ n, where M is an infinite transition probability matrix, i.e., M n [0, 1]. In this way, we can take into account all possible transitions between the excited states and the excited conjugated states. Let A = h z 2 +it and A = h z 2 it and consider the ground-states ψ 0 = 1 A α, ψ 0 = 1 A α.

21 ψ n = ( 2hz) n α(α + 1)... (α + n 1) 1 A α+n, ψ n = ( 2hz) n α(α + 1)... (α + n 1) 1 A α+n. K = 1 A 2α n 0 M nn z 2n A 2n The operator H is homogeneous with respect to the dilations (z, t) (λz, λ 2 t).

22 The term z 2n A 2n = ( z 2 h 2 z 4 + t 2 is not homogeneous with respect to the above dilations, unless n = 0. Hence K = ) n M 00 A 2α = M 00 ψ 0 ψ 0. We shall determine α such that H K = 0, for (z, t) (0, 0). H (ψ 0 ψ0 ) = 2αh A 2α 2h z 2 4αh z 2 AA = 0 for α = 1 2. Hence the fundamental solution has the form K = C (AA) 1/2 = C (h 2 z 4 + t 2 ) 1/2. (Folland)

23 The fundamental solution for λ = H i 2 λ[x 1, X 2 ] Look for a fundamental solution at the origin with K(z, t; 0, 0) = Cψ 0 ψ 0, ψ 0 = 1 A α, ψ 0 = 1 A β. λ is a linear combination of ZZ and ZZ λ = ZZ + ZZ + λ[z, Z] = (1 + λ)z Z + (1 λ)zz. A straightforward computation shows = λ (ψ 0 ψ0 ) h A α+1 A β+1 8αβh z 2 2[α + β +λ(α β)]h z 2 2it[β α λ(β + α)].

24 λ (ψ 0 ψ0 ) = 0 = α = 1 λ 2, β = 1 + λ 2 The fundamental solution becomes K(z, t; 0, 0) = CA 1 λ 2 A 1+λ 2 = (Beals, Gaveau, Greiner) C (h z 2 + it) 1 λ 2 (h z 2 it) 1+λ 2.

25 The Schrödinger kernel ih u Ψ + h 2 H Ψ = 0, Ψ(x, t, u 0 ) = Ψ 0 (x, t), An attempt to solve using supper-position of waves: Ψ = j Φ j (x, t)ϕ j (u), with ϕ j (u) = e i hu E j u 2 and h 2 H Φ j = E j Φ j Classical Mechanics: Eu = g (action) Eu 2 = gu(= f) modified action Let f j = E j u 2. Then Ψ = j Φ j (x, t)e if j hu.

26 A Green function P for the operator L is a distribution on R 2 x R t with the following properties 1) u P H P = 0, for u > 0, (7) 2) lim u 0 P (x, t, u) = δ(x)δ(t), (8) where δ stands for the Dirac distribution. The Green function P (x, t, u; x 0, y 0, u 0 ) is also referred to in Quantum Mechanics as a propagator. If Ψ is a wave function, which satisfies the equation ih u Ψ + h 2 H Ψ = 0, (9) with the initial condition Ψ 0 (x, t) = Ψ(x, t, u 0 ), then the solution at any instant of time u > u 0 is given by Ψ(x, t, u) = P (x, t, u; x 0, y 0, u 0 )Ψ 0 (x 0, t 0 ) dx 0 dt 0.

27 Theorem(Greiner) The Schrödinger kernel, i.e., the kernel of e ihu H at the origin is given by P (x, t, u) = 1 (2πhiu) 2 + e if(x,t,τ) hu 2hτ sinh(2hτ) dτ, where f = τg = iτt + hτ coth(2τh)(x x2 2 ) is the complex modified action function. This shows that the propagator P depends on each subriemannain distance d j, since the function f evaluated at its critical points produces all distances d j.

28 This is a well known fact of Quantum Mechanics stated by R. Feynman. He explained the double slit experiment assuming the fact that the electron takes all possible paths between any two points. Hence each path has a contribution to the propagator, which is defined as a path integral. For macro objects, like airplanes, planets or bullets, only the classical path is important and their motion is explained in the context of Riemannian geometry. The picture for the sub-atomic particles is completely different and the correct framework to study their motion is the subriemannian geometry.

29 The Green function at any point In the general case the complex action between the points (x 0, y 0 ) and (x, y) within time τ is g = it + h coth(2hτ)( x 2 + x 0 2 ) 1 2h x x 0 sinh(2hτ). Proposition. The complex action g given by (10) satisfies the Hamilton-Jacobi equation g τ (X 1g) (X 2g) 2 = 0.

30 Theorem. The Schrödinger kernel is given by = where P (x, t, x 0, t 0, u) h (2πhiu) 2 + e if(x,t,x0,t0,τ) 0 hu τ sinh(hτ)r r 1 cosh(hτ) r 0 r +1 dτ, f = τg = iτt + τh coth(2hτ)( x 2 + x 0 2 ) 1 2τh x x 0 sinh(2hτ). is the complex modified action function. (r = x, r 0 = x )

31 Theorem: At points (0, t) with t > 0, we have the following expansion P (0, t; u) = 1 4u 2 k=1 ( 1) k+1 ke l2 k (0,t)/2u as u 0 +, where l k (0, t) is the length of the k-th geodesic connecting the origin and the point (0, t).

32 Why choose in complex Hamiltonian mechanics θ = i? Harmonic oscillator leads to the Heisenberg operator by certain complex quantization. Consider a unit mass particle under the influence of force F (x) = sx. Newton s equation is ẍ = x. This is the equation which describes the dynamics of an inverse pendulum in an unstable equilibrium, for small angle x, see Figure 5.1. The potential energy is x U(x) = F (u) du = x The Lagrangian L : T R R is the difference between the kinetic and the potential energy L(x, ẋ) = K U = 1 2ẋ x2. x

33 Figure 5.1: The inverse pendulum problem. The momentum is p = L ẋ = ẋ. The Hamiltonian associated with the above Lagrangian is obtained using the Legendre transform: H : T R R H(x, p) = pẋ L(x, ẋ) = p p2 1 2 x2 = 1 2 p2 1 2 x2. Consider the following complexification x = x 1 + ip 2, p = p 1 + ix 2. (10) Hence H : T C C and H(x, p) = 1 2 p2 1 2 x2 = 1 2 (p 1 + ix 2 ) (x 1 + ip 2 ) 2 = 1 2 (p 1 + ix 2 ) i2 (x 1 + ip 2 ) 2 = 1 2 (p 1 + ix 2 ) (ix 1 p 2 ) 2 = 1 2 (p 1 + ix 2 ) (p 2 ix 1 ) 2.

34 Replacing θ = i, H(x, p; θ) = 1 2 (p 1 θx 2 ) (p 2 + θx 1 ) 2. (11) Quantizing, i.e., making the changes p 1 x1, p 2 x2, θ t, we get H X, where X = 1 ( ) 2 1( ) 2 x1 x 2 t + x2 + x 1 t 2 2 is the Heisenberg operator. Remarks: a) If we choose the complexification x = 2x 1 + ip 2, p = p 1 + 2ix 2 then we arrive at the operator X = 1 ( ) 2 1( ) 2. x1 2x 2 t + x2 + 2x 1 t 2 2 b) The expression θ = i was used in the literature for a long time without an explanation. We can see that this value for θ is related to quantum behaviors.