The adiabatic approximation

Transcription

1 Quantum mechanics 2 - Lecture 5 November 20, 2013

2 1 The adiabatic theorem 2 Berry s phase Nonholonomic processes Geometric phase 3 The Aharonov-Bohm effect 4 Literature

3 Contents 1 The adiabatic theorem 2 Berry s phase Nonholonomic processes Geometric phase 3 The Aharonov-Bohm effect 4 Literature

4 Adiabatic process gradual change of external conditions

5 Adiabatic process gradual change of external conditions internal time, T i external time, T e Adiabatic process T e T i

6 Basic strategy: 1 start external parameters constant 2 end external parameters slowly vary with time

7 Basic strategy: 1 start external parameters constant 2 end external parameters slowly vary with time Classical analog: L = const. T = 2π L = L(t) T = 2π L g L(t) g

8 Basic strategy: 1 start external parameters constant 2 end external parameters slowly vary with time Quantum example: r = const. electronic problem r const. ω vib

9 Theorem: the adiabatic approximation Suppose that Hamiltonian changes gradually from the initial form H i to the final form H f. If the particle was initially in the nth eigenstate of H i, it will be carried into the nth eigenstate of H f. You can learn the proof of the adiabatic theorem from Ref. [1].

10 Theorem: the adiabatic approximation Suppose that Hamiltonian changes gradually from the initial form H i to the final form H f. If the particle was initially in the nth eigenstate of H i, it will be carried into the nth eigenstate of H f. Example 1: Indiana Jones problem ψ i (x) = 2 a sin ( π a x ) ψ f (x) adiab. = 1 a sin ( π 2a x ) ψ f (x) non adiab. = ψi (x)

11 Theorem: the adiabatic approximation Suppose that Hamiltonian changes gradually from the initial form H i to the final form H f. If the particle was initially in the nth eigenstate of H i, it will be carried into the nth eigenstate of H f. Example 1: Indiana Jones problem it can be solved exactly (S. W. Doescher, M. H. Rice, Am. J. Phys. 37, 1246 (1969)): 2 ( nπ ) Φ n(x, t) = w sin w x e i(mvx2 2En i at)/2 w, where w(t) = a + vt and E i n = n 2 π 2 2 /2ma 2 general solution: Ψ(x, t) = c nφ n(x, t), c n = 2 π n=1 π α = mva/2π 2 measures the wall velocity 0 e iαz2 sin(nz) sin(z)dz

12 Example 1: Indiana Jones problem (cont.) w(t e) = 2a = T e = a/v T i = 2π ω = 2π E 1 = 4ma2 π Adiabatic T e T i a v 4ma2 π c n = 2 π π 0 α 1 sin(nz) sin(z)dz = δ n1

13 Example 1: Indiana Jones problem (cont.) w(t e) = 2a = T e = a/v T i = 2π ω = 2π E 1 = 4ma2 π = Ψ(x, t) = Adiabatic T e T i a v 4ma2 π π α 1 c n = 2 π 0 2 ( πx ) w sin e i(mvx2 2E1 i at)/2 w w sin(nz) sin(z)dz = δ n1 ground state of the instantaneous well of width w (apart from the phase factor)

14 Example 1: Indiana Jones problem (cont.) w(t e) = 2a = T e = a/v T i = 2π ω = 2π E 1 = 4ma2 π = Ψ(x, t) = Adiabatic T e T i a v 4ma2 π π α 1 c n = 2 π 0 2 ( πx ) w sin e i(mvx2 2E1 i at)/2 w w mvx 2 2 w < mva2 2 a = mva 2 1 sin(nz) sin(z)dz = δ n1

15 Example 1: Indiana Jones problem (cont.) w(t e) = 2a = T e = a/v T i = 2π ω = 2π E 1 = 4ma2 π = Ψ(x, t) = Adiabatic T e T i a v 4ma2 π π α 1 sin(nz) sin(z)dz = δ n1 c n = 2 π 0 2 ( πx ) w sin e i(mvx2 2E1 i at)/2 w w mvx 2 2 w < mva2 2 a = Ψ(x, t) = = mva ( πx w sin w ) e E i 1 at/ w

16 Example 1: Indiana Jones problem (cont.) Since phase factor in Ψ(x, t), θ(t) = π2 t 2maw 2 ( πx ) = Ψ(x, t) = w sin e iθ w

17 Example 1: Indiana Jones problem (cont.) Since phase factor in Ψ(x, t), θ(t) = π2 t 2maw 2 ( πx ) = Ψ(x, t) = w sin e iθ w for fixed well Ψ(x, t) = Ψ 1(x)e ie 1t/ for (adiabatically) expanding well: a w θ 1 t E 1 (t )dt 0

18 Example 2: Electron in magnetic field B(t) = B 0 [sin α cos(ωt)î + sin α sin(ωt)ĵ ] + cos αˆk

19 Example 2: Electron in magnetic field B(t) = B 0 [sin α cos(ωt)î + sin α sin(ωt)ĵ ] + cos αˆk The Hamiltonian: H(t) = e B m S = e B0 [sin α cos(ωt)σx + sin α sin(ωt)σy + cos ασz] 2m ( ) = ω1 cos α e iωt sin α 2 e iωt sin α cos α ω 1 = eb0 m

20 Example 2: Electron in magnetic field (cont.) normalized eigenspinors of H(t) (spins along the instantaneous direction of B(t)): ) ) χ +(t) = ( cos(α/2) e iωt sin(α/2) corresponding eigenvalues:, χ (t) = E ± = ± ω1 2 ( e iωt sin(α/2) cos(α/2)

21 Example 2: Electron in magnetic field (cont.) suppose electron starts with spin up along B(0): ( ) cos(α/2) χ(0) = sin(α/2) the exact solution of time-dependent S.E.: [ ( ) λt ω1 ω cos α χ(t) = cos i sin 2 λ [ ( ω λt +i sin α sin λ 2 λ = ω 2 + ω1 2 2ωω1 cos α transition probability to spin down χ(t) χ (t) 2 = )] e +iωt/2 χ (t), [ ω ω 1 sin α sin ( )] λt e iωt/2 χ +(t) 2 ( )] 2 λt 2

22 Example 2: Electron in magnetic field (cont.) the exact solution of time-dependent S.E.: [ ( ) λt ω1 ω cos α χ(t) = cos i sin 2 λ [ ( ω λt +i sin α sin λ 2 where λ = ω 2 + ω1 2 2ωω1 cos α transition probability to spin down χ(t) χ (t) 2 = )] e +iωt/2 χ (t), [ ω sin α sin λ ( )] λt e iωt/2 χ +(t) 2 ( )] 2 λt 2 A question What do you expect for this probability from the adiabatic theorem?

23 Example 2: Electron in magnetic field (cont.) T e characteristic time for changes in H(t) 1/ω T i characteristic time for changes in χ(t) 1/ω 1

24 Example 2: Electron in magnetic field (cont.) T e characteristic time for changes in H(t) (1/ω) T i characteristic time for changes in χ(t) (1/ω 1) 1 adiabatic regime T e T i = ω ω 1 = λ ω 1 = [ ( )] 2 χ(t) χ (t) 2 ω λt = sin α sin 0 ω 1 2

25 Example 2: Electron in magnetic field (cont.) T e characteristic time for changes in H(t) (1/ω) T i characteristic time for changes in χ(t) (1/ω 1) 1 adiabatic regime T e T i = ω ω 1 = λ ω 1 = χ(t) χ [ ( )] 2 (t) 2 ω λt = sin α sin 0 ω non-adiabatic regime ω ω 1 = λ ω = [ ( χ(t) χ (t) 2 ωt )] 2 = sin α sin 2

26 Example 2: Electron in magnetic field (cont.) 2 non-adiabatic regime ω ω 1 = λ ω = [ ( χ(t) χ (t) 2 ωt )] 2 = sin α sin 2

27 Contents 1 The adiabatic theorem 2 Berry s phase Nonholonomic processes Geometric phase 3 The Aharonov-Bohm effect 4 Literature

28 Nonholonomic processes A question What do you expect that happens with the plane of swinging?

29 Nonholonomic processes A question What do you expect that happens with the plane of swinging? Θ = A/R 2 = Ω Area A doesn t depend on the path shape.

30 Nonholonomic processes Equivalent problem is Foucault pendulum Ω = sin θdθdφ = 2π(1 cos θ 0)

31 Nonholonomic processes Equivalent problem is Foucault pendulum Ω = sin θdθdφ = 2π(1 cos θ 0) nonholonomic processes

32 Nonholonomic processes Equivalent problem is Foucault pendulum Ω = sin θdθdφ = 2π(1 cos θ 0) Problem we would like to answer How does the final state differ from the initial state, if the parameters in the Hamiltonian are carried adiabatically around some closed cycle?

33 Geometric phase Let us repeat: nth of H(0) nth of H(t) = where Ψ n(t) = e i[θn(t)+γn(t)] ψ n(t), θ n(t) = 1 γ n(t) = i t 0 t 0 E n(t )dt ψ n(t ) t ψn(t ) dt dynamic phase geometric phase

34 Geometric phase 1 one parameter R(t) = ψ n = ψ n(t), H = H(t) Rf ψ n = γ n(t) = i ψ n dr R i R Questions 1 What would R(t) be for the square potential well? 2 What s the value of γ n if the Hamiltonian returns to its initial form after time T?

35 Geometric phase 1 one parameter R(t) = ψ n = ψ n(t), H = H(t) Rf ψ n = γ n(t) = i ψ n dr R i R 2 N parameters R 1(t),..., R N (t) Rf = γ n(t) = i ψn R ψ n dr Ri Now, if H returns to its initial form after T : γ n(t ) = i ψn R ψ n d R 0

36 Geometric phase 1 one parameter R(t) = ψ n = ψ n(t), H = H(t) Rf ψ n = γ n(t) = i ψ n dr R i R 2 N parameters R 1(t),..., R N (t) Rf = γ n(t) = i ψn R ψ n dr Ri Now, if H returns to its initial form after T : ψn γ n(t ) = i R ψ n dr 0 Berry s phase M. V. Berry, Proc. R. Soc. Lond. A 392, 45 (1984).

37 Geometric phase 1 one parameter R(t) = ψ n = ψ n(t), H = H(t) Rf ψ n = γ n(t) = i ψ n dr R i R 2 N parameters R 1(t),..., R N (t) Rf = γ n(t) = i ψn R ψ n dr Ri Now, if H returns to its initial form after T : γ n(t ) = i ψn R ψ n d R 0 Berry s phase Berry s phase - depends only on the path taken dynamic phase - depends on the time elapsed

38 Geometric phase A question Do these phase factors have any physical significance?

39 Geometric phase A question Do these phase factors have any physical significance?

40 Geometric phase A question Do these phase factors have any physical significance? Ψ = 1 2 Ψ Ψ0eiΓ ) Ψ 2 = Ψ 0 2 cos 2 ( Γ 2

41 Geometric phase A question Do these phase factors have any physical significance? Ψ = 1 2 Ψ Ψ0eiΓ ( ) Ψ 2 = Ψ 0 2 cos 2 Γ chance for measuring Γ 2

42 Geometric phase Example 3: Electron in a magnetic field First, refer to Example 2 in adiabatic regime ω ω 1: λ = ω ω ( ) 2 ( ω cos α + ω 1 1 ω ) cos α = ω 1 ω cos α ω 1 ω 1 ω 1

43 Geometric phase Example 3: Electron in a magnetic field First, refer to Example 2 in adiabatic regime ω ω 1: λ = ω ω ( ) 2 ( ω cos α + ω 1 1 ω ) cos α = ω 1 ω cos α ω 1 ω 1 ω 1 Solution is then χ(t) e iω1t/2 e i(ω cos α)t/2 e iωt/2 χ +(t) [ ω ( ωt ) ] +i sin α sin e +iωt/2 χ (t) ω 1 2 A question What do we get for χ(t) when ω/ω 1 0?

44 Geometric phase Example 3: Electron in a magnetic field First, refer to Example 2 in adiabatic regime ω ω 1: λ = ω ω ( ) 2 ( ω cos α + ω 1 1 ω ) cos α = ω 1 ω cos α ω 1 ω 1 ω 1 Solution is then χ(t) e iω1t/2 e i(ω cos α)t/2 e iωt/2 χ +(t) [ ω ( ωt ) ] +i sin α sin e +iωt/2 χ (t) ω 1 2 dynamic phase θ +(t) = ω1t 2 geometric phase γ +(t) = (cos α 1) ωt 2 Berry s phase γ +(T ) = π(cos α 1)

45 Geometric phase Example 3: Electron in a magnetic field (cont.) Spin up: χ + = ( cos(θ/2) e iφ sin(θ/2) θ = θ(t), φ = φ(t) ),

46 Geometric phase Example 3: Electron in a magnetic field (cont.) Spin up: χ + = ( cos(θ/2) e iφ sin(θ/2) θ = θ(t), φ = φ(t) ), χ + χ + = i sin2 (θ/2) ˆφ r sin θ

47 Geometric phase Example 3: Electron in a magnetic field (cont.) Spin up: χ + = ( cos(θ/2) e iφ sin(θ/2) θ = θ(t), φ = φ(t) ), χ + χ + = i sin2 (θ/2) ˆφ r sin θ To apply the Stoke s theorem we need χ + χ + = i 2r 2 ˆr

48 Geometric phase Example 3: Electron in a magnetic field (cont.) Spin up: χ + = ( cos(θ/2) e iφ sin(θ/2) θ = θ(t), φ = φ(t) ), χ + χ + = i sin2 (θ/2) ˆφ r sin θ To apply the Stoke s theorem we need χ + χ + = i 2r 2 ˆr γ +(T ) = 1 1 ˆr d a 2 r 2 = d a = r 2 dωˆr = 1 2 Ω

49 Contents 1 The adiabatic theorem 2 Berry s phase Nonholonomic processes Geometric phase 3 The Aharonov-Bohm effect 4 Literature

50 Classical electrodynamics: E = ϕ A t, B = A, ϕ ϕ Λ t, A A + Λ

51 Classical electrodynamics: Quantum mechanics: E = ϕ A t, B = A, ϕ ϕ Λ t, A A + Λ H = 1 ( ) 2 2m i qa + qϕ

52 Aharonov & Bohm (1959) first predicted by Ehrenberg & Siday (1949) first experiment by Chambers (1960)

53 r < a B 0 r > a B = 0, A = Φ - magnetic flux Φ 2rπ ˆφ

54 r < a B 0 r > a B = 0, A = Φ - magnetic flux Φ 2rπ ˆφ H = 1 ( 2 + q 2 A 2 + 2i qa 2m )

55 E n = H = 1 ( 2 + q 2 A 2 + 2i qa 2m ) 2 2mb 2 ( n qφ ) 2, n = 0, ±1, ±2,... 2π

56 E n = H = 1 ( 2 + q 2 A 2 + 2i qa 2m ) 2 2mb 2 ( n qφ ) 2, n = 0, ±1, ±2,... 2π + E n = E n(φ), although Φ at r=b = 0

57 Thougth experiment Beam phase g q A d r = ± qφ 2

58 Thougth experiment Beam phase g q Phase difference A d r = ± qφ 2 = g = qφ measurable interference

59 Thougth experiment Beam phase g q Phase difference A d r = ± qφ 2 = g = qφ measurable interference Berry showed that: γ n = qφ = g There are electromagnetic effects where fields are zero.

60 Contents 1 The adiabatic theorem 2 Berry s phase Nonholonomic processes Geometric phase 3 The Aharonov-Bohm effect 4 Literature

61 Literature 1 D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., Pearson Education, Inc., Upper Saddle River, NJ, Y. Aharonov, D. Bohm, Phys. Rev. 115, 485 (1959). 3 R. G. Chambers, Phys. Rev. Lett. 5, 3 (1960). 4 H. Batelaan, A. Tonomura, Phys. Today, Sept. (2009). 5 I. Supek, Teorijska fizika i struktura materije, II. dio, Školska knjiga, Zagreb, 1989.