Dynamics of cold molecules in external electromagnetic fields. Roman Krems University of British Columbia
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1 Dynamics of cold molecules in external electromagnetic fields Roman Krems University of British Columbia
2 UBC group: Zhiying Li Timur Tscherbul Erik Abrahamsson Sergey Alyabishev Chris Hemming Collaborations: Alexei Buchachenko - Moscow Grzegorz Chalasinski - Warsaw Alex Dalgarno - Harvard John Doyle - Harvard Gerrit Groenenboom - Nijmegen Sture Nordholm - Sweden Maria Szczesniak - Michigan
3 Our research projects Collisions of molecules in electromagnetic elds eects of static electric elds on elastic and inelastic collisions eects of eld orientation on collision dynamics collision dynamics in laser elds Feshbach resonances electric-eld-induced Feshbach resonances resonance scattering mediated by second-order couplings scattering resonances in 2D laser-induced Feshbach resonances Ultracold Chemistry non-adiabatic eects in ultracold chemical reactions chemical reactions in external elds Statistical Properties of Ultracold Molecular Gases quantum diusion in binary mixtures of molecules collisional decoherence of wavepackets in ultracold gases
4 Our research projects Collisions of molecules in electromagnetic elds eects of static electric elds on elastic and inelastic collisions eects of eld orientation on collision dynamics collision dynamics in laser elds Feshbach resonances electric-eld-induced Feshbach resonances resonance scattering mediated by second-order couplings scattering resonances in 2D laser-induced Feshbach resonances Ultracold Chemistry non-adiabatic eects in ultracold chemical reactions chemical reactions in external elds Statistical Properties of Ultracold Molecular Gases quantum diusion in binary mixtures of molecules collisional decoherence of wavepackets in ultracold gases
5 Mechanisms to control collisions with electromagnetic elds Electric-eld-induced Feshbach resonances in ultracold collisions E d
6 Mechanisms to control collisions with electromagnetic elds Electric-eld-induced Feshbach resonances in ultracold collisions E d
7 Alkali atoms in a magnetic eld 2µ 0 B(Ŝ(a) z Ĥ = 1 d 2 2µRdR 2R + ˆl 2 (θ, φ) 2µR + 2 E cos θ S γ (a) Î (a) Ŝ(a) + γ (b) Î (b) Ŝ(b) + SM S V S (R) SM S + S M S SM S d S (R) SM S + M S ( ) µ (a) + Ŝ(b) I z ) B I (a)î(a) z + µ(b) I I (b)î(b) z R
8 Alkali atoms in a magnetic eld Good quantum numbers for the atoms: f = I + S and the projection m f - if there's no magnetic eld Only m f = m I + m S - in the presence of magnetic elds When the atoms come together in a magnetic eld, the only conserved quantum number is M = m fa + m fb which is the projection of F = f a + f b R
9 Alkali atoms in a magnetic eld Good quantum numbers for the atoms: f = I + S and the projection m f - if there's no magnetic eld Only m f = m I + m S - in the presence of magnetic elds When the atoms come together in a magnetic eld, the only conserved quantum number is M = m fa + m fb which is the projection of F = f a + f b R
10 The energy of Li and Cs in the states with m fli = 1 and m fcs = 1 Potential energy (K) Magnetic field (Gauss)
11 The energy of Li and Cs Total angular momentum projection M= Potential energy (K) Magnetic field (Gauss)
12 Cross section (Å 2 ) s-wave elastic scattering Magnetic field (Gauss) p-wave elastic scattering Magnetic field (Gauss) s p transition at E=30 kv/cm Magnetic field (Gauss)
13 Cross section (Å 2 ) s-wave elastic scattering p-wave elastic scattering Magnetic field (Gauss) s p transition at E=30 kv/cm Magnetic field (Gauss)
14 Alkali atoms in a magnetic eld 2µ 0 B(Ŝ(a) z Ĥ = 1 d 2 2µRdR 2R + ˆl 2 (θ, φ) 2µR + 2 E cos θ S γ (a) Î (a) Ŝ(a) + γ (b) Î (b) Ŝ(b) + SM S V S (R) SM S + S M S SM S d S (R) SM S + M S ( ) µ (a) + Ŝ(b) I z ) B I (a)î(a) z + µ(b) I I (b)î(b) z R
15 Alkali atoms in electric and magnetic elds 2µ 0 B(Ŝ(a) z Ĥ = 1 d 2 2µRdR 2R + ˆl 2 (θ, φ) 2µR + 2 E cos θ S γ (a) Î (a) Ŝ(a) + γ (b) Î (b) Ŝ(b) + SM S V S (R) SM S + S M S SM S d S (R) SM S + M S ( ) µ (a) + Ŝ(b) I z ) B I (a)î(a) z + µ(b) I I (b)î(b) z R
16 Cross section (Å 2 ) s-wave elastic scattering p-wave elastic scattering Magnetic field (Gauss) s p transition at E=30 kv/cm Magnetic field (Gauss)
17 Magnetic field (Gauss) s-wave elastic scattering Cross section (Å 2 ) p-wave elastic scattering 10 3 s p transition at E=30 kv/cm
18 Cross section (Å 2 ) Zero electric field Magnetic field (Gauss)
19 Cross section (Å 2 ) Zero electric field kv/cm Magnetic field (Gauss)
20 10 8 PRL 96, (2006) Cross section (Å 2 ) Electric Field (kv/cm)
21 Why electric elds for Feshbach resonances?
22 Why electric elds for Feshbach resonances? Electric elds can be tuned much faster than magnetic elds -Faster control over atom - atom interactions
23 Why electric elds for Feshbach resonances? Electric elds can be tuned much faster than magnetic elds -Faster control over atom - atom interactions Structure of the separated atoms is not modied -Fewer decoherence issues
24 Why electric elds for Feshbach resonances? Electric elds can be tuned much faster than magnetic elds -Faster control over atom - atom interactions Structure of the separated atoms is not modied -Fewer decoherence issues Novel three-state resonances -Two-dimensional control
25 Why electric elds for Feshbach resonances? Electric elds can be tuned much faster than magnetic elds -Faster control over atom - atom interactions Structure of the separated atoms is not modied -Fewer decoherence issues Novel three-state resonances -Two-dimensional control Anisotropic scattering at ultracold temperatures - New interesting physics?
26 Three-state Feshbach resonances The coupled Schrödinger equations for our system are (E H αα ) α + = V αβ β + (E H ββ ) β + = V βα α + + V βγ γ + (E H γγ ) γ + = V γβ β +
27 Three-state Feshbach resonances The coupled Schrödinger equations for our system are (E H αα ) α + = V αβ β + (E H ββ ) β + = V βα α + + V βγ γ + (E H γγ ) γ + = V γβ β + Divide the system into the open P -channel (the α- and β- channels together) - and the closed M-channel (the γ-channel).
28 The background scattering solution, for the isolated P-channel without coupling to the M-channel, is obtained from P P Ψ + = Ψ ( + P α = + ) β + (E H P P ) φ + P = 0 which is a system of two equations: (E H αα ) α + BG = V αβ β + BG (E H ββ ) β + BG = V βα α + BG
29 (E H αα ) α + BG = V αβ β + BG (E H ββ ) β + BG = V βα α + BG which can be re-written as ( E H αα V αβ 1 ) E + V H βα α BG + = 0 ββ ) β BG + = 0. ( E H ββ V βα 1 E + H αα V αβ Again, it's the same as (E H P P ) φ + P = 0
30 (E H αα ) α + BG = V αβ β + BG (E H ββ ) β + BG = V βα α + BG which can be re-written as ( E H αα V αβ 1 ) E + V H βα α BG + = 0 ββ ) β BG + = 0. ( E H ββ V βα 1 E + H αα V αβ Again, it's the same as (E H P P ) φ + P = 0 what is the inverse of the operator (E H P P )??
31 1 E + H P P = ( G + α(β) G + β(α) V βαg + α G + α(β) V αβg + β G + β(α) )
32 1 E + H P P = ( G + α(β) G + β(α) V βαg + α G + α(β) V αβg + β G + β(α) ) with the Green's functions dened as G + λ 1 E + H λλ V + λµν V λµg + µ V µν G + λ(µ) 1 E + H λλ V λµλ
33 1 E + H P P = ( G + α(β) G + β(α) V βαg + α G + α(β) V αβg + β G + β(α) ) with the Green's functions dened as G + λ 1 E + H λλ V + λµν V λµg + µ V µν G + λ(µ) 1 E + H λλ V λµλ This can be used to dene the scattering state in the open channels: Ψ + P = φ+ P E + H H P M H P P E H MM H 1 HP φ + P MP E + H H P P P M
34 If we know the scattering state, we know the T -matrix elements: T = χ P V 1 P P H P M H E H MM H MP (E + H P P ) 1 MP H P M Ψ+ P
35 If we know the scattering state, we know the T -matrix elements: T = χ P V 1 P P H P M H E H MM H MP (E + H P P ) 1 MP H P M Ψ+ P from where we obtain the resonant part of the scattering matrix element S r iγ = E E r + iγ 2 with Γ = 2π ( ) 1 ( ) φ m V γβ V E + H ββ V βα G + βα 1 + G + 2 α (E)V αα α 0 (E) α(e)v αβ and +P.V. 0 φ m V γβ βi BG 2 + E E i E r = ɛ m + i ( φ m V γβ de 1 E + H ββ V βα G + α (E )V αβ V βα (1 + G + α(e )V αα ) E E ) α 0 (E ) 2
36 Three-state Feshbach resonances (one scattering and two bound states)
37 Three-state Feshbach resonances (one scattering and two bound states) The coupled Schrödinger equations for our system are (E H αα ) α+ = V αβ β (E H ββ ) β = V βα α+ + V βγ γ (E H γγ ) γ = V γβ β.
38 Three-state Feshbach resonances (one scattering and two bound states) The coupled Schrödinger equations for our system are or (E H αα ) α+ = V αβ β (E H ββ ) β = V βα α+ + V βγ γ (E H γγ ) γ = V γβ β. γ = G γ V γβ β (E H ββ V βγ G γ V γβ ) β = V βα α+ So we have eliminated the γ state.
39 Inverting the last equation, gives: β = G β(γ) V βα α+ (E H αα V αβ G β(γ) V βα ) α+ = 0
40 Inverting the last equation, gives: β = G β(γ) V βα α+ (E H αα V αβ G β(γ) V βα ) α+ = 0 So the eective potential experienced in the α-channel, V eff = V αα + V αβ G β(γ) V βα
41 Inverting the last equation, gives: β = G β(γ) V βα α+ (E H αα V αβ G β(γ) V βα ) α+ = 0 So the eective potential experienced in the α-channel, V eff = V αα + V αβ G β(γ) V βα We can re-write the equation for α+ as (E H αα ) α+ = V αβ G β(γ) V βα α+
42 Inverting the last equation, gives: β = G β(γ) V βα α+ (E H αα V αβ G β(γ) V βα ) α+ = 0 So the eective potential experienced in the α-channel, V eff = V αα + V αβ G β(γ) V βα We can re-write the equation for α+ as (E H αα ) α+ = V αβ G β(γ) V βα α+ Multiplying on the left by G + α then gives α+ = φ + α + G + αv αβ G β(γ) V βα α+ where φ + α = a scattering state for the uncoupled α-channel: (E H αα ) φ + α = 0.
43 What is the free scattering state in the α-channel?
44 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0
45 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0 So we can write the φ + α in terms of the free state in the α-channel as φ + α = α 0 + G + αv αα α 0
46 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0 So we can write the φ + α in terms of the free state in the α-channel as φ + α = α 0 + G + αv αα α 0 The T-matrix element for α α scattering is T = α 0 V eff α+ = α 0 V αα α+ + α 0 V αβ G β(γ) V βα α+
47 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0 So we can write the φ + α in terms of the free state in the α-channel as φ + α = α 0 + G + αv αα α 0 The T-matrix element for α α scattering is or T = α 0 V eff α+ = α 0 V αα α+ + α 0 V αβ G β(γ) V βα α+ T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+
48 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0 So we can write the φ + α in terms of the free state in the α-channel as φ + α = α 0 + G + αv αα α 0 The T-matrix element for α α scattering is or T = α 0 V eff α+ = α 0 V αα α+ + α 0 V αβ G β(γ) V βα α+ T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ This is a very interesting expression!
49 What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0 So we can write the φ + α in terms of the free state in the α-channel as φ + α = α 0 + G + αv αα α 0 The T-matrix element for α α scattering is or T = α 0 V eff α+ = α 0 V αα α+ + α 0 V αβ G β(γ) V βα α+ T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ This is a very interesting expression! Let's see what happens to this expression as the total energy approaches the bound state β or the bound state γ:
50 T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ Let's see what happens to this expression as the total energy approaches the bound state β or the bound state γ.
51 T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ Let's see what happens to this expression as the total energy approaches the bound state β or the bound state γ. Remember the denitions: G + λ 1 E + H λλ V + λµν V λµg + µ V µν G + λ(µ) 1 E + H λλ V λµλ
52 T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ Let's see what happens to this expression as the total energy approaches the bound state β or the bound state γ. Remember the denitions: G + λ 1 E + H λλ V + λµν V λµg + µ V µν G + λ(µ) 1 E + H λλ V λµλ What happens to the Green's function G β(γ) as E ɛ γ?
53 What happens to the Green's function G β(γ) as E ɛ γ?
54 What happens to the Green's function G β(γ) as E ɛ γ? G β (γ) β 0 β 0 E ɛ β β 0 V βγ G γ V γβ β 0
55 What happens to the Green's function G β(γ) as E ɛ γ? G β (γ) β 0 β 0 E ɛ β β 0 V βγ G γ V γβ β 0 G γ γ 0 γ 0 E ɛ γ
56 What happens to the Green's function G β(γ) as E ɛ γ? G β (γ) β 0 β 0 E ɛ β β 0 V βγ G γ V γβ β 0 G γ γ 0 γ 0 E ɛ γ So, when E ɛ γ, G γ has a pole and the second and third terms in the expression T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ vanish.
57 What happens to the Green's function G β(γ) as E ɛ γ? G β (γ) β 0 β 0 E ɛ β β 0 V βγ G γ V γβ β 0 G γ γ 0 γ 0 E ɛ γ So, when E ɛ γ, G γ has a pole and the second and third terms in the expression T = α 0 V αα φ + α + α 0 V αα G + αv αβ G β(γ) V βα α+ + α 0 V αβ G β(γ) V βα α+ vanish. So, there is no resonant scattering!!!!
58 With some algebra, we can re-write the expression for the T -matrix as with T = T BG + Γ 2π(E ɛ β + iγ 2 ) + ΓA 2π(E ɛ β + iγ 2 ) ( (E ɛ γ )(E ɛ β + iγ 2 ) A) Γ 2 Im β 0 V βα G + αv αβ β 0 = 2π β 0 V βα φ + α 2 A β 0 V βγ γ 0 2 and (E) = Re β 0 V βα G + αv αβ β 0 = P.V. 0 de β V βα φ + α(e ) 2 E E
59 Let's see what happens to the expression T r = Γ 2π(E ɛ β + iγ 2 ) + ΓA 2π(E ɛ β + iγ 2 ) ( (E ɛ γ )(E ɛ β + iγ 2 ) A) as E ɛ β and E ɛ γ.
60 Open-shell molecules - alignment by magnetic fields
61 Magnetic trap
62 Trap loss...
63 How do electric fields affect spin relaxation? Induce couplings between the rotational levels ( N = 1) Increase the energy gap between the rotational levels R. V. Krems, A.Dalgarno, N.Balakrishnan, and G.C. Groenenboom, PRA 67, (R) (2003)
64 Spin relaxation is suppressed
65 Enhancement of spin relaxation First-order Stark effect T. V. Tscherbul and R.V. Krems, PRL 97, (2006)
66 Cold collisions in crossed fields
67 Collisions of molecules in a microwave cavity Molecular Hamiltonian: H mol = BN 2 Field Hamiltonian: H f = ω(aa n) Molecule - Field Hamiltonian: H mol,f = µ ω 2ɛ 0 V cos θ ( a + a ) Basis set: NM N n The matrix elements of of the molecule - eld Hamiltonian: n NM N H mol,f N M N n NM N cos θ N M N ( ) δ n,n +1 + δ n,n 1 NM N cos θ N M N δ M N,M N ( ) δ N,N +1 + δ N,N 1
68 30 Energy levels of a diatomic molecule in a laser field 21 photon number states, 5 molecular states Energy, cm Field strength, units of rotational constant
69 Energy levels of a diatomic molecule in a laser field 21 photon number states, 5 molecular states 2 1 Energy, cm-1 0 N=1 n = n - 2 N=0 n = n -1 N=0 n = n Field strength, units of rotational constant
70 References R. V. Krems, PRL 96, (2006). T. V. Tscherbul and R. V. Krems, PRL 97, (2006). T. V. Tscherbul and R. V. Krems, JCP 125, (2006). R. V. Krems et al., PRL 94, (2005). C. I. Hancox et al., PRL 94, (2005). R. V. Krems, PRL 93, (2004). Reviews R. V. Krems, Int. Rev. Phys. Chem. 24, 99 (2005). J. Doyle, B. Friedrich, R. V. Krems, and F. Masnou-Seeuws, Eur. Phys. J. D 31, 149 (2004).
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