Dynamics of cold molecules in external electromagnetic fields. Roman Krems University of British Columbia

Dynamics of cold molecules in external electromagnetic fields Roman Krems University of British Columbia

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

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

Mechanisms to control collisions with electromagnetic elds Electric-eld-induced Feshbach resonances in ultracold collisions E d

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

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

The energy of Li and Cs in the states 1 0.8 with m fli = 1 and m fcs = 1 Potential energy (K) 0.6 0.4 0.2 0-0.2-0.4-0.6-0.8-1 0 1000 2000 3000 4000 5000 6000 7000 Magnetic field (Gauss)

The energy of Li and Cs Total angular momentum projection M=2 1 0.8 0.6 Potential energy (K) 0.4 0.2 0-0.2-0.4-0.6-0.8-1 0 1000 2000 3000 4000 5000 6000 7000 Magnetic field (Gauss)

Cross section (Å 2 ) 10 8 10 6 10 4 10 2 10 0 s-wave elastic scattering 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss) p-wave elastic scattering 10 2 10 0 10-2 10-4 10-6 10-8 10 3 10 0 10-3 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss) s p transition at E=30 kv/cm 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss)

Cross section (Å 2 ) 10 8 10 6 10 4 10 2 10 0 10 2 10 0 10-2 10-4 10-6 10-8 s-wave elastic scattering p-wave elastic scattering 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss) 10 3 10 0 10-3 s p transition at E=30 kv/cm 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss)

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

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

Cross section (Å 2 ) 10 8 10 6 10 4 10 2 10 0 10 2 10 0 10-2 10-4 10-6 10-8 s-wave elastic scattering p-wave elastic scattering 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss) 10 3 10 0 10-3 s p transition at E=30 kv/cm 1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss)

1500 2000 2500 3000 3500 4000 4500 Magnetic field (Gauss) 10 8 10 6 s-wave elastic scattering 10 4 10 2 Cross section (Å 2 ) 10 0 10 2 10 0 10-2 10-4 10-6 10-8 p-wave elastic scattering 10 3 s p transition at E=30 kv/cm 10 0 10-3

Cross section (Å 2 ) 10 3 10 0 10-3 Zero electric field 1600 1800 2000 2200 Magnetic field (Gauss) 10 6 10 3 10 0 10-3

Cross section (Å 2 ) 10 3 10 0 10-3 10 6 Zero electric field 10 3 10 0 100 kv/cm 10-3 1600 1800 2000 2200 Magnetic field (Gauss)

10 8 PRL 96, 123202 (2006) Cross section (Å 2 ) 10 6 10 4 10 2 70 80 90 100 110 Electric Field (kv/cm)

Why electric elds for Feshbach resonances?

Why electric elds for Feshbach resonances? Electric elds can be tuned much faster than magnetic elds -Faster control over atom - atom interactions

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

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

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?

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 +

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).

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

(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

(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 )??

1 E + H P P = ( G + α(β) G + β(α) V βαg + α G + α(β) V αβg + β G + β(α) )

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

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 + 1 1 + E + H H P M H P P E H MM H 1 HP φ + P MP E + H H P P P M

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

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

Three-state Feshbach resonances (one scattering and two bound states)

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.

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.

Inverting the last equation, gives: β = G β(γ) V βα α+ (E H αα V αβ G β(γ) V βα ) α+ = 0

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

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+

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.

What is the free scattering state in the α-channel?

What is the free scattering state in the α-channel? It is the solution to (E K αα ) α 0 = 0

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

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+

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+

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!

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:

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.

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

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?

What happens to the Green's function G β(γ) as E ɛ γ?

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

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

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.

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!!!!

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

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.

Open-shell molecules - alignment by magnetic fields

Magnetic trap

Trap loss...

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, 060703(R) (2003)

Spin relaxation is suppressed

Enhancement of spin relaxation First-order Stark effect T. V. Tscherbul and R.V. Krems, PRL 97, 083201 (2006)

Cold collisions in crossed fields

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

30 Energy levels of a diatomic molecule in a laser field 21 photon number states, 5 molecular states 25 20 Energy, cm-1 15 10 5 0-5 -10 0 0.5 1 1.5 2 Field strength, units of rotational constant

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 - 1-2 0 0.5 1 1.5 2 Field strength, units of rotational constant

References R. V. Krems, PRL 96, 123202 (2006). T. V. Tscherbul and R. V. Krems, PRL 97, 083201 (2006). T. V. Tscherbul and R. V. Krems, JCP 125, 194311 (2006). R. V. Krems et al., PRL 94, 013202 (2005). C. I. Hancox et al., PRL 94, 013201 (2005). R. V. Krems, PRL 93, 013201 (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).