We have already demonstrated polarization of a singular nanodiamond (or bulk diamond) via Nitrogen-Vacancy (NV) centers 1

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1 We have already demonstrated polarization of a singular nanodiamond (or bulk diamond) via Nitrogen-Vacancy (NV) centers 1 Flip-flops Bath narrowing Experiment Source Power (dbm) Locking time ( sec) Simulation Rabi Frequency (MHz) Locking time ( sec) Fourier transform amplitude (b) Polarized nuclei (a) Unpolarized nuclei Frequency (MHz) 1 London, P.; Scheuer, J.; Cai, J. M.; Schwarz, I.; Retzker, A.; Plenio, M. B.; Katagiri, M.; Teraji, T.; Koizumi, S.; Isoya, J.; Fischer, R.; McGuinness, L. P.; Naydenov, B.; Jelezko, F., "Detecting and Polarizing Nuclear Spins with Double Resonance on a Single Electron Spin." Phys Rev Lett 2013, 111 (6). 0

2 Problem to be solved polarizing ND dust with random NV orientations However, when generalizing for nanodiamond (ND) dust, one encounters an additional problem how to account for the random ND (and therefore NV) orientation B 1 London, P.; Scheuer, J.; Cai, J. M.; Schwarz, I.; Retzker, A.; Plenio, M. B.; Katagiri, M.; Teraji, T.; Koizumi, S.; Isoya, J.; Fischer, R.; McGuinness, L. P.; Naydenov, B.; Jelezko, F., "Detecting and Polarizing Nuclear Spins with Double Resonance on a Single Electron Spin." Phys Rev Lett 2013, 111 (6). 1

3 Hamiltonian The Hamiltonian of an NV spin and a single nuclear spin, written in the principle axes of the NV center is: H = SD S + γ e Bcos θ S z + γ e Bsin θ S x + I θ + SA I In the following protocols we will focus on the weak-field regime (γ e B D). Therefore, the eigenstates of the NV spin will depend on the principal axes of the NV center (and not the magnetic field direction) D θ B 2

4 Hamiltonian H = SD S + γ e Bcos θ S z + γ e Bsin θ S x + I θ + SA I S=1 for the NV spin. However, in all following protocols, we will center on the transition between the states 0> and one of { 1>, - 1>} The following protocols will correspond to two regimes of interest: A the coupling between the NV and nuclear spins is larger than the Zeeman splitting A 1 > 1 > 0 > 3

5 Strong coupling regime (A ) H = SD S + γ e B cos θ S z + γ e B sin θ S x + I θ + SA I We introduce two π pulse external drivings, with frequencies as depicted in the illustration: 1. MW driving Ω mw cos(ω mw t) S x 2. RF driving Ω rf cos(ω rf t) S x The goal is to produce CNOT gates between the NV spin and nuclear spin to produce the polarization transfer 1 > 0 > ω e = D + γ e B cos θ A Ω mw cos[(ω e +A)t] Ω rf cos(at) 4

6 Strong coupling regime (A ) H = SD S + γ e B cos θ S z + γ e B sin θ S x + I θ + SA I Denote ψ, φ > the two system state, where ψ > is the state of the NV spin and φ > of the nuclear spin Effect of MW driving: a, b > a + b, b > (a, b 0,1 ) Effect of RF driving: a, b > a, a + b > (a, b 0,1 ) 1 > 0 > ω e = D + γ e B cos θ A Ω mw cos[(ω e +A)t] Ω rf cos(at) 5

7 Strong coupling regime protocol H = SD S + γ e B cos θ S z + γ e B sin θ S x + I θ + SA I The protocol is then as follows: Initialize the NV spin via optical pumping MW π pulse RF π pulse a 0 > +b 1 > c > +d +> NV init c 0, > +d 0,+> MW RF c 0, > +d 1,+> c 0, > +d 1, >= (c 0 > +d 1 >) 1 > 0 > ω e = D + γ e B cos θ A Ω mw cos[(ω e +A)t] Ω rf cos(at) 6

8 Strong coupling regime effect of random orientation H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ The scheme still has an angle dependency in ω e = D + γ e B cos θ, which could be quite large Therefore, to account for random ND orientation we take a very small magnetic field (B=0 1 or γ e B A). The probabilities of an electron, nuclear flips are then P x e = sin 2 (Ω mw t 1 sin θ), P x n = sin 2 (Ω rf t 2 sin θ) Overall probability for polarization transfer P = sin 2 (Ω mw t 1 sin θ) sin 2 (Ω rf t 2 sin θ) A 1 > ω e = D 0 > Ω mw cos[(ω e +A)t] Ω rf cos(at) 1 One can still differentiate between the 1> and -1> levels of the electron spin even in zero magnetic field, as there exists an energy difference of a few Mhz between the level from the strain in the diamond (which differs between diamonds) or by applying a homogeneous electric field 7

9 Strong coupling regime effect of random orientation H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ Consider N electron spins, each coupled to a single nuclei (i denoting the electron spin number, n number of polarization iterations: Average polarization for 1000 electron-nuclear spin pairs (all with same coupling) 8

10 Strong coupling regime multiple nuclear spins H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ Generalizing for multiple nuclear spins per NV spin works well for well distinguished coupling strengths (A 1 >A 2 >A 3 ) The polarization then proceeds for each nuclear spin separately, in a sequential fashion For relatively similar coupling strengths, direct polarization by the NV becomes more difficult, and polarization by nuclear-nuclear diffusion might be preferable 9

11 Strong coupling regime additional considerations H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ The CNOT gates rely on NV and nuclear spin coherence for the duration of the pulses. One then might need to take into account the NV spin s T 2, especially in the slow RF pulse. For smaller values of A, it might be necessary to apply dynamical decoupling schemes to prolong NV spin coherence This scheme works best for highly coupled (near to NV) nuclear spins. Further away spins could then be polarized by spin-diffusion The final achieved polarization in each ND will be along the NV center z axis in the half hemisphere dictated by θ < π/2. To bring all the nuclear spins in the same direction, we adiabatically ramp up the magnetic field, until it is the dominating parameter (γ e B > D) 10

12 Strong coupling regime additional considerations H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ The coupling to the NV will create a diffusion barrier because of energy mismatches between nuclear spins. However, when the NV is in the 0> state, the diffusion barrier vanishes The energy gap in the nuclear spins induced by the NV dipolar coupling has a direction dependency based on the anisotropic coupling, as well as the D orientation. In a ND there are 4 directions for D, which might cause a competition as to the polarization direction in the boundaries between. However, this should be solved by applying a small magnetic field so that only one direction is involved in polarization The dipolar coupling depends on the orientation between the NV and nuclear spins. Averaging the polarization given by the induced gap orientation does not average out to zero, but is not maximal The energy gap disappears when the NV is in the 0> state 11

13 Weak coupling regime ( A) H = DS z 2 + γ e B cos θ S z + γ e B sin θ S x + I θ + SA I In this regime there are two potential DNP protocols The solid effect Hartmann-Hahn (H.H.), e.g. NOVEL Both protocols involve MW driving Ω cos(ωt) S x In the rotating frame: H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + aσ z I x With x, z, in the principal axes of the magnetic field ( z is the magnetic field direction) 1 > 0 > ω e = D + γ e B cos θ Solid effect driving H.H. driving 12

14 Weak coupling regime NOVEL H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + Aσ z I x Choosing ω = D + γ e B cos θ for a specific value of (e.g. θ = 0) Applying first a π/2 pulse, waiting for rotation into spin-locking and then continuously driving in the H.H condition to achieve a polarization transfer As A << 1 MHz, one might need to apply dynamical decoupling schemes to ensure electron coherence during polarization transfer Can be applied also for θ 0, though the π/2 pulse will then be non-optimal 1 > 0 > ω e = D + γ e B cos θ H.H. driving 13

15 Weak coupling regime NOVEL H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + Aσ z I x Very angle dependent due to 2 causes: Angle dependence of the Rabi frequency: choosing θ = 0, considering small variations in θ, we get H eff = 1 2 γ eb θ 2 σ z + Ωσ x + I z + Aσ z I x Achieving the H.H condition requires γ e B θ 2 /2 <, therefore Δθ max 2γ N γ e 1 30 rad 1o Angle dependence in the π/2 pulse 1 > ω e = D + γ e B cos θ 0 > H.H. driving 14

16 Weak coupling regime Solid effect H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + Aσ z I x Choosing ω = D + γ e B cos θ ± δ for ω eff = = Ω/ δ 2 + Ω 2 gives δ 2 + Ω 2, and defining sin θ H eff = ω eff σ z + I z + A sin θ σ xi x, where x, z denote rotation by θ around the y axis Continuous driving when choosing ω eff will induce flip-flops between the NV center the nuclear spin Though A sin θ << 1 MHz, as the NV spin is in (or near) an eigenstate, the NV T 1 time should be long enough for the polarization transfer without (or with minimal) dynamical decoupling 1 > 0 > ω e = D + γ e B cos θ Solid effect driving 15

17 Weak coupling regime Solid effect H eff = ω eff σ z + I z + A sin θ σ x I x The angle dependence is weaker than in the NOVEL scheme as when choosing δ Ω, δ is always oriented along the NV z direction However, If B 0 there is still a strong angle dependence because of ω e = D + γ e B cos θ. Considering an angle pertubation, the Hamiltonian will be H eff = ω eff σ z + γ e B cos θ σ z + I z + a sin θ σ x I x Again leading to Δθ max 1 o In addition, as is smaller than the electron EPR linewidth, the case will be of the differential solid effect ω e ω e + ω e Differential solid effect 16

18 Weak coupling regime no magnetic field H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + Aσ z I x Proposal for reducing angle dependency: taking B 0 The solid effect will have almost no dependence on the angle for the polarization transfer or for the optical polarization 1 NOVEL/H.H. will also have a relatively weak angle dependence (caused only by the angle of the MW driving Ωσ x compared with the principal axes of D) However, there is also no energy gap between the nuclear spin states (except for that caused by the coupling to the NV spin, which cannot be used for solid effect polarization, as no flip-flop is induced) 1 > 0 > ω e = D 1 One can still differentiate between the 1> and -1> levels of the electron spin even in zero magnetic field, as there exists an energy difference of a few Mhz between the level from the strain in the diamond or by applying a homogeneous electric field 17

19 Weak coupling regime no magnetic field H eff = (D + γ e B cos θ ω) σ z + Ωσ x + I z + Aσ z I x Potentially, if the sample contains enough NV centers, one could use the energy gap induced by a stronger coupling of the nuclear spins to a second NV center For this to work, after the NV centers are initialized, they are brought to the 1> state with a MW pulse 1. This will induce an energy gap for nearby nuclear spins. The flipflop then will be from the 1> to the 0> state in the NV However, in order for nuclear spins for which a NV C 0.1 Mhz to have a stronger coupling to a nearby free electron spin (which will cause the energy gap) a very large NV density is required, much larger than typical densities NV 1 13 C 1 > A NV2 Induces flip-flop Creates energy gap NV 2 0 > ω e = D A NV2 1 This requires a π pulse for a known rabi frequency but with random orientation. Naively this would work but would be suboptimal. This could potentially be optimally performed by a composite pulse sequence 18

20 Strong coupling regime, no magnetic field additional considerations H = DS z 2 + AS z I z +Ω mw cos(ω mw t 1 ) S θ +Ω rf cos(ω rf t 2 ) S θ Larger NV center concentration (and the corresponding larger paramagnetic impurity concentration) decrease the nuclear T 1 The energy gap in the nuclear spins induced by the NV dipolar coupling has a direction dependency based on the anisotropic coupling, as well as the D orientation. In a ND there are 4 directions for D, which might cause a competition as to the polarization direction in the boundaries between. However, this should be solved by applying a small magnetic field so that only one direction is involved in polarization The dipolar coupling depends on the orientation between the NV and nuclear spins. It should be verified that the average induced gap orientation does not average out to zero As the NVs are very close together, one would need to take into account the coupling between the NVs as well 19

21 Thermal mixing/cross effect H = SD S + γ e Bcos θ S z + γ e Bsin θ S x + I θ + SA I Thermal mixing / cross effect both schemes achieve polarization transfer by simultaneous flipping of several electron spins, and require a large electron homogeneous/inhomogeneous broadening Thermal mixing Spin-temperature approximation - transferring of energy from the electron dipolar bath to the nuclear Zeeman terms Requires Δω > Cross effect Flip-flop of two electron spins and a single spin while conserving energy D Requires ω 2 ω 1 (large inhomogeneous broadening) At first glance, it seems that because of the angle dependence of the NV spins (γ e Bcos θ S z ) these mechanisms have no chance of working, as γ e B

22 Thermal mixing/cross effect However, in a specific nanodiamond, all NVs are in one of the four latticebase directions Therefore, the angle dependency will affect all NV spins in the nanodiamond in the same way No angle dependency in the interaction between neighboring NV spins Thermal mixing should work for nano-diamonds (Δω > for low magnetic fields). However, driving at the frequency required for thermal mixing (ω MW = ω e + for small values of ) will be challenging, as ω e = D + γ e B cos θ, and differs greatly with nano-diamond orientation D D D D D D B 21