The Effect of AC Zeeman Frequency Shifts Due to the NMR RF Field Used for AFP in EPR-Frequency-Shift 3 He Polarimetry

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1 The Effect of AC Zeeman Frequency Shifts Due to the NMR RF Field Used for AFP in EPR-Frequency-Shift 3 He Polarimetry Jaideep Singh, Peter Dolph, & Al Tobias University of Virginia Version 0.70 March 4, 009 Abstract We can determine the 3 He polarization inside polarized 3 He target cells by measuring the frequency shift in the EPR spectrum of alkali atoms. This shift is mostly due to the effective magnetic field associated with spin-exchange collisions with the polarized 3 He nuclei. Since we can only directly measure the EPR frequency due to the total magnetic field felt by the alkali atoms, we change the sign of the frequency shift by flipping the 3 He spins using the NMR technique of frequency sweep adiabatic fast passage (AFP). If all else is equal, then the difference in EPR frequencies for the two directions of the 3 He spins isolates the frequency shift due to the polarized 3 He nuclei: ν EPR ν EPR =Δν EPR = (Δν) He. What complicates this analysis is the fact that the oscillating NMR RF field used for AFP also causes a small frequency shift (Δν) rf, in this case, due to the AC Zeeman effect. If the magnitude of the NMR RF field changes during AFP, then its corresponding frequency shift is different for the two directions of the 3 He spins. Under this condition, the partially canceled frequency shift due to the NMR RF field also contributes to & contaminates the EPR frequency difference: Δν EPR = (Δν) He +(Δν) rf (Δν) rf)]. This could potentially be a large systematic error in the determination of the 3 He polarization. We ll (1) describe the problem in more detail, () calculate the absolute size of the frequency shifts due to the polarized 3 He nuclei & the NMR RF fields used for AFP and (3) present an the experimental study of this effect. Finally, we ll note that this effect can be completely eliminated by the measuring the EPR frequencies only when the NMR RF field is turned off. Contents 1 Description of the NMR RF Shift Effect 1 Calculating the Shift Due to Polarized 3 He 5 3 Calculating the Shift Due to the NMR RF Field: The AC Zeeman Effect 7 4 An Experimental Study of the NMR RF Shift Effect 11 1 Description of the NMR RF Shift Effect The result of a typical EPR measurement at UVa is shown in Fig. (1). At the start of an EPR polarization measurement, the total static magnetic field seen by the alkali atoms is B = B 0 + B He, which corresponds to an EPR frequency of: (ν EPR) off = ν 0 +(Δν) He (1) 1

2 khz, V = 300 mv set (Δν) = khz rf 9750 EPR Frequency (khz) khz 30.0 khz 30.0 khz 56.6 khz 30.0 khz 56.6 khz 56.6 khz 30.0 khz (Δν) = 66 khz He time (sec) EPR Frequency (khz) khz ON V set = 300 mv (Δν) = 0.4 khz rf 56.6 khz off time (sec) Figure 1: Typical EPR Measurement Sequence. The He spins are flipped using frequency sweep NMR-AFP. The top plot shows the entire data set which includes four 3 He spin flips. The shift due to the NMR RF field at 30.0 khz can be seen at 75 sec <t<175 sec, 50 sec <t<35 sec, and finally 45 sec <t<500 sec. Note that the sign of the RF shift is independent of the direction of the 3 He spins. The shift due to the NMR RF field at 56.6 khz is much smaller, but is visible at 0 sec <t<70 sec, see the bottom plot. Because the NMR RF coil resonates at about 0 khz, the amplitude of the RF field at 30.0 khz is much larger than the amplitude of the RF field at 56.6 khz for the same function generator set amplitude V set = 300 mv.

3 Under our typical conditions, B 0 = 13 G and, for the well state transition of 39 K, the EPR frequency is around 9.7 MHz. After measuring this EPR frequency for several seconds, the NMR RF field is turned on with a frequency of 56.6 khz. This NMR RF field causes an additional shift in the EPR frequency: (ν EPR) on = ν 0 +(Δν) He +(Δν) rf () After again measuring the EPR frequency for several seconds, the NMR RF frequency is swept over a period of 6 seconds from 56.6 khz, through resonance at 4. khz, to 30.0 khz. The rate is chosen to be fast enough to minimize the 3 He relaxation at resonance, but slow enough for the 3 He spins to follow the effective field in the rotating frame adiabatically. After the spins have been flipped, the total static magnetic field felt by the alkali atoms is B = B 0 B He which corresponds to an EPR frequency of: and then we turn the to measure: (ν EPR) on = ν 0 (Δν) He +(Δν) rf (3) (ν EPR) off = ν 0 (Δν) He (4) To conclude the EPR measurement, we follow the same steps already mentioned, but in reverse order: 1. turn the NMR RF on at 30.0 khz. measure the EPR frequency 3. sweep the NMR RF frequency from 30.0 khz to 56.6 khz 4. measure the EPR frequency 5. turn the 6. measure the EPR frequency In addition to this measurement sequence, Fig. (1) also shows the results of an analogous measurement sequence for which the NMR frequency sweep from 30.0 khz to 56.6 khz and then back to 30.0 khz. Using these measurements, there are two frequency differences that can be used to extract the 3 He polarization: ] Δ(ν EPR ) on = (νepr) on (νepr) on = (Δν) He + (Δν) rf (Δν) rf) (5) Δ(ν EPR ) off = (ν EPR) off (ν EPR) off = (Δν) He (6) If there is no shift associated with the NMR RF field ((Δν) rf =(Δν) rf = 0) or if the size of frequency shift associated with the NMR RF field stays the same after the 3 He spins are flipped ((Δν) rf (Δν) rf)] = 0), then the two frequency differences are identical. However, at least at UVa, this is not the case. Based on Figs. (1) & (), we ll make the following observations of the what we call NMR RF Shift Effect: 1. the size of (Δν) rf has no measurable dependence on the sign and magnitude of the 3 He polarization. the size of (Δν) rf at 13 Gauss appears to in the khz range for both 85 Rb and 39 K 3. the sign of (Δν) rf is positive for both the well state transition (I +1/) (I 1/) and the hat state transition +(I +1/) +(I 1/) 4. the size of (Δν) rf increases with the NMR RF amplitude and decreases with the NMR RF frequency Originally, we thought that (Δν) rf 0 was due to a direct coupling between the NMR RF coils and the photodiode used for EPR. We tested this hypothesis by increasing the distance between the EPR photodiode and the NMR RF coils. The fluorescence from the cell needed as an input to the EPR feedback circuit was captured by an optical fiber instead of the photodiode directly. Because we used a very long optical fiber, 3

4 EPR Frequency (khz) khz 30.0 khz 30.0 khz 56.6 khz 56.6 khz 35.0 khz 35.0 khz 56.6 khz 56.6 khz 5.0 khz 5.0 khz 56.6 khz (Δν) rf in khz (Δν) = 40 khz He time (sec) 5970 (Δν) = 1 khz rf (Δν) = 1 khz He EPR Frequency (khz) khz 30.0 khz 30.0 khz 56.6 khz (Δν) = 0 khz He time (sec) Figure : EPR in the High Energy or Hat State for 39 K (top) and 85 Rb (bottom) at 13 Gauss. In the top plot, we see that the shift depends on NMR frequency. This is because circuit used to drive the NMR RF coils resonates at about 0 khz. In the bottom plot, we see that the NMR RF shift in independent of the size of the shift due to the polarized 3 He gas. 4

5 the distance between the EPR photodiode and NMR RF coils was several meters. When we performed the EPR measurement under these conditions, we still observed a frequency shift when the NMR RF was turned on and off. Our new hypothesis (which we believe is correct) is that the NMR RF field directly interacts with the alkali atoms to produce a frequency shift via the AC Zeeman effect. Under our experimental conditions, B 0 > 10 Gauss, ν 0 > 5MHz,ν rf = (40 ± 0) khz, B rf < 0.5 Gauss, and, using the results of Sec. (3), the size of the shift is given by: (Δν) rf = F (F +1) m(m 1) 4ν EPR ] ν B B rf (7) where F = I ± 1/, I is the nuclear spin of the alkali atom, ν EPR is the frequency of the F, m F, m 1 transition, ν/ B is given by Eqn. (9), and B rf is the amplitude of the NMR RF Field in the lab frame. Tab. () lists the quantitative size of the shift for 85 Rb and 39 K under our conditions. Because, for incidental historical reasons, the NMR RF coils resonate at 0 khz (see Fig. (3)), the NMR RF field amplitude is higher at 30.0 khz than at 56.6 khz for the same set control voltage, which implies (Δν) rf (Δν) rf. Therefore, in order to avoid all these difficulties, it is much better to use Δ(ν EPR ) off than Δ(ν EPR ) on to calculate the 3 He polarization. To do this, all you have to do is take data with the NMR RF field off before and after the AFP frequency sweep. Calculating the Shift Due to Polarized 3 He More details about the EPR polarimetry method can be found in 1,, 3, 4]. To convert from the frequency shift (Δν) He to the effective field associated with the polarized 3 He, we make the following approximation: Up to fifth order in field, the derivative is: ν ± B = (g Iμ N g S μ B ) hi] ( ) 1 ν B He =(Δν) He (8) B 5 n=0 b n x n I] n (9) x = B (g I μ N g S μ B ) hν hfs (10) I] = I + 1 (11) b 0 = gi µn 1 ± g Sµ B (I] 1) gi µn =1 O(10 3 ) 1 g Sµ B (1) b 1 = (1 m) (13) ) b = 6 (1 3m +3m I] (14) 4 b 3 = 0 ( 1 4m +6m 4m 3) 6I] (1 m) (15) b 4 = 70 ( 1 5m +10m 10m 3 +5m 4) 5I] ( 1 3m +3m ) + 15I]4 (16) 8 b 5 = 5 ( 1 6m +15m 0m 3 +15m 4 6m 5) 105I] ( 1 4m +6m 4m 3) + 45I]4 (1 m) (17) 4 where I is the nuclear spin, g I is the nuclear g-factor, μ N is the nuclear magneton, h is the Planck constant, I] =I +1, and ν hfs is the zero-field hyperfine splitting between F = I +1/ andf = I 1/, see Tab. (1). For the special case of end transitions, F = I +1/ andm = I +1/ fori +1/ I 1/ &m = I +1/ 5

6 Isotope I upper F End g I ν hfs Transition MHz 6 Li 1 3/ s3/ s1/ Li 3/ s s Na 3/ s s K 3/ s s K 4 9/ s9/ s7/ K 3/ s s Rb 5/ 3 s3 s Rb 3/ s s Cs 7/ 4 s4 s Table 1: Hyperfine Structure Data for Alkali Atoms. F = I +1/ for the upper manifold and s = ±. for I +1/ I 1/ and the coefficients for the expansion of the derivative of the EPR frequency are: b 0 = gi µn 1+ g Sµ B (I) gi µn =1 O(10 3 ) 1 g Sµ B (18) b 1 = 4I (19) b = 6I (I 1) (0) b 3 = 8I ( 4I 6I +1 ) (1) b 4 = 10I (I 1) ( 4I 10I +1 ) () b 5 = 1I ( 16I 4 80I 3 +80I 0I +1 ) (3) where ± refers to the edge state m = ± ( I + 1 ) involved in the transition. Once we have BHe, we can find the He-3 polarization from the effective field due to the He-3 using: B He = μ 0 3 (κ P He 0 1+κ geo )He]g He μ N (4) where μ 0 is the magnetic permeability of free space, κ 0 is the empirical shift constant, κ geo is a geometric factor, He] is the He-3 number density, g He = is the He-3 g-factor, μ N is the nuclear magneton, P He = S z is the He-3 polarization. The values for κ 0 are given by 4, 5]: Rb : κ 0 = (T 00 o C) (5) K:κ 0 = (T 00 o C) (6) Na : κ 0 = (T 00 o C) (7) The geometric factor κ geo is the ratio of the classical magnetic field produced by the He-3 for the cell geometry averaged over the alkali sampling volume (ASV) to the classical magnetic field produced by a uniformly polarized spherical He-3 sample with the same density and polarization: 1 V κ geo = ASV ASV ẑ B c dv ẑ B Bsphere = μ 0 sphere 3 He]g P He Heμ N ẑ (8) where ẑ is the direction of the holding field. The classical field B c can be calculated using the method of magnetic scalar potentials (See section 5.9 of Blue Jackson 6]): B c = μ 0 ( Hc + M c ) H c = Φ M Mc =He] g Heμ N P c ( r) (9) 6

7 where P c ( r) is the possibly position dependent He polarization vector. For a uniformly polarized region bounded by a surface S, the magnetic scalar potential is given by: Φ M ( r) = 1 4π S ˆn M c ( r ) r r da (30) where ˆn is the normal unit vector pointing out of S at a position r on the surface S. For a region far from the polarized He-3, the magnetic scalar potential is given by: Φ M ( r) = V M c r 4πr 3 (31) where V is the total volume of the region containing the polarized He-3. The value of κ geo for a few different geometries are given here: 1. κ geo = 1 for a uniformly polarized sphere. κ geo =3/ for an infinitely long cylinder polarized along its axis 3. κ geo =3/4 for an infinitely long cylinder polarized perpendicular to its axis In quantitative terms, the effective field felt by the alkali metal due to the polarized He-3 vapor is: B He = μ 0 3 (κ P He 0 1+κ geo )He]g He μ N ]( κ0 1+κ geo 3 ) ] He] PHe = ( 14.5 milligauss) 6 1 amg 100% (3) (33) 3 Calculating the Shift Due to the NMR RF Field: The AC Zeeman Effect To calculate the shift due to the AC Zeeman effect, we ll follow and in some cases generalize the basic arguments found in problem.7 of 7]. Since we are only interested in a given Zeeman transition within the EPR spectrum, we ll focus only on the states involved in the transition. This results in a two state system ( m & m 1 ) with a Hamiltonian written as: H 0 = hω 0 F z (34) where F z is the z-component of the total atomic angular momentum, h is the Planck constant divided by π, and ω 0 is the EPR frequency for the transition for a magnetic field B 0. If another much smaller magnetic field B 1 B 0 is turned on, then we can write the additional Zeeman interaction as: H 1 = μ B 1 = gμ B F B1 (35) where g is the effective or Landé g-factor, μ B is the Bohr magneton, and F is the total atomic angular momentum. In general, g depends on the total magnetic field in the z-direction, the hyperfine manifold F, and the magnetic substate m. Its form can be obtained by considering the case when B 1 is in the z-direction (the same direction as B 0 ). In this case, the total Hamiltonian can be written in two ways: Taylor expanding ω = ω(b 0 + B 1 )aboutb 0 gives: H = H 0 + H 1 = hωf z = hω 0 F z gμ B F z B 1 (36) ω(b 0 + B 1 )=ω(b 0 )+(B 0 + B 1 B 0 ) ω B = ω 0 + B 1 ω B (37) 7

8 Putting this together gives the form of the effective g-factor: ( h hωf z hω 0 F z = gμ B F z B 1 g = where the derivative is evaluated at B 0. We ll now consider the case of an oscillating RF field in the x-direction: μ B ) ω B (38) H = hω 0 F z gμ B F x B rf cos(ω rf t) (39) where B rf is the amplitude of the RF field in the lab frame and ω rf is the angular frequency of the RF field. In general, the non-zero matrix elements of F z and F x =(F + + F )/ are: F (F +1) m(m 1) m F z m = m m 1 F z m 1 = m 1 m 1 F x m = m F x m 1 = (40) Using these formulas, we can rewrite the Hamiltonian using the Pauli spin matrices: H = hω 0 m + hω 0 σ z /+ hω rf σ x cos(ω rf t)/ (41) where m = m 1/ is the mean m associated with the transition and Ω rf is given by: Ω rf = ω B B rf F (F +1) m(m 1) (4) Since hω 0 m is an overall constant energy offset, we can safely drop it from the Hamiltonian. For convenience, we ll relabel two eigenstates of H 0 as a = m and b = m 1. The eigenstates of the full Hamiltonian H are denoted as 1 & and can be expanded as: 1 = a 1 exp( iω 0 t/) a + b 1 exp(+iω 0 t/) b = a exp( iω 0 t/) a + b exp(+iω 0 t/) b (43) where orthonormality enforces: a 1 + b 1 = a + b =1 a 1a + b 1b = 0 (44) Because we ve chose to expand 1 & in this way, they automatically satisfy the Schrodinger equations with a 1 = b =1&a = b 1 = 0 when there is no RF field Ω rf = 0. Our goal now is to obtain a solution correct to the lowest order for the case when 0 < Ω rf ω 0. Applying Schrodinger equation to 1, we get a pair of coupled equations: i h(ȧ 1 iω 0 a 1 /) exp( iω 0 t/) = +( hω 0 a 1 /) exp( iω 0 t/) + ( hω rf b 1 /) exp(+iω 0 t/) cos(ω rf t)(45) i h(ḃ1 + iω 0 b 1 /) exp(+iω 0 t/) = ( hω 0 b 1 /) exp(+iω 0 t/) + ( hω rf a 1 /) exp( iω 0 t/) cos(ω rf t)(46) After rearranging a few things we get: ȧ 1 = (iω rf b 1 /) exp(+iω 0 t)cos(ω rf t) (47) ḃ 1 = (iω rf a 1 /) exp( iω 0 t)cos(ω rf t) (48) Since Ω rf is small, it s not unreasonable to assume that a 1 1. In other words, we ll make the ansatz that a 1 =exp( iω! t). Using this form of a 1 allows us to directly integrate ḃ1 to give: b 1 = ḃ 1 dt = (iω rf /) exp( i(ω 1 + ω 0 )t)cos(ω rf t) dt (49) = (iω rf /4) exp( i(ω 1 + ω 0 ω rf )t)+exp( i(ω 1 + ω 0 + ω rf )t)] dt (50) exp( i(ω1 + ω 0 ω rf )t) = (iω rf /4) + exp( i(ω ] 1 + ω 0 + ω rf )t) (51) i(ω 1 + ω 0 ω rf ) i(ω 1 + ω 0 + ω rf ) = Ω rf exp( i(ω1 + ω 0 ω rf )t) + exp( i(ω ] 1 + ω 0 + ω rf )t) (5) 4 ω 1 + ω 0 ω rf ω 1 + ω 0 + ω rf 8

9 Using this solution for b 1 and the ansatz for a 1, we can now apply the equation for ȧ 1 : ȧ 1 = (iω rf b 1 /) exp(+iω 0 t)cos(ω rf t) (53) iω 1 exp( iω 1 t) = i Ω rf 8 exp(+iω exp( i(ω1 + ω 0 ω rf )t) 0t)cos(ω rf t) + exp( i(ω ] 1 + ω 0 + ω rf )t) (54) ω 1 + ω 0 ω rf ω 1 + ω 0 + ω rf ω 1 = Ω rf 8 cos(ω exp(+iωrf t) rft) + exp( iω ] rft) (55) ω 1 + ω 0 ω rf ω 1 + ω 0 + ω rf = Ω rf 16 exp(+iω exp(+iωrf t) rft) + exp( iω rf t)] + exp( iω ] rft) (56) ω 1 + ω 0 ω rf ω 1 + ω 0 + ω rf = Ω rf 1 + exp(+iωrf t) + 1+exp( iω ] rft) (57) 16 ω 1 + ω 0 ω rf ω 1 + ω 0 + ω rf (58) Our first approximation will be to drop the rapidly oscillating terms to get: ] ω 1 = Ω rf ω 1 + ω 0 8 (ω 1 + ω 0 ) ωrf (59) Our second approximation is that ω 1 ω 0, which allows us to solve for ω 1 : ] ω 1 = Ω rf ω 0 8 ω0 1+ ω ] 1 1+ ω 1ω 0 + ω ] 1 1 ω rf ω 0 ω0 (60) ω rf ] Ω rf ω 0 8 ω0 1 ω ( 1 ω 0 + ω )] rf ω rf ω 0 ω0 (61) ω rf ] = Ω rf ω 0 8 ω0 1+ Ω rf(ω0 + ω ] 1 rf) ω rf 8(ω0 ω rf) (6) For self-consistency, this approximation necessarily implies that Ω rf ω 0 which is roughly equivalent to our earlier assertion that B rf B 0. Our third and final approximation is that the frequency of the RF field is far from resonance, ω rf ω 0, which finally gives: ] ω 1 = Ω rf ω 0 8 ω0 1+ Ω rf(ω0 + ω ] 1 rf) ω rf 8(ω0 ω rf) Ω rf (63) 8ω 0 We can follow this same calculation for where a plays the same role as b 1, b plays the same role as a 1, and we must flip the sign of ω 0 : a = Ω rf exp( i(ω ω 0 ω rf )t) + exp( i(ω ] ω 0 + ω rf )t) (64) 4 ω ω 0 ω rf ω ω 0 + ω rf b = exp( iω t) (65) ω Ω rf (66) 8ω 0 Putting this altogether gives: 1 = exp( i(ω 0 /+ω 1 )t) a + b 1 exp(+iω 0 t/) b (67) = a exp( iω 0 t/) a +exp(+i(ω 0 / ω )t) b (68) b 1 = + Ω rf exp( i(ω 1 + ω 0 )t)cos(ω rf t) 4ω 0 (69) a = Ω rf exp( i(ω ω 0 )t)cos(ω rf t) 4ω 0 (70) ω 1 = ω = Ω rf 8ω 0 (71) 9

10 B 0 isotope I m ν 0 ν/ B (Δν) rf (Δν) He Gauss ±I]/ MHz khz/g khz/g khz/amg Rb 5/ 39 K 3/ Rb 5/ 39 K 3/ Table : EPR Frequency Shifts for the End Transitions Due to the NMR RF Field & Polarized 3 He. We ve assumed that B rf B 0 and ν rf ν 0. The (+) sign for m refers to the well ( hat ) state. For the shift due to the polarized 3 He, we ve assumed a spherical sample at a temperature of 00 o C and 100% polarization. The sign of the polarization is taken to be the same as the sign of the alkali m state. To lowest order, the energy levels are: and finally the frequency is: ω = 1 H 1 H h 1 H 1 = 1 i h 1 t H = i h t = ω 0 + ω 1 ω = ω 0 + Ω rf 4ω 0 = ω 0 + h(ω 0 /+ω 1 ) (7) h(ω 0 / ω ) (73) 1+ ( ω B ) B rf (F (F +1) m(m 1))] ω 0 (74) The AC Zeeman frequency shift due to an RF field for the transition F, m F, m 1 when B rf B 0 and ν rf ν 0 is, to lowest order, given by: (Δν) rf = F (F +1) m(m 1) 4ν 0 ] ν B B rf (75) where B rf is the magnitude of the RF field in the lab frame, ν 0 is the frequency of the transition when B rf = 0, and the derivative for F = I ± 1/ is given by (up to fifth order in field) Eqn. (9). For the special case of end transitions, F = I +1/ andm = I +1/ fori +1/ I 1/ &m = I +1/ for I +1/ I 1/ and the frequency shift is: (Δν) rf = I] ] ν 4ν 0 B B rf (76) 10

11 1.6 I RF, current in NMR RF coils (A) NMR RF frequency (khz) Figure 3: Current in NMR RF coils for a set amplitude of V set = 300 mv. The NMR RF coils resonate at about 0 khz. 4 An Experimental Study of the NMR RF Shift Effect To study the NMR RF Shift effect, we took data for: 1. fields of 13 Gauss and 3 Gauss. NMR RF frequencies of 5.0 khz, 30.0 khz, 35.0 khz, and 56.6 khz 3. NMR RF set amplitude from 5 mv up to 600 mv or whatever value the NMR RF amplifier would overload K in the well state At typical sequence is depicted in Fig. (4). For each data set, we would measure the EPR frequency, the current in the Helmholtz coils, and the current in the NMR RF coils. For all of the measurements, the current in the Helmholtz coils was very stable and the 3 He polarization had reached equilibrium. Therefore any change in the EPR frequency was mostly due to the NMR RF field. After collecting this data, we plot the shift vs. NMR RF set amplitude, see Fig. (5). The expected quadratic behavior is readily seen. We don t expect the NMR RF frequency to play a direct role in the shift. Therefore we measured the current in the NMR RF coils vs. NMR RF frequency for a fixed set amplitude V set, see Fig. (3). To make all the data comparable, we convert the shift into the amplitude of the RF field causing it using Eqn. (76). Since the same set amplitude doesn t correspond to the same current in the NMR RF coils for different NMR RF frequencies, we plot the data vs. the current in the RF coils instead. When we do this, if our calculation is correct, then all the data should lie on a line. To better than 5%, this is true and we get a slope of 18 mg/a. 11

12 (Gauss) B time (sec) I rf, measured in NMR RF Coils (A) time (sec) 14 Shift Due to NMR RF Field (khz) time (sec) Figure 4: Shift in the EPR Frequency for an NMR RF Field at 5 khz. The top plot shows the magnitude of the holding field as measured from the current in the Helmholtz coils. Over the course of the measurement, it was stable to a few milligauss. The middle plot shows the current in NMR RF coils. During the measurement, the RF field was turned off and on at linearly increasing amplitudes. The bottom plot shows the locked EPR frequency as the NMR RF field is varied. The measurement was done at the end of a spin-up, so we assume that the 3 He polarization was at equilibrium. The field due to the polarized 3 He is about 40 milligauss, which corresponds to a frequency shift of about 30 khz. The magnitude of the NMR RF field is increased linearly, whereas the shift is increasing in a quadratic manner. At around t = 700 sec, there was a glitch in the RF amplitude. In addition, the NMR RF amplifier overloaded at around t = 1175 sec. Both of these features can be seen in both the bottom two plots. 1

13 khz, 13 G 8 (khz) (Δν) rf khz, 3 G 5.0 khz, 13 G 30.0 khz, 13 G 30.0 khz, 3 G 35.0 khz, 13 G 35.0 khz, 3 G 56.6 khz, 13 G V set (mv) (G) B rf, calculated from (Δν) rf B 0 = 3 Gauss (190.8 ± 4.3) mg/a B 0 = 13 Gauss (181.6 ± 0.8) mg/a all data (181.9 ± 0.6) mg/a I rf, measured in NMR RF coils (A) Figure 5: Measurement of (Δν) rf under different conditions. We measured the NMR RF shift at two different fields, four different NMR RF frequencies, and several NMR RF field amplitudes. The top plot shows the shift plotted against set amplitude. Except for the 5.0 khz, 13 Gauss data set, there was a linear relationship between the current in the NMR RF coils and the set amplitude. For the 5.0 khz, 13 gauss data set, a glitch occurred during the measurement, see Fig. (4). For the bottom plot, we ve inserted the measured shift into Eqn. (76) and solved for the amplitude of the NMR RF field in the lab frame. Once this value for B rf is plotted against the measured current I rf in the NMR RF coils, the data cluster along a line. 13

14 References 1] S. R. Schaefer, G. D. Cates, Ting-Ray Chien, D. Gonatas, W. Happer, and T. G. Walker. Frequency shifts of the magnetic-resonance spectrum of mixtures of nuclear spin-polarized noble gases and vapors of spin-polarized alkali-metal atoms. Physical Review A, 39(11): , Jun ] N. R. Newbury, A. S. Barton, P. Bogorad, G. D. Cates, M. Gatzke, H. Mabuchi, and B. Saam. Polarization-dependent frequency shifts from Rb- 3 He collisions. Physical Review A, 48(1): , Jul ] A. S. Barton, N. R. Newbury, G. D. Cates, B. Driehuys, H. Middleton, and B. Saam. Self-calibrating measurement of polarization-dependent frequency shifts from Rb- 3 He collisions. Physical Review A, 49(4): , Apr ] M. V. Romalis and G. D. Cates. Accurate 3 He polarimetry using the Rb Zeeman frequency shift due to the Rb- 3 He spin-exchange collisions. Physical Review A, 58(4): , Oct ] Earl Babcock, Ian A. Nelson, Steve Kadlecek, and Thad G. Walker. 3 He polarization-dependent EPR frequency shifts of alkali-metal- 3 He pairs. Physical Review A, 71(1):013414, Jan ] John David Jackson. Classical Electrodynamics. John Wiley & Sons, New York, 3rd edition, ] Dmitry Budker, Derek K. Kimball, and David P. DeMille. Atomic Physics: An Exploration Through Problems and Solutions. Oxford University Press, Oxford, Great Britain, 1st edition,