Notes on Optical Pumping Procedure & Theory

Transcription

1 Notes on Otical Puming Procedure & Theory Pre-lab 1. Why is the exeriment called otical uming? What is umed? 2. What is the exerimental signature of having cancelled all magnetic fields in the samle cell? 3. What determines the maximum rate at hich the B field should be modulated? 4. What are the different roles of the RF and visible sources? 5. What are the advantages of studying 87 Rb instead of the isotoically more abundant 85 Rb? 6. What is the nature of the double-quantum transitions in the high-oer RF measurements? Procedure 1. The manual for this exeriment is not very ell ritten. So dealing ith this manual should be an interesting learning exerience that ill build character! 2. Turn on the electronics and ait for the cell to arm u: ~ ½ hour. The oven should already be set to 50 C don t change this setting. You can leave the black cover off the aaratus. Set the front anel settings to the values outlined in Section (g) on. 5-3 of the manual. Once the cell is armed u, look for inkish glo from the lam and a signal from the detector. Move the to lenses around to maximize the signal. Underneath the otical comonents is a cm scale that you should use to record the ositions of these various otical comonents, since other students may change these settings for their exeriment. Then check that the Linear Polarizer and ¼-ave late are roerly aligned. You can determine the degree of circular olarization of the light by lacing a second linear olarizer in front the detector and rotating its axis of olarization. The detector signal should change minimally as you rotate the linear olarizer. If you ish, you can use the rocedure on. 3-8 & 3-9 of the manual to maximize the circular olarization of the light. No remove second linear olarizer, ut the black cover on and shut its door to ut the aaratus in the dark. Note hat haens to the detector signal. 3. Start ith exeriment 4B, You should do a careful job of canceling the vertical and horizontal magnetic fields, as evidenced by a narro zero-field transition. At this time, make sure that the main horizontal-field coil is not connected to its current source. For the vertical coil, the red banana lug should be lugged into the red recetacle on its current source in order to cancel out the vertical comonent of earth s field. For the horizontal-field see coil, you ll have to figure this out for yourself. Use the comass to determine the direction of the horizontal comonent of the earth s field and then aly a current to the see coil of aroriate olarity so that the earth s field is oosed. Next aly a current to the main horizontal-field coil and see hat olarity is needed to roduce a horizontal field in the same direction as that of the see coil this ill be useful later hen you do the quadratic Zeeman Effect. No disconnect the main horizontal coil. It is a good idea to use a ortable DVM to measure the current through the see-field coil from the voltage across a 1-Ω resistor in series ith this coil. 1

2 Do a see of the horizontal field manually to observe the zero-field resonance. Note that the signal is negative being on resonance corresonds to minimum transmission of light to the detector. You ill need to adjust the vertical field to minimize the idth of the resonance. The best rocedure is to (1) adjust the horizontal see field so the detector outut is sitting halfay beteen its maximum and minimum values and then (2) adjust the vertical field until the outut is a maximum. Then reeat (1) & (2) until you cannot narro the resonance any further. Also use another DVM to measure the current through the vertical-field coil. The final ste is to reeat (1) but rock the S end of the aaratus E & W until you get a maximum signal. Here you are aligning the aaratus along the direction of the horizontal magnetic field in the lab. Once you have nulled out both comonents of the earth s field, you should calculate the magnitude and direction of the earth s field in the lab, using the aroximate coil calibrations given in the table on Do you get a reasonable result? Using the DVM and seeing manually, measure the-full-idth-at-half-maximum of the zero-field transition in Gauss it should be ~ a fe mg. No you can reeat this see using the Labvie rogram for automatic taking of the data, so you can generate a grah. 4. Next you must carefully calibrate the see-field and main-field coils as outlined on. 4-8 to Set the RF amlifier gain on 4 and the oscillator outut to near 0 db. For the loest-frequency 10-kHz Zeeman resonance, you can use the Labvie rogram to generate a grah like Figure 4B-2 on. 4-7 for your notebook. Hoever, for generating the frequency vs see-field-current grah shon on. 4-10, you should use the DVM to record the see-field current at each Zeeman resonance. On you are given the useful quantity: µ o /h = MHz/Gauss. For the main-field calibration, you should use to DVMs: one for see-field coil and another for the main-field coil. It is a good idea to adjust the see-field current so that its field and the horizontal comonent of the earth s field cancel. This ay your calibration of the main coil is almost decouled from your calibration of the see-field coil. Of course, you ill need to adjust the see-field current slightly aay from this zero-total-field state to attain a Zeeman resonance for a given frequency, because the main-coil current knob is rather coarse. You ill need to make u a table like the one on Your lot of main-coil field vs. current should ass close to (0,0). 5. For the quadratic Zeeman effect, kee the RF-amlifier gain on 4 and vary the outut of the oscillator. To avoid detecting the double-hoton resonances, kee oscillator outut as small as ossible, consistent ith having reasonable signal. Use the Labvie rogram to record the single-hoton resonances as a function of the see field, a lot that should like Fig. 4C-1 on You ill need to use both DVMs for the calibration: either manually recording the see-field current at each resonance or interolating off the grah you recorded ith Labvie. Use the MathCad rogram to obtain the exected values of B at each resonance and carefully comare them ith your data. To record qualitatively hat haens hen double-hoton absortion occurs, see the field ith higher oscillator outut like Fig. 4C-3 on For the Transient Effects, there are all sorts interesting exeriments to do using a digital oscilloscoe see 4D on

3 (1) After the RF modulation is turned on, the frequency of the transient oscillations ( Rabi oscillations ) should deend linearly on the amlitude of the RF magnetic field ( I RF, the amlitude of the RF current through the RF coil). See if you can confirm this. Although you ll robably be recording the oscillator outut voltage in db, you can devise an exeriment to determine I RF from the db oututs by looking at the discussion on You can use the knon geometry of this RF coil to calculate the field roduced by a given current and comare this result ith the actual RF field amlitude determined from the frequency of the Rabi oscillations. (2) You can determine ho the recovery rate for reestablishing uming (after the RF modulation is turn off) deends on the intensity of the light assing through the cell. Place an additional linear olarizer beteen the linear olarizer of the circular olarizer and the interference filter and rotate the axis of this ne olarizer to vary the light intensity. Also record the DC outut signal of the light detector to see if detector signal obeys Malus La. The folloing theory section ill give you a hint on ho to analyze your recovery-rate results. Theory of Otical Puming 1. From your Quantum Mechanics course (or the Otical Puming instruction manual), you kno that the total angular momentum of an electron (J) obeys J = L + S, here L is orbital angular momentum vector and S (± ½) is sin angular momentum vector. Recall that the values of J range over: L + S, L + S 1, L - S. For the ground state of Rb (L=0), one has 2 S J= (0 + ½) = ½ ; and for the first excited state, 2 P J= 1 - ½ = ½. 2. The recursor to Equation 2B-1 is: H = AI J - µ J B - µ I B (1) For the to Zeeman-Effect terms (electron and nucleus), the loest energy is hen moment µ is arallel to magnetic field B. The term AI J is the hyerfine couling beteen the electron and its nucleus. Because the electron has a negative charge, its moment µ J is oosite to its total angular momentum vector J. Thus µ J = - ( µ J / J )J. Thus Eq. 2B-1 should read: H = AI J + ( µ J / J )J B ( µ I / I )I B (2) here it is assumed for the nucleus that µ I is arallel to I, hich is the case for 87 Rb and 85 Rb. Next, Eq. 2B-5 should be corrected to read: W = + g J µ 0 B M J, here µ 0 is the Bohr magneton, M J is the comonent of J along B and g J = 2 for the ground state 2 S 1/2. Eq. 2B-7 should really read W = + g F µ 0 B here is the comonent of angular momentum vector F along B. Note that F is defined to be: F = J + I, and the values of F range over J + I, J+I 1, J - I. For 87 Rb, one has I = 3/2, giving F = J + I = 1/2 + 3/2 = 2 and F = J - I = 1/2-3/2 = 1 for both the 2 S 1/2 and 2 P 1/2 states. For F = 2, one has = -2, -1, 0, +1, +2; and for F = 1, = -1, 0, +1. Finally, Eq. 2B-8 is correct if you let M =. 3

4 3. Understanding Fig. 2B-3 the 2 S 1/2 ground state of 87 Rb: Note that the order of ith increasing energy is not monotonic (+1, 0,-1, -2, -1, 0, +1, +2). This can be understood by going to the large-b limit here g J µ 0 B M J >> AI J, here M J = ±1/2. In this limit, one can treat the vectors J and I as being indeendent, ith rojections along B of M J and M I, resectively. Then our Eq. (2) becomes H = A M J M I + g J µ 0 B M J ; M J M I e ignore the nuclear sin term because it s small. Thus one can categorize these levels by M J, M I here = M J + M I. Obviously, the four loest-energy levels at high B are associated ith M J = - ½. Since the A M J M I term is negative (due to M J = - ½), then the loest-energy hyerfine-slit level must have the largest ositive M I (=3/2), giving = -1/2 +3/2 = +1. The next largest energy level must have M I = ½, as shon in the table, etc. +1/2 +3/2 = 1/2 + 3/2 = +2 +1/2 +1/2 = +1/2 + 1/2 = +1 +1/2-1/2 = 1/2 = -1/2 = 0 +1/2-3/2 = 1/2 + -3/2 = -1-1/2-3/2 = -1/2 + -3/2 = -2-1/2-1/2 = -1/2 + -1/2 = - 1-1/2 +1/2 = -1/ = 0 4. A better understanding of Fig. 2D-3. Let s assume -1/2 +3/2 = -1/2 + 3/2 = +1 the circular olarization of the light is such that Δ = + 1. Further, let us focus only uon the to highest-energy F = 2 levels ( = +1 & +2) for the 2 S 1/2 and 2 P 1/2 states, as shon belo. Since Δ = + 1, the only excitation (uard solid arro) alloed by incident light is from the = +1 of 2 S 1/2 to = +2 of 2 P 1/2. We are ignoring the ossibility of loer-energy states of 2 S 1/2 feeding into the = +1 & +2 states of 2 P 1/2. No uard transition from = +2 state of 2 S 1/2 is alloed because there is no = +3 state for 2 P 1/2. For 2 S 1/2, all that the = +2 state of can do is relax back (donard gray arro) to the loer-energy = +1 state. Such sontaneous relaxation usually occurs due to scattering of the Rb atoms by (1) other Rb atoms, or (2) the alls of the cell. Sontaneous emission of hotons (donard oen arros) can occur hen an electron jums from the = +2 state of 2 P 1/2 to either the = +1 or = +2 state of 2 S 1/2. Such emission of P-to-S state hotons is not secifically directed toard the detector, so such emission looks like a reduction in transmission of light beteen the lam and the detector. b 11 b P-state = +2 = +1 F = 2 2 Level #2 S-state Level #1 relaxation 1 = +2 F = 2 = +1 Alying rf magnetic field For the S-state: b = robability er unit time for #1 #2 transition via P state b 11 = #1 #1 transition via P state 4

5 = #2 #1 relaxation 1 = robability of occuation of #1 state 2 = #2 state Note: = 1 Modifying Eq. 2D-1 for this simler case (here the b 11 transition can be ignored because it does not affect the value of 1 ), e get! (3) 1 = b Note: #2 #1 transition via P state is not ossible (b = 0), and #1 #2 relaxation in S state is not ossible ( = 0). Eliminating 2, e get ( 1 )!, and 1 = b ( b + ) 1! = (4) 1 + The solution to Eq. (4) is the folloing, here 1 = 1 at time t = 0: 1 [ (b + )t] b ex + = & b + If one takes the t limit of Eq. (5), then one obtains 1 = & b ( 1 ex[ (b + )t]) b = (5) b + b = (6) b + If = 0 (no relaxation), then 1 = 0 and 2 = 1. That is, the #2 state becomes fully occuied and the #1 state becomes emty strong uming situation. If < b, then Eq. (6) gives 1 < 1/2 and 2 > 1/2. The #2 level is still occuied, but the #1 level is not comletely deleted a eaker uming situation. If the magnetic field B = 0, then the #2 #1 relaxation become very raid because the #1 and #2 levels are degenerate. We can aroximate this case ith ( >> b ), and Eq. (6) no becomes b 1 and 2 = 0 1 = No suose e aly a rf transverse magnetic field that is in resonance for a articular value of B, as shon in the cartoon above as a double-headed arro connecting the #1 and #2 S states. If the rf-field-induced transition robabilities (er unit time) for #1 #2 are much larger than b and, then the derivation on ages 2-19 & 2-20 of the manual alies. In the t limit, one has 1 = ½ and 2 = ½. So let s summarize the results in the t limit: (a) If = 0, then 1 = 0 & 2 = 1, B 0 situation ith no relaxation 5

6 (b) If < b, then 1 < ½ & 2 > ½, B 0 situation ith eak relaxation (c) If =, then 1 = 1 & 2 = 0, B = 0 situation ith strong relaxation (d) If one alies a strong rf magnetic field in resonance, 1 = ½ & 2 = ½. No let us relate this summary to the transmission of light to the detector. First define the robability er unit time of hoton absortion (and re-emission), P abs : P abs = (b 11 + b ) 1 As e stated earlier, such emission of P-to-S state hotons is not secifically directed toard the detector, so such emission looks like a reduction in transmission of light beteen the lam and the detector. Thus a higher P abs corresonds to a loer transmission. Case (c) corresonds to maximum 1 (= 1) hich means P abs is a maximum, giving a minimum transmission (zero-field resonance). Cases (a) & (b) corresond to the uming situation here 1 < 1/2 and P abs is smaller, giving an increased transmission. Case (d), here a resonant rf magnetic field is alied, relates to an intermediate situation here 1 = ½, meaning that P abs could be beteen those values for case (c) and cases (a) & (b). Thus for case (d) the transmission at rf resonance should be beteen the zero-field case and the uming situation. Perhas this is the reason that the transmission reduction for the zero-field resonance is stronger than that for the rf field resonance. 2 here α = (b + ). Thus hen the rf field is suddenly turned off, the otical-uming state recovers at rate α hich is roortional to b. Try to observe this recovery-rate deendence on light intensity ( b ). Note that 2 in Eq. (5) has the form [ 1 ex( αt) ] 6