H Ml u H ‘ .t I “ "w ‘ ‘ I u l l M ‘IH r l \ i‘ in '1 mm 1 ‘ 1 i K ‘ WI 133 613 THS SOME ASPECTS OF THE QUANTUM MECHANICS OF THE DEUTERON Thesis for The Degree of M. S. MICHIGAN STATE COLLEGE Wen~Chin Yu 1949 —-———-——-"""—""- ‘A 3 1293 017749 llilllllllHHlllWWIHIUlllHIWll}11W “2W”! “1-— LIBRARY Mlchigan State . . University This is to certify that the thesis entitled SOME ASPECTS OF THE QUANTW MECHANICS OF THE DEU’IERON presented by Wen Chin Yu has been accepted towards fulfillment of the requirements for ___3Lo_s_o__degree in Physics Major ’profess -7 M TL Date Mamm— law; L “-796 .___________.._.,._.___,-, PLACE IN REFURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE 1/98 c-JCIRC/DateDue p65-p.14 SOME ASPECTS OF THE QUANTUM MECHANICS OF THE DEUTERON By 'Wen-Chin Yu A.Thesis Submitted to the School of Graduate Studies orllichigan State College of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Department of Physics 19h? rACKNOWLEDGEMENT I wish to express my appreciation to Dr. Chihiro Kikuchi for suggesting this problem and for his interest and patience during the preparation of this paper. W- Wv?’ l0 Lu 1% 01.9 TABLE 0! CONTENTS page I. Introduction . . . . . . . . . . . . 1 II. Binding Energy . . . . . . . . . . . 5 A. Experimental Dtermination . . . . . . 5 B. Theory of Nuclear Forces and the Binding Energy . . . . . . . . . . . . 8 III. Experimental Dtermination of Spin and Magnetic . . . . . . . . . . . . 15 A. The Deuteron Spin . . . . . . . . . 15 B. iMagnetic Mbment of the Deuteron . . . . 18 IV. Discovery of the Electric Quadrupole moment of the Deuteron . . . . . . . . . 22 ‘V. Electromagnetic Interaction of the Electrons with the Nucleus . . . . . . . . . 26 A. Electrostatic Interaction —- Electric Quadrupole Mbmgt . . . . . . . . . 27 B. Magnetostatic Interaction - Magnetic Dipole Moment Due to Nuclear Current Distribution . . . . . . . . . . 32 C. Angular Momentum Consideration for Non-central Forces . . . . . . . . 36 page VI. The Theoretical Interpretation of the IMagnetic Dipole and Electric Quadrupole moments of the Deuteron . . . . . . . f3 A. Necessity of Non-central Force . . . . 5 B. Properties of Pauli Spin Function . . . #5 C. The Evaluation of the Electric Quadrupole moment . . . . . . . . 52 VII. Conclusion and Discussion . . . . . . . 61 Bibliography 0 o o o o o o o o o o o o 62 SOME ASPECTS OF THE QUANTUM MECHANICS OF THE DEUTERON I. LNTRODUCTION— IMPORTANCE 91;; THE DEUTERON, PROBLEM _I_N_ NUCLEAR PHYSICS The deuteron plays in nuclear physics a role similar to that of the hydrogen atom in atomic physics. It consists of two elementary particles, one proton and one neutron. In investigating the quantitative theory of nuclear forces, its crucial test is the deuteron, the simplest stable combination of heavy particles (neutrons and protons) composing the nuclei. The deuteron tests the theory without aggravating the computational situation. In spite of the comparative simplicity of the dynamical system, many phenomena of nuclear physics are known to occur in conjunction with the deuteron, such as, photodis- integration, capture of neutrons by protons, scattering of slow and fast neutrons and protons, etc. The purpose of this paper, is to consider in detail a fan'of the experi- mental and theoretical investigations of the deuteron, namely, those associated with the binding energy, spin and magnetic dipole moment, and the electric quadrupole moment. Before proceeding further, let us briefly recall the discovery of the deuteron. The discovery of the isotope H2 of hydrogen was the result of very exact and careful measurement. At first it was believed that there was only one isotope of oxygen -- namely 016. Later, in 1929 1, oxygen was found to be a mixture of isotope 016 with slight traces of 017 and 018. Therefore it was concluded that hydrogen must also carry a trace of an isotope heavier than Bl. In 1931 Birge and Menzel 2 of the University of California estimated the relative abundance of isotope H2 to H1 as one part in 1.500. In 1932 Urey 3 of Columbia University, subjected a residue obtained by distilling liquid hydrogen to spectral analysis. The Balmer lines of B]- were obtained as usual, but each line was accompanied by a faint line on its short wave-length side. The Rydberg constant for an atom with a nucleus of mass M is related to that for an atom with a nucleus of infinite mass by Rm = Roe 7%? ’ (1'1) where m is the mass of the electron. Hence .112 = l‘2”‘2 + n) . (1.2) The subscript l and 2 refer to El and H2 respectively. If A1 is the wave-length of a certain line in the spectrum of H]- and *2 that of the corresponding line in the spectrum of H2, it is seen that 31/)‘2 =~ Rl/Rz . Replacing the left side of Eq. (1.2) by Al/ A2 and solving for M2 we obtain M1 m"1 (1.3) 112 _ . By measuring the difference )xl - A2 between a Balmer line and its faint companion, Urey found the latter was due to radiation from atoms of mass 2. Because the isotope H2 is of great interest and importance it was given the special name of deuterium and designated by the chemical symbol D. The nucleus of deuterium is called deuteron. The deuteron has about twice the mass of the proton, i.e., M = 2.01472 and carries the some electrical charge e. Its spin is l in units of 4h and obeys Bose statistics. Its magnetic moment is + 0.8565 :t. 0.0001. nuclear magnetons. These values refer to the ground state of the deuteron. In the quantum mechanical description of the deuteron, it is reasonable to assume the ground state to be an S-state, i.e., a state of zero orbital angular momentum, L: 0. Since a free neutron and a free proton each has a spin Q, the spin value of 1 seems to indicate that the proton and neutron have parallel spins in the deuteron. The deuteron also possesses a small electric quadrupole moment. This interesting and surprising discovery was made by Rabi 1* and his co-workers. According to their results, the deuteron is cigar-shaped, i.e., elongated in the direction on the Spin. The existence of the quadrupole moment seemed surprising, because it had been assumed that the ground state of the deuteron would be a pure S-state, for which the quadrupole moment would vanish. To resolve this difficulty, Rarita and Schwinger 5 introduced tensor, i.e., angle-dependent, forces. The effect of this type of force causes the deuteron to spend part of its time in the S-state (L==0) and part in the D-state (L==2). The quadrupole moment arises from the fact that the deuteron contains a small mixture of the D-state. One of the consequences of this theory is that the magnetic moment of the deuteron.wdll not be the emu of the magnetic moments of the free neutron and free proton. This is supported by recent measurements by.Arnold and Roberts 6. They showed that the deuteron moment is smaller than the sum of the magnetic moments of the free particles by +0.0228i0.0016 nuclear magnetons. Because the Rarita- Schwinger theory has been remarkably sucsessful in accounting for experhmental results, a detailed discussion of their theory will be given in a later section. In order to discuss this theory, we shall consider the deuteron binding energy from which the general characteri- stics of proton-neutron force can be deduced and then consider the concepts of magnetic and quadrupole moments, as they arise in classical electrodynamics. II. BINDING ENERGY In this section, we shall review the methods used in determining the deuteron binding energy, and point out its relation to nuclear forces. A. EXPERIMENTAL_DETERMINATION The binding energy of the deuteron was first measured by Chadwick and Golhaber in 1931. 7. They observed the photodisintegration of deuteron produced by 2.62 Mev. ‘Y- ray from.thorium C' by means of an ionization chamber. 112 + hi} ——’_1H1 + onl . (2.1) This reaction takes place when hy’ which is the energy of the incident )/-ray, is greater than the binding energy of the deuteron. The difference between hy’ and the binding energy appears as kinetic energy of the neutron and the proton. Because the momentum of the )/-ray is small, the momenta of the proton and neutron are almost equal but opposite. Further, since their masses are almost equal, they share the excess energy very nearly equally. The energy E of the proton can be determined by measuring its range. The binding energy is then hy’== 2E. In 1937 Chadwick, leather and Bretscher 8 used the 2.623 Mev. Y-ray from thorium C"to disintegrate deuteron in a cloud chamber and measured the range of disintegration protons. Using Blackett and Lee's 9 range-energy curve, they gave 0.185 Mev. as the proton energy and 2.62 - 2)(0.185 '= 2.25 i 0.05 Nev. for the binding energy. ‘Using Parkinson, 10 Herb, Bellamy, and Hudson's more recent values of proton ranges, they estimated the proton energy as 0.2 uev. and the binding energy as 2.22 Mev. Bethe 11 made corrections to these data and obtained a proton energy of 0.225 Mev. and the binding energy for the deuteron of 2.17.: 0.0h1lev. One of the direct measurements of the deuteron binding energy is that done by Stetter and Jentschke 12, who measured the ionization produced by the disintegration protons in an ionization chamber and obtained the value 2.189 :t 0.022 Mev. In 1940, K. Kimura 13 obtained the deuteron binding energy 2.189 - 0.007 Nev. by estimating Ithe energy of disintegration neutrons from.RaC-D. ’ A.more accurate determination was made by Wiedenbeck and Marhbfer 1h in 1945; (a similar method was also used by myers and L. C. van Atta 15). Let us describe the experiment in detail, shown in Fig. 1. They made use of the resonance absorption of neutron by rhodium or silver. It is well known from theory 16 and from experiments on the excitation of nuclei by x-rays 17, that the intensity of any isochremat in the thick target of continuousII-ray spectrum, increases linearly when the potential applied in WA 76/? COOLED X-RAY TARGEf ’32" 6040 CUP ***‘—“‘* i. ' ' r T? £25 02 O '3 h 06 3/1/54 owe ‘ _ - .1 l, _ _ __,_- ': \ :1? ‘ “““ ‘ I ’1 I' 1 u 1' r $ IVEU rRo/v COM/r41? l ‘ - _— I.) H. .i 1 ‘i 6215. .' “ .__.__._... -11..._.__. 4—1;}:3‘..m=—3J Sr‘hvmutic diagram of th(- apparatus used for -'.r-h~rrmninur Illt' threshold voltages. Phr- sr‘alv of thr- C a In I I I, ‘ , hurmnt: rs rnrlrr‘ntwl by lllt‘ total (hrr'knrws a3 )nl the haul ~§‘ir'lrllilg. accelerating the electrons is greater than the energy of the isochromat. Therefore, in a 'Y-n process, the number of neutrons of a given energy should be a linear function of the applied voltage V when .A Vévt+ —A——_-—l—-Vn , (2.2) where'Vt is the threshold potential, and.A the atomic weight of the neutron being considered. Thus the essential features of the experiment, shown in Fig. l, are the use of the high voltage Ikrays to produce disintegration and the detector which is sensitive to only one neutron "line" energy equal to the resonance energy of the detector. The experimental result will not be affected by the faster neutrons. Thus the activity vs. accelerating-potential curve should give a straight line intersecting the abscissa at the binding potential. The detector used was an argon- ether filled counter with a rhodium cathode. Small samples of deuterium.were bombarded for two minutes bin-rays produced by a beam.current of 100 microamperes striking a thick gold .target. The activity was taken as the number of counts above the background obtained during the two minutes after the irradiation was stopped. The activity was plotted as shown in Fig. 2. A straight line is obtained, which, when extrapolated to zero activity, gives for the deuteron binding energy the value 2.185;: 0.006 Mev. (5000 (2000 c cunts Firth". on O O O . / 6000 activity in / 3000 -j/ 2.0 1'13. 2. 210 2.2 2.4 2.6 2.8 3.0 3 2 electron beam energy mev loam: counting rate venue accelerating potential. Still another method is due to Stephens 18 who deduced the binding energy by mass considerations. The masses of deuteron and proton are known accurately from.mass spec- trographic data, and the mass difference of neutron and proton is known accurately from.measurements on the reaction chain 19 013(P. n)N13(/s*)013. From the equation [2M(P)+ {M(m) - M(p)} - M(D2)]= Binding Energy he obtains the result 2.187 1.0.011 Mev. B. THEORY‘QE NUCLEAR FORCES_AND THE BINDING ENERGY In order to account for the binding energy of the deuteron quantum.mechanically, one must know or guess something about the nature of the nuclear forces holding the neutron and proton together. Since nuclear effects cannot be accounted for by known forces, such as electric and gravitational forces, it is necessary to introduce an entirely new type of force. we shall assume it to be a short range central force. Let us now derive a few results of the type of force ‘we have postulated. The Schroedinger equation transformed to the center of mass of the proton-neutron system.is given by V2Y(/l.€.4’) + it; [E-v(/u]yrr/z.a.¢)=o, (2.3) where r is the distance between the proton and the neutron, and m.the reduced mass given by Mi-+IH2 M stands for the mass of either the proton or the neutron. The quantity E occurring in.Eq. (2.3) is a negative quantity and its absolute value is the binding energy as determined by experiment. Since E is a known quantity we can determine the unknown quantity V(r). For this purpose, consider the ground state. Since L=0 for the ground state,Y must be spherically symmetric. If we assume a central force, i.e., the interaction potential of neutron and proton is only a function of the distance r between the particles but independent of angles, then the wave function in Eq. (2.3) is also a function of r only. The Laplacian operator applied to the wave function 1V== U(r)/r gives er ‘v r r dr2 Eq. (2.3) then becomes 2 l d u M: u __ i.e., c132 .,_ 3!. [r - V(r)] u=0. (2.5) dr2 ‘52 An assumption about the shape of'V(r) is necessary. One possibility is the rectangular potential well. This shape, -10- (Fig. 3), certainly represents a short range force and leads to a differential eqwation which can easily be solved. Here, there are two parameters, the width and depth of the well. Since the Schroedinger equation with a given E I determines only one parameter, we expect to find only the relation between V and r0, e—flo ; - W) TV. -——)/L not definite values for them. ____i oL With E =- - E0, where E0 is 113. 3 ' positive, Eq. (2.5) becomes d2u u —-+—[V -E0]u==0 forrro. (2.6)b dr2 2 O . 3? must be continuous, finite and have a continuous first derivative everywhere. Therefore u = rY, which must have the same continuity conditions, must go to zero at r==0, and must not diverge faster than r as r-aoa. To satisfy the conditions at zero and infinity the solution of (2.6) must be u = A sin kr for r 4 r0, (2.7)a U =- Be'Sr for r > r0, (2.7)}, where 1: == JM(v - roT/ h , (2.8)a -11- Now if U and its derivative are continuous then the deriva- tive of log U must also be continuous. Applying these conditions at r = r0 we obtain the result k cot kro = - at . (2.9) Eq. (2.9) does not involve A and B but only the two unknowns ro and 70, E0 being 2.187 Mev. V0 and r0 are not restricted further. As seen from the above, E0 is small compared to '0 and can be neglected in Eq. (2.8)3. Eq. (2.9) can be put in a simple'and approximate form cot no == - 04/1: ass/m. (2.10) Thus cot kro is negative and small in absolute value. Therefore, kro is only slightly larger than 71’ /2. (We know the value of kro slightly larger than 371/2 is not admissible since there would be a radial node in the wave function «y at kro = 7T, indicating that this is not the lowest energy level. This result contradicts the hypothesis.) Using krom 7r/2 and again neglecting EC) in the expression for k, Eq. (2.10) gives v0.3 R5 712 412/411. (2.11) Actually (Vorg) is slightly greater than thequantity on the right. Thus 2 2 Vol-g < 7T Mfi OI’ kro 4 7T 0 (2012) (The expression Voro2 frequently occurs in nuclear calcula- tion; it is not necessary to know V0 and r0 separately.) -12- Let us consider another result which does not depend on the detailed form of the potential (as long as the force has a short range). Such is the case for the wave function U(r). When r is greater than the range of nuclear forces, it is given by U = ce'M‘. (2.13) For r less than the range of nuclear forces, the wave function should decrease as shown in the diagram (Fig. 1.). However, the wave function given above is sufficiently accurate over the whole region, as to be useful in many calculation. Since the fit 9" exponent o( r in the above I\ zit/”é 2 expression is a dimension- ) ~-‘_ or/\ less quantity, d has the we A ’ dimension of l/cm. Therefore we can take l/e( as a Fig. 1. measure of the size of the deuteron. Since we assume that a central force has short range, the "radius" of the deuteron is considerably larger than the range of nuclear forces. That is r0 << l/et . (2.11;) Most of the area under U(r) occurs for r > r0, The use of another shape for the potential function changes U(r) appreciably only for r < r0. Therefore whatever the shape of the potential, Ce'dr is close to the true wave function - 13 - over'most of space. Let us normalize “Y so as to determine the value of C. Then fy/Zdr = twjowuz dr = “T0210“ e‘zdr dr =-— 27102/04 = 1, 0 == /ok/27r . Therefore, the normalized wave function is 01‘ u(r).-.e _.L_ e'°‘1‘. (2.15) 27r Similarly if definite values are assigned to ro and V6 then .A and B of the true U(r) given by Eq. (2.7) can be found from.the continuity and normalization conditions. B is somewhat greater than C of the approximate U(r), thus B =.- .4 (1 + 5M0). (2.16) M In fact Eq. (2.16) is a good approximation. 'We shall conclude this section by showing that it is not possible for the deuteron to have bound excited states. Let us consider the possibility of the existence of excited states with.L==0. It will be assumed in this proof that the force between the neutron and the proton is the same for states of higher I. as for the case L==0. In order to prove our assertion we shall compute the minimum well depth V0 required to produce a bound state, i.e., one for which the binding energy no is Just zero. This required well -1),- depth will be found to be considerably larger than the actual well depth, as determined above from the binding energy of the ground state. Since the actual depth is less than the minimum value required for the binding of states of non-vanishing angular momentum, such bound states can not exist. The analytical proof is as follows. For I. = 0, the Schroedinger wave equation is 2 Lgirnp -V(r)]u - ““1 u=0. (2.17) dr £12 1‘2 Assume V: -vo, rzro, . E =-Eoo Set E0 = 0 for minimum well depth, and let L==1. The solution of Eq. (2.17) is given by U = M - cos kr, r 4 r0, (2.18)a kr U = 6"“ [:15 -r 1 ], (2.13),, where k2 = mvo - so)/ £2, (2.19)EL .22 = 1130/ 4.2. (2.19),, For simplicity set EO == 0, and use for the outside solution U =3 l/r, r 2 r0. (2.20) Using the definition of k with E0 =-- o, moroz/ 62 : n2. (2.21) The required depth V, .is almost four times as large as the actual depth in the ground state. Therefore no other bound state exists at I. =- 0. - 15 - III. EIPERD’IENTAI‘ DETERMINATION . Q; . TEE SPIN AND MAGNETIC MOMENT OF THE DEUTERDN A. THE_DEUTERON SPIN It is known that the electron possesses an intrinsic angular momentum, called the "spin", of value i and a magnetic moment of l Bohr’magneton. The simple relation between these quantitites has been pointed out by Dirac, who derived his equation by relativity arguments. ror protons and neutrons and nuclei consisting of these particles, however, the relation between the spin and magnetic moment is not a simple one. So far, there is no theory of general applicability which gives this relation- ship. The lack of such a theory makes the problem of measuring spins and magnetic moments of nuclei very important in nuclear physics. For this purpose a number of methods have been devised: hyperfine structure, mole- cular beams, band Spectra, etc. Since the spin of the deuteron has been determined by the band spectra method, this willa%iscussed in detail. The band Spectra method is very useful in determining nuclear spins, particularly for light nuclei which tend to form homonuclear molecules. The presence of a nuclear spin causes a change in the statistical weight associated with the given rotational state of a homonuclear diatomic molecule and thus causes a change in the expected intensities -16... found in the band spectrum. The reason for this is as follows. If the nuclear spin is zero, the statistical weight of any state for which the total angular momentum is 3’ would be 2J+l. The presence of a nuclear spin further increases the degeneracy of the states. A nucleus of total angular momentum I can have a component M in any prescribed direction, taking any of the values I, I - 1, ---I,a total of 21+l states. Let NF“ represent the wave functions of these states. If a molecule contains two identical nuclei, the wave function of each being represented by Y’Miu) and YMJB)’ the total number of ways in which these functions can be combined to form the product functions Yul (A) (”“2 (B) (3.1) is (2I+l)2. Since symmetric and antisymmetric states do not combine in diatomic molecules containing identical nuclei, it is necessary to separate the (ZI-rl)2 functions into symmetric and antisymmetric functions. We note that the product ftmctions given by Eq. (3.1) are symmetric when M1 = 112. Of the remaining (21.»1)2 -. (21+1) =- 2I(2I+l) functions of the form +111”) “I’M2(B) and YHZ(A)V’M1(B), one half will be symmetric and the other half antisymmetric. Thus the total number of symmetric functions is I(2I+1) + 21 +1 = (I+1)(2I+1) - 17 ; and (211-1)I, the number of antisymmetric functions. The ratio of these numbers is (I+1)(21-I-l) __ 1+1 . __ (3.2) I(2I+l) I If the nuclei obey Bose statistics, symmetric nuclear spin functions must be combined with rotational states of even.L and antisymmetric spins with those of odd L. Because of the statistical weights attached to the spin states, the intensity of even rotational lines will be (Ii-l)/I as great as that of the neighboring odd rotational lines. For Fermi statistics of the nuclei, the spin and the rotational states combine in a.manner to make the odd rotational lines more intense, in the ratio (I+—l)/I. Thus by determining which lines are more intense, even or odd, the nuclear statistics is determined and by measuring the ratio of intensities of adjacent lines, the nuclear spin is obtained as shown in Fig. 5. occupah'an num ber ) (11.. Infemiiyalfemdflon in Band spectra. Fig. 5 The deuteron spin was investigated in l93h by Mbrphy and Johnston 20, who measured photometrically the relative -13.. intensitites of 29 lines in the o< bands of the molecular spectrum of deuterium. These lines correspond to the tran- sition 3p 713 mu." 2p 0323/: analyzed by Dicke 21. The bands lie between 5939 and 6291 A and were photographed in the second order of a 21-foot grating. The experimental results show that the deuteron obeys Bose-Einstein statistics and that its spin is l. B. MAGNETIC MOMENroEm DEUTeRON Magnetic moment of the deuteron has been detemined by hyperfine structure and atomic beam methods. The latter method was used by Rabi and his co-workers 22, who measured the hyperfine structure separation by passing a beam of atomic hydrogen and deuterium through an inhomogeneous magnetic field. The magnetic moments are then calculated from the hyperfine structure separations. The disadvantage of the above method is that the result depends on the somewhat uncertain relation between the nuclear magnetic moment and the hyperfine structure. This criticism is avoided in the molecular-beam magnetic resonance method 23 in which the nuclear moment is obtained by observing the precessional frequency of nuclei in a uniform magnetic field. Consider a system with angular momentum J, in units of h and magnetic moment ,a. , in an external magnetic field Ho. The angular momentum vector - 19 - will precess about the Eb vector with the Larmor frequency given by =’“_Ho_ )2 In . (3.3) Of the quantities occurring in the equation, Eb may be measured by conventional procedures and J can be determined by some one of the methods mentioned in the previous section. For the determination of.v , Rabi 2“ has pointed out an ingeneous method. It is as follows. From.Eq, (3.3) we see that if a nucleus of magnetic moment,LL is placed in a uniform field Eb , it will continue to precess with a constant angular velocity.V and the angle which the angular momentum.vector’makes with Eb will remain constant. Now' suppose, in addition, that a small rotating field Hi is applied at right angles to the steady field E5. The nucleus will then also tend to precess also about this small rotating field, thereby changing the angle which J makes with Ho. If Bi rotates with the frequency 1/ this effect is cumula- tive and the change in the angle between Eb and I can be made large. If the frequency of revolution f of’Ei about BB is different from J , the net effect will be small. The smaller the ratio Hi/Bb the sharper this effect will be in its dependence on the exact agreement between the frequency of precession y’, and the resonance frequency f. The schematic diagram of the apparatus used in detecting these reorientations is shown in Fig. 6. .A stream of m .mp— ._5mso..oo_. :95. ms..— =020om3oa of 53:3 we .559 of 5.. .25 688B... .32. as: 3...: of was? 5.5.3:. “focus... of .3 :otoioha of not...» .0 2.02.3 £963 @3511. em :35: a.x_:om.:.mu 3.5. of .5 23...: .3 138:5. E of? Jesus... U of 3 common of E ceases—o cox—.1 1...... of use—e 392:3... “discus... doc—o2: omega .0 actoomoa of mo_=o..._oE 0.5 .3 223 of 38.3 Zeus... m~ o... .3 comm?— of 5 noise .533. c»: 2:. .18.: 2:55. Eoc of mac—m “5:5... 2.233. of 3 5.29%.... 2: sot—.5 E .m :2. emit .3 use so use... moacomoim :95 2: .3 3.8%.: a a: 63355... of snack... snowmen. 5.3. tow—:30 so: “to mEoEoE omen? cos 8:629» can 3:85... 3.3.5:... acts; 3.3.53... 9.: .o mfea of 382:: mo?.:o 1:8 0...: 2:. .8383... .o 2:...— uocmmEm _ — U _ HocmnE< -20- molecules coming from the source, 0, in a high vacuum apparatus, is defined by a collimating slit, S, and detected by some suitable device at D. The magnets.A and B produce inhomogeneous magnetic fields. When these magnets are turned on, molecules having magnetic moments will be deflected in the direction of the field gradient if the projection of the moment,‘;42, along the field is positive, and in the opposite direction, if,/cz is negative. The magnets A.and B are adjusted so that the number of molecules arriving at D is the same whether the magnets A and B are on or off. The molecules also pass between the poles of magnet O, which produces the uniform field 35. Instead of a rotating field, an oscillating field (which is equivalent to two fields rotating in opposite directions) at right angles to the steady field is produced by means of a loop of wire inserted between the poles of the C magnet. (The loop is not shown in the diagram.) If the frequency of the oscilla- ting field is just right to induce re¢orientations of the molecules, i.e., to cause,“z to change, magnet B will no longer be able to bring the molecules to the detector D, i.e., the deflection will be too large if,“ 2 increases and too small 1f.fi‘z decreases. Thus the condition of maximum.reorientations is determined by noting the minimum molecular*beam.arriving at D. -21- By the use of quantum mechanics, the reorientation process is more precisely described as one in which the system, originally in some state, with a magnetic quantum number, 111, makes a transition to another magnetic level m'. An exact solution for the transition probability for the case where El rotates and is arbitrary in magnitude was given by Rabi 21*. For the particular case J = i we have P(% _§)= ginzo Sin27rft(l+22-ZZ cos 9)%, (3.h) ’ 1+2 -2z cos 0 where P“? é) is the probability that the system, originally ’ in the state m=§, is found in the state m == -5 after a time t; z, the ratio of the frequency of revolution f to the frequency of precession y , given by y =,u(H§ + Hi?) 17: a, (3.5) tan 9 = Hl/Hoo (306) For an oscillating field, when Hl/H°<< l, in the neighbor- hood of f=fl this formula becomes 31:13:11: [(l-z)2+ A211), (3.7) Pat» .—_-. (l-z)2+A§- where A =Hl/éHO is one half the ratio of the amplitudes of the oscillating to the static fields. .To determine the magnetic moment of the deuteron 25, the molecular beam of HD or D2 is used. The molecule HD or D2 is in the state with rotational angular momentum, J, - 22 - equal to zero, the internuclear axis is oriented in every direction with equal probability. Therefore, if the applied steady field is large enough to decouple the particles from each other, the frequencies at which the protons or the deuterons make transitions will be effectively that of the free proton or the free deuteron. The experimental results are obtained in the following way. The radiofrequency oscillator is set at a definite constant frequency f and the molecular beam intensity is observed as a function of the homogeneous magnetic field B5. For certain values of the magnetic field the beam.intensity becomes a.minimum. This occurs whenever the product hf for the oscillator is equal to the difference in energy between two molecular states. The resonance curves for deuteron in the ED and D2 molecules are shown in Fig. 7, 8. By analyzing such curves, it is found that the magnetic moment of the deuterons is 0.855 x 0.006 nuclear magnetons. Iv. DISCOVERY 91 THE momma QUADRU'POLEMOMENT 9; THE DMRON In the last section, we outlined the general procedure for determining the magnetic moments of nuclei by the magnetic resonance method and in particular, we pointed out that the very deep minima at the center of the resonance curve corresponds to the transition frequency of the free nuclear particles, such as the free proton or the free .ouopu dosage»: cums on» no.“ gnu eonopsow on» no dogwood: mafiaonu hCoHno nogooaos am no“ Ergo oogomom é. .mrm. mmD§02_OJ!h”!hmZO<2 00mm 0050 Ome 9:5 u .q o : 9v a 5233.... a: so Ausuama m’rze .33- Heoogeaou one: 23 you one. scones: 3.3.9:." on» we “Savannah... Manson- uogoonoa an no» erase eonoaeuem .w mg 336 2. ad... ocuzosz oo.~ 000m 00o. O azozu30u¢u a s o T _ _ n. _ ”‘1 Museum "'39 - 23 - deuteron. However in attempting to account for the auxiliary minima flanking the central minimum, Rabi and his co-workersh’27 noticed a fundamental difference in the spectra of HZ and D2 . They found that the structure of the radio-frequency spectrum.of HQ molecule could be explained by the magnetic interactions of various angular momentum.vectors. They found, however, to their surprise, that a similar reasoning applied to the D2 spectrum led to results not in agreement with the observed facts. we shall explain below the main arguments which led Rabi and his co-workers to the conclu- sion that the discrepancy can be attributed to the electric quadrupole moment of the deuteron. Let us first consider the H22molecule. The angular momentum vectors associated with such a.molecule are as shown in the diagram. The quantitites f1 and a; are the spins of the two protons, and J is the angular momentum arising from.the rotation of the HQ molecule. It is readily seen that when such a molecule is placed in a magnetic field, its energy is given by —-’ *) E=-/u7,(01+?2)-H-,ufi?'H -/upH’ -/8). Substi- tution into Eq. (5.1) gives = °° fut? K v gj V THIN r/z 2K gunman” =f_gd_fi Eff" 7>(c.,sw)dr,, +f-éjfg A7110”? 77(Casw)dtn 7’ (:5 at?” +fjfjflnf" 2 o w) + (5.2) The first term is called the Coulomb energy and is Just the energy due to a charge Ze concentrated at the center of the nucleus. The second term is the electric dipole moment term. Emperiments show that the electric dipole moment does not manifest itself in atomic nuclei. Thus the contribution from this term vanishes. The third term is called the quadrupole interaction term. Let us examine this a little more closely. Since coscu contains both the electron and the nuclear angles, we shall separate them.by using the addition-formula 23 for the Legendre polynomials. -29- That is “P" (cos .0): x; (2— %)Lm+fi, ’P: ((05 0)?“ ((054)605 I“? 13). (5.3) Substituting this into the quadrupole twem, there results v 7:“ £K)g_;__:;; It: P2K(Co$ waltefflzfn 7;“ccosocosxd-purn. (5.1,) Since the nucleus is rotating rapidly about its spin axis, it is safe to assume that the nuclear charge distribution has axial symmetry. Then the charge density fn is inde- pendent of ,3 , and the only value of k for which the inte- gral does not vanish is for 1:240. Hence Va=_)}z‘§ £9 731co56)d‘( f/z: )0 amount?" __ _,_ 1° (3605254) 2 .. ‘ 4] 8 Ag are Pam‘s“ 013,4 (5.5) where eQ =-:: fijrnzm coszok- 1) find Tn , (5.6) which is the nuclear quadrupole moment, and 6‘12 o (507) fa (3 cosze - 1) 6. Te J 1.; It can be shown that this represents the second derivative of the potential, or the gradient of the electric field along the nuclear spin axis arising from the elctron charge distribution. The proof runs as follows. quf .Fe(3 00829 ' 1) d?e szeUZE " 1%) d ‘(e , r3 r5 6 - 3o - The potential due to the elctron charge distribution at points inside the nucleus is 're ' 3n) Fe d—EL ”If 1 53-51 Ifle" _—/z_—n) (€32: I/ze"/ln) Therefore .74 3..( - I/ze— 12.15 Hence, at the center of the nucleus 9: £132;- ( . ) [22% Illa-Ail.) /z,.= a 4%}, dz"- 5 9 To see how the shape of the nucleus affects the quadrupole moment, assume a nucleus to be a uniformly charged ellipsoid of revolution having total charge Ze with samiemajor and semieminor axes of a and b respectively. .Assume that the spin and the major axes coincide. Since the volmme of this ellipsoid of revolution is h/37rab2, the charge density is 326 #11 8th2 . fl — ) dTn=R «wan dz, r1210 cosZaL- 1):”: - r121 == 222 - 112, (5.10) Fig. 12 -31.. sq =.._ b 1’2 22/3 2” (222-32)_1§2.. R (it? (112 dz J J zoa ¢=0 a b 1-:2/a2 = 21f! 5 (222 - r2)R dR dz a 0 271’]a [b222(1-zz/a2) - b2(l - 22/a2)2/h] dz -a = 87Tab2(a2 - b2)/15, 32a 871‘ 2 2 2 eQ:———-— ab (a - b ) (5.11) [web2 is ’ Q .-.= E 2(82 - b2). (5.12) 5 This verifies a remark by Mattauch 30. Similarly in case the spin axis is the minor axis, the quadrupole moment turns out to be oz; 2(1)2 - a2). (5.13) From these results we see that in case the nucleus is lengthened along the spin axis the quadrupole moment is positive (Fig. 135) and negative when flattened (Fig. 131,). (I I I ZN 4. g a Fig. 13, b2< a2 Fig. 13b b2> a2 - 32 - The axis ratio a/b calculated according to Eq, (5.12) is plotted in Fig. 1A. B. MAMETOSTATIC INTERACTION .— MACMTIC f DIPOLE mom DUE gq NUCLEAR CURRENT DISTEBUTION In the preceding paragraph we considered the effect of stationary distribution of charge upon the energy of an atom. Let us now consider the magnetostatic interac- tion energy of the atom. By this we mean the contribution to the energy of the atom due to a stationary distribution of nuclear currents, that arise from the motion of charges within the nucleus. Our purpose is to determine the effect of the internal motion of nuclear charges on the magnetic moment of the nucleus. Let us suppose that the motion of particles with-in the nucleus gives rise to an effective current of density 3.. The vector potential 31 K due to this distribution is given by K =1] 35’“) (”n . (5.11.) Ir - rnl Expanding the denominator, we obtain the following. I) 7:”? , '0 1 K—; a = cml I} (72,»)714'0. \(r + all/HfIAKl?(An)71(cosw)dTn. (5.15) Fig. 15 I I f T V a Y . ‘r Y l )0 __.__._... --—. — ._- _- ._ _ __ _.___-__. l i 105 J 4; _---- gl-c ' ' l t A . lOO H + u t l l 1 l l l l l l l 1 1 l l l 40 50 60 70 80 90 100 110 l20 I30 140 ISO 160170 l80 I90 200 210 A—> Ratio of nuclear axes when viewed as ellipsoidal, as vulculated from the quadrupole moment. The arrows oorrespond to the limits of experimental error. A systematic dependence on mass number, A, appears to be indicated, but is uncertain herause of the large S(':)t((‘l'lllg and the small number of experi- mental points. -33.. The first term is zero. The second term is 1 l -9 d _ 1 —-y —’ "" --> .0. WIIrn|3(rn) 008a) (n-c'flzjewnflrfld‘fn, (5.16) where 'e’.""n = I?! (“I-‘3) coscu, and ‘6’ is a unit vector along the vector i". From.the continuity equation div 3 + 1% = o (5.17) C and from the fact that we are considering stationary distribution of charges and currents 2% -"-"-'- O) (5018) b we see that the current density satisfies the relation div 3:: o. (5.19) This means that 3 can be derived from the curl of some vector. Let us write therefore —+ J --' cvxm, (5020) A - l "‘ "’ “’ d -_—2—(e.r vxmd‘r . (5.21) 11) Ir) :1 n From the expansion formula 32 -D --) -" VX(UV)=VUXV+UVXV , we get gofiVXfi=VX(-é’o‘f;fi) "V‘E’o 1711))(3 Vx(g,oA_—’nfi) " (Gov-1:3)XE =VX(?. Ea) "' "é’XI—If, -34.. Amp: - 1 Jé’xfidrn:- 3’ xffidrn, (5.22) Ir)2 lrl2 since the volume integral of the first term vanishes. Comparing this with the vector potential of a dipole, .. M. (5.23) Adip — )rl3 , where 52’ is the magnetic dipole moment, we see that (Edrn=x7 (5.21.) is the contribution to the magnetic dipole moment due to the motion of charges in the nucleus. We shall now obtain a more convenient expression for this orbital magnetic moment. From Eq. (5.20) we have A .9 -5 -9 r J=rxvxm, x c g; f?x3’drn==ji7xvxfid‘rn. ' (5.25) c From the expansion formula 32 Grad (U . V) = V.VU +U.vV+ V curl U+ U x curl V, we obtain Grad (Y'dii): ?.vm +m.vr+iixvx?+ 1.131(va = m’.v'r’+m.vr+§ixvx'r’+?xvxfi. Then 'r’xvx'fi == Grad ("13.13) -'I'E.v'i"— mv?-'fiixvx? Grad. (120—13) ”m -m ~3KVX?. Substitute the above equation into Eq. (5.25) -35.. _’ 1 ”fix dr =-f2mdr . 3f ‘1 n 11 Comparing with Eq. (5.21»). we obtain gIE’xi'drn = 2,22, c "’ = 1 ?deT . ,u 20f 11 By means of 3" = fv , (5.26) where 3’13 the charge density and 5’ is the velocity, it follows that ,Z.’ =f£§£ri drI1 . (5.27) The integral of Eq. (5.27) gives the contribution of the motion of the charged nuclear particles to the nuclear magnetic moment. It can be written in the form ,7? .-.-. Z fig— L, (5.28) where the summation is over all particles and L is the orbital angular momentum of the particles. For the case of the deuteron which contains a proton and a neutron, 2M1° “20 2 Since the neutron has no charge the first term is zero. Therefore fl = .32— L2. (5.30) -35- Since L1= L2 = L/2 , (5.31) we obtain jZorb = 21:1 (533). (5.32) This is the contribution of the orbital motion of charged nuclear particles to the magnetic moment. This is also the quantumemechanical Operator for the orbital magnetic moment. 0. ANGULARMOMENTUM CONSIDERATION FOR NON-m_m Because we find it necessary to introduce non-central or angle-dependent forces to account for the experimentally observed electric quadrupole moment, we shall discuss the classical aspects of such forces. We shall show that, although the orbital angular momentum is a constant for central forces, it is not necessarily so for non-central forces. The fact that orbital angular momentum is conserved for central forces can be seen from the following considera- tions. Let us suppose that a particle is moving under the action of a central force fir). We can then write 111 a’ = ir’ = f(r) 3', (5.33) where 'r’ is the position vector, The quantity 'r’xf is known as the torque about the origin, exerted by the force 1'. Therefore, -37.. mfxa’ = ‘fxi" = i" x f(r) “r’, (5.3).) since the vector product of a vector into itself vanishes. But rxa = 2%.,- (Fur), d - v =._ ‘ E (m IXV) 0. (5035) So the orbital angular momentum is constant, that is mr x V = constant. Thus we see that a particle in a central force field will continue to move in a fixed invariable plane, that is, the orbit of the particle will not "wobble”. On the other hand, consider two spinning dipoles each of mass m and m' and of magnetic moments M and M', as shown in Fig. 16. Consider now the moment of the force about the center of mass. Let r be their distance from each other. The potential of the magnetic dipole 32 is U = - M . v l/r. (5.36) Since the first magnet gives rise to the field H = ( v v 1/r) . M, (5.37) Fig. 16 the potential energy of the dipole m' is “M. o H=-M' 0[M 0(VV1/r)]o (5038) The translational force on the dipole m' is computed by -33.. the formula F = M, 9 V H = M'.V[M.(VV1/1‘)] ’ (5039) and the torque on m' is given by LM' = M'x H = M' x (VVl/r).M. (5.1.0) But .1; = - i:— - (VL)?-l-_V? vv r Vr3 r3 r3 ’ i.e., vvi=2ii - i... (5.1.1) r r5 r3 Substitute Eq. (5.1.1) into Eq. (5.37). We have .._.. 232’-J_=2)££§-i. (5.42) ‘1'1' M ( 1,5 1.3} 1.5 1.3 Substitution of Eq. (5.13.1) into Eq. (5.39) gives —+ 3? if = ' . V . - — . . FR M [—1.5 M r r3] (5 (.3) Using the expansion formula -5 —’ —> VUl-V)= v/xv+,uvv, V(U°V) = V°vU + U-VV 1*le curl U + U x curl V, we obtain the first term of Eq. (5.13) ? 3" . 3" v{.13:.§u.r}={v(ll'r)};.:: + MrV i. =M°Vr2§+3M°r l 'r’+l-_V?} r5 (‘7‘? 1.5 -39- :3 {-J—Mw‘r’i‘ 1.2.1: MI ( .43) 1‘7 1’ r + r5 } 5 a and the second term of Eq. (5.1.3) 1 _-_- 1 M = " L V r3 V? r5 m0 (5oh3)b Substituting Eq. (5.1.3)a and Eq. (5.1.3)b into Eq. (5.1.3) we obtain 3!" = 3P 5 M.rn'r¥+M—-5—'Mr+M—-M°r '+3—M"1‘M° ‘5'“) M' :33 r 1‘ J 1‘5 Therefore the moment of the translational force on m' is E’xffi. = 3.1142 er' .,. 2. M'orrx M. (5.1.5) 1‘5 1‘5 If the moments are taken about the center of mass, then r' = 2R, r = - 212. Then moment of forces on m' about the center 0 is "" _ 6M’R RXM' 6M'.R RXFMg —— 32R5 + 3.55— RXM. (5oh6) Similarly, the moment of the translational force on 111 about 0 is -* 6M' 0R 6M0}! - x = RXM + RXMH e R PM 32125 '3—235 (5 117) Therefore the sum of the moments of the translational force on m and m' is -+ v MB 121M er -er = RXM+--—-RXM'. (5.1.8) 1" M 32125 32125 Thus the sum of the moment of the translational forces does not vanish. This shows that the orbital angular momentum is not a constant of motion. - #0 - Now let us go back to the spin moments of the dipoles m and m'. Let LM and LM' be the torques on m and m' respectively. Substitute Eq. (5.h1) into Eq. (5.40). 'We have for the torque on m! LM. M'XH== M'x[L¥F-L]-M 5 1.3 .13.. tx t _ M'XM BZRSM RMR 1:37. 9 (50119) H and similarly the torque on.m.is given by LM=-.12 MxRM'o'R - MXM' . (5.50) 32125 8R5 The sum of the torques is + = —-M'XRM-R+ 12 MXRM'oR. (5. 1) Hr 1M 7323 5 Therefore the sum.of moments of the translational force and the torques is 12x: 11' + (.42) x rm + LM +LM. = o. (5.52) This shows that the total angular'momentum.of the system remains constant. These results are true in quantum mechanics. ‘We shall show that when tensor, i.e. non-central, forces are intro- duced, the orbital angular momentum.L “111 no longer be a good quantum number, although the total angular:momentum is still a constant of motion. VI. THE THEORETICAL marmunou qg'rHE MAGNETIC DIPOLE AND ELECTRIC QQADRU‘POLE‘ MOMENTS 9;; THE DEUTERONS In the previous section, we outlined the experimental methods used in determining nuclear multipole moments, and develOped the concepts of multipole moments in classical electro dynamics. Our purpose now'will be to see how these ideas can be fit into a consistent quantum mechanical scheme. One suspects this to be possible because quantum.mechanics has been remarkably suCCessful in providing a qualitative interpretation of nuclear phenomena. It is interesting to note that the Rarita and Schwinger theory, which will be outlined below, was anticipated in 1938, in a paper by Yukawa, Sakata, and Taketani 33. Howe ever, the physiCal significance of their theory of the deuteron was not fully realized until after the discovery of the deuteron quadrupole moment 3h. In order to see what must be done, let us assemble the experimental data. The moments of the deuteron are ’“D =+O.8565 i 0.0001. nuclear magnetons. QD = 2.73 x 10'27 cm2. The magnetic moments of the proton and neutron are f‘r = +2.78% 1 0.0008, -42.. #11 =-_ - 1.9103 1 0.0012. We notice that the sum of proton and neutron moments is ,uP +5111 =+0.8793 1 0.0015, which differs from the deuteron moment by (”pI/“n’ - ,aP = 0.0228 1 0.0016. Thus, the magnetic moment of the deuteron is very nearly the sum of the proton and neutron moments. In addition, as has been pointed out previously, the deuteron spin is 1, which is Just the sum of the spin of the proton and the neutron. These results suggest that the ground state of the deuteron is a 381 state. The small deviation from additivity and the small quadrupole moment is due to the presence of a small amount of. 3D1 state. Let us examine the problem a little more carefully by applying the Lands formula of nuclear systems. Let S represent the intrinsic angular momenta of the particles, 3 the orbital angular momentum, and 3 the total angular momentum of the nucleus in the ground state. The eigenvalue of 3, is, of course, the experimentally observed nuclear spin I. Let the magnetic moments associated with the vector '3’ and I: be ’28 and ’21., respectively. Then, because of the Larmor precession of the vectors S and I: about 3’, the effective magnetic moment of the experimentally observable -43- value is #effzfll. cos (L, J)+/ds cos (S, J). (6.1) 1‘ Using the cosine law of trigonometry, we then get 2 2 2 +213 L2 ~23? - J2 . (6.2) “‘37— Introduce the gyromagnetic ratios given by flL =3 gL/ZN, .. (6.3)8 I“: = gJ/(N, (6.3)o where ’uN is the nuclear magneton. Then Eq. (6.2) becomes __ 82-12-12 1.2-3242 Thus the gyromagnetic ratio of the nucleus is _H_ _ g 1-; 1 Bar: - Nantes) 882233?in 38' 8L). _(605) In the last step we replaced 82 by S(S+l), etc., in accordance with the results of quantum mechanics. Under the section on non-central force, we showed that 3L: i. The quantity gs is Just the sum of the g's of the proton and neutron. Hence 33 = g1. + 82:1 207896 " 109103 = 0087930 (606) H -zm- From Eq. (6.6) we see that if the nucleus is in the S state, (J ='- S = l, L=O), the effective magnetic moment should be ”err = [5(3L+g3) +£(gs - ed] Jfln=esiflm (6.7) Thus, if the deuteron ground state were a pure S state, its magnetic moment should be Just the sum of the moments of the proton and the neutron. If the deuteron were a pure D state (S=l, J==l, L==2), its magnetic moment would be 1... .19) - 2(3) ’u-eff [é(gL+gs)+ 2(1)(2) (33 - 81.)] J’uN =[i(eL+gs) - (gs - 3L)]/£N = [,5,_-§_g91 - (1.8.92 - .5)];1N=.3101.,uN . (6.8) The experimentally observed value lies between these two values. Hence we are led to consider the possibility that the ground state is partly S and partly D states. If the percentage of the D state is represented by p, then the observed magetic moment will be given by ,uD = 0.8693(1 - p) + 0.3101111. (6.9) where the first term is the contribution to the magnetic moment from the 8 state and the second term, that from the D state. Inserting the observed value of ’“D in the above equation and solving for p we obtain p = 6%. It is a remarkable result of the Rarita and Schwinger theory 6 that Just this amount of D state admixture is necessary to account for the deuteron quadrupole moment. - 45 - B. PROPERTIES 92 PAULI (SPIN, FUNCTION In developing their theory, Rarita and Schwinger made an outright assumption that the interaction function between the proton and the neutron in the deuteron is given by V(r) = [1 - 532.11g r1-G'2+Y312] J(r), (6.10) 3(F'o r 0‘ °:r 312 = 1 122 2 12 -(rl oo-z . (6.11) 1'12 where 0'1 and 6'2 are the neutron and proton spin operators, reSpectively. Also g and Y represent the strengths of spin-spin and dipole-dipole interactions, respectively. The wave equation of the proton and neutron system is 2 2 [— 3?;- V; - fin— vg+v]yr=zyr. (5.12) This equation can be put into a.more convenient form by transforming it into a coordinate system.whose origin coincides with the center of’mass. The result is 2 -§_72y+vY=E)/a I (6.13) This equation differs a little from.wave equations treated in.many text books on quantum mechanics. The difference is that the potential function V contains an angle dependent term 312. Because of this term the orbital angular momentum. operatorL2 will no longer commute with theHamiltonian 2 . Has-1%- V2+V. (6.11.) -hé- That is [L2, H] 7‘- 0. (6.15) The physical meaning of this is that the orbital angular momentum.is not a constant of motion. Yet there are certain constants of motion. These can be determined by a.method given in Rojansky 35, i.e., by finding the dynamical variables which commute with the Hamiltonian. These are s2, J2, J‘ and the parity. The effect of the last quantum 2’ mechanical constant of motion is to prevent the mixing of states of odd and even orbital angular momentum, Therefore the ground state solution of the wave equation can be written in the form f(J’=l, S==1, mJ, parity even) =‘/’(3Sl) +Y’(3D1). (6.16) The notation used here is similar to that used in atomic spectroscopy. The symbol J represents the total angular momentum and is the observable spin of the nucleus. Let 3 us now proceed to determine the 381 and D1 wave functions. For this, we assume that these functions are of the form +1331) =—. c1 M14761“ , (6.17) 11(3111) = c2 w(r) 312751;, (6.18) where X? are the spin functions, given by xi = .1(1).((2), (6.19) O = _1_ .L(1),3(2)+o<(2)/9(1) , (6.20) )(1 ,2.) 1 X? = p(1)p(2). ‘ (6.21) -1”- The above are spin functions of total spin 1 with components in the direction of quantization equal to l, 0, -1, respec- tively. These remarks can be proved by showing that 32x? = 1(1+1)7c“i1 , (6.22) szx’l“ =-. 11X:Cl , (6.23) where 32=(°'1:0'2)2, 32: (12:942.. Using these results we see that the wave functions given in Eq. (6.17) are the wave functions of a state with J=S=l and L=O. To normalize the 331 wave function, i.e., to determine the constant 01 of Eq. (6.17), we write 2 m 2 __ °1)‘X1) d0. .. 1. Since (X1132 = l, Innis- 1. Therefore 01: l . . (6021+) The proof that Eq. (6.18) represents a state of {fr-1, S=l, L=2 is somewhat more involved. We shall show that it represents a D state. For this we need to prove that 1.2(slzx’f) == 2(2+1) 81sz . . (6.25) To demonstrate this, let us first show that m 172(1-2 8127(1) = o, (6.26) 01' 17262 31270;) = [(726.2 312)] x11“ . -43.. since X? do not depend upon the space coordinates. Using the relation 12: 1‘2 " (fl-0'2, and carrying through the differentiations, we have The Laplacian (72 can be written in the form 2 __ 1 a 2,2_ -L2 . Now apply the above operator to the wave function Slzch'. From Eq. (6.26) we have 1 _b_ 2 A. 2 _ 1.2 2 = :53]: (1‘ 3r) (_1‘ 812) :2- (1' 312) 00 Therefore 2 L 2 __ 1 a ;'2'(1'312’ —- par-‘1‘ A __ B 3 -> —-9 _ = 2(3) r2{30'1' ard’z er - (r1 . 0'2} = 2(3) r2 312 =. 2(2+1) r2 312. Consequently 112(r2 812) --- 2(2+1)(r2 312), 01‘ - 49 - 12(slzx1f) == 2(2+1)(slzx§_‘). (6.28) Comparing this with LZV’LZ, m) = 1(1+1)‘)"(1, m) , we see that 1= 2. This shows that 812%? is the wave function of a state with [=- 2. Let us normalize the 3D ‘wave function i.e., determine the constant c2 so that l 2 m.. m. ‘_ c2f(812xl 3127(1) “1‘1' The above integral can also be written in the form 2 m 2 m __ 6 (Thus in order to obtain the normalizing factor, we need to evaluate S 2 This can be done in the following manner. 12 ° Since '3(r' ° r0? .:r 1 2 S12== 2 ' 1 ’ r therefore (3,, + 1)2 _..—_ 2; (r1 . r)2 (r, - r)2. (6.29) r Using the relation given by Dirac 35 TAT-B=A°B-10'°AXB , Eq. (6.29) becomes (812+1)2._..__%[r2+i()':L ° rXr] [r2 + 10'2 ° rer=9 , 332:2: 8-2312. Therefore - 50 - 2 m m 2 In In __ 2 22 )X1 (8 - 2312)X1 cm=8e2 Jxlxl an _ 802 (MT), 2 l 1 c = o (6030) 2 Fr”. 2572 For the radial part of the wave function, write u(r)/r and u’(r)/r for 331 and 3D1 states, resPectively. Then from Eqs. (6.16), (6.17),),647)’, (6.18), (6.21.) and (6.30) we obtain - 2 Y=fi—l-f-{Eifl'+2 3/ amid-ilk? (6.31) where 361 ° r12‘7-2 ' 1’12 812 = __.___________.2 " 1 "V2 ' 1'12 The normalization condition is L (u2+w2) dr == 1. (6.32) Using the relation g—l.¢-2=23(s+1) -3=1, whenS=l into Eq. (6.10), we Obtain v = -(1 +Y812)J(r). When we substitute the above result into the deuteron wave Eq. (6.13), we obtain _ 1’12 2 r y... T v - [1+Y812]J(r))(’ . (6.33) The differential equations which u(r)/r and uJ(r)/r satisfy can be found by substituting Eq. (6.31) into the -51- deuteron wave Eq. (6.33). We get V2 31.91 X1 + S ZCAi-L x1111) 2 2 +fiz‘-[E+(1+Y812)J(r)] [ll-531+E-fs1z Mk!“ ‘0' (6‘3“ The quantity V 2u(r)/r can be evaluated easily, since u(r)/r depends upon r only. Thus V2 12 gégflzigf; %_cyfn(e,¢) 22 . (6.35), The term v2 ($129-$9- ). we note, can be written (7212 f+fl=vzlf%+}fe%n(o,¢)1rz, (6.35),, r dr where r21).(9, 4’) is the angular part of the Laplacian. To evaluate the last term, let us note that it contains the angular dependence of the wave function of the D state and that it is actually a linear combination of the spherical harmonics of the second degree. This can be seen by actually carrying out the operation of 812 on Xi, i.e., $12Xi = [3 sin2 9 e21¢X'1+3/2_ sin 9 cos 9 a” X: +(3 cos2 0 - 1)X1] . (6.36) The coefficient of the spin functions are Y2(2), Y2”), and Y2“) respectively. Since - 52 - n.(e,4>) Y1 + [(1151) Y£= 0, therefore n(e,4>)12 = - 2(2+1)1r2 = - 6Y2. Applying these results, we find 1249.4) 12 = - 681sz . (6.37) Hence Eq. (6.36) becomes 2 2 l d u l l g a) 6 [1‘ drag 2"? 12 1' dr + if; {h +(1 + rslz)J(r)} {Pi-13%;}; 812 gig” xii-“=0- . (6.38) Since Eq. (6.38) is identically zero the angle dependent and the angle independent parts must each be identically zero. This fact depends on the linear independence of the spherical harmonics. Making use of 813:: 8 - 2812, we can ’separate Eq. (6.38) into and 6.2“) _ £1.15... [3111-2 Y).T(r)]w=-23/2 H yu(r)w era r 62 22 (6.39)b The next step in the solatuon of the problem is to solve the differential equations Just obtained and to apply - 53 - the usual quantum mechanical continuity conditions to the wave functions. To do this, we find, unfortunately that the wave functions for points inside the potential well cannot be solved in terms of any elementary function. Rarita and Schwinger have given a power series solution, but this method involves tedious numerical work. Conse- quently, we shall develop a method which will enable us to ascertain the order of magnitude of the quadrupole moment arising from.the presence of h% D admixture. The differential Iqs. (6.39) can be written in the form 52:2” 12 u = -2\2 Vocab), (6.16), 2w 6w :12- r7 + 212w = -112 v0 u(r), (6.1.0)b where K23“ (V -‘EOI)=E2- 2 t P 0 £2 0L , (60M)a .2__11 _ _ ._ .12 . K 21? [(1 2)!)vo (r 00:? (1 2Y)vo , (6 1.1)b A2 23/2Y%§vo , (6.1.1)c £2 and E0, the experimentally determined binding energy 2.187 Mev. given earlier. - 54 - ror points outside the potential well (r > r0), v0 vanishes, and consequently, the differential equations assume the simple forms 2 1J3. _ 2 _—_—. dzw _ 6a) 2 __ -—-—-dr2 r2 - 0‘ O.) —-O. (6eh2)b The well-behaved solution of the first equation is U(r) 1‘0) = A @- °L(1‘ - r0). (60h3)a The solution of the second equation is a little more complicated; it can be shown to be w(r>ro) = c/f K5/2 (K'r), (6"‘3’11 where Kh(r) is known as the modified Bessel functions 36 of the second kind. That Ki (Kfir) is a solution of the 5/2 differential equation can be inferred from.the fact that the solution of aqazfliliriz x- }y=o37. (6.1.1.) is y=ffJS/2 (ax). Both J5/2 (x) and K5/2 (x) can be expressed in terms of elementary function. These are J5/2(x)=/7T'2; (- sin x - «3 cos x+i§ 8111 I). (6.651) __ 7T 2 -1 e 0 K5/2(x)— L2; (1 +3/x +3/x ) e (6 66) 'Using these results, we obtain the outside solutions u(r) = A 6 -oL(r - 1‘0), 1‘} 1‘0, (60h7)a -55... w (r) = Be- Och-TO )(l+3/°Lr+3/042r2), r>ro, (6.1.7)b The inside solutions are difficult to obtain because they satisfy a pair of coupled differential equations. Therefore, let us content ourselves by a very rough method of approximating the solutions. It is to be remembered that we are here primarily interested in determining the effect of the D state on the quadrupole moment and that the effect of the D state is very small, i.e., co sz dr = 0.01.. (6&8) From this fact we infer that w is small in comparison with 11. Therefore the solution of __52“ +K2 u = 0 cu"2 given by u = A0 sin Kr (6&9) is fairly close to the correct solution. Connecting this with the outside solution, we obtain Aosin Kro = A, AOK cos Kro == -o£A, 01‘ c013 K10 = - OL/Ko But I >> d , therefore Kro A; 7T/2 and A9—’AO . Experiment indicates that the ground state of the deuteron is practically a 8 state (actually 96%) . Conse- quently we may put fang dr = l. O -56- °° 2 _. " 2 2 2 zero -2oro) = B/rz. (6.52)-D Setting the logarithmic derivatives of the inside and outside solutions equal to each other, we obtain g—r [log 130"; log r + log J5/2 (K'r)]r_ _r=—[B-Zlogr]r= , 1.0 dr ! Eq. (6.53) reduces to (K'r) = 0. 5/2 The first root of this function 39 is h 49. The relation between the amplitude B and B0 is given by __ 2 Bo ffo 35/2 (K'ro) —- B/ro This gives us B in terms of 30' To evaluate Bo we need to consider a, (w _ 2 __ 2 2 dr 001+ ‘fw fir-Bel 1335/2 (K'r)dr+B 1;. E -53- 2% = 2 2 , 32 301 r J5/2 (K r) are + ”03 . (6.51.) To evaluate the remaining integral, we use the relation x I xJfilxldx =i 12(Jn2+J2+1) " 11X ‘Tn Jn+l 0 q=é x2{J§(x)- Jn-1(X)Jn+1 (26)}. Therefore f0“ .- :rg/zz (m) «u =- 2 r3 133;, (I: re), 2 ._ Bro From.Eq. (6.5h) we then obtain 32 2 r§J§/2 (K'r)+ B = 0.01., 1'3 J§/2(K'r) 31' = 6/5 (.01.) 2:3 . (6.55) ‘We are now ready to estimate the value of the quadru- pole moment. The quadrupole moment Q,is defined as the value of (322 - r2) averaged over the asymmetrical charge distribution in the magnetic sub-state m:=l, i.e., Q. =£<322 - r2), M; = 1. Using the wave function given by Eq. (6.31), the above expression becomes =iflféfizfr ('11-'17?) $.55 Snag)": ”22 ' “’2’ 1 1 =_.1.... °° 7' 2f- 312 "E‘z'hlfi léngaéz!‘ dr :2!) d0 sin 9 (291+— -59.. 7r 2 J: d4> {u2(3 cos2 9 - 1) XIII.)+T32-‘ (3 00829 - M11312“ Xi +3(0082 e - 1) 3&3 (Slzxi’2° (6.56) Angular integrations give zero for the first term. For the second and third terms we use Eq. (6.36) and the orthogonality conditions of the x-is. We get for the Q second term + _%-' dr :2 ua) and for the third term 5 2 0 - 5%)— !“ dr r2 602, upon performing the angular integration. 0 Thus we find 5f§ . ° = {—3-— 0 dr r2 (uw - —:-'=- 2). (6.57) The second tam involving 602 is the contribution to the quadrupole moment by the D state. Since this is mall, the quadrupole moment will be given practically by Therefore, oo 10 AoBofr sin Kr(fi" ifs/2m r))dr+lo£lo e r zlj'g' hereof? sin Kr(frJ5/2(K'r))dr+ 5- AB 10 26L The order of magnitude of the integral is -60.. f. 1%“ ARTS/235R (K' IchoBorg/2‘75/2m'flr z 1.5 This is clearly the upper limit of the value of the integral. Because the wave function vanishes at the origin, we would expect the correct value to be considerably smaller. There- fore, it will be neglected. Consequently, we obtain Q 91 5A3 ' (6.58) Squaring Eq. (6.58), we obtain .2 N 2 2 _6_ 3 Q _ (+0042 1"“0 5 (’01,) r° 3 = .-_2.i...1... r° . 50° °’~ l+okro Now 39 ,. 2 N .25 -13 (2.8)3x 10 500(1.6t.) 10 2.8x10 ' Q = 1.7X10-27 cm.2. This result is about half the experimentally observed quadrupole moment (2.73x10'27cm.). However, one should not attach too much significance to the numerical result because of the rough approximations that were made. Our calculations indicate only that a D state admixture of 1.7: leads to a positive quadrupole moment of the order of -61.. 10 cm.. VII. CONCLUSION AND DISCUSSION In this paper, a few of the quantum.mechanical problems in the theory of the deuteron were considered. In particular, it was shown that tensor forces can simultaneously account for the observed magnetic dipole ‘and electric quadrupole moments of the deuteron. It is to be noted, however, that the eXperiment, which we have discussed, does not give Q directly, but rather the product qQ. The actual value of quadrupole moment, therefore, depends to a large extent upon Nordsieck's 39 calculated value of q, which, in turn, depends upon the reliability of Wang's molecular wave functions. 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