Introductory Quantum Mechanics I
Introductory Quantum Mechanics I PH 451
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X111 3 THE NORMAL ZEEMANN EFFECT If we switch on a magnetic eld the still existing mdegeneracy in a central eld is lifted 2 EZ M H 2 CVr p 2f9Vr 1331 21710 2mo moc We have neglected here the term in A2 which is allowed in the case of suf ciently small ie easily technically achievable fields if we assume that the term linear in A has a nonzero contribution We use also n10 for the electron mass but remember that it is actually the reduced mass u which is basically m0 in the electron case to avoid confusion with the quantum number m The charge e of the electron is taken to be the elementary charge e0 Because of V A 0 the potential is not caused by monopoles we have 13212113 3 PI is Hermitian We assume a homogeneous magnetic eld given by the vector potential 1 g HgtltI 1332 2 hm 412 zHy sz gtf ltsz tz 1 ltny my gt13 2 IE Hx Hy Hz x y z A A a a A HZ H zz ltHx23 Hzx3gtjj 6y 6y T a a A A H H kk xyaz 6er a a yi izgx 62 6y f h Z a a A I we reco nizet at g 6x 62 y 3y2 2 6y 6x we see that the term 413 e A A A p mac corresponds to the potential energy of a magnetic moment 171 connected with the angular momentum in a magnetic field AU2 mHz iH 2moc One distinguishes between a canonical angular momentum L f X p here c c e a c a c e a p mv A and the mechanical angular momentum mr X I L I X A C C The magnetic moment is of course proportional to the mechanical angular momentum but we can neglect this in our approximation We put the zaXis in direction of ie Hx Hy 0 HZ H Schroedinger equation 122 m AVI 2LH Z 1 2 El 1333 me 0 0 The solution are the same as for the one without magnetic eld only the mdegeneracy has been lifted We have eh mH 1334 E quot1 2moc E nlm 414 For the energy perturbation by the magnetic eld we can write AE uBmH mZH the socalled BOHR MAGNETON the smallest unit of magnetic moment with L13 2 2m c 0 of an atom The wavefunction does not change the atom will not be deformed by the magnetic eld The energy depends now however on the direction of the angular momentum with respect to the eld ie on the magnetic azimuthal quantum number m The possible emission and absorption frequencies ofthe spectrum 1324 are according to the Bohr s frequency condition E E eH a nlmn l m h 2m 6 m m wo wLm m 1335 0 with the LARMOR FREQUENCY 6 ah l1 217106 The intensity and polarization of the spectral lines are connected with the radiation of a Hertz s dipole with the moment qltl7 im V640 1336 q charge of the moving particle witha 2 mo wLm m39 and ynlm 39739 WWW the transition matrix element with components f i E 415 If we introduce instead of X y z the quantities o 39 x i zy r51n3equot and zzrcosg we obtain with ynlm RnlYlm Sciz39yRn r I RM ltYlm sin 3 eii E Rn lrl RngtltYm 00s19l 7 gt Let us look at the angle dependent part The values are zero unless ll I mm 0I Selection rule AlI Am01 a y 0 2 7 0 7 For m 7 m we have 13 2 12 and thus l329andl3210 39 39 C0nS1OOCOSU01 1337 In this case the dipole moment oscillates in zdirection with the unshifted frequency 00 it does not radiate in zdirection but perpendicular to 2 it radiates linearly polarized light Formm lwehave 56ij7tO ijzz zOm w0wL 1335 Therefore we have Ft corzsfcosw0 wLtsina0 wLtO Dipole moment rotates counterclockwise around zaXis right hand screw Correspondingly we nd for m m l i i iO 3Ei 0aaO aL 1335 and 7t constcosa0 wL t sina0 wL t0 If we watch in zdirection ie in the direction of the magnetic eld longitudinal Zeemann e ect we see two circular polarized components with frequencies a 0 i a L If we watch however perpendicular to the magnetic eld transversal Zeemann effect we see one in z direction linearly polarized ncomponent with frequency 00 and two perpendicular to the magnetic eld linearly polarized ocomponents with equencies m0 i a L LORENTZ TRIPLET rn r e I I 7 r lsQ 1 yoUL wowL Q xIy Z x y 6 5 Q g 11 wo wL g It 00 x Z x x 0 E 10 UJL U0 U0CUL 0 Jr a Fig 13 2 Level diagram and line scheme for the normal Zeemann effect for the transition between a p and slevel 417 Transversal Zeemann effect Linear polarization because dipole does not radiate in direction of its axis Therefore only xaxis contributes if we look along the yaxis as in Fig 132 Polarization Abb 15 Abb 16 7T Phase d1fference of or 3 between V1brat1on 1n X and yd1rect1on x y 77 cosat cos wt smot x0 yo 2 2 x l cos2wts1n2wtl x0 y0 This is the equation of an ellipse with half axes X0 and y0 x0 yo 2 circular polarized Of course the situation is more complicated due to the spin If we denote the Hamilton operator of the electron without magnetic eld with H 0 A if H0 2 A 1338 2m then we obtain 418 a 1 6H A A lh Z HOW LZ h039zl 1339 6t 271100 From this equation we obtain neglecting again H2 this term determines weak diamagnetismthe term which represents the action of the magnetic eld as the potential energy of a magnetic dipole with moment n 2106 Z 7710 in the magnetic eld L h5 z 13310 A 1 0 Pay attention here that Oz 2 0 I the PAULI spin matrix and that the LANDE Factor for an electron is g20 Therefore we have the h in the expression There is another h connected with I because the eigenvalue of I is mh If we go from the operators to the expectation values which means we can replace the A with the vector arrow everything is right again ih yE xi ihi LAZ 6x 8y 6gp Remember Eigenvalue of I mh 419 Now we want to determine the stationary states We use the wavefunction E LPCZf lxyze ht 13311 E is the energy of the stationary state If we substitute 13311 into 1339 we nd A eH A A Holl LZ 13312 2moc If we choose the representation in which the matrix 639z is diagonal sZrepresentation then we have A 1 0 G 39 G 02W 2 13313 0 quot1 972 quotV z and the equation 13312 decomposes into two separate equations for l1 and Vz A HOV1 i h6z11 Eyl 1331401 2nc A eff HOV2 2 i Mybe EyZ 13314b nc The solution of these equations are obtained if we take into account that there are two solutions possible without a magnetic eld 420 1121 Z WSW E E2 for spin sZ 2 1331561 0 0 h 1121quot E E for spm s2 13315b lIZnlm 2 where nlm 13316 Since LAzVnlm 2 km Vnlm these solutions are simultaneously solutions of the equations 133 14a and b They belong however to different eigenvalues Ifwe substitute 13315 a and b into 13314 a and b we obtain two solutions 71H 1 Wm E 19 2 E3 e m1 s2 13317a 2moc 2 71H 1 Wm E it E31 e m l s2 1331719 2moc 2 We see that the inclusion of the spin leads only to an additional splitting even of the s term 1 0 m 0 This is the most important result of spin theory The magnetic moment of the spin interacts only weakly with the eld of the light wave Therefore only transitions are allowed where the spin does not change We should mention that the splitting into three lines of the normal Zeemann triplet or Lorentz triplet corresponds exactly to the classical theory of the Zeeman effect In the classical theory the Zeeman effect is explained by the precession of the orbit in the magnetic eld with the LAMOR frequency a L The quantum mechanical formula eH 2moc a 2 mo wLm39 mquot 13318 421 does not contain h and therefore the result must be in agreement with the classical one Homework 7727 mz7 317 5 1 m7 m 7 quot quot l a Z c a39 239 c L 75 m i 72 72 81quot SZ39Z Without eld H 0 With eld H i 0 Fig 13 3 The splitting of the s and pleVels in a strong magnetic eld with inclusion of the spin Let us dwell a little bit longer with spin 12 in a magnetic eld because I want to deal with rotations of the coordinate system and the question of diagonalization of H in this case The magnetic moment of an atom or ion in free space is given by 17 2 7h 2 gqu 13319 eh 2m0c uB BOHR MAGNETON where the total angular momentum hj is the sum of the orbital hi and spin g angular momenta The constant y is the ratio of the magnetic moment to the angular moment yis called the gyromagnetic or magnetogyric ratio For electronic systems a quantity g called g factor or spectroscopic splitting factor is de ned by 422 guB h 13320 For an electron spin g 2003 usually taken as 200 For a free atom the gfactor is given by the Lande equation JJ1SS1 LL1 2JJ 1 1 13321 The hydrogen atom in its ground state has no magnetic orbital momentum The spin however has a magnetic moment 77 guB 2JB JBB39 13322 The energy of a magnetic momentl in a magnetic eld H is Umg m H 13323 Since the atoms are supposed to be at rest we do not have a kinetic energy term and A A A gt H Umag H 13324 The eigenvectors are In lmszgt and we have to solve the Schroedinger equation lhE H Wnalymasz 0 13325 We want to make the problem simple and choose the coordinate system so that 1 00 H z 423 1 Pr 00HZ 33 ssl Fig 13 4 Eigenvalues of Z and 2 with the magnetic eld in zdirection Therefore the energy eigenvalue is from 13 323 Substitution of 13322 into 13324 E mag lt ZgtHZ 2 3ltSZgtHZ i HZ 13326 ltmzgt 2UBltSZgt ihuB 13327 because of 52gt i If I path l we get the result shown in many books that u uB The energy splitting is therefore i u HZ and the energy difference between the 1 spin and spin 5 level 2uHZ shown in Fig 135 424 Fig 13 5 Splitting of the spin states in a magnetic eld The Hamiltonian of the problem is given by 13324 and can now be written replacing the A h d operator S by the spin matrix 3 039 H H 13328 3 H H 13329 HF 13327 2 Rotation of the coordinate system should not change the energy eigenvalues since the hydrogen atom is rotationally symmetric We obtain uHZ O E 13330 0 uHZ 425 Fig 13 6 Rotation of the coordinate system As mentioned the energy eigenvalues do not change when we rotate the coordinate system but we have to replace HZ by HZ E 13331 0 HZ H 13332 u1HfH HZZ 0 gt E 2 2 2 13333 0 1Hx H y HZ Solution by coordinate transformation I I I Ex 9y aZ gt Ex9y9Z because the Hamiltonian 133 29 is diagonal in the X y z coordinate system as we know 426 1 0 Et 1 C1 0 C2 1 e 13 3 35 f1 F1 H 0 1 H 0 i m 1 0 13336 0 quot1 0 1 yi 0 z 0 1 211 1th 13337 Substitution of13335 into 13337 yields DEWHiOJWiciliwio O 1 O 1 1 0 1 O Wicioiwii 427 Obviously we have two equations ECl lLlZC1 C2 13338 EC2 lLlHx C1 IUHZCZ 13339 E 0 C1 H11 H12 C1 0 E c2 H21 H22 c2 H11 Hz H12 uHx iHy H22 IUHZ H21HxiHy 3 HZ Hx iHy OH Hx 1Hy H 2 Condition of existence of nontrivial solutions H11 E H12 H 21 H 22 E Det 428 H H 2 i H12H21 HIIHZZ J 0 EL2 MHj Hj 1le2 uH A result we have expected Determination of the eigenvectors written in the old unprimed coordinate system H11 ECl H1262 0 H2161 H22 ECz 0 i 2 H12 62 Eu H11 E ziyJHfHjHfziuH E12H11 2111743qu H12 Hx iHy E1 uHuHZ E2 uHuHZ 429 cl Hx iHy cf HHZ Hx z39Hyszinge i HZHCOSS c1 sin3 6 c2 1 0053 19 19 19 cos2 cos2 sin2 sm2 251n cos 2 2 2 2 2 2 1cosz sin2 20052 2 2 2 1cos3 1 cos2 sin2 a 2 sin2 2 2 2 100519 430 1 c058 Egt0 1cosG Elt0 sing cos 8 i2 3 2 8 8 CZ cos 6quot CZ sm 6quot 2 2 i l39 01 sm e 2 c1 2 005 6 2 2 i I 1392 02 smEe 2 02 2 005 6 2 2 Now we know how the matrix looks which will diagonalize H It has the eigenvectors as columns i 00336 2 sm3e 2 UT 2 if if sm e 2 003 6 2 2 2 Thus we have reduced the determination of the eigenvalues in the rotated system to the determination of a unitary transformation 431 XI THE HYDROGEN ATOM a Spherical coordinates Before we solve this problem let us talk about the Hamiltonian in spherical coordinates The relations between rectangular and spherical polar coordinates are given by x rsin cosq yzrsin sinq 1111 rcos19 1 Fig 11 1 The spherical coordinate system with position dependent unit vectors epe and eq The orthonormal basis vectors r and q satisfy the relations r39QZer39erpZerp 19 r r2619 6196p ed A A A A A A 1112 619 gtltep 26 280 er 2 ex Sin cosrpey 8111198111pez cosz A 619 x cos cosrp y cos sinrp z sinz 1113 gt b q 2 ex Sinrpey cosrp ex ersm cosg0e cos cosgo qpsmq ey 2 er s1n19s1ng0e cos s1ng0ez cosq 1114 62 zercos e sm Let us look how we got these relations A 3 6 1 quot1 NH ea i x 2 9 1 Fig112 1 1 A Cartesian and spherical 1 coordinate system unit Y vectors X From Fig 112 we see that 2 2 X y y 2 2 2 r 1x y Z tanzSl tanrp 1115 Z X 281 We have de ned a set of mutually perpendicular unit vectors r 13 and e in the sense of increasing r 19 and p respectively As the location of P is changed all three of these vectors change The Cartesian components of r and 19 are found from inspection of the gure to be er 2 ex sm cosgo ey sm smgo 62 cos1 L 19 2 ex sin1 l 90 cosg0 y sin1 l 90 sing0 z cos1 l 90 2 ex cos cosg0eycos19s1ng0 ez s1n15l We know that eipi r and 13 We nd q using the orthonormality relations erequque zo asin cosqbsin19ing0ccos 0 acos cosg0bcos singo csin 0 a2 b2 02 1 where Q a x b y 06 msin19cosltp n sin1 sinp 2 0051 B 39 h yusmg 0cos cosltp pzcos smq qsm19 we ave ambncl0 aobpcq0 a2b2c2 ambn l ambn am bn 282 a o7qjb p7qj0 plnq 1bmb 01mq olmq plnqm n cb olmq Zbnolnmq mpl nmqbln0 mp l 01 mq 01mq cbno mp olmq Z 2 2 b2 w b2b2 M 1 olmq olmq b2ltplnqgt2 1no mar 01 mq2 01 mq2 b2pl nq2 01 mg2 no mp2 2 01 mq2 b 01 mq xpl nq2 ol mq2 n0 mp2 283 pl 2 cos2 sinq nq sin2 19singo 01 cos2 cosgo mg 2 sin2 cosgo no 2 sin cos sin ocosgo mp sin cos sin ocosgo gtplnqsinq olmq cosq 110 mp O This yields cos b 0590 s1n g0cos go a sing0 c 0 Thus we have 1113 which can be written in matrix form s1n19 coup sin simp C0819 9 C0819 coup C0819 s1np Sim p sin p cos p 0 N lt N 1116 The inverse matrix is the transpose as you can show easily We are now ready to express the gradient operator V in spherical coordinates A a A a A a Vex ey ez 1117 8x 6y 82 We substitute for e y and 52 and collect terms 284 gt V r sin 19 cos sin 1 sin cos 6x 62 19 cos 19 cos cos 1 sin sin Sin 3cos 3 p g0 6x g0 The quantities within the square Brackets are easily recognized as 15 a 6r We prove this in the following way Let as take a function f Xyz and take the total differential df gdxgdygdz 6x 8y 82 Nowtake hm z zi Jriihi AHO Ar dr 0x dr 0y dr 02 dr i3ix3iy dr 3de 3ydr azdr or a 66x 66y 662 ar ax ar ay ar 62 ar sin coupisin19sinp3cost3 6x 6y 62 Analogously we obtain for the second square bracket 285 66 66y6g 619 6x619 6y619 Ear 6 6 6 rcos coup rcosz931np rs1n19 6x 6y 62 and for the third one i66x66y66z 6p 6x 6p 6y 6p 6z 6p rsjnzltsinpirsin coMpi 6x 5 3V er e e 1118 6r 619 r 5m 19 6p The derivatives of the unit vectors are easily obtained from 1112 gay 2 ex s1n19cosg0 ey sm slngo ez 00319 e 6 613 2 x cos sinq y cos cosq e 0031 0 6 A e 0 Q i a w x cosq y sinq r sin19 00519 e 286 A A x sin219cos 0 ysin219sin 0 zcos19sin19 e s1n19 e cos19 1 A 219 A 2 19 39 A 19 39 19 ex cos cosgD ey cos s1n 0 ez cos s1n x cosg0sin2 19 cos2 19 y sin 0sin2 19 cos2 19 x cosgD ysingo qed We can now use 1118 to express the kinetic energy in spherical coordinates 2 A 6 A 6 A 1 6 A 6 A 6 A 1 6 V rie ie 7 erie ie 7 6r 619 r s1n19 6gp 6r 619 r s1n19 6gp a2i 52 52 6r2 1126192 F281n2196 2 6 A 1 6 e 61quot Prsin196p A 1 6 A 1 6 e6 e r619 rs1n19 6p A 6A 1 6 er e6 6r r619 A 1 6 A 6 e19 e r619 6r A 1 6 A 6 A 1 6 A 1 6 e er e6 Prs1n196p 6r Prs1n196p r619 287 220 6A g e 0 6r 6r p 619 p A aer xcos cosg0 ycos19singo z Siml 19 er 2 exs1n19s1ng0exs1n19cosg0 e s1n15l 590 3amp9 q cos1 L 590 Product differentiation leads to the following result 2 62 1 62 1 62 2 2 2 6r r2619 rs1n196q A 6A1a A A 616 er el9 erel 6r r619 Tar r619 J 0 A GA 1 a A A a 1 a 6 e e e r 6r P rs1n196q war rs1n196q 1 0 0 2V A 1 ae a A A1 62 e eler r 619 6r H Jra ar H4 0 93 A 1 aa a A A 1 62 6 e e Prsmz 6p 6r Lil sm awar H 0 e 511119 A 1 6a 1a A A 1 a 16 6 e el Prsmz 6p r619 hrsm aq r619 H 0 e cos 19 288 6 1 6 1 6 6r r2 619 r s1n 19 61 1 6 1 6 cos19 1 6 s1n r 6r rsin19 6r rsin19 r 6r 1 a 2 a 62 2 a I Seepage 100 r261 E W 6r 2 V26723372 2 6r r6r r 619 rsm19 1 62 c0519 1 6 1 62 r 6r r2 111219 6 02 2 2 i sin19i 2 cos19isin19ai2 r 511119 619 619 r 511119 619 619 c0519 6 1 62 777 112511119 619 1126192 2 1 6 2 6 1 6 6 1 62 2V 2 2 2 s1n19 2 1119 r 6r 6r r s1n19619 619 r Sll l 1961p 289 V11 3 APPLICATION EXAMPLES FORTHE HARMONIC OSCELATOR A good example for aharmonic oscillator in one dimension is the Hcl molecule The hydrogen and chlorine nucleus oscillate around their equilibrium position hi Model for the Hcl molecule small circle H large circle cl waedenotewith L A Fig 75 c u rr position then we have as oscillator coordinate x X X From 2 k a 7 m mcl we can 39 k andthe atomic masses T may go 39 39 POTENTIAL wquy 3 rm 2 ho ENERGY STAN 5 2 6 Ener states ofthe harmonic oscillator 39 quot quot 39 cX th 39L39 quot 39 xrbelongstoxt Hcl gas absorbs light at amvelength of39 35 m The abmrbedene auses a transition n 0 to n 1 or n 1 to n 2 ofthe molecular oscillator the dipole strength ofwhich can be calculated using 6416 VIII 1 THE EQUIVALENCE OF OPERATOR AND MATRIX FORMULATION OF QUANTUM MECHANICS a Compilation of average values 1 multiplicative operators W ixvW dx W imww dx 6 ej 11 dx 2 differential operators mg ltEgt W Ta 3 Differential operators in position space can be multiplicative in momentum space and vice versa Position spaoe Momentum space 158 gt lt13XgtJpxwgwpch A s A e a M MAII dx ltxgt I Vplhawp dp The opmtm used in QM are linear upemtnrs because an opeiator F opeiates on I clwl czwz resulting in F I Fclll1 czwz 01Fw1 cngl2 811 b Two possibilities ofmeasurements 1mie 39 quot 39 are only possible statistically p PE p probablllty density av pa ket in pspace Spectral lines Fig 77 Spectral distributions cog 2 The oormoolc has nnly one value which can be predicted with certainty 517 F Fig 88 One value represented by a function e P Example Measurement ofthe velocity offree electron with sharp momentum or the measurement ofthe mvelength ofmonochromatic light The mountable attribute is obtained in this case from a pure mute 39 iclassical L W l 39 L L 39 W which 39 1 c complete nrtll unurmal iyrtem We start with eigenfunctions ofthe Hamilton operator given by 131W 2 Elll 812 are per de nition stationary solutions View ixe7E t 813 Forsimplicity r 439 L A 39 39 correspondingly to an energy continuum e eigenfunctions are nrtlmnnnnal ie we have Illflyde 5 814 Proof i and j are two eigenfunctions for E and E The time independent Schroedinger equation fortl and j respectively are fillj EjIlj 815 H hermi an operator Multiplication from the left with IV of the rst and j of the second equation integration and following subtraction of the rst equation from the second yields 112 i Id3x goll Aw 1JA1i 2 El EjId3xyiyj 816 The left side can be expressed as surface integral over the probability current h Jij 2WiV1Ij 1jV1i m Jiz39jdf Ei E1 d3XlIWJ 817 We extend the integration over the in nite space and correspondingly over a surface which will be in nite at the limit We demand that and disappear at in nity fast enough i e at least as lr for liilllfjadfzo 1 1 because jy 3 f N r2 product 3 1ntegral gt O for r gt no r r We obtain 161 E1 WWjd3x 0 gt E1 E gt wayJdbc 2 0 819 El 2 EgtJl1lde3x 0 We assume that the functions are normalized lIIfIde3x 51 The eigenfunctions of H form an orthonormal system ie a class of functions for which the above normalization condition is valid The system of eigenfunctions is complete Explanation The i form a class cE An arbitrary function f X square integrable within this class cE can be represented as linear combination gx Z Cklllk k1 so that fX lim gx Because of IlllIJa x 6y we have I 1fltxgt12dx Z 2 if i 2 C l 8111 162 This Bessel s equation is for all fx 6 CE a necessary and suf cient condition for completeness We can express ci by Cl ZJWX39fX39dX39 8112 andsubstitutein fxZc1x 8112 f x JZwxgt1x x39fx39dx39 01 fx jKxx39fx39dx39 8113 where KOBX39 5X X39 ZWXXWKX39 i d Expansion of Xt in a complete orthonormal system All eigenfunctions of 1 the class of functions cE form a complete orthonormal system Then we can expand i Et 11chka ch k eXp h k 8114 k k is normalized 163 Jwwdx1 is square integrable 3 Z C39ka is nite and because of the normalization k 2 Cka 1 8115 k We obtain the energy spectrum of ltEgt ZCCkEk 8116 k A We want to expand the matrix element of an arbitrary operator F the lnctions k used for the expansion are eigenfunctions of 1 but not of f7 A is A is is A ltFgtJ1 FIIdx 2 cl CJJIIiFldex 8117 1 where i J wf wjdx J ltx jltxdx exph Bi Ejy J jdx F 164 A icoit ltFgt chchj 6 J 8118 if with Fij and ij are obviously elements of a matrix We must demand that ltFgt is real as it represents an observable In the sum over i and j we have terms of the form iwijt iwijt cichUe cich e If this term is real we must have FF7k 0r FF J U U 8119 It means that the matrix Fij is identical with the transpose conjugate matrix Transpose of a matrix T T11 T 21 Tm T12 T22 722 T Rows and columns are interchanged 739ln T2n Tnn A square matrix is S YMMET RI C if it is equal to its transpose re ection on the main diagonal upper left to lower right leaves it unchanged it is ANT IS YMMET RI C if this operation reverses 165 sign N SYMMETRIC T T N ANTISYMMETRIC T T Note that the transpose of a COLUMN matrix is a ROW matrix i7 a1a2an COMPLEX CONJUGATE of a matrix T T5 T5 2 Ti 2 7 T 71 A matrix is REAL if all its elements are real and IMAGINARY if they are all imaginary REALT T IMAGINARYT T The HERMITIAN CONJU GATE or ADJOINT of a matrix indicated by a dagger T39 is the transposed conjugate T F A square matrix is HERMITIAN or SELFADJOINT if it is equal to its Hermitian conjugate if the Hermitian conjugation introduces a minus sign the matrix is SKEW HERMITIAN or AN TIHERMITIAN HERMITIAN T39 T SKEW HERMITIAN T39 T 166 E1 E2 F13 F 1721 122 F23 122 8120 F31 F32 E3 E3 F23 E3 A In the same sense one talks about the hermiticity of an operator F which generates this matrix If the operator f7 is hermitian one has IVElobe t0quotde or WWI11W Jlt wwdx J wwdx 8121 The proof of this relation is based on the expansion of the rst integrals and comparison of the terms with the same ij e Derivation of an operator with respect to position and momentum We ask ourselves what is the result if the momentum operator operates on an arbitrary operator F 3F1f 1 A7 Illf gllf 8122 x 167 gt 8123 lax 11212 8p 18p h 8 7513 1332 3213 8124 d a 1 A Quak gwkjnrwldx 8125 6 A A 6 F t clx F t dx flab ij 1 06th Substitution of the Schroedinger equation into 8125 yields d d 1 A 210 ijkwmdx 8126 J wk l Hdex J wzg amdx 168 a z H artI h m J wkmmdx iJHkaaxI x J HI1 FtIIldx at h W 11 isHermin39an i A A gJWkHFWIdx 6 z39 A or39k H F ail1k h Wk A a AA 2 JWkFtEW1dx JWkFHWldx This may be written as an equation for matrix elements since the functions Ill and k are quite arbitrary at any instant of time dt The le side 0f8 127 is the matrix whose elements are the time rate of change of the matrix 8127 elements of f7 and may be called the total time derivative of the matrix F The rst term on the right hand side is the matrix of the partial derivative of f7 with respect to time t and takes 169 into account the explicit time dependence of F The last term is that pa1t of the time derivative of the F matrix that arises from the change of time of the functions with respect to which the matrix is calculated 8127 is Heisenberg s form of the equation of motion of a dynamical variable Let us now assume thath7 has no explicit time dependence 3 0 l and that k and lare eigenfunctions of 1 d i A A A A A A EEC gHF FHk EkakFyldx ElekFyldx 8128 0 F EF FE dt kl kkl kll d Ed 2 0 1f Fkl Fkl k1 Fkk i e only ifFk1 is a diagonal dt matrix the time dependence disappears Solution of Schroedinger equation is equivalent to diagonalization of the matrix Hnn This result is not new but a consequence of the expansion W Z Ciel1k 2 Ck k exp 3 k k according to eigen lnctions of H 170 Ifthe k are not also eigenfunctions of f7 we have terms of the form 139 Ek Ez 1 Ede h 8129 Otherwise we would have only Fkk 3 H and F do not commute or d i M M F 2 HF FH 8130 dt kl h kl lt gt One could assume that in the case of nonstationary solutions Xt the factors ck would be functions of time ckt This is not the case ifthe operator does not contain the time explicitly via Vxt Proof A a 139 11 Z cktIIkxr with 5111 Hvk k h 6 d d h iatw 2k dtck wk 2k Cquot alth i PAHf Z ckEka k l7l h a h d h d E E 2 E jaill k dick1Ik l k ckdtwk k ck kwk h d h d 07Eckjllk439 7ckE1Ik ZkckEk1k Ef J izckEkqk because of i gwkIIqk O 2ickjlIk Ill and integrate k dt d d O ZliEckJWZykdx EckJWZykdx d d O 6 5 gt c t0 2k k lk k c does not depend on time Qed What does it mean if H and F commute H E means that the state k has a de ned energy value Ek k k k 172 FHWk EkFWk Ekfk39l k PIPka fkllk eka kakllk Ekfkllk k is simultaneously an eigenfunction of H and F In other words since the eigenvalues fk and Ek form diagonal matrices fkk and Ekk commute with each other Ek means measurement of a de ned Ek if we have the pure state described by k But simultaneously fk can be measured 3 0 8132 means possibility to measure Ek and 1 simultaneously with arbitrary accuracy The commutation relation is valid now only for k the eigen lnctions of the energy operator Since we can expand each arbitrary function in terms 0f 39 k 1I Z Z Cklllk k it is generally valid Classical analog We expect that is the quantum analog I have reversed here the order because of the factor i in the denominator With i in the nominator we would get of one of Hamilton s equations 173 dxc 6H6 dt 6px 0 where Hc is the classical Hamilton function of Xc ye zc px py pz characterizing the system The correspondence principle requires that hm 2 MO ih 6px Similarly for x we have in QM m lt139 192 dz if and classically dpxc 6H0 dt 6x6 The correspondence principle requires that Similar equation follow for y and z and their conjugate momenta All these conditions can be satis ed if we do the following 1 Let be a Hermitian operator identical in form with Hc but replacing all coordinates and 174 momenta by their corresponding operators 2 Postualte the fundamental commutation relations between the Hermitian operators representing coordinates and momenta 3713 971 213 ih xapy x7192 yapz yapx 27px Zapy 0 The three coordinates x y z commute with each other as well as the three momenta px py pZ Prescription 1 must be applied with care if Hc contains terms such as xcpx because x and pX would upon translation into noncommuting operators AC and x give rise to a nonHermitian H 1 AA A A The symmetrized operator 5xpx pxx can then be used instead It is Hermitian and leads to the correct classical limit Sometimes there may be several different ways of symmetrizing a term A 2 A2 A 2 A2 1 AA A A to make it Hermitian Thus the Hermitian operators 6 px pxx and 5xpx pxx2 both have the classical limit2 xcsz2 but they are not identical In practice it is usually possible to avoid such ambiguities f Extension of the expansion to continua As long as we deal only with discrete states harmonic oscillator square well potential the expansion is simple 1 chVk k i Ekt Illk k exp h for any square integrable and continuous function In most problems atom nucleus potentials we have however besides the discrete states a continuous spectrum 175 Fig 8 9 Example ofa core potential W Z Cielk J dECE ExeiEt 8134 k EX eigenfunctions for a continuous eigenvalue E g Addition and inu1tip1ication ofmatrices Addition A B c ABkAkBkCk Multiplication ofmatrices a multiplication with a constant C An cAik b nin1tip1ication oftwo inatn39ces ABlk ZAUBjk 8136 1 i is row index k is column index of product matrix AL I 0 column Scheme 1 J I H I Ax x1 x 8137 k H H ZAUBjkxk XI 2C1ka 8138 jk k Examples for multiplication All A12 Bll BIZ AllBll AIZBZI AllBIZ AIZBZZ A21 A22 321 BZZ 421311 AZZBZl AZIBIZ AZZBZZ AS SOCIATIVE LAW A BC AB C In general the commutative law is not valid AB39BA Example Aiijk Cik AB C BijJi Cki BA 2 C transpose of C The symbol is often omitted the convention is then that the sum is over the indices in the product which are the same in the above case therefore j The inverse matrix We substitute into 8138 x x 177 gt AUBjkxk x1 1x III I Unit matrix AB I B A391 InveIse matIiX A A391 A39IA I One nds the elements of A391 by solving a system of linear equations DetAij determinat of the matrix A detAij subdeterminant of element A A11 A12 39 A171 Delll A21 A22 I A2quot Aquot1 Fnz Am A11 A12 Alf A1quot A21 A22 AN A2quot H I I I I d A l et V An ATquot l A 1 AM Aw1 Ann erase the elements in ie the ith row and jth column 178 Azjl Det A det A Note the transpose Aji in detAji AB391 B1A1 Matrices with special properties 11k 2 Ala transpose matrix Aik conjugate matrix N gtilt A ik I Hermitian conjugate or adjoint matrix 0 A A Hermitian or self adjoint matrix N A A symmetric matrix Matrix of moment of inertia A 147k 2 Akl Akk O antisymmetric matrix angular momentum torque 179 A 1A1 unitary matrix A39A 1 a 0 0 l or 39 1 Special case A A I orthogonal matrix For example for example COSp sinp COSp sinp 1 O sinp COSp sinp COSp O 1 3 AA I h Application of matrix calculation in QM F and G be two operatom The functions form a complete orthogonal set jdx 5i 3 1131 byfax F G E G 1 180 Correspondingly j 11 wjdx FGU We expand Cw using k ll 1 ZV kGIg39 k because the expansion coef cients are obtained by multiplying from the left byIl and integrating over dX The term Ill61 jdx yields Gk J WZGWjdx J WZWkdxoGlg Glg39 ZlWi VkGIqdx 2 25kqu 2 FGU k k 181 e Rotational invariance of the Hamiltonian of the reduced one body problem CENTRAL POTENTIAL POTENTIAL WHICH DEPENDS ONLY ON THE DISTANCE FROM A CENTER Important Chapter before introduction of angular momentum Separation of variables Angular part Eigenstates of angular momentum operator Radial part The key point is the conservation of angular momentum What is conserved No torque which is true for central potentials Force and direction are parallel 7 f Torque thinking not useful in QM Answer in QM Angular momentum is conserved when there is rotational symmetry Clearly with angular momentum conservation the central potential problem becomes one of nding the simultaneous eigenstates of the Hamiltonian and the angular momentum The knowledge of the eigenstates of the latter is then used to simplify the simultaneous eigenstate problem Central potential problems important in classical mechanics Kepler39s Laws Same true for QM Major historically most important central force motion in QM is the Kepler motion found in hydrogen atoms few pedagogical examples of QM motion under central potential practice problems with only an occasional glimpse at real world applications Classical Kepler problem Two body problem Many QM central potential problems in the real world are also twobody problems We know how to solve only onebody Schroedinger equation exactly 4 We must leam how to convert the two body problem into a doable onebody problem Hamiltonian A2 A2 FIZZP rifll2p n2 11127 which is a sixdimensional Hamiltonian 301 The trick of the reduction of the two body problem is to replace the two initial bodies m1 and ml by two new bodies M p where l l l M m1 m2 and 11 m1 m2 As we have seen the massive particle moves free of any potential The light particle of reduced mass M moves within the central potential of interest Mathematically what we do is a coordinate transformation in three dimensional space taking us to the centerofmass reference frame de ned by MRmlr1m2r2 rr1r2 c p amp b a Fig 11 3 Transformation to the center of mass coordinate system a coordinate system with origin 0 xed in space b coordinate system with origin at the center of mass 0 302 In terms of these new variables A 2 F p2 2 M Z 11128 This is the Hamiltonian of two independently moving particles and therefore the reduction has been achieved Whenever the Hamiltonian is the sum of two independent parts the eigenvalue problem can be separated The Hamiltonian of the onebody Schroedinger equation for the particle u is A p2 h2 H E V I7l EV2 Vl 11129 Let us consider the symmetry of this Hamiltonian under rotation of the coordinate system Suppose we rotate the coordinate system about the zaXis by an angle 19 into a system x y Z I x xcos ys1n1 l I y xs1n1 1 ycos l I Z Z The inverse relations are x x39cos ly sin1 1 y x sin19y39 00519 22239 303 Proofofinvariance of H under this rotation r39 x392 y392 2392 2 x2 y2 22 2 r 4 potential does not change under rotation V2 a 8x 2 a 6y 2 a 6232 6x6x 6 6x8y6x 6 6y2 ltax ay39xa axgtltayay39gtlta aygt12 a az2 cos 196 ax sin 196 6y2 sin196 ax cos 196 EMT a azf V2 df zgdxgdygd2 6x 6y 62 dfgdxgdygdz dx 6x dx 6y dx 62 dx 3df l lf ilfflf l As X is a function of X and y we can replace the total differential d by the partial one a 304 4 the Hamiltonian kinetic plus potential energy remains unchanged under rotations If there is an invariance there should be a conservation law and we will see that there is an operator connected with rotation that commutes with the rotationally invariant Hamiltonian To nd the operator consider an in nitesimal rotation about the zaXis that is put 0051921 811119 2 19 in the transformation equations x39 x 19y y39 y 19x 11130 239 Z Let 1xyz denote an eigenfunction of H HWCX 1 Z Z EWCX 1 Z 11131 Now impose the condition that HIIx39y39z39 Ell268 Or substituting for X y and z from 11130 we have PAD4x 19yy 19xz EWX 19yy 19xz Taylor expansion of III and keeping terms only to rst order in19 and then subtract 11131 yields 305 A a a a a H E x 6y y aCIxyz x 6y y aCIxyz 6 6 A H xay yax wxyz Wx19ygty19xgtz Wxgtygtz a W 19ya W19xa WO 6x 8y 82 terms in 19y2 and 19x2and higher powers 6 6 Wxyz 19x a y wa y z y 6x We divide by 19 which is in nitesimal small but not zero and get the above result after subtracting 1JX y z 3 3 i A X Z 11132 8y 8x h A LZI zcomponent of the angular momentum 306 In this way we see that Lz commutes with H LZH 0 as far as operation on III is concerned However the functions 1JX y 2 being eigenfunctions of H form a complete and any wave lnction can be expanded in terms of them It follows that the commutation relation 2191 0 1111111 holds true in general Similarly by considering rotations about the XaXis and yaXis we can easily show that A A A A H L HLy 0 1111111 angular momentum conserved quantity we have been looking for Infinitesimal rotation about zaXis ii 19 Wltxlaylaz39 1 lIxyz 11135 2 I is recognized as the generator of in nitesimal rotation about the zaxis in the same spirit that we recognize the Hamiltonian H as the generator of translation in time Coming back to the treatment of the central potential Hamiltonian how do we take advantage of the conserved quantities namely AWLv and I discovered above Slight problem these operators do not form a commuting set and therefore cannot ALL have simultaneous eigenvalues with H and cannot ALL be used as labels for the eigenstates of H There is however the operator 2 which commutes with all the if i x y z and with H 307 Accordingly we seek simultaneous eigenvalues of H L2 and L2 the last one by convention Note that since we are going to calculate eigenstates in 3 dimensions we must have three labels for them the above three eigenvalues provide these labels Now let us label the states as in m lgt n quantum number related to energy I orbital quantum number eigenvalue ofi is ll1h m eigenvalue ofiz is mh I I 1 I 2 I 1 I I Z m Bound state problem 4 no generality is lost assuming discrete energy levels Eigenvalue equation A Hlnlmgt Elnlmgt in spherical coordinate representation 2 A i 10 a 2mm h2ltr19 DHnlmgtltr19 DW w rzwjFfnlmgt ZHE hZ r19 lnlmgt 11136 Since the Hamiltonian contains additive terms that act on separate subspaces the angular and radial spaces respectively the Schroedinger equation for the eigenfunction ra awlnalanO WWW clearly is separable in angular and radial parts the angular part being the eigenfunction of 2 Accordingly we factorize 1Jnlm in the form Z 11137 308 where the quantum number n as the quantum number related to energy will emerge from the solution of the radial equation 1 d d 211 hzll1 72 F2 EVrW Rnlr 11138 This follows from the solution of AJZlEm19p 1 lh2Ylm19p 11139 After this introduction we will deal with the hydrogen atom and achieve a deeper understanding before we deal with the angular momentum operator on a higher level 309 XIV 3 GENERAL CONSIDERATIONS OF THE VECTOR COUPLING The procedure of products of angular momentum eigenvectors developed for the Zeeman effect can be applied to arbitrary angular momenta J1J2M1M2gtJ1M1gtJ2M2gt 1431 In the language of representation theory this means One can build the new reducible subspace J1J2 from the subspaces J1 and J2 of dimensions N1 l and N2 l respectively Da J1J2 M1 M2gt Da YAIIM2iJ1 J2 M1 M2gt W Rotations transform the product space in itself Since the space is reducible it can be decomposed into irreducible spaces of the vectors l J M gt In order to determine alle possible values of J we start from the following relation We have seen before that the functions J 1 M1gt and J2 M 2gt are eigenfunctions of jlzj1z and JAZZsz respectively Then the product of the two functions which we want to write J 1 J 2 M1 M 2gt is an eigenfunction of the zcomponent of the total angular momentum A A A JZ J12 J22 with the eigenvalue M M M 2 We label the eigenfunctions of j 2 and jz with IJ M gt We can write it as a linear combination ofthe products J1J2M1M2gt J1 J2 JMgt Z ZlJ1J2AIlM2gtltJ1J2A4M2lJMgt 1433 M17J1 M27J2 471 The coef cients ltJ1 J2 M1 M2 J Mgt are real numbers our CLEBSCHGORDON coef cients which we have introduced in the discussion of the Zeeman effect In order to de ne completely the Clebsch Gordon coef cients we must x the phases of the vectors lJ J gt We do this by requiring that the component of lJ M gt along J1J2J1J J1 34 T be real and positive J1J2J1J J1 2 0 Real They are zero for M M1 M so that the double sum 1433 is in reality a single sum We can see that by taking A JZ JMgt MJMgt and le J22 J1J2ZVI1M2gtZ M leJ1J2MM2gt If we apply to both sides of the equation J1 J2 IJMgt Z ZlJtaJ2MM2gtltJDJ2MM2IJaMgt M1 J1 M2 J2 the operator JZ 2 J12 J22 and multiply from the left with lt12M1M2 we should obtain 472 MMIM2 J1J2M1M2JMgt O onlyif MM1 M2 The expression 1433 allows an inverse transformation JlJ2 J J1J2M1M2gt Z ZlJMgtltJMlJ1J2M1M2gt ll142 MEJ The sum over M contains only the one term with M M1 M 2 The condition that the functions Ji M igt and J M gt have to be orthogonal yield the following orthogonality conditions for the Clebsch Gordon coef cients J1 3912 Z ZltJ1J2M1M2JMgtltJ1J2M1M2J M gt5JJ5W M17J1 M27J2 J1J2 J ZltJ1J2M1M2 JMgtltJ1J2MM239 JlJ17J2l M7J J 5M1Mi5M2Mi J1 J2 2J1 Z ZltJ19J29M1 M2J9MgtltJ19JiaM1aMiJaMgt JzJi MzMi M17J1MZ7JZ 2J1 1 If we continue now analogously as in the case of J and J z with 473 17L17 WW ZWFW IiZWFW PuinWWlt 39019210u suuetuele Xpwul 9111 II 1 IZW 3W pup 11M 2 WI H zr4zW 141W WWW Zf IrX V IIIV ZF IFIRFIZW IW ZF WK Z Zr Ir C 04 r zr4zW Ir4IW mm FTXM JW Zr r rZW W Zr rZ Z Zr Ir W i 1 WV ZMNW ZI TXI IV TrfIVif zfzW aIW m IW ZI IIXM JW Zr IrFZrZ Z Zr 1 C lt14 1 zrzW IIW m2W 9 er JW ZI II F I Z Z Zr 1 1 14 XIV i 10 uquo 9AA W 1 ltIIV 1 We generate the function by operating Jli 0n J1J2M1M2gt and JZi 0n J1J2MIM2gt The coef cients are Jul i M1J1 M11J1J2M11M2gt and We i MZXJZ M2 1gtJ1J2MlM2 1 gtJMJiM1ltJ1J2M1M2 JMi1gt J14rM1J1T M11ltJ1J2M1 1M2 JMgt 1ltJ1gtJ2gtM1gtM2 When M J the top sign on the left hand side vanish es How does it look Using the upper Sign above and M1 J and M J we can determine M2 M1M2M1 M2MM11JJI1 and 475 mltJ1J2J1J J1 1lJ1J 1gt JJ2 JJ11J2 J J1ltJ1J2J1J J1lJJgt So J1J2J1J J1 1lJ1J 1gt J2 JJ11J2 J J1 x2J One can proceed in this way for other values of M1 and M2 and get all coefficients as multiples of one of them for example ltJ1 J 2 J1 J J1 J J gt With the normalization condition for the vector IJJgt J1J2J1J J1lJJgt J1 J2 Z ZltJ1J2M1M2J9Jgt2l M1 J1 M2 J2 and the phase condition they are completely determined One can see from M 2 J J that the coefficients are nonzero only for values of 39Jz J 39 J1 J2 M2 can have all values between J2 and J2 from where we get J1 39 J2 J J2 J We could also get the ClebschGordon coefficients as multiples of ltJ1 J 2 J J 2 J 2 J J gt ieM1JJ2 withtheresult JzJ1 J J1J2 476 Since the reduction of the product space is independent of it we conclude lJI39J2l J JzJ1 but this is the condition we would expect for J based on the J L S example We want to demonstrate this condition using the example J11 J232 J123252 5 I J 24 J1 Al J2 J2 A J 3 1 J25 J2 L 1 F1g 147 J Vector addition As one can see J J1 J2 form a triangle with integer or halfinteger sides the area of this triangle disappears in the limits Jmax and ijn More often than the ClebschGordon coefficients the 3J symbols introduced by Wigner are used The de nition is as follows J J J m m2 m3 1J1 J2 M32J3 1 2 J1J2 M1 M2J3 M3 2 3 The 3J symbols are tabulated in Edmonds BI 5353a Angular momenta in Quantum Mechanics 477
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