Ⅵ ELEMENTARY APPLICATIONS
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34. The harmonic oscillator

A SIMPLE and interesting example of a dynamical system in quantum mechanics is the harmonic oscillator. This example is of importance for general theory, because it forms a corner-stone in the theory of radiation. The dynamical variables needed for describing the system are just one coordinate g and its conjugate momentum p. The Hamiltonian in classical mechanics is H == (p*+mra%?), (1) where m is the mass of the oscillating particle and w is 27 times the frequency. We assume the same Hamiltonian in quantum mechanics. This Hamiltonian, together with the quantum condition (10) of § 22, define the system completely. The Heisenberg equations of motion are By = [p, H] = —maq,. It is convenient to introduce the dimensionless complex dynamical variable 4 == (2mfiw)?(p+imag). (3) The equations of motion (2) give ty = (2mhw)*(—mw*g, twp) = tony This equation can be integrated to give ; m= noe, (4) where 7 is a linear operator independent of t, and is equal to the value of 7, at time t= 0. The above equations are all as in the classical theory. We can express g and p in terms of 7 and its conjugate complex 7 and may thus work entirely in terms of 7 and 7. We have hon = (2m)~(p+-imag)(p—imeg) = (2m)~[p?-+-m?w*g? +-imeo(gp —P9)} = H—thw (5) and similarly hwiyn = H+thew. (8) Thus 99-7 = 1. (7) § 34 THE HARMONIC OSCILLATOR 137 Equation (5) or (6) gives H in terms of 7 and 4 and (7) gives the commutation relation connecting y and 7. From (5) heoinh = TH hoa and from (6) hong = H7+ es. Thus FH—Hy = hoi. (8) Also, (7) leads to An — qq = ny" (9) for any positive integer n, as may be verified by induction, since, by multiplying (9) by 7 on the left, we can deduce (9) with n+1 for n. Let H’ be an eigenvalue of H and |H’> an eigenket belonging to it. From. (5) hw |\n7q| > = = (f’—tho) is the square of the length of the ket 7|H’>, and h ‘ - ‘ ence > 0, the case of equality occurring only if 7|/H’> = 0. Also (H’|H’> > 0. the case of equality occurring only if 7/H’> = 0. From the form (1) of H as a sum of squares, we should expect its eigenvalues to be all positive or zero (since the average value of H for any state must be positive or zero). We now have the more stringent condition (10). From. (8) HG\H'> = (GH —het)|H"> = (H'—hw)q|H). (11) Now if H’ + dw, 7|H'> is not zero and is then according to (11) an eigenket of H belonging to the eigenvalue H’—tw. Thus, with H’ any eigenvalue of H not equal to diw, H’—fw is another eigenvalue of H. We can repeat the argument and infer that, if H’—hw ¥¢ fho, H’—2haw is another eigenvalue of H. Continuing in this way, we obtain the series of eigenvalues H’, H’—hw, H'—2hw, H’—dhw,..., which cannot extend to infinity; because then it would contain eigen- values contradicting (10), and can terminate only with the value jiw. Again, from the conjugate complex of equation (8) Hn|\H'> = (nH+hwn)|H"> = (H'+hew)n|H">, showing that H’+%w is another eigenvalue of H,,with 7|H’> as an eigenket belonging to it, unless y/H’> = 0. The latter alternative can be ruled out, since it would lead to 0 = hwin|H'> = (H+ fhw)|H"> = (H+ thw) |H, 138 ELEMENTARY APPLICATIONS § 34 which contradicts (10). Thus H’--iw is always another eigenvalue of H, and so are H’+ 2iw, H’-++-3hw and soon. Hence the eigenvalues of H are the series of numbers Mio, Bho, Bw, Tio, ... (12) extending to infinity. These are the possible energy values for the harmonic oscillator. Let |0> be an eigenket of H belonging to the lowest eigenvalue 1 and form the sequence of kets 10>, 9/0, nP10>, —-8]0>, (14) These kets are all eigenkets of H, belonging to the sequence of eigen- values (12) respectively. From (9) and (13) n"|0> = na"4]0 (15) for any non-negative integer n. Thus the set of kets (14) is such that n or 4 applied to any one of the set gives a ket dependent on the set. Now all the dynamical variables in our problem are expressible in terms of » and 7, so the kets (14) must form a complete set (otherwise there would be some more dynamical variables). There is just one of these kets for each eigenvalue (12) of H, so H by itself forms a complete commuting set of observables. The kets (14) correspond to the various stationary states of the oscillator. The stationary state with energy (n-+-4)hw, corresponding to y”|0), is called the nth quantum state. The square of the length of the ket 4”|0> is (O|7"9" 0) = 2X0] "49-40 with the help of (15). By induction, we find that <0|q""|0> = ni} (16) provided |0> is normalized. Thus the kets (14) multiplied by the coefficients n!-? with n = 0,1, 2,..., respectively form the basic kets of a representation, namely the representation with H diagonal. Any ket |z> can be expanded in the form ke) = Se, 10>, ~ a7) where the 2,’s are numbers. In this way the ket |x) is put into correspondence with a power series > x, 7” in the variable 7, the various terms in the power series corresponding to the various stationary states. If jx> is normalized, it defines a state for which § 34 THE HARMONIC OSCILLATOR 139 the probability of the oscillator being in the nth quantum state, ie. the probability of H having the value (n+4)ha, is P= n! |x|, (18) as follows from the same argument which led to (51) of § 18. We may consider the ket |0) as a standard ket and the power series in 7 as a wave function, since any ket can be expressed as such a wave function multiplied into this standard ket. We get a kind of wave function differing from the usual kind, introduced by equations (62) of § 20, in that it is a function of the complex dynamical variable y instead of observables. It was first introduced by V. Fock, so we shall call the representation Fock’s representation. It is for many purposes the most convenient representation for describing states of the harmonic oscillator. The standard ket [0 satisfies the condition (13), which replaces the conditions (43) of § 22 for the standard ket in Schrédinger’s representation. Let us introduce Schrédinger’s representation with g diagonal and obtain the representatives of the stationary states. From (13) and (3) (p—imag)|0> = 0, so = 0. With the help of (45) of § 22, this gives a ? f i head |0>+-mwg' = 0. (19) The solution of this differential equation is = (marfhiyremoa"?h, (20) the numerical coefficient being chosen so as to make |0> normalized. We have here the representative of the normal state, as the state of lowest energy is called. The representatives of the other stationary states can be obtained from it. We have from (3) = (2mhiw)-"?¢q'|(p +imagq)”|0> = (mfiny-nPin( —h jgi-t+ men’) = i*(2fen)-*( mean) z+ mcf) emo, (21) This may easily be worked out for small values of‘n. The result is of the form of e~”@7l2% times a power series of degree n in g’. A further factor n!-? must be inserted in (21) to get the normalized representa- tive of the nth quantum state. The phase factor i” may be discarded. 140 ELEMENTARY APPLICATIONS § 35 35. Angular momentum Let us consider a particle described by the three Cartesian coordi- nates x, y, z and their conjugate momenta p,, p,, p,. Its angular momentum about the origin is defined as in the classical theory, by or by the vector equation m= XX p. We must evaluate the P.B.s of the angular momentum components with the dynamical variables x, p,, etc., and with each other. This we can do most conveniently with the help of the laws (4) and (5) of § 21, thus and similarly, (23) [m,, Pa] = Py [m,, Py] = —DPz; (25) [m,, P2| = 0, (26) with corresponding relations for m, and m,. Again [m,; m,| = [zp,—xp,,m,| = 2[ Pz; m,|—[x, Mp] P, (27) [m., mz = My [m,, my} = ™M,- These results are all the same as in the classical theory. The sign in the results (23), (25), and (27) may easily be remembered from the rule that the + sign occurs when the three dynamical! variables, con- sisting of the two in the P.B. on the left-hand side’and the one forming the result on the right, are in the cyclic order (xyz) and the — sign occurs otherwise. Equations (27) may be put in the vector form mxm = iim. (28) Now suppose we have several particles with angular momenta m,,m,,.... Each of these angular momentum vectors will satisfy 9 and any one of them will commute with any other, so that m. m, is the total angular momentum, r MxM => m,xm, = > m,xm,+ > (m,xm,-+m,x m,) rs r r, say, satisfying M,|S> = M,|S) = M,|8 = 0, and hence ,|S> = r,|S> = 7,|S> = 0. This shows that the ket |S) is unaltered by infinitesimal rotations, and it must therefore be unaltered by finite rotations, since the latter can be built up from infinitesimal ones. Thus the state is spherically symmetrical. The converse theorem, a spherically symmetrical state has zero total angular momentum, is also true, though its proof is not quite so simple. A spherically symmetrical state corresponds to a ket |S whose direction is unaltered by any rotation. Thus the change 144 ELEMENTARY APPLICATIONS § 35 in |S) produced by a rotation operator 7,,7,, or 7, must be a numerical multiple of |S), say e|S> = Cy|S>, ry|S> = Cy |S, 7,|S> = c,|8>, where the c’s are numbers. This gives M,|S> = ttie,|S>, MM, |8> = ihe,|S>, M,|S> = ike,|S). (33) These equations are not consistent with the commutation relations (29) for M,, M,, M, unless c, = ¢, = c, = 0, in which case the state has zero total angular momentum. We have in (33) an example of a ket which is simultaneously an eigenket of the three non-commuting linear operators M,, M,, M,, and this is possible only if all three eigenvalues are zero. 36. Properties of angular momentum There are some general properties of angular momentum, deducible simply from the commutation relations between the three compo- nents. These properties must hold equally for spin and orbital angular momentum. Let m,, m,, m, be the three components of an angular momentum, and introduce the quantity 8 defined by B= mh+m2 +m. Since f is a scalar it must commute with m,, m,, and m,. Let us suppose we have a dynamical system for which m,, m,, m, are the only dynamical variables. Then 8 commutes with everything and must be a number. We can study this dynamical system on much the same lines as we used for the harmonic oscillator in § 34. Put M,—tM, = 7. From the commutation relations (27) we get m,-+mZ— tm, My—M, Mz) = B—m2--tim, (34) and similarly qq = B—m2—im,. (85) Thus AN—H = BZhm,. (36) Also M,N—NM, = tim, —im, = —tin. (37) We assume that the components of an angular momentum are observables and thus m, has eigenvalues. Let mj be one of them, and |m;> an eigenket belonging to it. From (34) = = (B—m?+-hm,) and is thus greater than or equal to zero, the case of equality occur- ring if and only if y|{m,)> = 0. Hence B—m2-+him!, > 0, or B+-40" > (mi—4h)?. (38) Thus B+it? > 0. Defining the number & by be = (BEE)! = (m3--m}-m2+ Hey, (39) so that k > —4$h, the inequality (38) becomes k-+-Hi > |i —4h| or k+h > m, > —k. (40) An equality occurs if and only if y]m,> = 0. Similarly from (35) mt nii|m,> = (B—m,2—Fim) mim’), showing that p—mZ—im, > 0 or k > m, > —k—Ah, with an equality occurring if and only if q]m> == 0. This result combined with (40) shows that &£ > 0 and k>m, > ~k, (41) with m, = kif aim = 0 and m, = —k if ylm> = 0. From (37) m,\m,> = (ym,—fin)|m,> = (m,—h)y|m,). Now if m, 4 —k, y/mj> is not zero and is then an eigenket of m, belonging to the eigenvalue m,—h. Similarly, ifmj,—h + —k, m,—2h is another eigenvalue of m,, and so on. We get in this way a series of eigenvalues mj, m,—h, m,—2#,..., which must terminate from (41), and can terminate only with the value —k. Again, from the conjugate complex of equation (37) mi, ilmgy = (Fm, HHA) |my> = (mi, -A)Alm,, showing that m}+-# is another eigenvalue of m, unless 7|/m,> = 0, in which case m, = &. Continuing in this way we get a series of eigen- values m,,m,+h,m,+2h,.... which must terminate from (41), and can terminate only with the value &. We can conclude that 2% is an integral multiple of 4 and that the eigenvalues of m, are k, k—-h, k—2h, ..., —k+h, —k. (42) 146 ELEMENTARY APPLICATIONS § 36 The eigenvalues of m, and m, are the same, from symmetry. These eigenvalues are all integral or half odd integral multiples of %, accord- ing to whether 2k is an even or odd multiple of %. Let |max> be an eigenket of m, belonging to the maximum eigen- value k, so that qimax) = 0, (43) and form the sequence of kets jmax>, ylmax>, 7?|max>, ..., »?/*|max). (44) These kets are all eigenkets of m,, belonging to the sequence of eigen- values (42) respectively. The set of kets (44) is such that the operator 7 applied to any one of them gives a ket dependent on the set (y applied to the last gives zero), and from (36) and (43) one sees that 7 applied to any one of the set also gives a ket dependent on the set. All the dynamical variables for the system we are now dealing with are expressible in terms of 7 and 7, so the set of kets (44) is a complete set. There is just one of these kets for each eigenvalue (42) of m,, so m, by itself forms a complete commuting set of observables. It is convenient to define the magnitude of the angular momentum vector m to be k, given by (39), rather than f!, because the possible values for k are 0, fh, h, 3h, Qh, on, (45) extending to infinity, while the possible values for £? are a more complicated set of numbers. For a dynamical system involving other dynamical variables besides Mz, M,, and m,, there may be variables that do not commute with f. Then f is no longer a number, but a general linear operator. This happens for any orbital angular momentum (22), as x, y, z, Dz, Py, and p, do not commute with 8. We shall assume that 8 is always an observable, and & can then be defined by (39) with the positive square root function and is also an observable. We shall call & so defined the magnitude of the angular momentum vector m in the general case. The above analysis by which we obtained the eigenvalues of m, is still valid if we replace |m,> by a simultaneous eigenket |k’m,> of the commuting observables k and m,, and leads to the result that the possible eigenvalues for k are the numbers (45), and for each eigenvalue k’ of k the eigenvalues of m, are the numbers (42) with k’ substituted for k. We have here an example of a phenomenon which we have not met with previously, namely that with two commuting observables, the eigenvalues of one depend on what eigenvalue we § 36 PROPERTIES OF ANGULAR MOMENTUM 147 assign to the other. This phenomenon may be understood as the two observables being not altogether independent, but partially functions of one another. The number of independent simultaneous eigenkets of & and m, belonging to the eigenvalues k’ and m, must be indepen- dent of mj, since for each independent jk’m}> we can obtain an independent |k’m;>, for any mz in the sequence (42), by multiplying \b’m,> by a suitable power of y or 7. As an example let us consider a dynamical system with two angular momenta m, and m,, which commute with one another. If there are no other dynamical variables, then all the dynamical variables com- mute with the magnitudes /, and k, of m, and mg, so k, and k, are numbers. However, the magnitude K of the resultant angular momentum M = m,-+m, is not a number (it does not commute with the components of m, and m,) and it is interesting to work out the eigenvalues of K. This can be done most simply by a method of counting independent kets. There is one independent simultaneous eigenket of m,, and m,, belonging to any eigenvalue m,, having one of the values k,, ky —h, ky —26h,..., —k, and any eigenvalue m,, having one of the values k,, k,—f, k,—2h,..., —k,, and this ket is an eigenket of M, belonging to the eigenvalue M, = m,,+mj,. The possible values of M’, are thus h,-+-k,, hk, tky—h, ky +k,—2h,...,—k,—ky, and the number of times each of them occurs is given by the following scheme (if we assume for definiteness that k, > k,), Tey} Keng bey hhg Fi, Boy + ig — 2h ooey ley leg, bey — leg,» 1 2 3 Qhg tl k+1 (46) Jey len, —hey + lig Ih...) — ho — hog ky 1 Qk, .. 1 Now each eigenvalue K’ of K will be associated with the eigenvalues K’, K'—h, K'—%,..., ~K’' for M,, with the same number of indepen- dent simultaneous eigenkets of K and M, for each of them. The total number of independent eigenkets of J, belonging to any eigenvalue Mi must be the same, whether we take them to be simultaneous eigenkets of m,, and mg, or simultaneous eigenkets of K and M,, i.e. it is always given by the scheme (46). It follows that the eigenvalues for K are lett, ey-tky—h, fey hog — 2H, ws ey — eg, (47) and that for each of these eigenvalues for K and an eigenvalue for 148 ELEMENTARY APPLICATIONS § 36 M, going with it there is just one independent simultaneous eigenket of K and M,. The effect of rotations on eigenkets of angular momentum variables should be noted. Take any eigenket |) of the z component of total angular momentum for any dynamical system, and apply to it a small rotation through an angle 6¢ about the z-axis. It will change into (1-+8¢r,)|Mz> = (1—i8$.M,/)|M,) with the help of (32). This equals (1—i86.Mi/%) |My == e~BOMIh| > to the first order in 8¢. Thus |M{> gets multiplied by the numerical factor e-®¢MJk, By applying a succession of these small rotations, we find that the application of a finite rotation through an angle ¢ about the z-axis causes |M{> to get multiplied by e~**™%"_ Putting 6 = 2m, we find that an application of one revolution about the z-axis leaves |M“> unchanged if the eigenvalue M;, is an integral multiple of f and causes |M{> to change sign if 1, is half an odd integral multiple of h. Now consider an eigenket |K’> of the magnitude K of the total angu- lar momentum. If the eigenvalue K’ is an integral multiple of #, the possible eigenvalues of M, are all integral multiples of# and the applica- tion of one revolution about the z-axis must leave |K’> unchanged. Conversely, if K’ is halfan odd integral multiple of 4, the possible eigen- values of M, are all half odd integral multiples of # and the revolution must change the sign of |K’>. From symmetry, the application of a revolution about any other axis must have the same effect on |’) as one about the z-axis. We thus get the general result, the application of one revolution about any axis leaves a ket unchanged or changes tts sign according to whether it belongs to eigenvalues of the magnitude of the total angular momentum which are integral or half odd integral multiples of h. A state, of course, is always unaffected by the revolu- tion, since a state is unaffected by a change of sign of the ket corre- sponding to it. For a dynamical system involving only orbital angular momenta, a ket must be unchanged by a revolution about an axis, since we can set up Schrédinger’s representation, with the coordinates of all the particles diagonal, and the Schrédinger representative of a ket will get brought back to its original value by the revolution. It follows that the eigenvalues of the magnitude of an orbital angular momentum are always integral multiples of k. The eigenvalues of a component § 36 PROPERTIES OF ANGULAR MOMENTUM 149 of an orbital angular momentum are also always integral multiples of #. For a spin angular momentum, Schrédinger’s representation does not exist and both kinds of eigenvalue are possible. 37. The spin of the electron Electrons, and also some of the other fundamental particles (pro- tons, neutrons) have a spin whose magnitude is 4%. This is found from experimental evidence, and also there are theoretical reasons showing that this spin value is more elementary than any other, even spin zero (see Chapter XI). The study of this particular spin is there- fore of special importance. For dealing with an angular momentum m whose magnitude is 3%, it is convenient to put m = Hie. (48) The components of the vector « then satisfy, from (27), Oy Fg 0, 0y = Qoz 0,0 y-—8,0, = to, (49) Og Sy—FySz, = 2to,. The eigenvalues of m, are $7 and —4%, so the eigenvalues of a, are 1 and —1, and o2 has just the one eigenvalue 1. It follows that o? must equal 1, and similarly for o2 and a3, i.e. B= =of =i, (50) We can get equations (49) and (50) into a simpler form by means of some straightforward non-commutative algebra. From (50) 2 2. OF, 9, 0% = 0 or 0,(Fy F,— 0, Fy) + (oy 0,—G, 5y)o, = 0 or Oy Fg tz Fy = with the help of the first of equations (49). Thismeansa, oa, = —oy oz. Two dynamical variables or linear operators like these which satisfy the commutative law of multiplication except for a minus sign will be said to anticommute. Thus o, anticommutes with c,. From sym- metry each of the three dynamical variables c,, o,, o, must anti- commute with any other. Equations (49) may now be written 0,0, = 0, = —G, Oy, Oz F_ = Wy = —O,5,, (51) Og Fy = 1G, = — dy Fz, and also from (50 GO, 6,0, == t. 52 Yo) 7 3595.57 L 150 ELEMENTARY APPLICATIONS § 37 Equations (50), (51), (52) are the fundamental equations satisfied by the spin variables o describing a spin whose magnitude is 4%. Let us set up a matrix representation for the o’s and let us take o, to be diagonal. If there are no other independent dynamical variables besides the m’s or o’s in our dynamical system, then o, by itself forms a complete set of commuting observables, since the form of equations (50) and (51) is such that we cannot construct out of o,, o,, and o, any new dynamical variable that commutes with o,. The diagonal elements of the matrix representing o, being the eigenvalues 1 and —1 of o,, the matrix itself will be ( 1 0 0 i}: L ay a) et a, be represented by ): a This matrix must be Hermitian, so that a, and a, must be real and a, and a, conjugate complex numbers. The equation o,¢, = —o,9, gives us ay Ge) f% a —d, —a Mz —a,)’ so that a, = a, = 0. Hence o, is represented by a matrix of the form 0 a, a, Of The equation of = 1 now shows that a,a, = 1. Thus a, and ag, being conjugate complex numbers, must be of the form e’“ and e-* re- spectively, where « is a real number, so that o, is represented by a matrix of the form 0 “ ete OP Similarly it may be shown that a, is also represented by a matrix of this form. By suitably choosing the phase factors in the representa- tion, which is not completely determined by the condition that o, shall be diagonal, we can arrange that o, shall be represented by the matrix 01 1 of The representative of o, is then determined by the equation = 0,0, We thus obtain finally the three matrices (to) (Fo op a Cy § 37 THE SPIN OF THE ELECTRON 151 to represent o,, o,, and a, respectively, which matrices satisfy all the algebraic relations (49), (50), (51), (52). The component of the vector o in an arbitrary direction specified by the direction cosines 1, m, n, namely Ic,-+-mo,-+no,, is represented by n l—im lism on} (8 The representative of a ket vector will consist of just two numbers, corresponding to the two values +1 and —1 for oj. These two num- bers form a function of the variable of whose domain consists of only the two points +land —1. The state for which oc, has the value unity will be represented by the function, f,(o{) say, consisting of the pair of numbers 1, 0 and that for which o, has the value —1 will be represented by the function, fs(o,) say, consisting of the pair 0, 1. Any function of the variable oj, i.e. any pair of numbers, can be expressed as a linear combination of these two. Thus any state can be obtained by superposition of the two states for which o, equals +1 and —1 respectively. For example, the state for which the component of o in the direction |, m, n, represented hy (54), has the value +1 is represented by the pair of numbers a, 6 which satisfy n a) a\ fa itim —n ) ~ (; or nat+(l—im)b = a, (+tim)a—nb = b. a l—im I+” Th ~ = = . “ 6 in ~ Tim This state can be regarded as a superposition of the two states for which o, equals +1 and —1, the relative weights in the superposition process being as jal? : [|]? = [l—im|?: (l—n)? = l+n:l—n. (55) For the complete description of an electron (or other elementary particle with spin 44) we require the spin dynamical variables o, whose connexion with the spin angular momentum is given by (48), together with the Cartesian coordinates x, y, z and momenta p,, Dy, p, The spin dynamical variables commute with these coordinates and momenta. Thus a complete set of commuting observables for a system consisting of a single electron will be z, y, z, ¢,. In a repre- sentation in which these are diagonal, the representative of any state 152 ELEMENTARY APPLICATIONS § 37 will be a function of four variables 2’, y’, 2’, oj. Since o, has a domain consisting of only two points, namely 1 and —1, this function of four variables is the same as two functions of three variables, namely the two functions Ca'y"2" |) == <8", y', 2’, +1), kaly'2'|>_ == . (56) Thus the presence of the spin may be considered either as introducing a new variable into the representative of a state or as giving this representa- tive two components. 38. Motion in a central field of force An atom consists of a massive positively charged nucleus together with a number of electrons moving round, under the influence of the attractive force of the nucleus and their own mutual repulsions. An exact treatment of this dynamical system is a very difficult mathe- matical problem. One can, however, gain some insight into the main features of the system by making the rough approximation of regard- ing each electron as moving independently in a certain cenéral field of force, namely that of the nucleus, assumed fixed, together with some kind of average of the forces due to the other electrons. Thus our present problem of the motion of a particle in a central field of force forms a corner-stone in the theory of the atom. Let the Cartesian coordinates of the particle, referred to a system of axes with the centre of force as origin, be x, y, z and the corre- sponding components of momentum p,, p,, Pp, The Hamiltonian, with neglect of relativistic mechanics, will be of the form H = 1/2m.(p2+p2+p2)+V, (57) where V, the potential energy, is a function only of (2?+y?+2*). To develop the theory it is convenient to introduce polar dynamical variables. We introduce first the radius 7, defined as the positive square root r= (aL y?+a2)i, Its eigenvalues go from 0 too. If we evaluate its P.B.s with p,, p,, and p,, we obtain, with the help of formula (32) of § 22, - or 2 y 2 [7, De = ex = r [7, Pyl = , [7, Pe] = 7’ the same as in the classical theory. We introduce also the dynamical variable p, defined by Dp == Tey t+ YPy t+ ePe)- (58 § 38 MOTION IN A CENTRAL FIELD OF FORCE 153 Its P.B. with r is given by rr, Pr] = [r, rp, | —_ [7, CPs YPy } zp,| = alr, Pal+yl7, Py] +2[r, p,] == e.a/rty.y/rte.2z/r =r. Hence [7,p,] = 1 or 1),—P,T = th. The commutation relation between r and p, is just the one for a canonical coordinate and momentum, namely equation (10) of § 22. This makes p, like the momentum conjugate to the r coordinate, but it is not exactly equal to this momentum because it is not real, its conjugate complex being By = (Pet +iyytp,2)0* = (epz+ypy+ep,— 3ih)r == (rp.—dvh)r“ = p,—2ihir-1. (59) Thus p,—tir— is real and is the true momentum conjugate to r. The angular momentum m of the particle about the origin is given by (22) and its magnitude & is given by (39). Since r and p, are scalars, they commute with m, and therefore also with &. We can express the Hamiltonian in terms ofr, p,, and k. We have, if ¥ denotes a sum over cyclic permutations of the suffixes x, y, z, LYS k(k-+-h) =a = > (xp,—yp.)* xyz = > (Cy XPy+YDz YP2—ZPy YPz—YPz LP y) = > (x2 p32 + y*p2— xp, Py Y—YPy Pn ED — UP Dy C— cyz —2hap,) = (22+-y2-+2%)(p2-++p? +p) — (ap, +YyPy+2p,)(Pet+Py Y +P, 2+ 2m) 7? (pt pi + p2)—1D,(B,7 + 21) = P(pitpi+pe)—Tpr. from (59). Hence H=-— (oer SE a7. (60) ~~ 9m 7 This form for H is such that k commutes not only with H, as is necessary since & is a constant of the motion, but also with every dynamical variable occurring in H, namely r, p,, and V, which is a 154 ELEMENTARY APPLICATIONS § 38 function of r. In consequence, a simple treatment becomes possible, namely, we may consider an eigenstate of k belonging to an eigen- value k’ and then we can substitute b’ for & in (60) and get a problem in one degree of freedom r. Let us introduce Schrédinger’s representation with x, y, z diagonal. Then pz, Py, p, are equal to the operators —i% 6/éx, —ih a/éy, —th d/0z respectively. A state is represented by a wave function y(xyzt) satis- fying Schrédinger’s wave equation (7) of § 27, which now reads, with Hf given by (57), ea = oO =| ! a (61) Qm\ ax? ' dy? ' az? We may pass from the Cartesian coordinates w,y,z to the polar coordinates 7,9,6é by means of the equations x = rsin@cos¢, y = rsin @sin ¢d, (62) z= rcos6, and may express the wave function in terms of the polar coordinates, so that it reads 4(rédt). The equations (62) give the operator equation a. de 6 by a Ge HA Ye 220 ar er aut or by | ar bz roe ay tr oz’ which shows, on being compared with (58), that p, = —i#a/ér. Thus Schrédinger’s wave equation reads, with the form (60) for H, Op [ie 1 @ k(k-+h)\ | Q i at (= (—3 Bret Here]! a (68) Here & is a certain linear operator which, since it commutes with r and 6/ér, can involve only @, ¢, @/€0, and 8/86. From the formula k(k-+h) = m-m2 +m, (64) which comes from (39), and from (62) one can work out the form of k(k-+-%) and one finds h(k-+h) 1@.,8@ 1 @ i nd 0a Sk OE (8) This operator is well known in mathematical physics. Its eigen- functions are called spherical harmonics and its eigenvalues are n(n-+-1) where nm is an integer. Thus the theory of spherical har- monies provides an alternative proof that the eigenvalues of k are integral multiples of %. § 38 MOTION IN A CENTRAL FIELD OF FORCE 155 For an eigenstate of & belonging to the eigenvalue nfi (n a non- negative integer) the wave function will be of the form | . = r2y(r8)S,,(66), (66) where S,,(@¢) satisfies K(k+-R)S, (86) = n(n+ 1)hS,,(89), (67) ie. from (65) S,, is a spherical harmonic of order ». The factor r-t is inserted in (66) for convenience. Substituting (66) into (63), we get as the equation for x 0x (#8 & nlnt+))\ , he (sa Bre pe V}x. (68) If the state is a stationary state belonging to the energy value H’, x will be of the form x(rt) = xo(ryentettt and (68) will reduce to 2 2 By = [E(t +7) (69) Qm\ drt This equation may be used to determine the energy-levels H’ cf the system. For each solution y, of (69), arising from a given n, there will be 2n+-1 independent states, because there are 2n--1 indepen- dent solutions of (67) corresponding to the 2n+1 different values that a component of the angular momentum, m, say, can take on. The probability of the particle being in an element of volume dadyéz is proportional to |b/?dzdydz. With % of the form (66) this becomes r-?|y/2|S,,|2dadydz. The probability of the particle being in a spherical shell between r and r+dr is then proportional to |x|?dr. It now becomes clear that, in solving equation (68) or (69), we must impose a boundary condition on the function y at r = 0, namely the function must be such that the integral to the origin i ly|? dr is 0 convergent. If this integral were not convergent, the wave function would represent a state for which the chances are infinitely in favour of the particle being at the origin and such a state would not be physically admissible. The boundary condition at r = 0 obtained by the above considera- tion of probabilities is, however, not sufficiently stringent. We get a more stringent condition by verifying that the wave function obtained by solving the wave equation in polar coordinates (63) really satisfies the wave equation in Cartesian coordinates (61). Let us take the case 156 ELEMENTARY APPLICATIONS § 38 of V = 0, giving us the problem of the free particle. Applied to a stationary state with energy H’ == 0, equation (61) gives V4 = 0, (70) where V? is written for the Laplacian operator 6?/éx?+-6?/ay? + 27/é2", and equation (63) gives ( a ses = 0. (71) r or? hr? A solution of (71) for k=0 is 4 =r. This does not satisfy (70), since, although V?r-! vanishes for any finite value of 7, its integral through a volume containing the origin is —47 (as may be verified by transforming this volume integral to a surface integral by means of Gauss’s theorem), and hence V2r-k == — 4 8(x)5(y)5(2). (72) Thus not every solution of (71) gives a solution of (70), and more generally, not every solution of (63) is a solution of (61). We must impose on the solution of (63) the condition that it shall not tend to infinity as rapidly as r-1 when r -> 0 in order that, when substituted into (61), it shall not give a 5 function on the right like the right-hand side of (72), Only when equation (63) is supplemented with this condi- tion does it become equivalent to equation (61). We thus have the boundary condition rf > 0 or y> 0 as r—-> 0. There are also boundary conditions for the wave function at r = oo. If we are interested only in ‘closed’ states, i.e. states for which the particle does not go off to infinity, we must restrict the integral to infinity i |x(r)|? dr to be convergent. These closed states, however, are not the only ones that are physically permissible, as we can also have states in which the particle arrives from infinity, is scattered by the central field of force, and goes off to infinity again. For these states the wave function may remain finite as r -> 00. Such states will be dealt with in Chapter VIII under the heading of collision problems. In any case the wave function must not tend to infinity as r > 00, or it will represent a state that has no physical meaning. 39. Energy-levels of the hydrogen atom The above analysis may be applied to the problem of the hydrogen atom with neglect of relativistic mechanics and the spin of the § 39 ENERGY-LEVELS OF THE HYDROGEN ATOM 157 electron. The potential energy V is nowt —e?/r, so that equation (69) becomes a n(nt+l) , 2me* 1 2mH’ ie re + Fe 7 (Xo Fe Xo (73) ; A thorough investigation of this equation has been given by Schré- dinger.t We shall here obtain its eigenvalues H’ by an elementary argument. t is convenient to put Xo = (rye, (74) introducing the new function f(r), where a is one or other of the square roots a == t4)(—H2/2mH’). (75) Equation (73) now becomes @ 2d n(nm+il) , Ime? 1 _ fe adr 7 or Flr) = 0. (78) We look for a solution of this equation in the form of a power series f(r) = 2 C,r, (77) in which consecutive values for s differ by unity although these values themselves need not be integers. On substituting (77) in (76) we obtain ¥ o,{s(s—1)r*-2— (28/a)r8-1—n(n-+ 1 prs? + (2me?/h?)rs-1} = 0, 8 which gives, on equating to zero the coefficient of r°-*, the following relation between successive coefficients c,, ca[s(s—1)—n(n-+1)] = ¢,_,[2(s—1)/a—2me?/h?]. (78) We saw in the preceding section that only those eigenfunctions x are allowed that tend to zero with r and hence, from (74), f(r) must tend to zero with r. The series (77) must therefore terminate on the side of small s and the minimum value of s must be greater than zero. Now the only possible minimum values of s are those that make the coefficient of c, in (78) vanish, ie. n-+1 and —n, and the second of these is negative or zero. Thus the minimum value of s must be n+1. Since n is always an integer, the values of s will all be integers. + The e here, denoting minus the charge on an electron, is, of course, to be dis- tinguished from the e denoting the base of exponentials. } Schrédinger, Ann, d. Physik, 79 (1926), 361. 158 ELEMENTARY APPLICATIONS § 39 The series (77) will in general extend to infinity on the side of large s. For large values of s the ratio of successive terms is Cy ar t= Cy-4 8a according to (78). Thus the series (77) will always converge, as the ratios of the higher terms to one another are the same as for the series 1 /2r\s S siz] ; (79) & which converges to e?7/@, We must now examine how our solution xy, behaves for large values of r. We must distinguish between the two cases of H’ positive and H' negative. For H’ negative, a given by (75) will be real. Sup- pose we take the positive value for a. Then as roo the sum of the series (77) will tend to infinity according to the same law as the sum of the series (79), ie. the law e2/¢. Thus, from (74), yo will tend to infinity according to the law e/@ and will not represent a physically possible state. There is therefore in general no permissible solution of (73) for negative values of H’. An exception arises, however, when- ever the series (77) terminates on the side of large s, in which case the boundary conditions are all satisfied. The condition for this termina- tion of the series is that the coefficient of c,_, in (78) shall vanish for some value of the suffix s—1 not less than its minimum value n+ 1, which is the same as the condition that s me? 2 =0 a for some integer s not less than n+-1. With the help of (75) this condition becomes 4 me Ht’ — 282K?’ I (80) and is thus a condition for the energy-level H’. Since s may be any positive integer, the formula (80) gives a discrete set of negative energy-levels for the hydrogen atom. These are in agreement with experiment. For each of them (except the lowest one s = 1) there are several independent states, as there are various possible values for n, namely any positive or zero integer less than s. This multi- plicity of states belonging to an energy-level is in addition to that mentioned in the preceding section arising from the various possible § 39 ENERGY-LEVELS OF THE HYDROGEN ATOM 159 values for a component of angular momentum, which latter multi- plicity occurs with any central field of force. The x multiplicity occurs only with an inverse square law of force and even then is removed when one takes relativistic mechanics into account, as will be found in Chapter XI. The solution x, of (73) when H’ satisfies (80) tends to zero exponentially as r > co and thus represents a closed state (corre- sponding to an elliptic orbit in Bohr’s theory). For any positive values of H’, a given by (75) will be pure imaginary. The series (77), which is like the series (79) for large r, will now have a sum that remains finite asr ->00. Thus x, given by (74) willnowremain finite as r > co and will therefore be a permissible solution of (73), giving a wave function ¢ that tends to zero according to the law r-* as r->co. Hence in addition to the discrete set of negative energy-levels (80), all positive energy-levels are allowed. The states of positive co energy are not closed, since for them the integral to infinity | lyql? dr does not converge. (These states correspond to the hyperbolic orbits of Bohr’s theory.) 40. Selection rules If a dynamical system is set up in a certain stationary state, it will remain in that stationary state so long as it is not acted upon by outside forces. Any atomic system in practice, however, frequently gets acted upon by external electromagnetic fields, under whose influence it is liable to cease to be in one stationary state and to make a transition to another. The theory of such transitions will be de- veloped in §§ 44and 45. A result of this theory is that, to a high degree of accuracy, transitions between two states cannot occur under the influence of electromagnetic radiation if, in a Heisenberg representa- tion with these two stationary states as two of the basic states, the matrix element, referring to these two states, of the representative of the total electric displacement D of the system vanishes. Now it happens for many atomic systems that the great majority of the matrix elements of D in a Heisenberg representation do vanish, and hence there are severe limitations on the possibilities for transitions. The rules that express these limitations are called selection rules. The idea of selection rules can be refined by a more detailed application of the theory of §§44 and 45, according to which the matrix elements of the different Cartesian components of the vector D are associated with different states of polarization of the 160 ELEMENTARY APPLICATIONS § 40 electromagnetic radiation. The nature of this association is just what one would get if one considered the matrix elements, or rather their real parts, as the amplitudes of harmonic oscillators which interact with the field of radiation according to classical electrodynamics. There is a general method for obtaining all selection rules, as follows. Let us call the constants of the motion which are diagonal in the Heisenberg representation «’s and let D be one of the Cartesian components of D. We must obtain an algebraic equation connecting D and the «’s which does not involve any dynamical variables other than D and the a’s and which is linear in D. Such an equation will be of the form YF-Do, = 0, (81) where the f,’s and g,’s are functions of the a’s only. If this equation is expressed in terms of representatives, it gives us X fla’) flo )ge(") = 0, which shows that == 0 unless E fila! gp(o") = 0. (82) This last equation, giving the connexion which must exist between « and «” in order that <«'|D|a”> may not vanish, constitutes the selection rule, so far as the component D of D is concerned. Our work on the harmonic oscillator in § 34 provides an example of a selection rule. Equation (8) is of the form (81) with 7 for D and H playing the part of the «’s, and it shows that the matrix elements (H'|q|H"> of 7 all vanish except those for which H”’—H’ = hw. The conjugate complex of this result is that the matrix elements (H’|n|H’> of » all vanish except those for which H”—H’ = —iiw. Since q is a numerical multiple of »—4, its matrix elements (H’|q|H”) all vanish except those for which H’—H’ = tiw. If the harmonic oscillator carries an electric charge, its electric displacement D will be pro- portional to g. The selection rule is then that only those transitions can take place in which the energy H changes by a single quan- tum fiw. We shall now obtain the selection rules fer m, and k for an electron moving in a central field of force. The components. of electric dis- § 40 SELECTION RULES 161 placement are here proportional to the Cartesian coordinates 2, y, 2. Taking first m,, we have that m, commutes with z, or that M,2—zm, = 0. This is an equation of the required type (81), giving us the selection rule , ” m,—M, = 0 for the z-component of the displacement. Again, from equations (23) we have [ms [m,, al] = [m,, y| =e or mxz—2m,7m,+omi—a = 0, which is also of the type (81) and gives us the selection rule a” "2 tt A yp Ff2 = m,2—In,m,+-mPe—h 0 or (m,—m,—h)(m,—m, +h) = 0 for the z-component of the displacement. The selection rule for the y-component is the same. Thus our selection rules for m, are that in transitions associated with radiation with a polarization corresponding to an electric dipole in the z-direction, m, cannot change, while in transi- tions associated with a polarization corresponding to an electric dipole in the x-direction or y-direction, m, must change by -h. We can determine more accurately the state of polarization of the radiation associated with a transition in which m, changes by +4, by considering the condition for the non-vanishing of matrix elements of a-+iy and z—iy. We have [m,,2+iy] = y—-w = —iatity) or m,(a-+iy)—(a+iy)(m,+h) = 9, which is again of the type (81). It gives m,—m,—h = 0 as the condition that shall not vanish. Similarly, m,—m,+h = 0 is the condition that shall not vanish. Hence = 0 or = i-+}, which determines 162 ELEMENTARY APPLICATIONS § 40 the state of polarization of the radiation associated with transitions for which m, = m,—h, has the following three components 4{ + } = U(a+tibjeiot+ (a—ibjeo} = acoswt—bsinut, Hdmi ly|m,—h> + } (83) = fi{—(a+ibje!+ (a—ibje} = asinwt+6 cos wt, { +m, —h|z|my} = 0. From the form of these components we see that the associated radia- tion moving in the z-direction will be circularly polarized, that moving in any direction in the «y-plane will be linearly polarized in this plane, and that moving in intermediate directions will be elliptically polarized. The direction of circular polarization for radia- tion moving in the z-direction will depend on whether w is positive or negative, and this will depend on which of the two states m, or m, = m,—h has the greater energy. We shall now determine the selection rule for k. We have [k(k-+-h), 2] = [mi,z]+[m2, z] YM. My, YELM, HM ye = 2(m,x—m, y+ itz) = 2(m,x—ym,) = 2am —M,Y¥). Similarly, [k(k-+-h), c] = 2(ym,—m, 2) and [k(k-+-4), y] = 2(m,2—am,). Hence [k(e-Lh), [e(e-+A), 2] = 4k(k+-h), m,x—m, y+tiz] “. am, [k(k-+h), x] —2m,[k(k+h), y]+ 2ih[k(k-+H), 2] z)—4m,,(m, z2—xm,)-+ Wh(k-h)z—2k(k-+hy} 2b my, ym, 2)m,—4(m3-+m2-+-m2)z4 + 2fk(k+-h)e—zk(ke+h)}. From (22) MyzL+-MyY+M,% = 0 (84) and hence [k(h-+h), [k(k-+-h), 2] = —2{k(k+hje+ek(kh+hy}, which gives k2(ke--i)22—~ Qhe(k-+ hiek(k--h) 2k? (lo-fi)? QWAAk(k+Aje+ek(k+h)} = 0. (85) § 40 SELECTION RULES 163 Similar equations hold for « and y. These equations are of the re- quired type (81), and give us the selection rule h'?(Ke’ -+-5)2— 2h (Ie! +B) k" (kl +h) + b"2(k" Bs)? — 282k’ (ke! +h) —QH2k" (k” Lh) = 0, which reduces to (ho Lk" + 2h) (k' +k Vk’ — kh’ +h) (kb’ —k” —h) = 0. A transition can take place between two states &’ and k” only if one of these four factors vanishes. Now the first of the factors, (k’+-k’ +24), can never vanish, since the eigenvalues of & are all positive or zero. The second, (k'--4"), can vanish only if k’ = O0and k” = 0. But transitions between two states with these values for & cannot occur on account of other selection rules, as may be seen from the following argument. If two states (labelled respectively with a single prime and a double prime) are such that &’ = 0 and hk” = 0, then from (41) and the corresponding results for m, and m,, mi, = ml = m, = 0 and mZ = mj) = mi = 0. The selection rule for m, now shows that the matrix elements of a and y referring to the two states must vanish, as the value of m, does not change during the transition, and the similar selection rule for m, or m, shows that the matrix element of z also vanishes. Thus ransitions between the two states cannot occur. Our selection rule for k now reduces to (h’—k’ +h)(k’—k’ fh) = 0, showing that k must change by --h. This selection rule may be written hy? 2h’ kh’ +-h2—H? = 0, and since this is the condition that a matrix element ¢k’ |z|k"> shall not vanish, we get the equation h?z—Qkhzek+zkh?*—h2 = 0 or [k, [k, 2] = —2, (86) a result which could not easily be obtained in a more direct way. As a final example we shall obtain the selection rule for the magni- tude K of the total angular momentum M ofa general atomic system. Let x,y,z be the coordinates of one of the electrons. We must obtain the condition that the (K’, K”) matrix element of 2, y, or z sball not vanish. This is evidently the same as the condition that the (K’, K”) matrix element of A,, A,, or A, shall not vanish, where A,, Ag, and Ag 164 ELEMENTARY APPLICATIONS § 40 are any three independent linear functions of x, y, and z with numeri- cal coefficients, or more generally with any coefficients that commute with K and are thus represented by matrices which are diagonal with respect to K. Let Ay = e+ M, y+ M2, A, = M,z—M,y—ta, d, = M,x—M,2—thy, A, = M,y—M,x—thz. We have M,d,+M,r, +r, = > (Mf, M,2z—M, M,y—hM, x) Eye => (M,M,—M,M,—ihMjze=0 (87) Lye from (29). Thus A,, A,, and A, are not linearly independent functions of z, y, and z. Any two of them, however, together with A, are three linearly independent functions of x, y, and z and may be taken as the above A,, As; Ag, Since the coefficients M,, M,, M, all commute with K. Our problem thus reduces to finding the condition that the (K’, K”) matrix elements of Ay, A,, A,, and A, shall not vanish. The physical meanings of these )’s are that A, is proportional to the component of the vector (x, y,2) in the direction of the vector M, and A,, A,, A, are proportional to the Cartesian components of the component of (xz, y, 2) perpendicular to M. Since A, is a scalar it must commute with K. It follows that only the diagonal elements (K'|A,|K’> of A, can differ from zero, so the selection rule is that K cannot change so far as A, is concerned. Apply- ing (30) to the vector A,,A,,A,, we have eo Ayo These relations between M, and A,,A,, A, are of exactly the same form or Ny, as the relations (23), (24) between m, and x,y,z, and also (87) is of the same form as (84). The dynamical variables ,, i,, A, thus have the same properties relative to the angular momentum M as x,y,z have relative to m. The deduction of the selection rule for k when the electric displacement is proportional to (x,y,z) can therefore be taken over and applied to the selection rule for K when the electric displace- ment is proportional to (A,,A,,4,). We find in this way that, so far as Ap Ay: A, ave concerned, the selection rule for K is that it must change by -Lh. Collecting results, we have as the selection rule for K that it must change by 0 or +#. We have considered the electric displacement § 40 SELECTION RULES 165 produced by only one of the electrons, but the same selection rule must hold for each electron and thus also for the total electric dis- placement. 41. The Zeeman effect for the hydrogen atom We shall now consider the system of a hydrogen atom in a uniform magnetic field. The Hamiltonian (57) with V = —e®/r, which describes the hydrogen atom in no external field, gets modified by the magnetic field, the modification, according to classical mechanics, consisting in the replacement of the components of momentum, p,, Py: Pz bY Detele. Ag, Pypele-Ay, Patele-A,, where A,, A,, A, are the com- ponents of the vector potential describing the field. For a uniform field of magnitude # in the direction of the z-axis we may take A, = —}y, A, = 34s, A, = 0. The classical Hamiltonian will then be oe? ~~ 2m r This classical Hamiltonian may be taken over into the quantum theory if we add on to it a term giving the effect of the spin of the electron. According to experimental evidence and according to the theory of Chapter XI, theelectronhasa magnetic moment —eh/2mc.c, where o is the spin vector of §37. The energy of this magnetic moment in the magnetic field will be eh#/2me.o,. Thus the total quantum Hamiltonian will be a 2 2 = (e354) +(0y-+5