初歩的な応用
Ⅵ ELEMENTARY APPLICATIONS
初歩的な応用
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