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\begin{document}
E. Wigner Phys. Rev. {\bf 51,} 106 \hfill {\large \bf 1937}\\
\vspace{2cm}
\begin{center}
{\bf \Large On the Consequences of the Symmetry of the Nuclear Hamiltonian on the Spectroscopy of Nuclei}
\end{center}
\begin{center}
E. WIGNER\\
Princeton University, Princeton, New Jersey \\
(Received October 23, 1936)
\end{center}
\begin{abstract}
The structure of the multiplets of nuclear terms is investigated, using as
first approximation a Hamiltonian which does not involve the ordinary
spin and corresponds to equal forces between all
nuclear constituents, protons and neutrons. The multiplets turn out to
have a rather complicated
structure, instead of the $S$ of atomic spectroscopy, one has three
quantum numbers $S,~ T,~ Y.$ The
second approximation can either introduce spin forces (method 2), or
else can discriminate between
protons and neutrons (method 3). The last approximation discriminates
between protons and neutrons in method 2 and takes the spin forces into
account in method 3. The method 2 is worked
out schematically and is shown to explain qualitatively the table
of stable nuclei to about Mo.
\end{abstract}
\section*{1}
~~~Recent investigations\footnote{M. A. Tuve, N. P. Heydenburg and
L. R. Hafstad, Phys. Rev. {\bf 50,} 806 (1936); G. Breit,
E. U. Condon and R. D. Present, Phys. Rev. {\bf 50,} 825 (1936).}
appear to show that the forces between all pairs of constituents makes
it desirable to treat the protons and neutrons on an equal footing. A scheme
for this was devised in his original paper by W.
Heisenberg\footnote{W. Heisenberg, Zeits. f. Physik {\bf 77,} 1 (1932).}
who considered protons and neutrons as different states of the same particle.
Heisenberg introduced a variable $\tau$ which we shall call the
isotopic spin, the value $-1$ of this variable can be assigned to
the proton state of the particle, the value $+1$ to the neutron state.
The assumption that the forces between all pairs of particles
are equal is equivalent, then to the assumption that they do not depend on
$\tau$ or that the Hamiltonian does not in involve the isotopic spin.
In addition to this isotopic spin $\tau$, we must keep, of
course, the ordinary spin variable $s$ also;
$s$ also can assume the two values $+1$ and $-1.$ It has
been pointed out lately\footnote{J. H. Bartlett, Phys. Rev. {\bf 49,}
102 (1936); W. Elsasser, J. de phys. et rad. {\bf 7,} 312 (1936),
and especially B. Cassen and E. U. Condon, Phys. Rev. {\bf 50,} 846
(1936).} that the Pauli principle requires that the wave function
\begin{equation}
\Psi(r_1 s_1 \tau_1,~ r_2 s_2 \tau_2,~ \ldots r_n s_n \tau_n)
\end{equation}
be antisymmetric with respect to the simultaneous
interchange of Cartesian, spin and isotopic spin
variables of any pair of heavy particles. This fact is
quite analogous to the similar statement for ordinary spin.
Of course, if Eq. (1) is to represent the state of a
given nucleus, say with $n_P$ protons and $n_N$ neutrons, it
must vanish at every place where the sum of the $\tau$'s
\begin{equation}
\tau_1 + \tau_2 + \ldots + \tau_n \ne n_N - n_P
\end{equation}
is not equal to the ``isotopic number'' of this element.
All wave functions which are finite for several sums of
the $\tau$'s, refer to states which can be different elements
with finite probabilities. No such states are known to
be of any importance and the mathematical apparatus
of the isotopic spin is, hence, somewhat redundant. It
will turn out that it is very useful in spite of this.
In addition to the assumption of the approximate
equality of forces between all pairs of particles, it
appears to be a useful approximation to neglect the
forces involving the ordinary spin. The Hamiltonian
depends then on the space coordinates alone. By
keeping both, one or none of these assumptions, one
comes to four possible schemes;
{\small
(1) Take into account forces depending on space co-
ordinates alone.
(2) Take into account forces depending on space and
ordinary spin coordinates, assuming, however, interactions
between all kinds of pairs to be equal.
(3) Neglect ordinary spin forces, take into account forces
depending on space coordinates and isotopic spin, i.e.,
discriminate between proton-proton, proton-neutron and
neutron-neutron interactions.
(4) Take all kinds of interaction into account.}
The first is the roughest method, the last the most
exact and it is probable that (2) is more accurate for
light elements, (3) for heavy
elements. On the other hand, of course, one can obtain
most results from symmetry considerations for 1,
fewest for 4. Approximation (1) is identical with the
``all orbital forces equal''
model\footnote{E. Feenberg and E. Wigner, Phys. Rev. This issue.}.
The statement that an operator involves only one or
another set of variables needs further amplification. As
used in the ordinary theory of spectra, this expression
means that the operator {\it can be written} in terms of
these variables alone. It did not mean that it cannot be
written in some other way as well. Thus, e.g., the
inter-change $P$ of the space coordinates acts only on space
coordinates, although it can be written by Dirac's
identity,
$$
P = - \frac{1}{2} - \frac{1}{2} (s_1 \cdot s_2)
$$
entirely in terms of spin operators for antisymmetric
functions. We shall keep this definition for the forces
depending on Cartesian and ordnary spin coordinates
for nuclei also.
The operators which involve $\tau$ are, however,
somewhat specialized to begin with. Using
Heisenberg's notation for isotopic spin operators
\setcounter{equation}{3}
\begin{equation}
\tau = \tau_{\zeta} = \Biggl| \Biggl|
\begin{array}{rr}
-1&0\\
0&1
\end{array} \Biggr| \Biggr|, \quad
\tau_{\xi} = \Biggl| \Biggl|
\begin{array}{rr}
0&i\\-i&0
\end{array} \Biggr| \Biggr|, \quad
\tau_{\eta} = \Biggl| \Biggl|
\begin{array}{rr}
0&1\\1&0
\end{array} \Biggr| \Biggr|
\end{equation}
the conservation law for electric charge requires that
all operators commute with
$$
\tau_{\zeta 1} + \tau_{\zeta 2} + \ldots + \tau_{\zeta n} = n_N - n_P = 2T_{\zeta}.
\eqno(3)
$$
In addition to this, one hardly would say that
\setcounter{equation}{4}
\begin{equation}
\tau_{\xi 1} \tau_{\xi 2} + \tau_{\eta 1} \tau_{\eta 2} + \tau_{\zeta 1}
\tau_{\zeta 2} = - 1 - 2PQ.
\end{equation}
($P$ interchange of space, $Q$ interchange of spin
coordinates) does not involve the Cartesian or spin
coordinates, since Eq. (5) is a rather artificial
expression, $\tau_{\xi}$ and $\tau_{\eta}$ having no immediate physical
significance. We shall assume hence for
approximation (3) only such operators which are
equivalent to operators acting on the Cartesian
coordinates alone, but in a different way for protons
and neutrons. This is equivalent to using only
operators involving the space coordinates and the $\tau_{\zeta}$'s.
If we do this, the results of method (3) must become
equivalent to the usual theory (without $\tau$'s) which
neglects the spin. As a matter of fact, for approximation (3), the
introduction of $\tau$ is entirely useless and it is taken up
here only in order to establish the transition from
approximation (1) to (3).\\
\section*{2}
~~~The interaction in the electronic shells of atoms is a
sum of terms containing two particles only and the
momenta is no higher than the second power. The
reason for the first is, that the interaction occurs
through a field and this gives in first approximation
only interaction between two particles. The reason that
one can stop with the second power of the momenta is
that these always enter in the combination $p/mc$ which
is a small quantity.
An advantage of introducing the variable $\tau$
is\footnote{W. Heisenberg, Zeits. f. Physik {\bf 77,}
1 (1932).},\footnote{J. H. Bartlett, Phys. Rev. {\bf 49,}
102 (1936); W. Elsasser, J. de phys. et rad. {\bf 7,} 312 (1936),
and especially B. Cassen and E. U. Condon, Phys. Rev. {\bf 50,} 846
(1936).} that one can take over these assumptions to nuclei. If one
does not use the variable $\tau$ the interchange of two
particles if expressed as a power series of the
momenta is an infinite series\footnote{J.A. Wheeler, Rhys. Rev. {\bf 50},
643 (1936).}
$$
\operatornamewithlimits{\sum}_{n_1 n_2 n_3} \frac{(x_2 - x_1)^{n_1} (y_2
- y_1)^{n_2} (z_2 - z_1)^{n_3}}{n_1! n_2! n_3!}
$$
$$
\times \left( \frac{\partial}{\partial x_1} - \frac{\partial}{\partial x_2}
\right)^{n_1} \left( \frac{\partial}{\partial y_1} - \frac{\partial}{\partial
y_2} \right)^{n_2} \left( \frac{\partial}{\partial z_1} - \frac{\partial}{\partial
z_2} \right)^{n_3}.
$$
However, it can be expressed by means of Dirac's
identity also entirely without the momenta by means
of Eq. (5). It must be admitted, however, that the spin
cannot be considered to be small as in the atomic
theory. We shall determine here all interaction forms
between two particles which do not contain higher
than first power terms of momenta\footnote{ Some of these
were given previously by Cassen and Condon, reference 3.
The expressions given here are invariant only under Galilei
transformations. G. Breit has shown that, in order to ensure
relativistic invariance, correction terms must be added to the
expressions derived here.} as far as the dependence
on $s$ and $\tau$ goes. Nothing can be said, of
course, on the dependence on the distance, and this
factor will be omitted hence. It seems to be of lesser
importance for the present.
The interaction must have spherical symmetry,
depending on the differences of coordinates and
momenta only, be invariant under inversion,
substitution of $-t$ for $t$ and also be
symmetric in the particles. The first requirements
determine the dependence on $s,~x$ and $p$. From the two
triples of spin operators, one can form two invariants
$$
(i)\quad 1; ~\mbox{and}~ (i') \quad \frac{1}{2} + \frac{1}{2} (s_{x 1} s_{x
2}
+ s_{y 1} s_{y 2} + s_{z 1} s_{z 2}) = Q_{12}
$$
three axial vectors with $Z$ components
$$
(v) \quad s_{z 1} + s_{z 2}; \quad s_{z 1} - s_{z 2}; \quad
- s_{x 1} s_{y 2} - s_{y 1} s_{x 2},
$$
respectively, and one axial tensor, with components
$$
s_{x 1} s_{y2} + s_{y1} s_{x2}; \quad s_{y1} s_{z2} + s_{z1} s_{y2};
\quad s_{z1} s_{x2} + s_{x1} s_{x2};
$$
$$
s_{x1} s_{x2} - s_{y1} s_{y2}; \quad s_{x1} s_{x2} + s_{y1} s_{y2}
- 2s_{z1} s_{z2}.
$$
The first two of these, (i) and (i$'$), can be used as they
stand, cannot be combined with first power
expressions of $p$, however, since these change sign
under the $t' = - t$ substitution. The last one ($t$) gives the
familiar expression
$$
(i'') \quad (s_1 \cdot {\bf r}_{12})(s_2 \cdot {\bf r}_{12} - {\bf 3}(s_1
\cdot s_2){\bf r}_{12}^2
$$
if combined with the similar tensor of the
coordinates\footnote{(ii) has the property that it is identical with
$Q_{12}(ii)$. It is an interaction which shows saturation.}.
It cannot be combined with the $p$ either.
The middle one must be combined with the vector
$p_1 - p_2$ which gives a useless axial invariant and tensor
and an ordinary vector. This combined with the
distance vector gives the familiar
$$
(ia)(ib)(ic) \Biggl|
\begin{array}{ccc}
s_x&s_y&s_z\\
x_1-x_2&y_1-y_2&z_1-z_2\\
p_{x1}-p_{x2}&p_{y1}-p_{y2}&p_{z1}-p_{z2}
\end{array} \Biggr|.
$$
Here $s_x,s_y,s_z$ can be the components of one of the
three vectors ($v$). On the whole, we have 6 invariants.
These invariants can be multiplied with one of the six
expressions in $\tau$ which commute with
$\tau_{\zeta 1}+ \tau_{\zeta 2}$. These are, first of all
$$
(\tau_0) \quad 1 \quad \mbox{and} \quad (\tau'_0) ~ - \frac{1}{2} - \frac{1}{2}
(\tau_1 \cdot \tau_2) = P_{12} Q_{12}
$$
which give the same interaction between all pairs of
particles. In addition to these, we have
$$
(\tau_1) \quad \frac{1}{2} + \frac{1}{2} \tau_{\zeta 1} \tau_{\zeta 2} \quad
\mbox{and} \quad (\tau'_2) \quad \frac{1}{2} (\tau_{\zeta 1}= \tau_{\zeta
2}).
$$
The first of these gives ordinary interaction but only
between like particles; the second gives a negative
interaction for proton pairs, a positive
for neutron pairs, none for unlike particles. These
interactions are symmetric in the particles and can be
combined with $(i), (i$'$), (i$''$)$ and $(ia)$, giving in the
whole 16 different forms.
Finally we have
$$
(\tau_2) \quad \tau_{\zeta 1} - \tau_{\zeta 2} \quad \mbox{and} \quad (\tau^{\prime}_2)
\quad \frac{1}{2} (\tau_{\xi 1} \tau_{\eta^2} - \tau_{\eta 1} \tau_{\xi 2}),
$$
which can be combined with ($ib$) and ($ic$) giving 4
more types of interaction.
In approximation (1) we can have only $(i)(\tau_0)$ and
$(i')(\tau_0')$, i.e., ordinary and Majorana exchange forces.
In approximation (2), all 8 forms derived from $(\tau_0), (\tau_0')$
and $(i), (i'), (i'')$ and $(ia)$. These are, in addition to
the previous ones, spin-spin $(i'')(\tau_0)$, spin-orbit
$(ia)(\tau_0)$ ordinary forces, Heisenberg forces $(i)(\tau'_0)$.
Furthermore spin-spin exchange forces $(i')(\tau_0')$ and
spin-orbit exchange forces $(ia)(\tau_0')$ of the Heisenberg
type. The Majorana exchange forces of these types are
identical with the ordinary forces. Finally, we have the
spin-exchange forces $(i')(\tau_0)$
of Bartlett\footnote{J. H. Bartlett, Phys. Rev. {\bf 49,}
102 (1936); W. Elsasser, J. de phys. et rad. {\bf 7,} 312 (1936),
and especially B. Cassen and E. U. Condon, Phys. Rev. {\bf 50,} 846
(1936).}.
In approximation (3) we must permit according to
the preceding section, in addition to those of 1, only
$(i)(\tau_1)$ and $(i)(\tau_1')$, allowing for different interactions
between different kinds of pairs. The coefficient of
$(i)(\tau_1')$ is certainly very small, the proton-proton
interaction being very nearly equal to the neutron-neutron interaction.
In approximation (4), all 20 types become possible.
\section*{3}
~~~~We next go over to approximation (1), and try to
define the analog of the multiple! system. This can be
defined in two ways: either by considering the
functional dependence of the wave functions on the
spins or else by considering their dependence on the
space coordinates. We shall first consider the spin
function\footnote{The content of this section is based on the
fundamental mathematical works of E. Cartan, Bull. Soc. Math. de France
{\bf 41,} 43 (1913). J. de Math. {\bf 10,} 149 (1914); I. Schur, Berl.
Ber., pp. 189, 297, 346 (1924) and particularly, H. Weyl,
Math. Zs. {\bf 23,} 271 (1925). I attempted to compile in this
section-often without giving rigorous proofs-those results
which suffice for the discussion of the physical problems in
question.}.
The great difference between the ordinary spin and
the spin considered here is that we have, for every
particle, two spin coordinates s and â, giving in the
whole four different sets of values
$-1, -1; -1, 1; 1, -1; 1, 1.$ Instead of two two-valued
spins, one can introduce one four-valued spin $\eta$, which
has the values 1, 2, 3, 4 for the four different doublets
of values of $s$ and $\tau$, respectively. This $\eta$ plays the
same role which the two-valued spin plays in the
ordnary spin theory. However, because of the four-valuedness
of $\eta$, instead of the representations of the
two-dimensional unitary group (or the equivalent
three-dimensional rotation group), the representations of the
four-dimensional unitary group will characterize the
multiplet systems.
Since the Hamiltonian does not contain the spin
coordinates, any transformation which affects only
these, will bring a characteristic function into a
characteristic function. We can consider first, the
permutations of the $\eta_i$ and second, simultaneous
unitary transformations of all the $\eta$:
\begin{equation}
R_u \psi (\eta 1, \ldots, \eta n)= \sum_{\vartheta} u_{\eta 1} \vartheta
1 u_{\eta 2} \vartheta_2 \ldots u_{\eta n} \vartheta_n \times \psi (\vartheta_1,
\ldots, \vartheta_n).
\end{equation}
We can first define something analogous to the $Z$
component of the spin momentum by considering the
$u$'s of the type
\begin{equation}
u(\varphi_1, \varphi_2,\varphi_3,\varphi_4) =
\Biggl| \Biggl|
\begin{array}{cccc}
e^{i \varphi 1}&0&0&0\\
0&e^{i \varphi 2}&0&0\\
0&0&e^{i \varphi 3}&0\\
0&0&0&e^{i \varphi 4}
\end{array} \Biggr| \Biggr|
\end{equation}
These operations all commute and, hence, a system of
functions of the $\eta$ can be found, the members of which
are merely multiplied by constants if an $R_u$ with $u$ of
the form (7) is applied to them\footnote{The $F_{\mu_1 \mu_2 \mu_3 \mu_4}(\eta_1
\ldots \eta_0)$ are zero for every set $\eta_1 \eta_2 \ldots \eta_n$
of the $\eta$, except for those sets in which $\mu_1$ of them have the
value 1, exactly $\mu_2$ of them have the value 2, and $\mu_3$ of them
are 3. Then $\mu_4$ of them will be equal to 4. Otherwise they can
be arbitrary and will still satisfy Eq. (8).}
\begin{equation}
R^{u(\varphi_1 \varphi_2 \varphi_3 \varphi_4)}F^{\nu}_{\mu_1 \mu_2 \mu_3 \mu_4}(\eta_1
\ldots \eta_n) = e^{i(\mu_1 \varphi_1 + \mu_2 \varphi_2 + \mu_3 \varphi_3
+ \mu_4 \varphi_4)}F^{\nu}_{\mu_1 \mu_2 \mu_3 \mu_4}.
\end{equation}
The $\mu$ must be integers in this equation, they will be
called diagonal quantum numbers. The $\nu$ serves only to
discriminate between different functions of the $\eta$ with
the same diagonal quantum numbers.
Since $u(\varphi, \varphi, \varphi, \varphi)$ with four equal $\varphi$
is only multiplication with $e^{i \varphi}$, because of (6), $R_u$ is
multiplication with $e^{i n \varphi}$. It is, on the other hand,
multiplication with $e^{i (\mu_1 \varphi + \mu_2 \varphi + \mu_3 \varphi
+ \mu_4 \varphi)}$ which shows that all
possible systems of diagonal quantum numbers satisfy the equation
\begin{equation}
\mu_1 + \mu_2 + \mu_3 + \mu_4 = n,
\end{equation}
where $n$ is the number of variables $\eta$. There is a simple
connection between the diagonal quantum numbers
and the $Z$ component of the spin momentum $S$. One
obtains it by considering a rotation of the spin
coordinates around $Z$ by $\varphi$, the matrix of which is of
the form (7) with \mbox{$\varphi_1 = \varphi_2 = - \varphi_3 = - \varphi_4 = -\frac{1}{2} \varphi.$}
Under the influence of the corresponding $R_u$, the wave
function will be multiplied by $e^{iS_z \varphi}$ which gives
$$
S_z = \frac{1}{2} \sum_k s_{zk} = \frac{1}{2} (\mu_4 + \mu_3 - \mu_2 - \mu_1).
\eqno(10a)
$$
The $1/2$ before the $s_{zk}$ enters because the usual definition
of the Pauli-matrices is $1/2$ of that given in (4).
Similarly, we have
$$
T_{\zeta} = \frac{1}{2} \sum_k \tau_{\zeta k} = \frac{1}{2} (\mu_4 - \mu_3
+ \mu_2 - \mu_1) \eqno(10b)
$$
and we define also a
$$
Y_{\zeta} = \frac{1}{2} \sum_k s_{zk} \tau_{\zeta k} = \frac{1}{2} (\mu_4 -
\mu_3 - \mu_2 + \mu_1). \eqno(10c)
$$
The quantum numbers $S_z, T_{\zeta}, Y_{\zeta}$ can be called magnetic
quantum numbers. They determine, together with $n$,
the $\mu$ uniquely. Their importance for spectroscopic
considerations is the same as that of the single ordinary
magnetic quantum number in atomic spectroscopy;
they can be easily found simultaneously for all states
of a multiplet.
Several states with different magnetic quantum
numbers form sets ``multiplets'' which always have
common energy. These sets contain in atomic
spectroscopy {\it one} state with every magnetic quantum
number from a maximum, $S$, to $-S$. We must find the
corresponding sets for four-valued spin.
The multiplet will be denoted by the highest set
$\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1$ of $\mu$ which occurs
in it. The set $\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1$ is called
higher than the set $\mu_4 \mu_3 \mu_2 \mu_1$ if
either $\Lambda_4 > \mu_4$, or if $\Lambda_4 = \mu_4$ but
$\Lambda_3 > \mu_3$, or
finally, if $\Lambda_4 = \mu_4,~ \Lambda_3 = \mu_3$ but
$\Lambda_2 > \mu_2$. The reason for
several states with different diagonal quantum numbers
being united into the same multiplet, is that they are
transformed into each other by $R_u$, the $u$ of which have
not the form (7). Instead of the $\Lambda$, we can use for the
characterization of a multiplet also
$$
S = \frac{1}{2} (\Lambda_4 + \Lambda_3 - \Lambda_2 - \Lambda_1), \eqno(11a)
$$
$$
T = \frac{1}{2} (\Lambda_4 - \Lambda_3 + \Lambda_2 - \Lambda_1), \eqno(11b)
$$
$$
Y = \frac{1}{2} (\Lambda_4 - \Lambda_3 - \Lambda_2 + \Lambda_1), \eqno(11c)
$$
which together with $\Lambda_4 + \Lambda_3 + \Lambda_2 + \Lambda_1 = n$
completely determine the $\Lambda$.
The character of the multiplets is for the fourfold
spin not as simple as for the twofold spin. While the
latter ones can be represented by the points on a line
from $-S$ to $S$, the former ones must be represented at
least in a three-dimensional space, giving the possible
$S_z, T_{\zeta}, Y_{\zeta}$ values and {\it their multiplicities.} This is necessary because it is not true any more that every
combination of $S_z T_{\zeta} Y_{\zeta}$ occurs only once. The
multiplicity of every $S_z T_{\zeta} Y_{\zeta}$ is the same as that
of any permutation of these numbers and also that of
$-S_z - T_{\zeta} Y_{\zeta}; - S_z T_{\zeta} - Y_{\zeta}; S_z - T_{\zeta}
- Y_{\zeta}.$ The figure of the multiplet
has, therefore, tetrahedral symmetry in the $S_z T_{\zeta} Y_{\zeta}$
space. Using $S_z T_{\zeta} Y_{\zeta}$ has the advantage over using the
$\Lambda$ that the multiplets for $n=1,~5,~9,~13,$ are represented
by the same figures. The quantum numbers are all
half-integers, the occurring $S_z T_{\zeta} Y_{\zeta}$
combinations form a face centered lattice for which
$S_z + T_{\zeta} + Y_{\zeta}$ is of the
form $2k - 1/2$. These figures, reflected in any of the
planes $T_{\zeta} Y_{\zeta},~ Y_{\zeta} S_z,~ S_z T_{\zeta},$
give the multiplets existing for
$n=3,~7,~11,~ \ldots,$ the face centered lattice being
characterized by $S_z + T_{\zeta} + Y_{\zeta}$ having the form
$2k+1/2.$
The quantum numbers $S_z,~ T_{\zeta},~ Y_{\zeta}$ are integers for even
$n.$ Their sum is even or $n=4,~8,~12,~\ldots,$ odd for
$n=2,~6,~10,~\ldots.$
In order to find the figures for the multiplets, one
must know how many states with a certain
$\mu_4 \mu_3 \mu_2 \mu_1$ combination are present in the multiplet
$(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$. We shall denote this
number by
\setcounter{equation}{11}
\begin{equation}
\left(
\begin{array}{c}
\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1\\
\mu_4 \mu_3 \mu_2 \mu_1
\end{array} \right)
\end{equation}
The calculation of the quantities (12) is important for
the following section also. The simplest
interpretation of (12) is obtained by considering the
subgroup of the unitary group formed by the matrices
(7). The symbol (12) denotes how often the
(one-dimensional) representation $e^{i(\mu_1 \varphi_1 + \mu_2 \varphi_2
+ \mu_3 \varphi_3 + \mu_4 \varphi_4)}$
occurs in the representation of the total unitary group
which is designated by
$(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$.
The symbols (12) are defined only if $\Lambda_1 + \Lambda_2
+ \Lambda_3 + \Lambda_4 = \mu_4 + \mu_3 + \mu_2 + \mu_1,~ \Lambda_4 \geq \Lambda_3
\geq \Lambda_2 \geq \Lambda_1 \geq 0$
and since a permutation of the $\mu$ does not change the
value of (12), we can assume also $\mu_4 \geq \mu_3 \geq \mu_2 \geq \mu_1 \geq 0.$ The value of (12) is 0, unless $\Lambda_4 \geq \mu_4$ since
$\Lambda_4$ was
the greatest $\mu$ of the multiplet. In addition to this, (12)
vanishes unless\footnote{Similar formulas hold also for symbols of
the kind (12)
with more than four $\Lambda$'s. They can be proved by an argument
similar to that of the next section.}
\begin{equation}
\begin{array}{c}
\Lambda_4 \geq \mu_4; \quad \Lambda_4 + \Lambda_3 \geq \mu_4 + \mu_3;\\
\Lambda_4 + \Lambda_3 + \Lambda_2 \geq \mu_4 + \mu_3 + \mu_2.
\end{array}
\end{equation}
The last of these can be written also as $\Lambda_1 \leq \mu_1$. There
are several ways of evaluating (12). One of them is to
consider the matrices $\mu$ which correspond to a
three-dimensional unitary transformation of the $\eta$ values $1, ~
2,~ 3$ only. The representation $(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$ contains all those representations $(\Lambda_{3}' \Lambda_2'\Lambda_1'$)
of the three dimensional unitary group exactly once for which
\begin{equation}
\Lambda_4 \geq \Lambda'_3 \geq \Lambda_3 \geq \Lambda'_2 \geq \Lambda_2 \geq \Lambda'_1
\geq \Lambda_1.
\end{equation}
Thus the quartets of diagonal quantum numbers of
$(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$ are those of all
$(\Lambda'_3 \Lambda'_2 \Lambda'_1)$ satisfying
(14), together with the last diagonal quantum number
$\Lambda_4 + \Lambda_3 + \Lambda_2 + \Lambda_1 - \Lambda'_3 - \Lambda'_2
- \Lambda'_1$. After this, we can reduce
the representation of the three-dimensional unitary
group to a two-dimensional: in $(\Lambda'_3 \Lambda'_2 \Lambda'_1)$
those $(\Lambda^{\prime \prime}_2 \Lambda^{\prime \prime}_1)$
will occur for which
$$
\Lambda_3' \geq \Lambda_2^{\prime \prime} \geq \Lambda_2' \geq \Lambda_1^{\prime
\prime} \geq \Lambda_1'. \eqno(14a)
$$
Finally $(\Lambda_2^{\prime \prime} \Lambda_1^{\prime \prime})$
contains the pairs of diagonal quantum
numbers \mbox{$\Lambda_2^{\prime \prime}, \Lambda_1^{\prime \prime};~
\Lambda_2^{\prime \prime} - 1, \Lambda_1^{\prime \prime} + 1;~ \Lambda_2^{\prime
\prime} - 2, \Lambda_1^{\prime \prime} + 2; \ldots; \Lambda_1^{\prime \prime},
\Lambda_2^{\prime \prime}.$}
For instance, in order to find the multiplet (3 1 1 0) we
can calculate $(3~ 1~ 1~ 0) = (3~ 1 ~1)0 + (2~1~1)1 + \underline{(1~1~1)2}
+ (3~1~0)1 + \underline{(2~1~0)2} + \underline{(1~1~0)3.}$
It suffices to obtain those
quadruplets of $\mu$ which are in a descending order. The
others can be obtained then by permutation. We can omit hence the
underlined ones. To reduce further
$$
\begin{array}{ll}
(3~1~1)=(3~1)1 + \underline{(2~1)2} + \underline{(1~1)3}&\\
(2~1~1) = (2~1)1 + \underline{(1~1)2},&\\
(3~1~0) = \underline{(3~1)0} + (2~1)1 + \underline{(1~1)2}&
+ (3~0)1\\
& + \underline{(2~0)2} + \underline{(1~0)3}.
\end{array}
$$
\begin{figure}
\centerline{\resizebox{12cm}{!}{\includegraphics{fig1.gif}}}
\caption{}
\end{figure}
\begin{figure}
\centerline{\resizebox{12cm}{!}{\includegraphics{fig2.gif}}}
\caption{Every set of figures represents a multiplet, the $STY$ sign of which,
together with one corresponding $\Lambda_4 + \Lambda_3 + \Lambda_2 + \Lambda_1$,
is given on top. Every circle represents a $S_z T_{\zeta} T_{\zeta}$ state,
$T_{\zeta}$ is given below the figure, $Y_{\zeta}$ and $S_z$ are the coordinates
of the circle, the origin of the coordinate system being at the center of
the figure, the $Y_{\zeta}$ axis runs to the right, the $S_z$ axis downward.
The numbers in the circles give the number of states $S_z T_{\zeta} Y_{\zeta}$
in the multiplet. The distance between two adjoining circles on a horizontal
or vertical is 2. The multiplets with half integer $STY$ correspond to elements
with masses $4n+1$. The multiplets for masses $4n+3$ are obtined from these
by reversing the direction of the $Y_{\zeta}$ axis. The sign of $Y$ must
be changed also. The 2 at the center of this figure should be replaced by
a 3.}
\end{figure}
This gives the $\mu$ systems 3 1 1 0, 2 2 1 0, 2 1 1 1,
2 1 1 1, 2 1 1 1 and their permutations. In the, language of the
magnetic quantum number expressed, the multiplet
$(S,T,Y) = (3/2, 3/2, 1/2)$
contains $S_z = 3/2,$ \mbox{$T_{\zeta} = 3/2,$} \mbox{$Y_{\zeta} = 1/2$}
once, $S_z = 3/2,$ $T_{\zeta} = 1/2,$ $Y_{\zeta} = - 1/2$ once,
$S_z = T_{\zeta}=Y_{\zeta}=1/2$
three times. In addition to these, all permutations of
these and those triplets in which any two of the
$S_z T_{\zeta} Y_{\zeta}$
are replaced by their negative values. The multiplet is
shown, along with some other ones, in Fig. 2, it is the
third one.
According to the general theory\footnote{The content of this section is based on the
fundamental mathematical works of E. Cartan, Bull. Soc. Math. de France
{\bf 41,} 43 (1913). J. de Math. {\bf 10,} 149 (1914); I. Schur, Berl.
Ber., pp. 189, 297, 346 (1924) and particularly, H. Weyl,
Math. Zs. {\bf 23,} 271 (1925). I attempted to compile in this
section-often without giving rigorous proofs-those results
which suffice for the discussion of the physical problems in
question.} the wave
functions of the multiplet $\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1$
belong with
respect to interchange of the $\eta$ to the representation
of the symmetric group which is characterized by the
partition \mbox{$\Lambda_4 + \Lambda_3 + \Lambda_2 + \Lambda_1 = n$.}
It has been shown by Slater\footnote{J. C. Slater, Phys. Rev.
{\bf 34,} 1293 (1929).} for atomic spectra that
the knowledge of the structure of multiplets enables
one to determine the numbers and characters of the
terms which arise from any configuration. The same is
true in principle for nuclear spectra. The difference is
that instead of the two-dimensional plot of occurring
$L_z S_z$ values, one should prepare a four-dimensional
plot of $L_z S_z T_{\zeta} Y_{\zeta}$ values.
Or perhaps for every $L_z$ and $S_z$ a
two-dimensional plot of the occurring $T_{\zeta} Y_{\zeta}$ values. For
every $L_z$ these plots must be decomposed into $S_z T_{\zeta} Y_{\zeta}$
combinations which form multiplets. After this, the $L_z$
values for every multiplet must be grouped together
into sets ranging from $-L$ to $L$ thus obtaining the
azimuthal quantum member.
The most practical procedure along these lines
which I could find was one using the $\Lambda$ and the
diagonal quantum numbers. A state with $\mu_1$ protons
and $\mu_2$ neutrons with spin $-1/2$ and $\mu_3$ protons and
$\mu_4$ neutrons with spin $1/2$ is a state with the diagonal
quantum numbers $\mu_1 \mu_2 \mu_3 \mu_4$. One first makes a plot of
the occurring $\mu_1 \mu_2 \mu_3$ values for every $L_z$ and $\mu_4$.
For this purpose, one draws an equilateral triangle with the
altitude
%\begin{figure}
%\centerline{\resizebox{10cm}{!}{\includegraphics{fig1.gif}}}
%\caption{
%\end{figure}
$\mu_1 + \mu_2 + \mu_3 = n - \mu_4$. The point which has the
distances $\mu_1,~ \mu_2,~ \mu_3$
from the three sides of the triangle, respectively, will
correspond to a state with the diagonal quantum numbers
$\mu_1,~ \mu_2,~ \mu_3$
and $\mu_4 = n - \mu_1 - \mu_2 - \mu_3.$ Every plot forms a
representation of the three dimensional unitary group. The
irreducible representations of this group are rather simple,
their plots are shown in Figs. 3a and 3b: the combination
$\mu_1 \mu_2 \mu_3$ occurs in $\Lambda_1' \Lambda_2' \Lambda_3'$
\begin{equation}
\left(
\begin{array}{c}
\Lambda'_3 \Lambda'_2 \Lambda'_1\\
\mu_3 \mu_2 \mu_1
\end{array} \right)
= 1 + \mbox{Min}~ (\Lambda_3' - \mu_3, \Lambda_3' + \Lambda_2' - \mu_3 -
\mu_2, \Lambda_3' - \Lambda_2', \Lambda_2' - \Lambda_1')
\end{equation}
times, where $\mbox{Min}~ (\alpha, \beta, \ldots)$ is the smallest
of the numbers $\alpha, \beta, \ldots$ if they are positive and
equals $- 1$ if any of them is negative.
One can decompose the $\mu_1 \mu_2 \mu_3$ plots for every $L_z$ into
irreducible plots, characterized by ($\Lambda_3' \Lambda_2' \Lambda_1'$) (the primes on
the $\Lambda$ are omitted in the figure). Having obtained the number
of ($\Lambda_3' \Lambda_2' \Lambda_1'$), one unites these for every
$L_z$ separately into total multiplets $(\Lambda_4 \Lambda_3 \Lambda_2
\Lambda_1)$ according to
\begin{equation}
(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1) = \sum (\Lambda_3' \Lambda_2' \Lambda_1').
\end{equation}
The limits of summation are given in (14). Finally the $L_z$
values for every $(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$ are united
separately to total
azimuthal quantum numbers $L$. On the whole, the procedure
is much more cumbersome than the analogous one for
atomic spectra. It has the disadvantage also, that one first
obtains the highest multiplicities, which have the highest
energies.\\
\section*{4}
~~~~An alternative method which leads much more rapidly to
the goal is to consider, for the time being, only the
dependence of the wave function on space coordinates. This
method was worked out for atomic spectra by the present
author and Delbr\"uck\footnote{E. Wigner, Zeits. f. Physik {\bf 43,}
627 (1927); M. Delbr\"uck,
Zeits. f. Physik {\bf 51,} 181 (1928). Eq. (19) of the former and
(14) of the latter give an explicit expression for (20) in the
case all $\lambda$ and $\mu$ are 1 or 2. Only this case occurs in
ordinary spectroscopy. A similar expression would be too
complicated if the $\lambda$ and $\mu$ can be 3 and 4 also.} before the
spin theory of Pauli was
known. One considers first again the possible configurations
i.e. the distributions of the particles into the different states,
without, however, taking into
consideration the spin. For every such state one determines
the $L_z$ (or $m_L$ as it is often called) as the sum of the $l_z$
of all the particles. Next one determines how many terms with a
certain partition $\lambda_{\rho} + \lambda_{\rho - 1} + \ldots
+ \lambda_1$ this configuration gives.
Finally, one considers the states corresponding to every
partition separately and unites states with $L_z$ from $-L$
to $L$ into a total multiple! with azimuthal quantum number $L$.
These steps performed, one has all one is interested in for
approximation (1): the multiplicities and azimuthal quantum
numbers of every term. For later work, it is still necessary to
know how to complete the wave functions depending on
coordinates alone, by functions depending on the spins, to
total wave functions. Since the total wave function must be
antisymmetric, the space-coordinate wave function and the
spin function used for the completion must have adjoint
characters, i.e., belong to two such representations the
matrices of which are equal for even, oppositely equal for
odd permutations. The adjoint partition to the partition
$4 + 4 + \ldots + 1$ with $n_4$ fours, $n_3$ threes, $n_2$
twos and $n_1$ ones is the partition $\Lambda_4 + \Lambda_3 + \Lambda_2 +
\Lambda_1$ where
\begin{equation}
\begin{array}{ll}
\Lambda_4 = n_4 + n_3 + n_2 + n_1, & \Lambda_3 = n_4 + n_3 + n_2,\\
\Lambda_3 = n_4 + n_3 & \Lambda_1 = n_4.
\end{array}
\end{equation}
Thus, e.g., the partition $4+3+2+2+2+1$ for the spacial wave
function is equivalent to the partition $6+5+2+1$ for the $\Lambda$ or
to the $STY$ set $(4~ 1~ 0).$
The step which involves the difficulty for this procedure
is to determine how many terms with a certain partition
$\lambda_{\rho} + \lambda_{\rho - 1} + \ldots + \lambda_1$ a configuration
gives in which there are $\mu_1$
particles in the first, $\mu_2$ in the second, etc., $\mu_{\nu}$ in the
$\nu$-th state. We consider the wave function
\begin{equation}
\psi_1 (x_1) \psi_1(x_2) \ldots \psi_1(x_{\mu 1}) \psi_2(x_{\mu 1 + 1}) \ldots
\psi_2(x_{\mu 1 + \mu 2}) \cdot \psi_3(x_{\mu 1 + \mu 2 +1}) \ldots \psi_{\nu}(x_n)
\end{equation}
together with those arising from (18) by a permutation of the
$x$. There are $n!/\mu_1! \mu_2! \ldots \mu_{\nu}!$ of these.
Under a permutation of
the $x$ they naturally transform among themselves, the
corresponding (reducible) representation of the symmetric
\begin{figure}
\centerline{\resizebox{10cm}{!}{\includegraphics{fig3.gif}}}
\caption{Every circle corresponds to a triple $\mu_1 \mu_2 \mu_3$ these
numbers being the distances of the circle from the three
sides of the equilateral triangle, surrounding the figure. The
number in the circle tells how often the corresponding set of
$\mu$'s occurs in the representation characterized by
$(\Lambda_3 \Lambda_2 \Lambda_1)$.
Fig. 3a holds for $\Lambda_3 - \Lambda_2 \geq \Lambda_2 - \Lambda_1$,
Fig. 3b for $\Lambda_3 - \Lambda_2 \leq \Lambda_2 - \Lambda_1$
(the first is actually the multiplet (9 4 2) the second
one (8 6 1)). In both cases the boundary hexagon contains one's, the
next two's and so on until the hexagon reduces to a triangle.
The $\mu$ sets within the triangle occur all equally often in the
representation: $\Lambda_2 - \Lambda_1 + 1$ times in the first,
$\Lambda_3 - \Lambda_2 + 1$ times in the second case.}
\end{figure}
group will be denoted by $[\mu_{\nu} \mu_{\nu - 1}
\ldots \mu_1]$. Upon
decomposing this into irreducible representations
\begin{equation}
[\mu_{\nu} \mu_{\nu - 1} \ldots \mu_1] = \sum_{\lambda} \left(
\begin{array}{c}
\lambda_{\rho} \lambda_{\rho - 1} \ldots \lambda_1\\
\mu_{\nu} \mu_{\nu - 1} \ldots \mu_1
\end{array}
\right) (\lambda_{\rho} \lambda_{\rho - 1} \ldots \lambda_1),
\end{equation}
the coefficients (some of which occurred previously in (12))
tell us how many terms with the partition $\lambda_{\rho} + \lambda_{\rho
- 1} + \ldots + \lambda_1$
the configuration gives. We shall arrange the $\mu$ again in a
descending order $\mu_{\nu} \geq \mu_{\nu - 1} \geq \ldots \geq \mu_1 \geq 0$.
Only those partitions $\lambda_{\rho} + \lambda_{\rho - 1} + \ldots +
\lambda_1$ will correspond to
real terms in which none of the $\lambda$ is greater than 4. This
must hold then, because of
(13), for the $\mu$ as well: no orbit can be more than four times
occupied. One can even omit for the calculation of the
coefficients all the fourfold occupied states, i.e., drop all the
$\mu = 4$, since because of (13), the $\lambda$ above a 4 must be a 4 also.
In the $p$ shell, there are only three states $l_z = - 1,~ 0,~ 1$
and the coefficients of interest are therefore of the form (15)
(if the partition contains only two addends, a 0 can be
affixed for $\lambda_1$) and they are explicitly given in (15). One
must only arrange the $\mu$ in descending order.
In this case the calculation is especially simple. For
instance, for 3 particles we have the ten configurations of
Table I. The figures below the
$l_z$ values give the number of particles ($\mu$) in this state, the
figure below $L_z$ is the total $L_z$ the last columns give the
number of terms with the partitions 3 or $(3+0+0)$ and $2+1$
(i.e., $(2+1+0)$) which this configuration gives. This gives $F$
and $P$ terms with the partition (3) and $D$ and $P$ terms with
the partition $(2+1)$. Table
II, reference 4, was prepared in this way. The adjoint
partitions to (3) and $(2 + 1)$ are $(1 + 1 + 1)$ and $(2 + 1)$,
respectively, the $STY$ characterization is $\left(\frac{1}{2} ~ \frac{1}{2}-
\frac{1}{2}\right)$ for the
former, $\left(\frac{3}{2}~ \frac{1}{2} ~\frac{1}{2}\right)$ for
the latter. There is in addition to these, one
$S$ term of the multiplicity $(\frac{3}{2}~ \frac{3}{2}~ \frac{3}{2})$.
In the general case the explicit formulas for the
\begin{equation}
\left(
\begin{array}{ccccc}
\lambda_{\rho} & \lambda_{\rho - 1} & \lambda_{\rho - 2} & \ldots & \lambda_1\\
\mu_{\nu} & \mu_{\nu - 1} & \mu_{\nu - 2} & \ldots & \mu_1
\end{array} \right)
\end{equation}
are too complicated. A useful way of evaluating (20) starts
from another interpretation of (20) than given in (19). For
Frobenius' reciprocity theorem\footnote{G. Frobenius, Berl. Ber. 501
(1898). The reader will find
a straight forward proof in H. Weyl's {\it Gruppentheorie und
Quantenmechanik} (Leipzig 1928), first edition, p. 254. The
proof in the English translation by H. P. Robertson (London,
1931), pp. 332-338, is more abstract.} one considers the subgroup
which contains the permutations of the first $\mu_1$ elements\\
TABLE I. {\it The ten configurations for 3 particles.}
\begin{center}
\begin{tabular}{c|r|l|c|c|c}
\hline \hline
&&&&&\\
$l_z = - 1~0~~~1$&$L_z~~$&(3)~~~~(2 + 1)&$l_z = - 1~0~~~1$&$L_z$& (3)~~~~(2 + 1)\\
&&&&&\\
\hline
3~ 0~ 0&$-3$&1&1~ 0~ 2&1&1~~~~~~~ 1\\
2~ 1~ 0&$-2$&1~~~~~~~1&0~ 3~ 0&0&1~~~~~~~ --\\
2~ 0~ 1&$-1$&1~~~~~~~1&0~ 2~1&1&1~~~~~~~ 1\\
1~ 2~ 0&$-1$&1~~~~~~~1&0~ 1~ 2&2&1~~~~~~~ 1\\
1~ 1~ 1&0&1~~~~~~~2&0~ 0~ 3&3&1~~~~~~~ --\\
&&&&&\\
\hline \hline
\end{tabular}
\end{center}
among themselves, the next $\mu_2$ elements among
themselves, etc. and all the products of these permutations.
Then (20) gives the number, how often the unit representation in which
every element of the subgroup is represented by the
``matrix'' (1), occurs in $(\lambda_{\rho} + \lambda_{\rho - 1} + \ldots
+ \lambda_1)$, if this is consider as a representation of the subgroup. This
is, however, also the number of times the adjoint representation
$(\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1)$, again considered as a representation
of the subgroup, contains the ``antisymmetric'' representation, in which
every even permutation is represented by (1), every odd by $(-1)$.
If we denote this by
\begin{equation}
\left[
\begin{array}{c}
\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1\\
\mu_{\nu} \mu_{\nu - 1} \ldots \mu_1
\end{array} \right] =
\left(
\begin{array}{c}
\lambda_{\rho} \lambda_{\rho - 1} \ldots \lambda_1\\
\mu_{\nu} \mu_{\nu - 1} \ldots \mu_1
\end{array} \right)
\end{equation}
(21) expresses Frobenius' theorem. The $\Lambda$ are defined in (17).
The expressions (21) can be calculated by recursion formulas. If $\mu_{\nu}
= 4$, we have
$$
\left[
\begin{array}{ccc}
& \Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1&\\
4& \mu_{\nu - 1} \ldots& \mu_1
\end{array} \right] = \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3 - 1& \Lambda_2 - 1& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right] \eqno(22a)
$$
By means of this formula, one can get rid of all 4 among the $\mu$. If $\mu_{\nu}
= 3$
$$
\left[
\begin{array}{ccc}
& \Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1&\\
3& \mu_{\nu - 1} \ldots& \mu_1
\end{array} \right] = \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3 - 1& \Lambda_2 - 1& \Lambda_1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right] $$
$$
+ \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3 - 1& \Lambda_2& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
+ \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3& \Lambda_2 - 1& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right] $$
$$
+ \left[
\begin{array}{cccc}
\Lambda_4& \Lambda_3 - 1& \Lambda_2 - 1& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]. \eqno(22b)
$$
After a sufficient number of reductions of this type, there will be
only 2's and 1's among the $\mu$. If one of the [ ], occurring in
the right side of (22b) contains a negative number in the upper row,
or a number which is greater than the preceding one, the
whole [ ] is zero. To get rid of the 2's
$$
\left[
\begin{array}{ccc}
& \Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1&\\
2& \mu_{\nu - 1} \ldots& \mu_1
\end{array} \right] = \left[
\begin{array}{cccc}
\Lambda_4& \Lambda_3& \Lambda_2 - 1& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
$$
$$
+ \left[
\begin{array}{cccc}
\Lambda_4& \Lambda_3 - 1& \Lambda_2 & \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
+ \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3 & \Lambda_2& \Lambda_1 - 1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
$$
$$
+ \left[
\begin{array}{cccc}
\Lambda_4& \Lambda_3 - 1& \Lambda_2 - 1& \Lambda_1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
+ \left[
\begin{array}{cccc}
\Lambda_4-1 & \Lambda_3& \Lambda_2 - 1& \Lambda_1 \\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]
$$
$$
+ \left[
\begin{array}{cccc}
\Lambda_4 - 1& \Lambda_3-1 & \Lambda_2& \Lambda_1\\
\mu_{\nu - 1}& \mu_{\nu - 2} \ldots&& \ldots \mu_1
\end{array} \right]. \eqno(22c)
$$
If all the $\mu = 1$, which will be true after some reductions of this
type, one can use a formula analogous to the previous
ones. It is quicker to notice that (23) is the dimension of $(\Lambda_4 \Lambda_3
\Lambda_2 \Lambda_1)$ and hence
\setcounter{equation}{22}
$$
\left[
\begin{array}{c}
\Lambda_4 \Lambda_3 \Lambda_2 \Lambda_1\\
11 \ldots ~~~1
\end{array} \right] = \frac{(\Lambda_4 + \Lambda_3 + \Lambda_2 + \Lambda_1)!
(\Lambda_4 + 3 - \Lambda_1)(\Lambda_4 + 2 - \Lambda_2)(\Lambda_4 + 1 - \Lambda_3)}
{(\Lambda_4 + 3)!(\Lambda_3 + 2)!(\Lambda_2 + 1)! \Lambda_1!}
$$
\begin{equation}
\cdot (\Lambda_3 + 2 - \Lambda_1)(\Lambda_3 + 1 - \Lambda_2)(\Lambda_2 +
1 - \Lambda_1).
\end{equation}
One will use the formulas (22), (23) very rarely.
In most practical cases, the calculation of the coefficients (21) is greatly
facilitated by special conditions.
\section*{5}
~~~We can go over, now, to approximation (2). In this approximation the Hamiltonian
will be invariant with respect to all operations involving $\tau$ only. Since
$\tau$ has, mathematically, the same properties as the ordinary spin variable,
we shell have the analogs of the quantum numbers $L$ and $S$ of ordinary
spectroscopy. Instead of the azimuthal quantum number of the common spectroscopy,
we have the total quantum number $J$, since the spin forces are taken into
account in approximation 2; instead of the total spin, we have an
isotopic spin $T$. Of course, the interaction of $J$ and $T$ in the
higher approximation 4 will be entirely different from the
interaction between azimuthal and spin quantum numbers
of ordinary spectroscopy. Above all, the resulting angular
momentum $J$ will be a good quantum number in all
approximations.
The existence of a total isotopic spin means that terms
with different $\zeta$ components of the isotopic spin have the
same energy in approximation 2. These are, of course,
terms of different isobaric nuclei, and a total isotopic spin $T$
will be a term with the same binding energy for all nuclei
with isotopic numbers from $- T$ to $T$. This shows that to
every term of an element with a certain isotopic number $T_{\zeta}$
terms of all elements with smaller isotopic numbers will
correspond. The element with the smallest isotopic number
($n_P = n_N$ for even masses, $n_P = n_N \pm 1$ for odd masses) has the
greatest number of terms. In approximation 4, the equality
of these term values will cease to hold and the Coulomb
energy, already, will introduce a splitting.
If one is interested in the number of terms of
approximation (2), arising from a certain configuration, one
can use the ordinary Hund-Russell-Saunders method to
determine these. The only difference is that the ``orbits''
contain the ordinary spin quantum number already and one
has, therefore, for instance, six $p$ states, with $Z$ components
of the angular momentum $3/2, ~1/2,~ 1/2,~-1/2,~ -1/2,~ -3/2. $
Everyone of these six states can be doubly occupied, with a
particle $\tau = 1$ and $\tau = - 1$ (neutron or proton). The half
sum of the $\tau$ is denoted by $T_{\zeta}$ and the different $T_{\zeta}$ from $-T$ to $T$ united into a multiplet. The number of terms,
arising even from a simple configuration, is very great, however.
It is more important, perhaps, to consider the terms into
which a term of approximation (1) splits if we introduce the
spin forces and thus go over to (2). The transition from
approximation (1) to (2) can be performed in two steps: first
disregarding $Y_{\zeta}$ every multiplet goes over into several
multiplets which still have an $S$ and $T$. One obtains these by
simply projecting every point of Fig. 1 into the $S_z T_{\zeta}$ plane.
This is done in Fig. 4 for the multiplet ($3/2,~ 3/2, ~1/2$) as an
example. We see that it gives one term with
$S=3/2,$ $T=3/2,$ one with $S=3/2,$ $T=1/2,$ one with $S=1/2$
$T=3/2,$ and one with $S=1/2,$ $T= 1/2.$ The second step, then,
is to combine the $S$'s with the azimuthal quantum number $L$
to $J$'s, according to the vector addition model.
It would be very important to know experimentally the
relative separation of the terms which arise from the same
approximation 1 term, since this would allow us to tell
which of the 6 possible interactions, given in Section 2,
describes the spin forces.
Fig. 5 shows what can be expected in approximation (2).
Every figure corresponds to a set of isobars. The abscissa is
the isotopic number, the ordinate the total energy. Every
line corresponds to a term of approximation (2) all lines
arising from the same term of approximation (1) are
grouped close together. The $STY$ symbol of this term is
given on the right, it is, of course, the same for all the
group. The $T$ of the term is represented by the length of its
line, so that the term exists for elements with those isotopic
numbers $T_{\zeta}$ over which the line extends. The number on the
left of the line is the $S$ characterizing its spin after the first
step in the transition from approximation (1) to (2) is
performed. This $S$ will be the total angular momentum $J$ of
the nucleus, if the azimuthal quantum number was zero,
which will be very frequently the case. The energy of the
approximation (1) term is estimated on the basis of Eq. (8),
reference 4. This estimate gives the same value for all terms
of the same configuration with the same multiplicity $STY$
which is, of course, only approximately true. We are
interested, of course, only in the lowest term of every
multiplicity. The distances between the lines of the same
group have no significance.
We see that in several cases the approximation (2) terms
extend over several isobars and the question of the most
stable isobar will be decided, hence, only in the next
approximation. We may assume that the most important
term in the next approximation is the Coulomb
energy\footnote{The most stable isobar has the smallest mass, not the
greatest binding energy. For the consideration of the
stability, therefore, $T_{\zeta}$ times the mass difference between
neutron and H1 should be added to the total energy. This
will cause the lines of Fig. 4 to slope upward to the right.
This slope is soon overcompensated, however, by the opposite effect of the Coulomb energy.}. This will
\begin{figure}[h]
\centerline{\resizebox{9cm}{!}{\includegraphics{fig4.gif}}}
\caption{Determination of the spin angular momentum $S$
and isotopic spin $T$ of the terms arising from the multiplet
$(3/2,~ 3/2, ~1/2)$ if the spin forces are introduced. The spin
angular momenta $S$ must be added, subsequently, to the
orbital angular momentum $L$, according to the vector
addition principle, in order to obtain the total angular
momentum $J$.}
\end{figure}
decrease the binding energy of the nuclei with negative $T_{\zeta}$
compared with the binding energy of nuclei with positive $T_{\zeta}$
and cause the horizontal lines of Fig. 5 to slope downward
to the right. The slope will be very roughly proportional to
the $2/3$ power of the charge.
This slope will have the most noticeable effect for
isobars with masses $4n+2$. While for small charge, the point
$a$ most stable, beginning at.O$^{18}$, the point $b$ will become
stable.
We can proceed even to higher elements, by successively
increasing the slope of the lines more and more. For
elements $4n$, if the slope becomes $3/2,$ in the arbitrary units
of the figure, the point $b$ will become most stable. This
happens to be at A$^{40}$. The point $c$ never will become most
stable, since $b$ reaches $a$ before $c$ does. This seems to be the
explanation why no nuclei of mass $4n$ with odd number of
protons and neutrons exist. There are, however, radioactive
nuclei of this type.
The situation is very similar for nuclei with masses $4n+2$.
Here the critical slope is 2, when the point $c$ reaches $a$.
Again, point d is not the most stable for any slope, and there
are (apart from o) no stable nuclei with odd neutron and
proton number, for elements $4n+2$ either. The slope 2 seems
to be reached at Ti$^{50}$, later, of course, than slope $3/2.$
It should be mentioned that the whole Fig. 5 will be
compressed in energy scale as we proceed to higher
elements, because the exchange integrals decrease. It has
been shown by Bethe and
\begin{figure}
\centerline{\resizebox{9cm}{!}{\includegraphics{fig5.gif}}}
\caption{The different kinds of multiplets are shown for
elements with mass numbers of the form $4n,~ 4n \pm 1,~ 4n+2.$
The ordinate is the energy in arbitrary units. Only one term
for every multiple! system is given, with a position on the
energy scale corresponding to long range forces. The
abscissa is the difference between the number of neutrons
and protons, divided by 2. The circles correspond to stable
nuclei, the squares to unstable nuclei.}
\end{figure}
Bacher\footnote{H. A. Bethe and R. F. Bacher, Rev. Mod. Phys. {\bf 8,}
82 (1936), Section VI.} that this happens in discontinuous steps,
corresponding to the completion of shells. There is,
therefore, an increased probability for the slope to pass a
certain amount in the regions where shells are completed.
In case of elements with mass numbers $4n+1$ and $4n+3$
we obtain the same picture. First, the point $a$ most stable,
at a slope $3/2$ the point $b$ reaches $a$. This seem to happen at
Cl$^{37}$ and Ca$^{43}$, respectively, quite in the neighborhood of
A$^{40}$. For the slope 2, the point $a$ passes $b$ and Ti$^{49}$
and V$^{51}$ become
the stable isobars. These are near indeed to Ti$^{50}$. This
explanation of the places where the isotopic number of
stable isobars shifts to higher values works rather better than
could be expected and the agreement is beyond doubt,
partly accidental.\\
\section*{6}
~~~~As a last point, I should like to establish the connection
between approximations 1 and 3, i.e., determine the terms
into which an $STY$ term of approximation (1) splits if one
introduces, as a perturbation, the difference between
proton-proton, proton-neutron, neutron-neutron interactions,
neglecting, however, spin forces.
The operator
\begin{equation}
\frac{1}{4} (s_{z1} (1 + \tau_{\zeta 1}) + s_{z2} (1 + \tau_{\zeta 2}) +
\ldots + s_{zn} (1 + \tau_{\zeta n})) = \frac{1}{2} (S_z + Y_{\zeta}) = S_{z
N}
\end{equation}
gives the $Z$ component of the neutron spin angular
momentum, since $1 + \tau_{\zeta}$ gives 0, if applied to a proton
state. Similarly
$$
\frac{1}{4} (s_{z1} (1 + \tau_{\zeta 1}) + s_{z2} (1 + \tau_{\zeta 2}) +
\ldots + s_{zn} (1 + \tau_{\zeta n})) = \frac{1}{2} (S_z - Y_{\zeta}) = S_{z
P} \eqno(24a)
$$
gives the $Z$ component of the proton spin angular
momentum. If we go through all the points of Fig. 1 for a
certain $T_{\zeta}$ and insert their $S_{zN}$ and $S_{zP}$ values
into a table, one
can unite the points of the table in the normal way to a
$S_N S_P$
multiplet\footnote{Instead of this, one can simply turn the
corresponding level in the diagrams of Fig. 1 by 45$^{\circ}$.
}. The azimuthal quantum number is, of course,
unchanged by the transition from approximation (1) to (3).
If the proton-proton forces are assumed to be equal to the
neutron-neutron forces, the $S_N S_P$ term will coincide in case
of equal number of protons and neutrons with the $S_P S_N$
term, if $S_N \ne S_P$. In the work of reference 4, these terms were
given as one term. The values of the multiplicities $2S_N + 1,~ 2S_N + 1$
are given in Fig. 1 below every $T_{\zeta}$.
We may consider, as an example, the $(3/2 ~3/2 ~1/2)$
multiplet, given in Fig. 1. Below $T_{\zeta} = 1/2$ we have the three
pairs 23, 41, 21. This means that for an element of the mass
$4n+1$ with $T_{\zeta} = 1/2$ (e.g., C$^{13}$), certain states in which the
neutrons are in the doublet, the protons in the triplet state,
exactly coincides in approximation (1) with a state in which
the neutrons are in the quartet, the protons in the singlet
state, and with another state in which the neutrons are in
doublet, the protons in the singlet state. It may be added that
the binding energy of these states is equal in approximation
(1), to the binding energy of a hypothetical B$^{13}$ nucleus
($T_{\zeta} =3/2$) in which the neutrons are in the triplet,
the protons
in the doublet state. In reality, all these states will be
unstable for isobars with the mass 13, because of the
comparatively high position of the ($3/2~ 3/2 ~1/2$) multiplet in
Fig. 5a. The example is thus, perhaps, not a very fortunate
one but it illustrates the kind of regularities to be expected
more clearly than a simpler case.
\end{document}
%ENCODED April 2003 BY NIS;
%EDITED April 2003 BY NIS;