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G. Breit, Phys. Rev., {\bf Vol. 50,} 825 \hfill {\large \bf 1936}\\
\vspace{2cm}
\begin{center}
{\Large \bf The Possibility of the Same Form of Specific Interaction
for All Nuclear Particles}\\
\end{center}
\vspace{0.5cm}
\begin{center}
G. BREIT AND E. FEENBERG\footnote{Now at the Institute for Advanced study},\\
University of Wisconsin, Madison Wisconsin \\
(Received August 19, 1936)
\end{center}
\begin{abstract}
Experimental evidence pointing to a universal form of interaction for all
nuclear particles is summarized. The form given by Eq. (1) is discussed. It is
found to be satisfactory for H$^3$, He$^3$, He$^4$
considered together with proton and neutron scattering. In order that heavy
nuclei should not be too stable it is necessary to require the inequality (7.2)
and in order that there be no very heavy and
electrically neutral nuclei the inequality (7) has to be satisfied. The form
of the interaction energy restricted by these conditions is as satisfactory as
nuclear theories postulating different forms for like
and unlike particles. The simplest form, satisfying all present requirements
is obtained from Eq. (1) by letting $g_1 = g_2 = 0$.
\end{abstract}
The interaction between protons and protons which
acts in addition to the Coulombian repulsion is
generally supposed to be the same as that between
neutrons and neutrons. The evidence for this lies in the
fact\footnote{E. Feenberg and J. Knipp, Phys. Rev. {\bf 48,} 906 (1935);
S. S. Share, Phys. Rev. {\bf 50}, 488 (1936).} that the mass defect of He$^3$
is smaller than that of H$^3$ by approximately the amount required by the
Coulombian repulsion between the protons in He$^3$.
Further strong support for regarding the specific like-particle interaction
as the same for protons and neutrons is found in the close
agreement\footnote{G. Breit, E. U. Condon and R. D. Present, Phys. Rev. this
issue; Fermi and Amaldi, Ricerca Scient. {\bf 1,} 1 (1936);
Fermi, ibid., July (1936).} of the values
obtained for its magnitude from experiments on the
scattering of protons in
hydrogen\footnote{M. A. Tuve, N. P. Heydenburg and L. R. Hafstad, Phys.
Rev. this issue.} with that derived
from the mass defect of H$^3$.
The accuracy of the determination of the like-particle interaction
from the proton-proton scattering
experiments is great. Numerical comparison of its
magnitude with that which follows from experiments
on the scattering and recombination of slow neutrons
in hydrogen shows that in $^1S$ states the specific
interaction between all nuclear particles are the same
to within the accuracy of present experiments. The
equality of the interactions is obtained independently
of what range is used for the spacial extension of the
force within wide limits and is, therefore, probably
real rather than accidental. It thus becomes reasonable
to attempt to regard the specific nuclear forces as the
same between all nuclear particles independently of
whether they are neutrons or protons. In the present note the
qualitative consequences of this view are discussed.
In the early development of nuclear theory it was
supposed that the principal interactions within the
nucleus take place between neutrons and protons
because by doing so the approximate equality of the
number of neutrons with the number of protons
automatically received a simple explanation.
According to present evidence, mentioned above, this
explanation does not apply and a new one must be
made. This is offered by the exclusion principle which
favors the addition of unlike particles analogously to
the way in which the building up of electronic shells in
an atom proceeds by adding electrons with opposite
spin directions. In addition it is favorable for stability
to have proton-neutron pairs in symmetrical $(S, D, G, \ldots)$
triplet states because the attraction is then greater
than in symmetrical singlet states. For like particles
the exclusion principle rules out the symmetrical
triplet state. This circumstance is important in the
building up of the first shell up to He$^4$.
The universal interaction energy between any pair
of heavy elementary particles will be assumed to be of
the form
\begin{equation}
V_{ij} = \left\{ (1 - g - g_1 - g_2) P_{ij}^M + g P_{ij}^H + g_1 \cdot
1 + g_2 P_{ij}^S \right\} J(r_{ij}),
\end{equation}
where $g, g_1, g_2$ are constants and $P^M, P^H$ are the
Majorana and Heisenberg exchange operators while $P^S = P^M P^H$
is the operator proposed by Bartlett which
exchanges the spins without affecting the
coordinates\footnote{Another formulation of the symmetrical Hamiltonian
using a fifth ``character'' variable is given by Condon and Cassen in this issue.}.
For this interaction\\
\begin{center}
Table I. {\it Two-particle systems.}
\end{center}
{\footnotesize
\begin{center}
\begin{tabular}{c|c|c|c}
\hline \hline
\multicolumn{2}{c|}{Even $L$}&\multicolumn{2}{c}{Odd $L$}\\\cline{1-4}
\hline
triplets&singlets&triplets&singlets\\
$^1S,~ ^3D, \ldots$&$^1S,~ ^1D, \ldots$&$^3P,~ ^3F, \ldots$&$^1P, ~^1F, \ldots$\\
$J(r)$&$(1 - 2g - 2g_2)J(r)$& $(1 + 2g_1 + 2g_2)J(r)$&$(- 1 + 2g + 2g_1)J(r)$\\
Absent for equiva-&&&Absent for equiva-\\
lent like particles&&&lent like particles\\
\hline \hline
\end{tabular}
\end{center}}
the potential energies in different states are as in Table I
where $L$ is the orbital angular momentum in units of $\hbar$. Two
isolated like particles moving in each other's field cannot have
different principal quantum numbers and are in equivalent
states if $L$ is evert and the spins are parallel or when $L$ is odd
and the spins are antiparallel. Triplets of even $L$ and singlets
of odd $L$ are thus ruled out by the exclusion principle. In
proton-proton scattering experiments only states with $L = 0$
matter and, therefore, the effective interaction energy is that
corresponding to $^1S$ namely $(1 - 2g - 2g_2)J(r)$. In the
scattering of slow neutrons by hydrogen a similar $^1S$ state is
responsible for the large collision cross section according to
Wigner's hypothesis. Both sets of scattering experiments thus
have to do with $(1 - 2g - 2g_2)J(r)$ while $J(r)$ determines the
binding energy of the deuteron.
\section{Light Nuclei}
~~~~In H$^3$, He$^3$, He$^4$ like particles are in nearly spacially
symmetric states with respect to each other and the effective
interaction potential is thus again $(1 - 2g - 2g_2)J(r)$. This is
consistent with the agreement between the values of the like
particle interaction energy as obtained from the mass defects
of the isotopes of hydrogen and helium with that derived from
the scattering of protons in hydrogen. Unlike particles are
partly in triplet and partly in singlet states with respect to each
other. However, the space wave function
is nearly symmetric and thus the effective interaction potential
between protons and neutrons is $[1 - \frac{1}{2} (g + g_2)]J(r)$ which is
the weighted mean of the potentials in singlets and triplets
with statistical weights $\frac{1}{4}$ and $\frac{3}{4}$. In these questions
the sum of $g + g_2$ occurs as a whole and one cannot distinguish between $g$
and $g_2$ separately. The calculations on mass defects that have
been made without using $g_1, g_2$ thus apply directly to (1) and
are consistent with it. The way in which this comes about may
be seen for H$^3$ as follows. The wave equation is
$$
(T + \Sigma_{i > j} V_{ij}) \psi = E \psi,
$$
where $T$ is the operator representing the sum of kinetic
energies of the three particles. Since $\psi$ is antisymmetric in the
neutrons it is of the form
$$
\psi = u((12_+),3)S^0((12_-),3) + v((12_-),3)S'((12_+),3),
$$
where the two neutrons are denoted by 1, 2 and a plus suffix
indicates that the function is symmetric in the particles while a
minus suffix similarly indicates antisymmetry. The functions
$u,v$ contain only the Cartesian coordinates while the functions
$S^0, S'$ contain only the spin coordinates. Considering the state
with total spin angular momentum $\frac{1}{2}$ in the $z$ direction one
may use
$$
\begin{array}{l}
S^0 = 2^{-\frac{1}{2}}[(+-+)- (-++)],\\
S' = 6^{-\frac{1}{2}}[(+-+)+(-++)-2(++-)],\\
\end{array}
$$
where the $+$ and $-$ signs correspond to positive and negative
orientations of the spin axis in an arbitrary fixed direction.
Each () stands for a product of spin functions referring to the
particles in the order 1, 2, 3. One obtains
\begin{equation}(H_0+H'-E) \left( \begin{array}{c}
u\\
v\\
\end{array} \right) =0,
\end{equation}
where
$$
H_0 = T + \left(
\begin{array}{c}
1 - 2g - 2g_2,0\\
0,-1 + 2g_1 + 2g_2
\end{array} \right) J_{12} + \left(
\begin{array}{c}
1 - \frac{\displaystyle 1}{\displaystyle 2}(g + g_2), 0\\
0,1 - (3/2)(g + g_2)
\end{array}
\right) (J_{13} + J_{23}), \eqno(2.1)
$$
$$
H' = \left(
\begin{array}{c}
X, Y\\
Y, Z
\end{array} \right), \eqno(2.2)
$$
$$
X =(1 - \frac{1}{2} g - g_1 - g_2)[J_{23}(P_{23}^M - 1)+J_{13}(P_{13}^M
- 1)], \eqno(2.3)
$$
$$
Y = 3^{\frac{1}{2}} 2^{-1}[(gP_{13}^M + g_2)J_{13} - (gP_{23}^M + g_2)J_{23}],
\eqno(2.4)
$$
$$
Z = (1 - (3/2)g - g_1 - g_2)[J_{23}(P_{23}^M - 1)+J_{13}(P_{13}^M - 1)]. \eqno(2.5)
$$
Neglecting $H'$ one obtains an approximation to $\psi$,
\begin{equation}
\psi_0 = \left( \begin{array}{c}
u\\
0
\end{array}
\right),
\end{equation}
with $u$ defined by
$$
\{ T +(1 - 2g - 2g_2) J_{12} + (1 - \frac{1}{2} g - \frac{1}{2} g_2)(J_{13}
+ J_{23}) - E \} u = 0 \eqno(3.1)
$$
In the approximation of using $\psi_0$ the energy can be calculated without
exchange operators by using $(1 - 2g - 2g_2)J,~ (1 - \frac{1}{2} g - \frac{1}{2}
g_2)J,$ respectively, for the potential energies between like and unlike
particles. The function $u$ is symmetric in the neutrons but has,
in general, no symmetry property for interchange of a proton and a neutron.
If, however, $g + g_2$ is small then $u$ becomes symmetric in all three
particles. In this case the first-order perturbation due to $H'$ vanishes as is
obvious from (2.3). The second-order perturbation may be
estimated as $-(\psi_0 |H'^2|\psi_0)/E_m$ where $E_m$ is an average energy
of perturbing levels above that of the normal state. According to
(2.2) and (3)
$$
(\psi_0|H'^2|\psi_0) = (u|X^2 + Y^2|u).
$$
The calculation of this matrix element is readily made using (2.3) and (2.4)
and gives for a completely symmetric $u$
$$
(\psi_0|H'^2|\psi_0) =(3/2)(g + g_2)^2 \int u^2 (J_{13}^2 - J_{13} J_{23})
d \tau_1 d \tau_2. \eqno(3.2)
$$
Thus the estimate of the second-order perturbation energy depends only on
$g + g_2$ and one may expect the energy
to depend on $g - g_2, g_1$ only in a secondary way. Since $g + g_2$ is the only
combination of the $g$'s which matters for the mass defect of the deuteron and
the present proton-proton and proton-neutron scattering experiments the
inclusion of $g_1$ and $g_2$ makes little practical difference in the comparison
of the mass defect of H$^3$, H$^2$, He$^3$ with the scattering experiments.
Similarly in He$^4$
$$
H_0 = T + \left( \begin{array}{c}
1 - 2g - 2g_2, 0\\
0, - 1 + 2g_1 + 2g_2
\end{array} \right)
(J_{12} + J_{34}) + $$
$$
+ \left(
\begin{array}{c} 1 - \frac{\displaystyle 1}{\displaystyle 2} g - \frac{\displaystyle
1}{\displaystyle 2} g_2, 0\\
0, 1 - (3/2)g - (3/2) g_2
\end{array} \right)
(J_{13} + J_{23} + J_{14} + J_{24}), \eqno(4.0)
$$
$$
H' = \left(
\begin{array}{c}
X,Y\\
Y,Z
\end{array}
\right) \eqno(4.1)
$$
$$
X = [J_{13}(P_{13}^M - 1) + J_{14} (P_{14}^M - 1) + J_{23} (P_{23}^M -
1) + J_{24} (P_{24}^M - 1)] (1 - \frac{1}{2} g - g_1 - g_2),
$$
$$
Y = 3^{\frac{1}{2}} 2 ^{-1} [J_{23} (g P_{23}^M + g_2) + J_{14}(gP_{14}^M
+ g_2) - J_{13} (g P_{13}^M + g_2) - J_{24} (g P_{24}^M + g_2)],
\eqno(4.2)
$$
$$
Z = (1 - (3/2) g - g_1 - g_2)X/(1 - \frac{1}{2} g - g_1 - g_2).
$$
These equations differ from Eqs. (2.1) to (2.5) essentially only through the
presence of more terms in sums over
like and unlike particles. The wave Eq. (2) is the same in form but now.
$$
\begin{array}{l}
S^0 = \frac{\displaystyle 1}{\displaystyle 2} [(+-+-) + (-+-+) - (+--+)
- (-++-),\\
S' = 12^{- \frac{1}{2}} [2(++--) + 2 (--++) - (+-+-) - (-+-+) - \\
~~~~~~~~- (- ++-) -(+--+)].
\end{array}
$$
Particles 1 and 2 are alike and so are 3 and 4. With a completely symmetric $u$
$$
(\psi_0|H'^2|\psi_0) = 3(g + g_2)^2 (u|J_{13}^2 + J_{13} J_{24} - 2 J_{13}
J_{23}| u).
\eqno(4.3)
$$
Thus here also only $g + g_2$ enters in the
calculation of the normal state. With wave functions
$u = \mbox{exp} \{- (\nu/2)(r_{12}^2 + r_{13}^2 + r_{23}^2)\}$ for H$^3$
and \mbox{$\mbox{exp} \{-(\nu/2) (r_{12}^2 + r_{13}^2 + r_{14}^2 + r_{23}^2 +
r_{24}^2 + r_{34}^2)\}$} for He$^4$ and with
$$
J(r) = Ae^{- \alpha r^2}
\eqno(4.4)
$$
one obtains for H$^3$
$$
(\psi_0 |H'^2| \psi_0) = \frac{3}{2} (g + g_2)^2 A^2 \left[ \left( \frac{3
\nu}{3 \nu + 4 \alpha} \right)^{\frac{3}{2}} - \left( \frac{\nu}{\nu +
\alpha} \right)^{\frac{3}{3}} \left( \frac{3 \nu}{3 \nu + \alpha}
\right)^{\frac{3}{2}} \right] \eqno(4.5)
$$
and for He$^4$
$$
(\psi_0|H'^2|\psi_0) = 3 (g + g_2)^2 A^2 \left\{ \left( \frac{\nu}{\nu
+ \alpha} \right)^{\frac{3}{2}} + \left( \frac{2 \nu}{2 \nu + \alpha} \right)^3
\right.$$
$$
\left. - \frac{128 \nu^3}{[3(\nu + \alpha)^2
+ 10 \nu (\nu + \alpha) + 3 \nu^2]^{\frac{3}{2}}}
\right\}. \eqno(4.6)
$$
The contributions to (3.2) and (4.3) in the present
approximation are due entirely to $Y^2$ and thus, according to
(2.4) and (4.2) originate in interactions between unlike
particles. They are thus present even if the form of the interaction
is not made to be the same for like and unlike particles.
Formulas (4.5) and (4.6) are then correct provided for $A$ the
value corresponding to the neutron-proton potential in the $^3S$
state is used. The magnitude of the perturbation can be
estimated using the approximation
\begin{equation}
\Delta E = - (\psi_0 |H'^2| \psi-0)/E_m,
\end{equation}
where $E_m$ is a properly taken mean of the perturbing
energy levels. For lack of better knowledge of the
relative importance of different levels one may use the
binding energy for $E_m$ and thus obtain most probably
an overestimate of the absolute value of $\Delta E$. Using the
units $\hbar (mM)^{\frac{1}{2}} c^{-1}$ for length and $mc^2$ for energy and
letting $\alpha = 16$ the values of $\nu$ which minimize $(0|H_0|0)$
are $\nu \sim 10$ for both H$^3$ and He$^4$. Eq. (5) gives then
$\Delta E \sim - mc^2$ for H$^3$ and $-0.4 mc^2$ for He$^4$. It should be
noted that these estimates refer only to the effect of $Y$
and do not represent the total effect of terms in $(g + g_2)^2$
in an expansion of the energy starting from a value
corresponding to a symmetric wave function. Thus if the Hamiltonians
$$
\begin{array}{l}
T + (1 - g - g_2) (J_{12} + J_{13} + J_{23}),\\
T + (1 - g - g_2) (J_{12} + J_{13} + J_{23} + J_{24} + J_{34}),
\end{array}
$$
are used respectively for H$^3$ and He$^4$ in order to define
unperturbed energy values and wave functions then the
values of $(\psi_0|H'2|\psi-0)$ given by Eqs. (4.5) and (4.6)
should be multiplied by 2.
Calculations have already been made for
Hamiltonians having different symmetries\footnote{E. Feenberg and S. S. Share,
Phys. Rev. {\bf 50,} 253 (1936); E. Feenberg, Phys. Rev. {\bf 49,} 273 (1936).}.
According to these the difference in energy values of H$^3$
corresponding to
$$
\begin{array}{c}
H = T + J^{\ast}_{23} + J^{\ast}_{13}\\
\mbox{and} \quad H = T + \frac{\displaystyle 2}{\displaystyle 3}
(J^{\ast}_{13} + J^{\ast}_{13} + J^{\ast}_{23})
\end{array}
$$
is about $2mc^2$. The quantity $\frac{2}{3} J^{\ast}_{12} - \frac{1}{3}J^{\ast}_{23}
- \frac{1}{3} J^{\ast}_{13} = H''$
may be considered as a perturbation which when
applied to the symmetric form gives the unsymmetric
form. For a symmetric initial wave function
$$
(0|H^{\prime \prime 2}|0) = \frac{2}{3} (0|J^{\ast 2}_{13} - J^{\ast}_{13} J^{\ast}_{23}|0).
$$
Equating $2mc^2$ and $(0|H''^2|0)/E_m^{\prime}$ it is found that
$E_m^{\prime} \sim 100 mc^2$. The energy levels which matter in this
connection are those corresponding to pure singlet
states in neutrons while the effect of $Y$ is concerned
with perturbations by states in which the neutrons are
in a triplet condition. Nevertheless it is probable that
there is a qualitative similarity between $E_m$ and $E_m^{\prime}$
Using $E_m = E_m^{\prime}$ the estimate of perturbations due to triplet
neutron states in H$^3$ drops to $0.14 mc^2$. It is thus seen
that the second-order perturbations which depend
essentially only on $(g + g_2)^2$ are themselves likely to be
small. Ordinarily the effect of these perturbations is
neglected and the calculations are made using $(1 - \frac{1}{2} g) A_{\pi
\nu} e_{- \alpha r^2}$ for the average neutron-proton interaction
and $A_{\nu \nu} e^{- \alpha r^2}$ for the interaction between like particles.
The value obtained for $A_{\nu \nu}$ agrees with that from
proton-proton scattering experiments and is equal to
$(1 - 2g) A_{\pi \nu}$. The mass defects of the
isotopes of hydrogen and helium are thus in agreement
with the interaction energy (1), but give at present
practically no information about $g, g_1, g_2$ except for
determining $g + g_2$.
\section{Heavy Nuclei}
~~~~Calculations of any exactness are difficult for heavy
nuclei. It is, however, possible to establish some
inequalities which are necessary for stability of
existing nuclei by means of the statistical model. The
density matrices for protons and neutrons will be
written $\rho_{\pi} (x,x')$ and $\rho_{\nu}(x,x')$. These functions are
supposed to involve only the space coordinates. For
simplicity it will be supposed that each space state is
filled by two like particles having opposite spin
directions. The density matrix contains therefore two
identical terms for each occupied space state. For
diagonal elements the abbreviations
$$
\rho_{\pi}(x) = \rho_{\pi}(x,x), \quad \rho_{\nu}(x) = \rho_{\nu}(x,x)
$$
will be used.
For the estimates made below the wave function is
approximated by a product of two determinants, one
for the neutrons and the other for the protons.
Substitution of such a wave function into the
variational integral gives an upper limit to the energy.
The expression thus obtained is
\begin{equation}
E = T + W^{\nu} + W^{\pi} + W^{\pi \nu}
\end{equation}
where $T$ is the average value of the kinetic energy
operator (a sum of the kinetic energies of the
individual particles). The quantities $W$ represent the
contributions due to (1). The interactions between
neutrons give $W^{\nu}$, those between protons give $W^{\pi}$, and
those between neutrons and protons give $W^{\pi \nu}$. It is found that
$$
W^{\nu} = (1 - \frac{1}{2}g - (3/2)g_1 - 2 g_2) E_{ex}^{\nu} + (1 - \frac{1}{2}
- \frac{1}{2} g + (3/2) g_1 + g_2) E^{\nu},
\eqno(6.1)
$$
$$
W^{\pi} = (1 - \frac{1}{2} g - (3/2) g_1 - 2g_2) E_{ex}^{\pi} + (- \frac{1}{2}
- \frac{1}{2} g + (3/2)g_1 + g_2)E^{\pi}, \eqno(6.2)
$$
$$
W^{\pi \nu} = (2 - g - 2g_1 - 2g_2) E_{ex}^{\pi \nu} + (2 g_1 + g_2)E^{\pi
\nu}, \eqno(6.3)
$$
where
$$
E^{\alpha \beta} = \frac{1}{2} \int \rho_{\alpha}(x) J(x - x') \rho_{\beta}(x')
dx dx';
$$
$$
E_{ex}^{\alpha \beta} = \frac{1}{2} \int \rho_{\alpha}(xx') J(x - x') \rho_{\beta}
(x'x) dx dx', \eqno(6.4)
$$
$$
E^{\alpha \alpha} = E^{\alpha}; \quad E_{ex}^{\alpha \alpha} = E_{ex}^{\alpha};
\quad \alpha, \beta = \mu ~\mbox{or} ~\nu.
$$
If the density of particles is so high that a large
number of them are on the average within the range of
$J$ then the quantities $E_{ex}$ show saturation while the
quantities $E$ are proportional to the square of the
number of particles. This circumstance may lead to
instability. Thus consider $W^{\nu}$. If the coefficient of $E^{\nu}$ in
it is positive, the nuclear energy can be lowered by
increasing the number of neutrons $N$. In particular it
would be possible to have nuclei consisting entirely of
large numbers of neutrons. For such nuclei the kinetic
energy would be proportional to $N^{5/3} r_0^{-2}$ where $r_0$ is
the nuclear radius while the potential energy would
vary as $N^2$ and would be independent of $r_0$, if $r_0$ is
sufficiently smaller than the range of force. Keeping
$r_0$ constant the kinetic energy varies more slowly with
$N$ for high $N$ than the potential energy and therefore a
negative potential energy will lead to infinite stability
for very high $N$. A positive coefficient of $E^{\nu}$ in Eq.
(6.1) would require, therefore, the existence of
infinitely heavy nuclei of high stability. This is
contrary to experience and hence\footnote{Condition (7) is equivalent
to requiring that for like particles the interaction be of the form
$(a + b P^M)J$ where $b \ge 2a$
as is clear from the fact that Eq. (1) is equivalent to an
interaction energy $(1 - g - g_1 - 2g_1) P^M J + (g_1 - g)J$ for like
particles.'}
\begin{equation}
1 + g \ge 3g_1 + 2g_2.
\end{equation}
For a nucleus in which the number of neutrons $N$ is
equal to the number of protons $Z$ the density matrices
$\rho_{\nu}, \rho_{\pi}$ may be taken to be equal. Then
$$
W = W^{\pi \nu} + 2W^{\nu \nu} = 4(1 - \frac{1}{2} g - (5/4)g_1 - (3/2)g_2
E_{ex} + (-1 - g + 5g_1 + 3g_2) E, \eqno(7.1)
$$
where the superscripts $\pi, \nu$ on the $E$'s are dropped.
The trial wave function can be made to correspond to
an $r_0$ smaller than the range of $J$. By increasing $N$ and
$Z$ simultaneously, the governing term in $W$ becomes
that due to $E$ which varies as $N^2$. The only other term
which varies as rapidly is that due to the Coulomb
energy. Since in the present case $N=Z$ the whole
energy expression can be considered as a
function of $N$. In order that it be impossible to have stable
nuclei of this type with very high $N$ and $Z$ it is necessary to
require that the coefficient of $N^2$ in the energy expression be
positive. This condition is
$$
1 + g + \frac{6e^2}{5r_0|J(0)|} \ge 5g_1 + 3g_2. \eqno(7.2)
$$
The nuclear radius ro must be chosen smaller than the
range of nuclear forces which is of the order $e^2/mc^2$. If
$r_0$ is made equal to $e^2mc^2$ the last term on the left side
of (7.2) is $1/58$ for $|J(0)| = 70 mc^2$. Since $r_0$ must be
made still somewhat smaller in order to bring about
the full activity of $J(0)$ this term should be considered
as somewhat greater than $1/58,$ say $1/20.$ It is
nevertheless numerically insignificant. The conditions
(7.1) and (7.2) may be summarized, using the
empirical value of $g + g_2 = 0.2$, as
$$
1.2 = 1 + g + g_2 \ge 3(g_1 + g_2);
$$
$$
1.2 = 1 + g + g_2 \ge 5g_1 + 4g_2. \eqno(7.3)
$$
On these grounds therefore the interaction in triplet
states of odd $L$ can be expected to lie between $-0.2 J$
and $-J$ for $g + g_2 = 0.20$ and between $0.1J$ and $-J$
for $g + g_2 = 0.25$ (provided that $g_1 + g_2$ is not negative). It
is these states that matter in the photoelectric disintegration
of the deuteron by high energy $\gamma$-rays. For high
particle densities with $N>Z$ the value of the variational
integral is
$$
E = t - N (1 - \frac{1}{2}g - (3/2)g_1 - 2g_2)|J(0)| - Z (3 - (3/2)g -
(7/2)g_1 - 4g_2)|J(0)|
$$
$$
+ \frac{1}{4} (N^2 + Z^2) (1 + g - 3g_1 - 2g_2)|J(0)| - NZ (g_1 + \frac{1}{2}
g_2)|J(0)| + \frac{3}{5} \frac{Z^2 e^2}{r_0}. \eqno(7.4)
$$
In order that it be impossible to obtain in this
expression contributions varying quadratically with $N$
and $Z$ for a given ratio $N/Z$ it is necessary to require
that
$$
1 + g \ge 3g_1 + 2g_2 + \frac{2NZ}{N^2 + Z^2} (2g_1 + g_2) - \frac{12}{5}
\frac{Z^2}{Z^2 + N^2} \frac{e^2}{r_0|J(0)|}. \eqno(7.5)
$$
For $Z=0$ this gives (7) and for $Z=N$ it gives (7.2).
Neglecting the relatively insignificant Coulomb energy
all conditions implied by (7.4) are contained in (7.3).
These conditions are obtained above by considering a
higher particle density than that which exists in actual
nuclei. This circumstance does not normally spoil the
argument because the energy is lower than the value of
the variational integral. The above discussion does not
say much about the reaction rate or the values of $N$ and
$Z$ at which transformations would occur. It does not
make much difference, however, if the possibility of
very heavy neutral nuclei is allowed but instead it is
required that there should be no neutral nuclei of mass
200 having a greater stability than a collection of equal
numbers of protons and neutrons in the form of alpha-particles.
The kinetic energy is statistically
$11.3 mc^2 N^{5/3}(e^2/mc^2 r_0)^2$. The requirement is then
$$
1 + g - 3g_1 - 2g_2 \le - \frac{45e^4N^{-1/2}}{|J(0)|r_0^2 mc^2} + \frac{4}{N}
(1 - \frac{1}{2} g - (3/2) g_1 - 2g_2) - \frac{54mc^2}{N|J(0)|}. \eqno(7.6)
$$
By using $|J(0)| = 70 mc^2$ and $r_0 = e^2/mc^2$ the right side is
$-0.09.$ This condition differs very little from (7). If
instead of requiring a smaller stability than that
corresponding to groups of alpha-particles one requires
a smaller stability than that corresponding to $17mc^2$
mass defect per particle the right-hand side of the last
inequality is changed only by $-0.001.$ Similarly if
instead of (7.2) it is required that a nucleus with $N=Z$
should not have a greater stability than that
corresponding to a mass defect of $17mc^2$ per particle,
which corresponds to the stablest nuclei, one obtains
$$
1 + g - 5g_1 - 3g_2 \ge - \frac{45e^4 N^{-1/2}}{|J(0)|r_0^2mc^2} + \frac{8}{N}
\left( 1 - \frac{1}{2} g - \frac{5}{4} g_1 - \frac{3}{2} g_2 \right)
$$
$$
-\frac{6e^2}{5r_0|J(0)|} - \frac{68mc^2}{N|J(0)|}. \eqno(7.7)
$$
For nuclei with $Z=N=100$ this condition is only slightly
weaker than (7.2) since the right side of the last inequality is
$-0.09$ using the
same constants as previously. The inequality (7.2) is thus also
nearly the condition for not having too much stability in
nuclei with ordinary numbers of particles.
Actual heavy nuclei are far from being in a state of high
particle density such as was just used. For Hg$^{200} E^{\nu}/E_{ex}^{\nu}$ can
be estimated to be approximately 4 using the statistical model
and $r_0 = 0.8 \times 10^{-12}$ cm with $\alpha^{-1/2} = 2.2 \times 10^{-13}$ cm.
For a very high
particle density this ratio would be $N/2 = 60$. Eq. (7.4) is thus
not a guide for the actual energy dependence. A too large
numerical value of a negative coefficient of $E$ in (7.1) would
be harmful to stability but it is impossible to tell without
further calculation whether higher approximations reduce the
energy sufficiently to allow appreciable positive values of
$1 + g - 5g_1 - 3g_2$. If $g_1 = g_2 = 0$ and if $E/E_{ex} = 4$ the potential
energy occurring in (7.1) is $-6gE_{ex}$ and is positive.
Nevertheless this does not exclude the possibility of a simple
theory with $g_1 = g_2 = 0$ because (1) the energy corresponding to
an assumed interaction is lower than the value of the
variational integral with an approximate wave function;
(2) $E^{\pi}/E_{ex}^{\pi}$ is $< 4;$ (3) exact values of $E/E_{ex}$
are hard to obtain since the uncertain nuclear radius is involved.
In all of the above discussion it was supposed that $J(r)$ has
the same sign throughout and complications arising from a
reversal of sign near $r=0$ were not taken up. In proton-proton
scattering experiments the kinetic energy is of the order of 1
Mev while in the nucleus the kinetic energy of individual
particles is of the order of 30 Mev. The kinetic energy inside
the ``potential well'' representing the interaction of two
particles is changed, however, only from $\sim 20$ Mev to $\sim 35$
Mev or perhaps $\sim 50$ Mev in the comparison of scattering
experiments with conditions inside the nucleus. This
corresponds to a decrease of the wave-length inside the deep
part of the ``well'' by a factor of about 1.3 which existing that existing
heavy roughly the same features of $J(r)$
come into consideration in existing existing heavy nuclei as in proton-proton
scattering experiments. In the deduction of the
inequalities (7.3), however, the size of the nucleus was taken
to be smaller than that of an actual nucleus and, therefore, the
kinetic energy per particle was increased to a maximum of
about $(8 \times 10^{-13}/2.8 \times 10^{-13})^2$ 30 Mev $\sim 200$ Mev
for $N=100$.
Since some of the nuclear particles collide while traveling in
opposite directions the relative kinetic energy is changed by a
factor of about 10 in the deep part of the ``well'' which
corresponds to a factor 3 in the wave-length. A reversal of
sign of $J$ within about $1/3$ of $e^2/mc^2$ will thus begin to affect the
conditions (7.6) and (7.7) without seriously affecting the
scattering experiments and the mass defect calculations of
light nuclei. The conditions (7.3) for infinite $N$ and $Z$ are
changed to their opposites for such a reversal.
Although it appears from the above that there is nothing
against considering the main part of the interaction of nuclear
particles as being the same in form for all particles it is not
likely that the interaction law is identical in all
approximations. Spin-orbit interactions and spin-spin
interactions involve the magnetic moments which are
different for protons and neutrons. In such, essentially
relativistic approximations it appears necessary to consider
the interactions to be different between different kinds of
nuclear particles. It is also unlikely that the Coulombian
interaction which already destroys the symmetry has no more
intimate connection with the specific forces than that given
by an additive term in the Hamiltonian.
For the sake of simplicity the same form of $J(r)$ was used
in Eq. (1) for $P^M, P^H, 1, P^S$. It is obviously possible to use
different space functions as multipliers of these operators.
There appears to be at present no call for such a
generalization.
We are very grateful to Professor E. Wigner for
discussions of nuclear stability questions related to the
conditions (7).
\end{document}
%ENCODED DECEMBER 2002 BY NIS;