Current State of the Theory of Intranuclear Forces

I. Golovin, Moscow

This review considers works by various authors, carried out over the last four years and aiming to clarify the character of the forces of interaction of the particles that make up the atomic nucleus, and the general features of a theory that would make it possible to describe the nucleus as a quantum-mechanical system.

§ 1. Constituent Elements of the Nucleus

Before the discovery of the neutron in 1932, theory was powerless to explain even the most striking regularities in the structure of nuclei throughout the periodic system. To explain the magnitude of the mass \(A\) and charge \(Z\) of the nucleus, it was necessary to regard them as being built from \(A\) protons and \(A - Z\) electrons. But by including electrons as constituent parts of the nucleus, we encounter a number of difficulties.

1. If the classical radius of the electron is adopted, then the volume of the nucleus is insufficient to contain all the necessary electrons.

2. For electrons confined in so small a region of space as the volume of the nucleus, and therefore possessing, according to the uncertainty relation, a very large kinetic energy (\(\sim 4 \cdot 10^{10}\ \mathrm{eV}\)), only a—still not yet created—relativistic quantum mechanics would be applicable.

3. Because of the small dimensions of the nucleus, the accelerations of the electrons would be so considerable that the concept of their electromagnetic mass would disappear.

4. Since the spin of the electron and of the proton is equal to \(1/2\), the spin of nuclei would have to be integral or half-integral and obey Bose or Fermi statistics depending on whether it consists, respectively, of an even or an odd number of particles.

Meanwhile, the nitrogen nucleus, which according to this conception contains 14 protons and 7 electrons, has spin 1 and obeys Bose statistics. The same contradiction is obtained also for the nucleus \(\mathrm{H}_1^2\).

1. The magnetic moments of all nuclei are, in order of magnitude, 1000 times smaller than the magnetic moment of the free electron.

2. Regarding the nucleus as being built from protons, electrons, and \(\alpha\)-particles, from the curve of mass defects we obtain that all elements, beginning with Sn,

must be radioactive, whereas radioactivity begins 32 elements later.

Thus an electron can be placed in the nucleus only by depriving it of the basic properties that it possesses in the free state. But this is equivalent to the assertion that there are no electrons in the nucleus.

When the neutron is introduced as a structural unit of the nucleus, all the above-mentioned difficulties disappear. For the questions of nuclear theory considered below it is immaterial whether the neutron is a complex or a simple particle, provided only that the binding energy of the neutron from its components is considerably greater than the energies with which one has to deal in the nucleus. Wave mechanics does not allow, in the presence only of Coulomb attraction, the formation from a proton and an electron of a system denser than the hydrogen atom. To construct a neutron from a proton and an electron one would have to assume that, at distances \(\sim 10^{-13}\,\text{cm}\) between the proton and the electron, there act, besides the Coulomb forces, still other, more considerable attractive forces. At present no objections are seen to the now accepted point of view, according to which the proton and the neutron are equally simple particles, capable, as a result of \(\beta\)- and, respectively, positron decay, of transforming one into the other. Thus, despite the phenomena of \(\beta^{+}\)- and \(\beta^{-}\)-decay, there are no positrons and electrons in nuclei, just as there are no light quanta in excited atoms.

To explain the magnitudes of the magnetic and mechanical moments of nuclei one has to ascribe to the neutron a spin equal to \(\frac{1}{2}\), and a magnetic moment equal to one third of the proton moment and directed in the opposite direction with respect to the mechanical moment. This fact, that the spin of the neutron is equal to \(\frac{1}{2}\), is the weightiest argument in favor of the point of view that regards the neutron as a simple particle. Indeed, the spin of the proton, like that of the electron, is equal to \(\frac{1}{2}\). The orbital moment of the electron can only be an integer. Therefore, for a neutron constructed from a proton and an electron, the total moment should likewise have come out an integer, and not \(\frac{1}{2}\).

§ 2. The Character and First Problems of Nuclear Theory

The first approximation to nuclear theory may be nonrelativistic. Indeed, according to experiments on neutron scattering, the diameter of light nuclei is of the order of \(5\cdot 10^{-13}\,\text{cm}\) and increases with increasing nuclear mass.1 Therefore, according to the uncertainty relation, the mean velocity of the particles in the nucleus is \(\sim 6\cdot 10^{9}\,\text{cm/sec}\), which gives a relativistic correction to the mass of \(\sim 2\%\). The energy significance of spins in this approximation is sufficiently accounted for from the point of view of the Pauli principle. Magnetic interactions cannot appreciably affect the energy of the nucleus. The most exact formulation of the nonrelativistic problem will therefore be the compilation of the Schrödinger equation for the nucleus. For

nuclei containing a large number of particles, it will be necessary to apply a statistical method that most accurately reproduces this rigorous wave problem.

This theory must first of all explain the following most striking regularities:

1. The number of neutrons \(n_1\) in a series of stable nuclei is equal to the number of protons \(n_2\) from the beginning of the periodic system up to \(\mathrm{Ca}_{20}^{40}\), and thereafter the ratio \(\frac{n_1}{n_2}\) continuously increases.

2. The radius of nuclei increases approximately as \(\sqrt[3]{n}\), where \(n = n_1 + n_2\), i.e. the density of nuclei remains approximately constant.

3. Nuclei with odd \(n_1\) and \(n_2\) simultaneously are almost absent.

4. The binding energy of nuclei increases almost linearly with their mass; moreover, the mass-defect curve has, at the beginning of the periodic system, kinks at all elements with mass \(4n\), and, beginning with \(\mathrm{Ne}_{10}^{20}\), at every element with even mass (Fig. 1).

Fig. 1. Mass defects of nuclei as a function of the number of particles

§ 3. Construction of the Hamiltonian for the nucleus

The Schrödinger equation for the nucleus was proposed by Heisenberg2. Initially he proceeded from the analogy arising between the bond in the ion \(\mathrm{H}_2^+\) and the bond of a proton–neutron pair, if the neutron is regarded as a system built of a tightly bound proton and electron. Then one must assume that the binding forces of the proton with the neutron have an exchange character. If it is assumed that exchange by an electron takes place, then all particles in the nucleus will be equivalent and can only be in the neutron or proton state. For the further conclusions on the range of questions outlined above it is not essential what the mechanism of interaction of the particles is. It is most correct not to try to reduce these exchange forces to electrical interactions, but to regard them as a new type of force, specific to the interaction of the neutron, as an uncharged elementary particle, with the proton.

Let the potential function \(J_0(r)\) determine the dependence of the magnitude of the exchange forces on the distance between particles, and let the function \(K(r)\) determine the law of the interaction forces of two neutrons. We shall denote by \(D\) the mass defect of the neutron, composed of a proton and an electron. For the time being, only two assumptions can be made about the form of the functions \(J_0\) and \(K\): 1) by analogy with the molecule \(\mathrm{H}_2\), let us take it that for nuclear values of \(r\) the function \(K(r) < J_0(r)\); 2) experiments on the scattering of \(\alpha\)-particles show that these forces are no longer noticeable at \(r = 10^{-12}\ \mathrm{cm}\), whereas the strength of nuclei compels one to suppose that \(J_0(r)\) at intranuclear distances is measured in millions of volts. Therefore \(J_0(r)\) has the form of a steep potential well with a mean width less than \(10^{-12}\ \mathrm{cm}\). In view of the fact that the classical radius of protons is considerably smaller than intranuclear distances, it is admissible that Coulomb repulsive forces act between them.

However, at present this picture of the nucleus due to Heisenberg, which he created even before the discovery of the positron, cannot be considered entirely satisfactory. Indeed, the transformation of a proton into a neutron with emission of a positron is just as probable as the reverse process, the transformation of a neutron into a proton with emission of an electron. Therefore, if there are interaction forces between two neutrons determined by the potential function \(K(r)\), then exactly the same forces, due to “orbital” positrons, must also exist between any two protons. The only asymmetry in the properties of protons and neutrons is created by the electric charges of the protons.

In setting up the Schrödinger equation, Heisenberg characterizes each particle by five quantities: three spatial coordinates \((x, y, z) = r\), the spin \(\sigma_k^z\) in the direction of the \(z\)-axis, and the number \(\beta_{\mu k}\), which can take only two values \(+1\) and \(-1\). For \(\beta_{1,k}=+1\) the \(k\)-th particle described is in the state of a neutron; for \(\beta_{-1,k}=-1\) this particle is in the state of a proton. To write the Hamiltonian the matrices may be used:

\[ \beta_k^\xi = \begin{vmatrix} 0 & 1\\ 1 & 0 \end{vmatrix}; \quad \rho_k^\eta = \begin{vmatrix} 0 & -i\\ i & 0 \end{vmatrix}; \quad \rho_k^\zeta = \begin{vmatrix} 1 & 0\\ 0 & -1 \end{vmatrix} \tag{3, 1} \]

The index \(k\) on the operator means that it acts on the \(\beta\)-coordinate only of the \(k\)-th particle. The eigenfunction of a nucleus consisting of \(n\) particles can be written in the form

\[ \psi = \psi(x_1,\ldots x_n, \beta_{\mu,1}, \beta_{\nu,2}\ldots \beta_{\tau,n}), \]

where \(\mu,\nu,\ldots \tau\) are equal to \(+1\) or \(-1\). Here \(x_k\) denotes all three spatial coordinates \(x_k, y_k, z_k\) and the spin \(\sigma_k^z\) of the \(k\)-th particle. For the use of the operators (3.1) we establish the following rule. The action of the \(k\)-th operator \(\rho_k^\xi, \rho_k^\eta\), or \(\rho_k^\zeta\) on the function \(\psi\) will be represented as follows:

take it to be a one-column matrix \(\binom{\psi}{0}\), if the \(k\)-th \(\beta\)-coordinate is \(\beta_{+1,k}\), and in the form \(\binom{0}{\psi}\), if the \(\beta\)-coordinate is \(\beta_{-1,k}\). As a result of the action of the operator, the matrix \(\binom{\psi}{0}\) will pass into a matrix of the form \(\binom{0}{\psi}\) (or conversely), or will remain unchanged. In the case of a change in the form of the matrix we change the sign \(\beta_{h,k}\) to the opposite one.

Using this rule, we obtain the identities:

\[ \rho_k^\xi f(x_1\ldots x_n;\ldots \beta_{1,k}\ldots) = f(x_1\ldots x_n;\ldots \beta_{1,k}), \]

\[ \rho_k^\xi f(x_1\ldots x_n;\ldots \beta_{-1,k}\ldots) = - f(x_1\ldots x_n;\ldots \beta_{-1,k}\ldots) \]

and, correspondingly, denoting for brevity only the \(k\)-th \(\beta\)-coordinate:

\[ \rho_k^\xi f(\beta_{1,k}) = f(\beta_{-1,k}); \qquad \rho_k^\xi f(\beta_{-1,k}) = f(\beta_{1,k}), \]

\[ \rho_k^\eta f(\beta_{1,k}) = i f(\beta_{-1,k}); \qquad \rho_k^\eta f(\beta_{-1,k}) = - i f(\beta_{1,k}). \]

One may even dispense with the matrix notation and regard these as identities defining the properties of the operators (3,1).

Let us now turn to the construction of the Hamiltonian. The kinetic term has the usual form

\[ \sum_{k=1}^{n}\frac{\hbar^2}{2m_k}\nabla_k^2. \]

The term giving the exchange energy must exchange the coordinates of corresponding pairs of particles if they are in states with different names (one in the proton state, the other in the neutron state) and must give zero for any pair of particles in the state with the same name. The operator

\[ A=\frac{1}{2}\left(\rho_k^\xi\rho_l^\xi+\rho_k^\eta\rho_l^\eta\right) \]

satisfies the stated conditions.

Indeed,

\[ Af(\beta_{1,k},\beta_{-1,l}) = f(\beta_{-1,k},\beta_{1,l}); \]

\[ Af(\beta_{1,k},\beta_{1,l}) = 0 \quad \text{etc.} \]

The term giving the attraction of neutrons must turn into zero if the particles are in different states, or both particles are in the proton state. Therefore it must be written in the form:

\[ -\frac{1}{4}\sum_{k>l} K(r_{kl})(1+\rho_k^\zeta)(1+\rho_l^\zeta). \]

The remaining operators are constructed similarly. The entire Hamiltonian takes the form:

\[ \begin{aligned} H=&-\frac{\hbar^2}{2}\sum_{k=1}^{n}\frac{1}{m_k}\nabla_k^2 -\frac{1}{2}\sum_{k>l}J_0(r_{k,l})(\rho_k^\xi\rho_l^\xi+\rho_k^\eta\rho_l^\eta)-\\ &-\frac{1}{4}\sum_{k>l}K(r_{kl})(1+\rho_k^\xi)(1+\rho_l^\xi)+\\ &+\frac{1}{4}\sum_{k>l}\frac{e^2}{r_{kl}}(1-\rho_k^\xi)(1-\rho_l^\xi) +\frac{1}{2}D\sum_{k=1}^{n}(1+\rho_k^\xi) \end{aligned} \tag{3, 2} \]

§ 4. Physical consequences

Let us see what consequences are obtained by a purely mathematical route from expression (3,2). Comparison of them with experiment will show how close to reality the hypotheses used in constructing the operator are.

First of all, let us verify whether the numbers of protons \(n_2\) and neutrons \(n_1\) are integrals of motion2. The operator of the number of neutrons is \(\frac{1}{2}\sum_k(1+\rho_k^\xi)\), that of the number of protons: \(\frac{1}{2}\sum_k(1-\rho_k^\xi)\).

From the relation

\[ (\rho_k^\xi+\rho_l^\xi)(\rho_k^\xi\rho_l^\xi+\rho_k^\eta\rho_l^\eta) = (\rho_k^\xi\rho_l^\xi+\rho_k^\eta\rho_l^\eta)(\rho_k^\xi+\rho_l^\xi)=0 \]

it follows that these operators commute with \(H\). Thus, in agreement with experiment, \(n_1\) and \(n_2\) do not change in the nucleus with time, provided that its energy remains constant.

If the last three terms in the Hamiltonian are neglected, it will be symmetric in protons and neutrons. Such a neglect is legitimate for light nuclei, since for them one may assume that \(J_0(r)\) at intranuclear distances is considerably greater not only than \(K(r)\), but also than \(\frac{e^2}{r}\). The symmetry of the Hamiltonian means that it does not depend on the sign of \(n_1-n_2=\sum \rho_k^\xi\), and therefore the binding energy of the nucleus, determined by the equation \(H\psi=E\psi\), reaches an extremum at \(n_1=n_2\). In the approximation under consideration, when \(n_2=0\), the energy \(E=0\). Therefore, as is easy to see, at \(n_1=n_2\) the binding energy (by binding energy we shall mean the positive work necessary to separate the nucleus into individual protons and neutrons) reaches a maximum. This conclusion is in full agreement with the experimental fact asserting that, in stable light nuclei, the number of protons is close to the number of neutrons.

If \(J_0(r)\) is regarded as decreasing much more rapidly with distance than \(\frac{1}{r}\), then, as the number \(n\) of particles in the nucleus increases, the Coulo-

new term \(\displaystyle \sum_{k>l}\frac{e^2}{r_{kl}}(1-\rho_k^2)(1-\rho_l^2)\) will acquire an ever greater value. Here the Coulomb forces are repulsive forces. Therefore it will be energetically more advantageous to increase the number of neutrons in comparison with the number of protons. This also agrees excellently with experiment.

It is now possible to give a qualitative picture of radioactive decay\(^{3}\). Let us imagine a nucleus consisting of neutrons alone. In view of the fact that the forces of interaction between oppositely charged particles are considerably greater than the forces of attraction between neutrons, it is energetically advantageous to replace part of the neutrons by protons. Such a nucleus will therefore be \(\beta\)-active. The binding energy upon the appearance of protons will then increase until the gain in energy from the increase in the number of exchange bonds is no longer compensated by Coulomb repulsion. A further increase in the number of protons will already lead to a decrease in the binding energy. Therefore the curve of the dependence of the energy of the nucleus on the number of neutrons at a given nuclear weight has a minimum corresponding to the greatest binding energy. The nuclei marked in Fig. 2 by circles are \(\beta^{+}\)-active; those marked by crosses are \(\beta^{-}\)-active. If near the minimum on the curve, in a region of width \(m_e c^2\), several nuclei fit, then they will be stable isobars, since with an electron or positron a greater energy would be emitted than the gain in energy obtained in the radioactive transition.

Fig. 2.

However, such a smooth curve can be drawn only for nuclei with odd atomic weight. For nuclei with even weight, account of the Pauli principle is essential. In this case one cannot draw a single smooth curve for all isobars. Two smooth curves are obtained: one, lying somewhat higher, for nuclei with an odd number of pro-

tons and, correspondingly, neutrons; the other—below the first, for nuclei with an even number of protons and neutrons. This changes the conditions for the stability of nuclei. Indeed, if the difference between the minima of the two curves is greater than \(m_e c^2\), then there are no stable isobars with an odd number of particles of both kinds at all. At the same time, such a nucleus as, for example, the nucleus \(P\) in Fig. 3 will be stable, since decay would transform it into the nucleus \(P'\), which is energetically unfavorable. The nucleus \(P''\) can emit both positrons and electrons, transforming respectively into the nuclei \(A\) or \(B\). Experiment shows that nuclei with an odd number of protons and neutrons exist only at the beginning of the periodic system, up to \(N^{14}_{7}\). This indicates that the difference between the minima of the two curves in heavy nuclei is greater than \(m_e c^2\).

Fig. 3.

Analogous energy considerations also explain the course of natural radioactive decay. In fact, \(\beta^-\)-decay begins in all cases with a nucleus possessing an even charge, and two successive \(\beta^-\)-decays always occur one after another. Thus the decay also ends on a nucleus with an even charge. This is explained by the fact that if the first decay, leading to an odd number of protons in the nucleus and therefore forcing the new proton to occupy a higher energy level, was energetically advantageous, then the second decay, leading to the possibility of forming a new \(\alpha\)-particle in the nucleus, will also be advantageous. To be sure, after the first \(\beta^-\)-decay a situation may arise in which, in view of the resulting excess of repelling protons, \(\alpha\)-decay will also prove advantageous. Then a branching appears, which occurs in all four radioactive series and begins at RaC, AcC, ThC and \(\mathrm{Th}^{233}_{90}\). The anomaly in the actinium series can be explained in the same way.

§ 5. The Schrödinger Equation for the Deuteron

Let us now write the Schrödinger equation for the simplest nuclei. The simplest system, consisting of two particles, can be in four states

\[ \psi_1(x_1,x_2;\beta_{1,1},\beta_{-1,2});\quad \psi_2(x_1,x_2;\beta_{-1,1},\beta_{1,2}); \]

\[ \psi_3(x_1,x_2;\beta_{1,1},\beta_{1,2});\quad \psi_4(x_1,x_2;\beta_{-1,1},\beta_{-1,2}). \]

We are interested only in the first two states, which describe the deuteron. In the given problem it turns out to be possible to represent these functions in the form of a product:

\[ \varphi_1(x_1,x_2)f(\beta_{1,1},\beta_{-1,2}),\quad \varphi_2(x_1,x_2)f(\beta_{-1,1},\beta_{1,2}). \]

Indeed, according to (3,2), we obtain the Schrödinger equations in the form

\[ \left. \begin{aligned} \Omega_1\varphi_1(x_1,x_2)f(\beta_{11},\beta_{-12}) &-J_0(r_{12})\varphi_1(x_1,x_2)f(\beta_{-11},\beta_{12})=0,\\ \Omega_2\varphi_2(x_1,x_2)f(\beta_{-11},\beta_{12}) &-J_0(r_{12})\varphi_2(x_1,x_2)f(\beta_{11},\beta_{-12})=0, \end{aligned} \right\} \tag{5,1} \]

where, for brevity, we have denoted:

\[ \Omega_i=-\frac{\hbar^2}{2}\sum_k\frac{1}{m_k}\nabla_k^2+D-E_i. \]

Putting \(E_1=E_2=E\), we obtain \(\varphi_1=\varphi_2\), since a unique solution is required. Equations (5,1) show that \(\varphi\) is symmetric in the spatial coordinates of the particles. Moreover, it is known from experiment that the spin of the deuteron is equal to unity. Therefore the deuteron function is also symmetric in the spins. For the wave function of the nucleus to satisfy the Pauli principle, it is necessary that it be antisymmetric in the \(\beta\)-coordinates. Here the antisymmetric function must be a monomial constructed from the quantities \(\beta_{\mu k}\). To obtain a function satisfying this requirement, we subtract the second equation (5,1) from the first. Then we find that \(\varphi\) is determined by the equation

\[ \Omega_i\varphi+J_0\varphi=0. \tag{5,2} \]

At the same time the condition must be fulfilled

\[ f(\beta_{1,1},\beta_{-1,2})=-f(\beta_{-1,1},\beta_{1,2}), \]

which is easily satisfied by a suitable choice of the function \(f\). The complete eigenfunction takes the form

\[ \varphi(x_1,x_2)\,[f(\beta_{11},\beta_{-12})-f(\beta_{-11},\beta_{12})]. \]

It is antisymmetric in the particles. It is essential that the binding is given by equation (5,2) only under the condition that \(J_0\) is a negative quantity. In what follows we shall use only the positive quantity \(J(r)\), equal to \(-J_0(r)\).

§ 6. The Significance of the Shape of the Well

For the part of the deuteron eigenfunction depending only on the relative coordinates, as a result of separating off the coordinates of the center of gravity, we obtain from (3,2) the following equation:

\[ \left[-\frac{\hbar^2}{m}\nabla^2-J(r)\right]\psi(r)=E_D\psi(r). \tag{6,1} \]

For integrating it, it is necessary to know the form of the function \(J(r)\). From experiment one can only estimate the mean depth and width of the potential well. Therefore the function \(J(r)\) has to be chosen arbitrarily. It must, however, depend on two parameters \(a\) and \(\rho\), characterizing the depth and width of the well.

If we integrate equation (6,1) and set \(E_D\) equal to its experimental value \((2.1\cdot 10^6 \mathrm{eV})\), then we obtain an equation relating the parameters \(a\) and \(\rho\).

In Fig. 4 are given the results of exact integration (analytical, or, when such integration is impossible, numerical) for the functions

\[ \mathrm{I}\qquad J(r)= \frac{4a}{\left(1+e^{-r/\rho}\right)\left(1+e^{r/\rho'}\right)}, \]

\[ \mathrm{II}\qquad J(r)=a'e^{-r/\rho'}, \]

\[ \mathrm{III}\qquad J(r)=a'' \quad \text{for } r<\rho''; \]

\[ J(r)=0 \quad \text{for } r>\rho'', \]

\[ \mathrm{IV}\qquad J(r)=a_0 e^{-r^2/\rho_0^2}. \]

Fig. 4.
Dependence of the parameters \(a\) on \(\rho^{-1}\)

It is easy to see that the parameters \(a\) and \(\rho\) have here different meanings. Thus, \(a''\) is the mean (here constant) depth of the well

Fig. 5a.

\[ a''=a=a'=a_0 \]

\[ \rho''=\rho=\rho'=\rho_0 \]

Fig. 5b.

\[ a'=0.9a=0.63a''=a_0 \]

\[ \rho''=2.2\rho=1.4\rho'=0.9\rho_0 \]

But \(a,\ a'\), and \(a_0\) give the maximum depths of the wells. Obviously, the mean depths have a different value. If the scales are changed by taking

\[ \left. \begin{aligned} a''&=0.9a=0.63a'=a_0,\\ \rho''&=2.2\rho=1.4\rho'=0.9\rho_0 \end{aligned} \right\}, \tag{6, 2} \]

then all the curves of Fig. 4 coincide with the curve for a rectangular well. In Fig. 5 the relative dimensions of the wells corresponding to these scales are shown. This figure does not contradict taking \(a''\), \(\rho''\), determined by the equalities (6,2), as the mean depth and width of the wells. If the equalities (6,2) are satisfied, then the deuteron energy does not depend on the shape of the well, but only on its mean depth and width.

Since up to now there is no exact solution even for the triton, it is unknown how strongly the exact shape of the well affects the binding energy of nuclei. However, the calculations carried out show that the shape of the well is of secondary importance; the binding energy of a nucleus depends essentially only on its mean depth and width.

§ 7. Wigner’s Calculation\(^4\)

For the nuclei \(\mathrm{H}_1^3\) and \(\mathrm{He}_2^3\) one already obtains a system of three simultaneous equations, the exact solution of which is impossible. Therefore one has to choose an approximate method, as close as possible to the solution of the exact system of equations proposed by Heisenberg.

Wigner makes the following simplification. He solves the problem of the nucleus \(\mathrm{He}_2^4\), assuming that the forces of interaction of the proton with the neutron are not exchange forces, but potential forces. He calculated the energy of the nucleus \(\mathrm{He}_2^4\) by means of the variational principle:

\[ \delta E_{\mathrm{He}}=\delta\frac{\int \psi H\psi\,dv}{\int \psi^2\,dv}=0, \]

where in the Hamiltonian the following expression was taken as the potential function:

\[ J(r)=\frac{4.56a}{\left(1+e^{\frac{r}{\rho}}\right)\left(1+e^{-\frac{r}{\rho}}\right)}. \]

For a well width \(2\rho=1.2\cdot10^{-13}\ \mathrm{cm}\) and a depth \(63\cdot10^6\ \mathrm{eV}\), taken in accordance with curve 1 of Fig. 4, i.e. when using the exact solution for the deuteron, he obtained the binding energy of \(\mathrm{He}_2^4\) to be 14 times greater than the binding energy of \(\mathrm{H}_1^2\). This well explains the experimental fact of the enormous difference in the binding energy of \(\mathrm{He}_2^4\) and \(\mathrm{H}_1^2\) \((27\ \text{and}\ 2\cdot10^6\ \mathrm{eV})\);

which had previously remained unclear. In this calculation only the interaction of protons with neutrons was taken into account. Allowance for Coulomb repulsion would reduce the binding energy by 2–3 million volts. This calculation, carried out by the Ritz method in the first approximation, must at present nevertheless be regarded as the most accurate, since the value of the well width and the binding energy agree comparatively well with the experimental data.

§ 8. The inapplicability of Wigner forces for explaining the structure of heavy nuclei

Let us try to calculate the dependence of the binding energy on the mass of the nucleus. In making an approximate calculation, we shall average the energy over all values of the \(\rho\)-coordinates. Having performed this averaging, Heisenberg obtained, according to (3,2), the interaction energy of two particles equal to

\[ U(r)=-2\,\frac{n_1 n_2}{n(n-1)}\,J(r) -\frac{n_1(n_1-1)}{n(n-1)}\,K(r) +\frac{n_2(n_2-1)}{n(n-1)}\,\frac{e^2}{r}, \]

where the nucleus must be regarded as constructed from particles of one kind and obeying Fermi statistics. Introducing the particle density \(\rho(r)\) and using the Thomas–Fermi method, we obtain:

\[ E=\frac{h^2}{M}\frac{4\pi}{5} \left(\frac{3}{8\pi}\right)^{\frac{5}{3}} \int \rho(r)^{\frac{5}{3}}\,d\tau+ \]

\[ +\frac{1}{2}\int\!\!\int \rho(r)\rho(r')\,U(|r-r'|)\,d\tau\,d\tau' -n_1D. \]

We find the density \(\rho(r)\) from the condition of a minimum of \(E\), assuming that \(\int \rho(r)d\tau=n\). The above averaging already excludes the exchange character of the forces and reduces them to the ordinary potential interaction \(U(r)\). In order to take into account the incompressibility of nuclei, i.e. the experimentally known proportionality between their volume and mass, it is necessary to introduce a finite radius of the particles and the maximum nuclear density \(\rho_0\) which would be obtained if all the particles of the nucleus were placed tightly against one another. As a result of the transformations we obtain:

\[ E-\frac{n}{2}\frac{dE}{dn}= \]

\[ =\frac{h^2}{M}\frac{2\pi}{15} \left(\frac{3}{8\pi}\right)^{\frac{5}{3}} \left[ \int\left(\frac{1}{\rho}-\frac{1}{\rho_0}\right)^{-\frac{2}{3}}\rho\,d\tau -2\int\frac{\rho}{\rho_0} \left(\frac{1}{\rho}-\frac{1}{\rho_0}\right)^{-\frac{5}{3}}d\tau \right]. \tag{8,1} \]

For light nuclei, for which one may assume \(\frac{1}{\rho}\ll\frac{1}{\rho_0}\), the energy \(E\)

increases more rapidly than \(\mathrm{const}\cdot n^{2}\), whereas experiment gives a linear increase of the energy with mass. It is true that by choosing the constants \(a\), \(b\), and \(A\) one can make the curve given by (8,1)

\[ E=a(n-b)^2+A \]

coincide with portions of the mass-defect curve (Fig. 1), but this is achieved only by an artificial introduction of the constants \(a\), \(b\), and \(A\). Thus potential forces, even with the additional introduction of a finite particle radius, do not give the basic regularity—the proportionality between the energy and the mass of nuclei.

§ 9. Majorana Forces

Majorana showed that exchange forces make it possible to explain the proportionality between the mass, volume, and energy of nuclei without introducing any additional hypotheses, such as, for example, a finite particle radius.^6

For what follows it is convenient to pass to another notation for the exchange forces, observing that charge exchange is completely equivalent to the simultaneous exchange of the spatial coordinates and spin of unlike particles (a proton and a neutron). Instead of \(\zeta\)-coordinates we shall introduce the division into protons and neutrons. With such a description of the nucleus, instead of the operator

\[ \frac{1}{2}J(r_{kl})\bigl([[unclear: expression involving }\zeta_k,\zeta_l\text{]]}\bigr) \]

one may use the operator

\[ -J(r_{kl})P'_{kl}, \tag{9, 1} \]

where \(P'_{kl}\) is the operator that simultaneously permutes the spin and spatial coordinates of the \(k\)-th neutron and the \(l\)-th proton.

If it is assumed that the binding of nuclei is due to Heisenberg forces, then the \(\alpha\)-particle turns out to be an unstable system, decaying into two deuterons, whereas experiment indicates the exceptional stability precisely of the \(\alpha\)-particle. Indeed, in the lowest energy state, in which the spatial part of the eigenfunction of \(\mathrm{He}_2^4\) is symmetric with respect to the interchange of any two particles, the eigenfunction must be antisymmetric in the spins of both protons and, correspondingly, of both neutrons. The interaction energy of all four particles, according to (9,1), is equal to

\[ \int \psi^*\bigl[J(r_{11})P'_{11}\psi+J(r_{12})P'_{12}\psi+J(r_{12})P'_{21}\psi+ \]

\[ +J(r_{22})P'_{22}\psi\bigr]\,d\omega . \]

If one adopts the orientation of the spins shown in the drawing (Fig. 6), then the first and fourth terms are equal to zero. Indeed, the operator \(P'_{11}\) changes the spin of the 1st neutron and, correspondingly, of the 1st proton to the opposite ones. But, for the parallel arrangement of the spins of like particles, \(\psi\) is equal to zero, in accordance with what was said above about its symmetry properties. Therefore only those protons and neutrons that have identically oriented spins will be attracted. Thus in the \(\alpha\)-particle there will be only two bonds and, if one neglects the hypothetical forces of attraction between like particles mentioned up to now, these will be two deuterons, repelling one another by the electric charges of the protons entering into them.

Majorana introduced another type of exchange forces, which, in the old description of the nucleus, give an exchange term of the form

\[ \frac{1}{4}\,J(r_{kl})(p_k^\xi p_l^\xi + p_k^\eta p_l^\eta)[1+(\sigma_k\sigma_l)], \]

and in the new one

\[ -J(r_{kl})P_{kl}, \tag{9,2} \]

where the operator \(P_{kl}\) exchanges only the spatial coordinates of the \(k\)-th neutron with the coordinates of the \(l\)-th proton.

If these Majorana forces act in nuclei, then in the \(\alpha\)-particle we already obtain four exchange bonds, which makes it possible to explain its exceptional stability.

Fig. 6.

§ 10. Approximate Formulation of the Problem

The exact formulation of the problem for Majorana forces leads to a system of equations analogous to Heisenberg’s system. For simplification we shall assume (in the zeroth approximation) that the interaction forces of a given particle with all the others can be replaced by some average conservative field, in such a way that the energy of interaction of each particle with the rest of the nucleus may be regarded as a function of the coordinates of this particle alone. Then we shall represent the total wave function of the nucleus in the form of a linear combination of products of the proper functions of the individual particles. The most convenient form of such a combination is ...

\[ \Psi=\frac{1}{\sqrt{n_1!n_2!}} \left| \begin{array}{ccc} \psi_1(x_1\sigma_1) & \ldots & \psi_{n_1}(x_1\sigma_1)\\ \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot\\ \psi_1(x_{n_1}\sigma_{n_1}) & \ldots & \psi_{n_1}(x_{n_1}\sigma_{n_1}) \end{array} \right| \times \left| \begin{array}{ccc} \psi'_1(x'_1\sigma'_1) & \ldots & \psi'_{n_2}(x'_1\sigma'_1)\\ \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot\\ \psi'_1(x'_{n_2}\sigma'_{n_2}) & \ldots & \psi'_{n_2}(x'_{n_2}\sigma'_{n_2}) \end{array} \right| \tag{10, 1} \]

where \(\psi_i\) describe neutron states, and \(\psi'_i\) proton states. We shall regard the functions \(\psi_i\), and correspondingly \(\psi'_i\), as mutually orthogonal. Analogously to the Hartree–Fock problem, the form of the functions \(\psi_i\) and \(\psi'_i\) can be found from the variational principle:

\[ \delta \int \Psi^* \left\{ \sum \frac{\hbar^2}{2m_k}\nabla_k^2 -\sum J(r)P_{kl} +\sum \frac{e^2}{r_{kl}} -E \right\} \Psi\,dv=0. \]

Here only the exchange forces between unlike particles and the Coulomb repulsion of protons are taken into account.

However, for purposes of a first orientation, solving the resulting system of equations is too cumbersome. We shall therefore turn to approximate methods.

§ 11. First Approximation to the Formulation Problem (Thomas–Fermi Method)\(^{6,7}\)

As Dirac\(^{8}\) showed, for systems composed of a very large number of particles obeying Fermi statistics, when the quantity \(\hbar^3\) may be neglected in comparison with the phase region occupied by the system, the Hartree–Fock solution becomes equivalent to the Thomas–Fermi method. Therefore, for nuclei consisting of a sufficiently large number of particles, one may hope to obtain agreement with experiment by using this method.

The total energy of the nucleus is

\[ E=E_{\mathrm{kin}}+E_{\mathrm{pot}}+E_{\mathrm{coul}}, \]

where

\[ E_{\mathrm{kin}} = \frac{\hbar^2}{M}\frac{4\pi}{5} \left(\frac{3}{8\pi}\right)^{\frac{5}{3}} \int dr\left\{[\rho_N(r)]^{\frac{5}{3}}+[\rho_P(r)]^{\frac{5}{3}}\right\}, \tag{11, 1} \]

\[ E_{\mathrm{pot}} = -\int \Psi^*\sum_{k,l}J(r_{kl})P_{kl}\Psi\,d\omega = \]

\[ = -\int \rho_N(r,r')J\left(|r-r'|\right)\rho_P^*(r,r')\,dr\,dr' \tag{11, 2} \]

Here \(\rho_N(r)\) and \(\rho_P(r)\) are the Thomas–Fermi number densities of protons and neutrons, while \(\rho_N(r,r')\) and \(\rho_P(r,r')\) are the corresponding mixed Dirac densities; \(M\) is the mass of the proton or neutron. In Thomas–Fermi:

\[ \rho_N(r)=\frac{2}{h^3}\int_0^{P_N(r)} dp;\qquad \rho_P(r)=\frac{2}{h^3}\int_0^{P_P(r)} dp, \tag{11,3} \]

where \(P_N(r)\) and \(P_P(r)\) are the maximum momenta of neutrons and protons at the point \(r\).

The element of the density matrix may be written in the form:

\[ \rho(r,r')=\frac{1}{h^3}\int\!\!\int (p|\rho|p') e^{-\frac{ip'r'}{\hbar}} e^{\frac{ipr}{\hbar}}\,dp\,dp'. \tag{11,4} \]

In the case where Thomas–Fermi statistics is applicable, this formula can be simplified. Indeed,

\[ -i\hbar \frac{\partial \rho}{\partial r}=p\rho-\rho p; \]

regarding \(\rho\) as varying slowly with \(r\) and neglecting the quantity \(\hbar\), we obtain

\[ p\rho-\rho p=0. \]

Consequently, in our approximation \((p|\rho|p')=\rho(p)\delta(p-p')\). According to (11,4) we obtain

\[ \rho(r,r')=\frac{1}{h^3}\int \rho(p)e^{\frac{ip(r-r')}{\hbar}}\,dp. \]

Since the eigenvalues \(\rho(r,r')\) are equal to 0 or 2, we have

\[ \rho(r,r')=\frac{2}{h^3}\int_0^{P\left(\frac{r+r'}{2}\right)} e^{\frac{ip(r-r')}{\hbar}}\,dp, \]

where the momentum \(P\) is related to the density \(\rho\) by formulas (11,3).

Using these relations, we obtain an expression for the exchange energy in terms of the ordinary density \(\rho(r)\):

\[ E_{\text{pot}}= -\int\!\!\int dr\,dr'\,\frac{4}{h^6} \int_0^{P_N} dp\int_0^{P_P} dp'\, e^{\frac{i}{\hbar}(p-p')(r-r')}J(|r'-r|). \tag{11,5} \]

If the function \(J\) is known, then this formula makes it possible to cal-

to regard \(E_{\mathrm{pot}}\) in the approximation corresponding to the applicability of the Thomas–Fermi method.

From formula (11,2), for the limiting case of a very large number of particles, under the condition that \(n_1 \gg n_2\), we obtain:

\[ E_{\mathrm{pot}}=-2n_2J(0), \]

where it is assumed that \(n_1\) is so large that \(\rho_N(r,r')\), as one moves away from the point \(r-r'=0\), falls to zero faster than \(J(r-r')\).

Thus, the use of the function in the form (10,1) leads, in the case of a large number of particles, to proportionality between the exchange energy and the number of protons (provided that \(n_1>n_2\)). This indicates saturation of the bonds. A calculation carried out for Heisenberg forces gives an exchange energy smaller by a factor of two.

According to formula (11,5), the potential energy may be written in the form:

\[ E_{\mathrm{pot}}=-\int dr\, f[\rho_N(r),\rho_P(r)], \tag{11,6} \]

where \(f\) is a function symmetric in \(\rho_N\) and \(\rho_P\), vanishing when \(\rho_N=0\) or \(\rho_P=0\), and, for large numbers of particles \(n_1\) or \(n_2\), tending to \(2\rho_NJ(0)\) if \(n_1<n_2\), and to \(2\rho_PJ(0)\) for \(n_1>n_2\). Without specifying the form of this function, one can already draw a number of conclusions about the structure of the nucleus.

If the Coulomb energy is neglected, then the total energy of the nucleus is

\[ E=\int\left[\frac{h^2}{M}\frac{4\pi}{3}\left(\frac{3}{8\pi}\right)^{\frac{5}{3}} \left(\rho_N^{\frac{5}{3}}+\rho_P^{\frac{5}{3}}\right) -f(\rho_N,\rho_P)\right]dr. \tag{11,7} \]

The densities \(\rho_N\) and \(\rho_P\), as functions of the coordinates of the nucleus, are found from the condition of the minimum of \(E\) with \(\int \rho_Ndr=n_1\) and \(\int \rho_Pdr=n_2\). This gives

\[ \rho_N(r)=\mathrm{const}=\frac{n_1}{V}; \qquad \rho_P(r)=\mathrm{const}=\frac{n_2}{V}, \]

if \(V\) is the volume of the nucleus. This means that the density inside the nucleus is constant. In order to obtain a finite mass of the nucleus, the density must be cut off discontinuously at its surface.

Thus

\[ E=\frac{h^2}{M}\frac{4\pi}{5}\left(\frac{3}{8\pi}\right)^{\frac{5}{3}} \left(n_1^{\frac{5}{3}}+n_2^{\frac{5}{3}}\right)V^{-\frac{2}{3}} - Vf\left(\frac{n_1}{V},\frac{n_2}{V}\right). \]

From the condition that the energy \(E\), as a function of the volume \(V\) and of the ratio \(\dfrac{n_1}{n_2}\) for a given \(n\), be minimal, we obtain that \(\dfrac{n_1}{n_2}\) is constant for nuclei of any mass. This would also occur in nature in the absence of the Coulomb repulsion of the protons, which we have not taken into account here. Moreover, it follows from this that the volume of nuclei \(V\) is proportional to the number of particles \(n\). The latter is in good agreement with experiment. Consequently, according to (11,7), the binding energy of nuclei increases in proportion to their volume, and therefore also to their mass. We were unable to explain this experimental fact by means of Wigner forces. Thus the theory of exchange forces qualitatively reproduces the basic regularities in the structure of nuclei.

§ 12. Quantitative Results of the Solution by the Thomas–Fermi Method

If one prescribes a definite form of the law \(J(r)\), then one can calculate the form of the function \(f\) and draw quantitative conclusions. Taking

\[ J=a'e^{-\frac{r}{\rho'}}, \]

Heisenberg calculated \(f\). Then, using formula (11,7) and the condition of minimum for \(E\), he chose the constants \(a'\) and \(\rho'\) in such a way that the correct binding energy would be obtained for the nuclei \(A_{18}^{40}\) and \(Se_{34}^{80}\). He found the following values

\[ a'=25\cdot 10^6 \mathrm{eV};\qquad \rho'=8\cdot 10^{-13}\ \mathrm{cm}, \]

which is equivalent to a rectangular well with the parameters

\[ a''=17\cdot 10^6 \mathrm{eV};\qquad \rho''=10\cdot 10^{-13}\ \mathrm{cm}. \]

These numbers agree poorly with Wigner’s data and with the results of later investigations. In Heisenberg’s calculation the binding energy is expressed by the formula:

\[ \frac{E}{Mc^2} = 0.00347 n_2 - 0.0364 n_1 + 0.01211\frac{n_1^2}{n_2} + \]

\[ + n_2^{\frac{5}{3}} \left(3.19-0.715\frac{n_1}{n_2}\right)10^{-4} + 0.049. \tag{12,1} \]

The curve of the dependence of the energy \(E\) on the mass of the nucleus agrees satisfactorily with experiment only for heavy nuclei, since the Thomas–Fermi method is not applicable to light nuclei.

Formula (12,1) makes it possible to give a quantitative justification of the arguments of § 4. Indeed, the condition for \(\alpha\)-decay is given by the formula:

\[ -2\left(\frac{\partial E}{\partial n_1}+\frac{\partial E}{\partial n_2}\right)+\Delta<0, \]

where \(\Delta\) is the mass defect of the \(\alpha\)-particle. The conditions for \(\beta\)- and positron decay are:

\[ -\frac{\partial E}{\partial n_1}+\frac{\partial E}{\partial n_2}+m_e c^2<0 \]

and

\[ -\frac{\partial E}{\partial n_2}+\frac{\partial E}{\partial n_1}+m_e c^2<0. \]

Unfortunately, the decay boundaries given by these formulas agree very poorly with the experimental data.

Thus, in view of the qualitative success, the further task is the quantitative refinement of the theory of exchange forces.

§ 13. Allowance for the surface effect. Heavy nuclei

The most striking inconsistency in the theory set forth above is the artificial discontinuity of the density at the surface of the nucleus. In reality such a jump of the density to zero would be possible if the nucleus were in an infinitely deep rectangular potential well, for the occurrence of which, however, there are no reasons. This conclusion is not changed for any form of \(J(r)\) taken with the sign corresponding to attraction. The necessity of cutting off the density appeared only as a consequence of the approximate nature of the method of calculation. Namely: the Thomas–Fermi method would be quite rigorous only for infinitely large nuclei. It assumes densities varying so slowly that over an elementary volume \(v_0\) (small in comparison with the whole volume occupied by the system, but containing a sufficiently large number of particles in the classical picture) the eigenfunction of the system may be represented in the form of a monochromatic wave

\[ \psi=\frac{1}{\sqrt{v_0}}e^{i(pr)/\hbar}. \]

Hence, according to the formula of wave mechanics,

\[ E_{\text{kin}}=\frac{1}{v_0}\int_{0}^{P}\frac{v_0\,dp}{h^3}\int_{v_0}dr\,2\,\frac{\hbar^2}{2M}\,|\operatorname{grad}\psi|^2 \tag{13,1} \]

one obtains the density of kinetic energy that appears in the Thomas–Fermi theory,

\[ E_{\text{kin}}=\frac{1}{v_0}\int_{0}^{P}\frac{v_0\,dp}{h^3}\int dr\,\frac{p^2}{Mv_0} =\frac{4\pi}{5}\frac{P^5}{Mh^3} =\frac{4\pi}{5}\frac{\hbar^2}{M}\left(\frac{3\rho}{8\pi}\right)^{5/3}. \]

If one introduces a refinement, assuming that the density changes appreciably also over the extent of the elementary volume \(v_0\), i.e. sets

\[ \psi=\frac{1}{\sqrt{v_0}}\,[1+(ar)]\,e^{i(pr)/h}, \]

then, by formula (13.1), the density of kinetic energy is (for the details see Weizsäcker’s paper \(^{9}\))

\[ E_{\mathrm{kin}}=\frac{4\pi}{5}\frac{P^5}{Mh^3}+\frac{1}{4\pi^2 hM}\int [a(p)]^2\,dp = \]

\[ =\frac{4\pi h^2}{5M}\left(\frac{3\rho}{8\pi}\right)^{5/3} +\frac{h^2}{32\pi^2 M}\frac{(\operatorname{grad}\rho)^2}{\rho}. \tag{13.2} \]

The second term is substantial in the surface transition layer, where the particle density falls to zero. Owing to the change of density at the surface, there is also a potential effect analogous to the surface tension of liquids. Both effects are of the same order of magnitude. Weizsäcker \(^{9}\) takes into account only the kinetic one, since the whole method is not so accurate that it would make sense to complicate the calculations. In that case, according to formula (13.2), the energy density in the nucleus is equal to

\[ F=\frac{\xi}{2}\left[(\operatorname{grad}\psi_P)^2+(\operatorname{grad}\psi_N)^2\right]+ \]

\[ +\frac{\eta}{2}\left(\psi_P^{10/3}+\psi_N^{10/3}\right)-f(\psi_P^2,\psi_N^2), \tag{13.3} \]

where

\[ \xi=\frac{h^2}{4\pi^2 M}; \qquad \eta=\frac{8\pi h^2}{5M}\left(\frac{3}{8\pi}\right)^{5/3}; \qquad \psi_P=\sqrt{\rho_P}; \qquad \psi_N=\sqrt{\rho_N}, \]

and \(f\) is the function computed by Heisenberg from formula (11.5) under the assumption that \(J=a'e^{-r/\rho'}\). Formula (13.3) was used by Weizsäcker, thus refining Heisenberg’s calculation presented above. The kinetic energy in this calculation must come out larger than in Heisenberg’s, and therefore the potential well will be deeper, which brings the values of the constants \(a'\) and \(\rho'\) closer to the Wigner ones.

As in the preceding problems, one must now solve the variational problem

\[ E=\min \int F\,dr \]

under the condition

\[ \int\left(\psi_N^2+\psi_P^2\right)d\tau=n. \]

For simplification Weizsäcker sets \(\psi_P=\psi_N=\psi\) and uses the Ritz method. For nuclei heavier than \(O^{16}_8\) it proves possible to reduce the problem to a one-dimensional model. Therefore Weizsäcker takes

\[ \psi= \left\{ \begin{array}{ll} \dfrac{\psi_0}{2}e^{\lambda x}, & \text{for } x\leq 0,\\[6pt] \psi_0\left(1-\dfrac{1}{2}e^{-\lambda x}\right), & \text{for } x\geq 0. \end{array} \right. \tag{13,4} \]

This one-dimensional Weizsäcker model gives inside the nucleus a constant density \(\psi_0^2\), calculated by the method of § 11—the “Majorana density,” and at the surface a smooth decrease of it to zero (Fig. 7).

Fig. 7.
[Diagram with a plateau labeled \(\psi_0^2\) and width \(2r\).]

In (13,4) \(\lambda\) is a variational parameter. Weizsäcker calculated the Coulomb energy by the formula for the energy of a uniformly charged sphere,

\[ E_{\text{Coul}}=\frac{3}{5}\frac{Z^2e^2}{r}, \]

where \(r\) is the nuclear radius obtained from the calculation.

The values of the parameters \(\alpha'\), \(\rho'\), the Majorana density \(\psi_0^2\), and the thickness of the transition layer \(d=\dfrac{1}{\lambda}\), found from this calculation, are given in Table 1. The table gives the energies per one particle in the nucleus.

Thus the depth of the potential well obtained by Weizsäcker is too large, since Wigner’s already correctly reproduced the energy of \(He^4_2\), and therefore Weizsäcker’s values would give too large a binding energy.

§ 14. Allowance for the Surface Effect for Light Nuclei

For light nuclei the calculation of §§ 11 and 12 gave quite incorrect results, since for them the surface effect is especially essential. The latter finds its expression in the fact that, if one moves from \(O^{16}_8\) to lighter nuclei, the binding energy per particle decreases continuously. Flugge somewhat modified the calculation

TABLE 1

$\psi_0$	$a'$ $(10^6\ \mathrm{eV})$	$\rho'$ $(10^{-13}\ \mathrm{cm})$	$d=\dfrac{1}{\lambda}$ $(10^{-13}\ \mathrm{cm})$	Nucleus	$r$ $(10^{-13}\ \mathrm{cm})$	$E_{\mathrm{vol}}$ $(10^6\ \mathrm{eV})$	$E_{\mathrm{surf}}$ $(10^6\ \mathrm{eV})$	$E_{\mathrm{coul}}$ $(10^6\ \mathrm{eV})$	$E_{\mathrm{theor}}$ $(10^6\ \mathrm{eV})$	$E_{\mathrm{exp}}$ $(10^6\ \mathrm{eV})$
1.4	185	1.03	1.85	${}^{16}_{8}\mathrm{O}$	4.06	$-14.8$	6.4	1.2	$-7.6$	$-7.7$
—	—	—	—	${}^{100}_{42}\mathrm{Mo}$	—	$-14.8$	3.3	2.9	$-8.7$	$-8.2$
—	—	—	—	${}^{200}_{80}\mathrm{Hg}$	9.44	$-14.8$	2.6	4.2	$-8.1$	$-7.7$

TABLE 2*

Element	$a_0=100$, $\rho_0$	$a_0=100$, $R$	$a_0=156$, $\rho_0$	$a_0=156$, $R$	$a_0=193$, $\rho_0$	$a_0=193$, $R$	$a_0=218$, $\rho_0$	$a_0=218$, $R$
${}^{4}_{2}\mathrm{He}$	1.76	2.72	1.27	2.32	1.10	2.20	1.00	2.08
${}^{6}_{3}\mathrm{Li}$	1.64	3.20	1.17	2.78	1.04	2.62	0.94	2.47
${}^{12}_{6}\mathrm{C}$	1.70	3.65	1.21	3.13	1.05	2.94	0.95	2.81
${}^{16}_{8}\mathrm{O}$	1.74	3.87	1.19	3.35	1.04	3.17	0.94	3.02
${}^{28}_{14}\mathrm{Si}$	1.67	4.58	1.19	3.92	1.03	3.67	0.93	3.48

* $\rho_0$ and $R$ are expressed in $10^{-13}\ \mathrm{cm}$, and $a_0$ in $10^6\ \mathrm{eV}$.

Weizsäcker, thereby making it possible to apply formula (13.3) also to light nuclei. In Weizsäcker’s model, the density of the nuclear core is determined only by the surface layer, by which the parameters $a'$ and $\rho'$ are determined. Therefore the surface—

of the surface layer to the volume of the nucleus is proportional to \(1/R\), where \(R\) is the radius of the nucleus, and for light nuclei one may entirely neglect the region of constant Majorana density. Flügge represents the density distribution in the nucleus, for example, by means of a Gaussian curve:

\[ \psi=-Ae^{-\frac{r^2}{R^2}}, \]

where, for simplicity, it is assumed that \(\sqrt{\rho_p}=\sqrt{\rho_M}=\psi\), which is quite accurately satisfied for many light nuclei. The coefficient \(A\) is determined from the normalization condition:

\[ A^2=\left(\frac{2}{\pi}\right)^{\frac{3}{2}}\frac{Z}{R^3}, \]

where \(Z\) is the number of particles of each kind. \(A^2\) has the meaning of the density at the center of the nucleus. The calculation by Flügge cited below showed that the nuclear radius varies approximately in proportion to \(\sqrt[3]{Z}\). Therefore, in agreement with experiment and with Majorana’s theory (§ 11), the density of nuclei is obtained as almost independent of their masses. The further calculation is carried out according to the usual variational scheme. The binding energy of the nucleus is equal to

\[ E=\min 4\pi\int_0^\infty (E_{\mathrm{kin}}+f)\,r^2\,dr, \tag{14, 1} \]

where \(E_{\mathrm{kin}}\) is taken according to formula (13,2), while \(f\) is calculated by formulas (11,5) and (11,6) for the chosen \(J(r)\). The variational parameter is the nuclear radius \(R\). The minimum condition is

\[ \int_0^\infty \frac{\partial}{\partial R}(E_{\mathrm{kin}}+f)\,r^2\,dr=0. \tag{14, 2} \]

Choosing a definite form for the exchange potential \(J\), depending on two parameters \(a_0\) and \(\rho_0\), we thus obtain two equations, (14,1) and (14,2), containing the quantities \(E\), \(R\), \(a_0\), and \(\rho_0\). From experiment only the binding energy \(E\) is known sufficiently well. Therefore there is not the necessary number of conditions for calculating the constants of interest to us. By assigning different values to the depth of the potential well, we can obtain only the dependence of \(R\) and \(\rho_0\) for different nuclei on \(a_0\). If Weizsäcker’s theory contains no internal contradiction, then \(\rho_0\) should not change from element to element for a given \(a_0\).

Flügge obtained the results (for \(J=a_0 e^{-\frac{r^2}{\rho_0^2}}\)) given in Table 2.

The pair of values closest to Weizsäcker’s is

\[ a_0=193\cdot 10^6\ \mathrm{eV}; \qquad \rho_0=-1.04\cdot 10^{-13}\ \mathrm{cm}. \]

Thus the Thomas–Fermi theory, supplemented by a correction for the surface effect, gives the experimental values of the binding energy throughout the entire periodic system, beginning with lithium, for \(a_0 \approx 190\cdot 10^6\ \mathrm{eV}\) and \(\rho_0=1.04\cdot 10^{-13}\ \mathrm{cm}\). Only for \(\mathrm{He}_2^4\) are results obtained that do not agree with all the others. This is quite understandable. The Fermi statistical method is no longer applicable to \(\mathrm{He}_2^4\), since all four particles may be in the lowest energy state. However, with Wigner’s values an even greater discrepancy is obtained here than in the Majorana–Heisenberg calculation. As we shall see below, despite the fact that Wigner’s values lie between Majorana’s and Weizsäcker’s, the latter two calculations contain inaccuracies of one and the same character.

§ 15. Oscillator Model of Light Nuclei3

The Thomas–Fermi method is the zeroth approximation to the solution of the Hartree–Fock problem. Taking the surface effect into account increases the accuracy of the calculation. It may be regarded as a kind of “first approximation” in the Hartree–Fock problem. Further refinement of the approximate problem posed in § 10 can be achieved by passing directly to the Hartree–Fock method.

Flügge’s calculations showed that for light nuclei there is no point in speaking of a region of constant Majorana density. Since the width of the well of the exchange potential turned out to be of the same order of magnitude as the radius of light nuclei, when a particle is removed from the center of the nucleus it will be acted upon by an ever-increasing force of attraction toward the center. Heisenberg considers it possible to approximate the potential of this force by a parabola. Then the behavior of each particle in the nucleus can be described by the eigenfunction of a three-dimensional oscillator. The approximate application of the Hartree–Fock method amounts to using the formula of this theory for the total energy of the system:

\[ E=\sum_{i=1}^{n_1+n_2}\int \left(\psi_i^{*}\frac{\hbar^2}{2m_i}\nabla^2\psi_i\,dr\right) + \]

\[ +\frac{e^2}{2}\int \frac{\rho_p(r)\rho_p(r')-\left|\rho_p(r,r')\right|^2}{|r-r'|} \,dr\,dr' +E_{\mathrm{pot}}. \tag{15,1} \]

The last term gives the energy of interaction of protons with neutrons, computed by formula (11,3). Heisenberg used an exchange potential in the form \(J=a_0 e^{-\frac{r^2}{\rho_0^2}}\) and carried out the calculation for three

nuclei \( \mathrm{He}^{4}_{2} \), \( \mathrm{O}^{16}_{8} \), and \( \mathrm{Ca}^{40}_{20} \), for which, in the oscillator model, closed shells are obtained.

The mixed density

\[ \rho_p(r,r')=\rho_N(r,r')=\sum_{\substack{\text{over all}\\ \text{occ. states}}}\psi_n^*(r)\psi_n(r') \]

we compute directly, substituting the oscillator eigenfunctions in the corresponding states. Since the characteristics of the “quasi-elastic force” are unknown, the oscillator frequency \(\nu\) remains as yet undetermined. We use it as a variable parameter in finding the minimum of \(E\). Computing

\[ \delta E = 0 \]

and equating \(E\) in (15,1) to its experimental value, we obtain two equations from which, for any \(\nu\), one can calculate the corresponding pair of values \(a\) and \(\rho\).

The curves shown in Fig. 8 indicate that the application of this oscillator model to light nuclei gives considerably better results than Majorana’s method, but results that all deteriorate with increasing atomic weight. This is also understandable. In heavier nuclei one can already speak of a region of constant density, where, owing to the short-range character of the forces between protons and neutrons, no force will prevent displacement from the center. But for light nuclei Heisenberg’s calculation gives results closer to the exact deuteron curve than Weizsäcker’s method (§ 13) (Flügge’s curve approximately coincides with Heisenberg’s curve for \(\mathrm{Ca}^{40}_{20}\)). For heavy nuclei the curves calculated by this oscillator model, as was to be expected, approach Majorana’s curve, since both these methods coincide in the limit for infinitely heavy nuclei.

Fig. 8.

§ 16. Calculations of the nuclei \( \mathrm{H}^{3}_{1} \), \( \mathrm{He}^{3}_{2} \), and \( \mathrm{He}^{4}_{2} \)

The values of the parameters \(a\) and \(\rho\) obtained in the calculations of Heisenberg (§ 12) and Weizsäcker (§ 13) are highly unreliable.

The latest works—by Flügge and the oscillator model of Heisenberg—give only the dependence of the parameters on one another. Wigner’s calculation, as was indicated, gave results in comparatively good agreement with the experimental data. But it was carried out for potential forces. In addition, this calculation was performed only for the nuclei \(\mathrm{H}^{1}_{2}\) and \(\mathrm{He}^{4}_{2}\). If the theory being applied is correct, then for the same values of the parameters \(a\) and \(\rho\) one should correctly obtain the binding energy of \(\mathrm{H}^{3}_{1}\) and \(\mathrm{He}^{3}_{2}\). According to the theorem proved by Eckart \(^{12}\), a calculation carried out for exchange forces always gives a smaller binding energy, for the same well parameters \(a\) and \(\rho\), than a calculation with Wigner forces. Unfortunately, Eckart’s theorem does not make it possible to estimate this difference quantitatively. It is therefore essential to carry out the calculation also for the nuclei \(\mathrm{H}^{3}_{1}\) and \(\mathrm{He}^{3}_{2}\), taking into account as well the exchange character of the forces.

It is quite evident that the most accurate results are most easily obtained by solving the problem of the simplest nuclei.

The deuteron problem was treated by us in detail in §§ 5 and 6.

The triton problem can be solved only approximately, by means of the variational method. The Schrödinger equation for the triton in Cartesian coordinates has the form:

\[ \left[-\frac{\hbar^{2}}{2m}\nabla^{2}+V(r_{1},r_{2},r_{3})\right]\psi_{0}=E\psi_{0}, \tag{16, 1} \]

where \(r_{1}\) and \(r_{2}\) are the distances from the proton to the two neutrons, and \(r_{3}\) is the distance between the neutrons. Separating the coordinates of the center of gravity, for the part of the function depending only on the relative coordinates, we obtain the equation:

\[ H\psi=\left[-\frac{\hbar^{2}}{m}(\nabla_{1}^{2}+\nabla_{2}^{2}+\nabla_{1}\nabla_{2})+ \right. \]

\[ \left. +\,V(r_{1},r_{2},r_{3})\right]\psi=E_T\psi, \tag{16, 2} \]

where:

\[ \begin{aligned} \nabla_i^{2} &= \frac{\partial^{2}}{\partial x_i^{2}}+\frac{\partial^{2}}{\partial y_i^{2}}+\frac{\partial^{2}}{\partial z_i^{2}},\\ r_i^{2} &= x_i^{2}+y_i^{2}+z_i^{2} \end{aligned} \qquad i=1,2 \]

\[ \nabla_{1}\nabla_{2} = \frac{\partial^{2}}{\partial x_{1}\partial x_{2}} + \frac{\partial^{2}}{\partial y_{1}\partial y_{2}} + \frac{\partial^{2}}{\partial z_{1}\partial z_{2}}, \]

and

\[ r_{3}^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}. \]

According to the concepts developed above,

\[ V(r_1,r_2,r_3)=P_1 I(r_1)+P_2 I(r_2)+P_3 I(r_3), \]

where \(P_i\) is the operator of permutation of the coordinates of the corresponding particles, and \(I(r_i)\) is the potential function describing the interaction of like particles. The latter, obviously, is small.

The calculation of the binding energy of the triplon thus leads to the solution of the variational problem

\[ E_T=\min \frac{\int \psi^*H\psi\,dv}{\int |\psi|^2\,dv}, \tag{16, 3} \]

where the operator \(H\) is defined by equation (16,2). If \(E_T\) is set equal to its experimental value \(8\cdot 10^6\ \mathrm{eV}\), then relation (16,3) gives an equation connecting the parameters \(a\) and \(\rho\). The point of intersection of this curve \(a=F(\rho)\) with the deuteron curve (Fig. 4) will determine the sought values of these parameters. A correct theory, when solving the problem of any other nucleus, must give the corresponding curve, intersecting the first two at one and the same point.

The solution of the variational problem for the triplon, \(\mathrm{He}_2^3\) and \(\mathrm{He}_2^4\), and a comparison of the exchange forces with Wigner’s, were given in Finberg’s work4. Unfortunately, in calculating \(a=f(\rho)\) in the deuteron problem he already uses approximate integration and therefore obtains too slow an increase of the well depth \(a\) when its width \(\rho\) is decreased. According to the uncertainty relation, the mean kinetic energy \(E_k\) of a particle confined to a region of width \(\rho\) is determined by the formula

\[ E_k\rho^2 \gtrless \frac{\hbar^2}{2m}. \]

In order for the particle to remain in the nucleus, the depth of the potential well \(a\) must be greater than \(E_k\), i.e., if \(c>1\), it is necessary that \(a\), when \(\rho\) is decreased, increase no more slowly than according to the condition

\[ a\rho^2=c\,\frac{\hbar^2}{2m}=\mathrm{const}. \]

Meanwhile, in Finberg’s work,

\[ a\rho^{\frac{3}{2}}=\mathrm{const}. \]

Therefore, for narrow potential wells Finberg obtains repulsion instead of binding.

For the deuteron he solves the variational problem

\[ E_D=\min \int \psi^*H\psi\,dv, \]

using the function

\[ \psi=\left(\frac{\nu}{\pi}\right)^{\frac{3}{4}} r e^{-\frac{\nu r^2}{2}}, \]

where \(\nu\) is a variational parameter. In exactly the same way he posed the problem of the triton, \(\mathrm{He}_2^3\), and \(\mathrm{He}_2^4\), using the functions

\[ \psi(1,2,3)=u(12)u(13)v(23) \]

for a nucleus consisting of three particles, and

\[ \psi(1,2,3,4)=u(13)u(14)u(23)u(24)v(12)v(34) \]

for the nucleus \(\mathrm{He}_2^4\), with

\[ u=Ne^{-\frac{\nu r^2}{2}};\qquad v=Ne^{-\frac{\mu r^2}{2}}. \]

Fig. 9.

Fig. 10.

Here \(\nu\) and \(\mu\) are new variational parameters. He takes Coulomb repulsion into account with the help of unperturbed wave functions. All calculations are carried out in the first approximation both for Majorana forces and for Wigner forces. The exchange potential is taken in the form \(J=a_0 e^{-r^2/\rho_0^2}\).

The results of his calculations are as follows: for the deuteron, assuming \(E_D=2\cdot10^6\ \mathrm{eV}\), he obtains the relation of \(a_0\) to \(\rho_0\) given by the curve in Fig. 9.

The lower curves in Fig. 10 show the course of the binding energy of the triplon, calculated by Wigner (solid line) and Majorana (dashed line). The upper curves give the binding energy of the nucleus \(\mathrm{He}_2^4\). Thus, for neither nucleus do the curves reach the experimental value of the binding energy (\(8\cdot 10^6\ \mathrm{eV}\) for \(\mathrm{H}_1^3\) and \(27\cdot 10^6\ \mathrm{eV}\) for \(\mathrm{He}_2^4\)). For \(\mathrm{H}_1^3\) one obtains even a qualitatively incorrect dependence of the binding energy on the width of the well. In fact, Thomas\(^{14}\) proved for a broad class of functions \(J(r)\) that, with a sufficiently good choice of the varied function, one can obtain such a relation of \(E_T\) to \(\rho\) that, as \(\rho\) tends to zero, the binding energy of the triplon will increase without bound. The opposite result of Finberg’s calculation—repulsion at small \(\rho\)—is explained primarily by the use of an inaccurate solution for the deuton.

With the aid of the formulae used by Eckart in deriving his theorem, one can estimate that the discrepancy between the Wigner and Majorana energies increases as the potential well is narrowed. Finberg’s calculation confirms this estimate and may serve as a guide to the order of magnitude of this discrepancy.

Greater success in the solution of this problem was achieved by Massey and Mohr\(^{15}\). Proceeding from the hypothesis that in the nucleus there exist only Wigner attractive forces between unlike particles, they carried out a series of calculations with various trial functions and exchange potentials \(J(r)\). The proper function of the triplon was taken in the form

\[ \Psi=\psi(r_1)\psi(r_2)\{1+f(r_1,r_2,\theta)\}, \]

where \(r_1\) and \(r_2\) are the distances from the proton to each of the neutrons, and \(\theta\) is the angle between \(r_1\) and \(r_2\). All calculations in which \(\psi\) was not an exact solution of the diplon led to completely unsatisfactory results, similar to Finberg’s calculations. With \(\psi\) taken from the exact solution of the diplon problem for a rectangular potential well and with the function

\[ f=\gamma r_1 r_2 \cos\theta, \tag{16, 4} \]

where \(\gamma\) is a variational parameter, somewhat better results were obtained. But the experimental value of the binding energy was not reached. As the well width tended to zero, the binding energy was found to increase, but asymptotically to approach the value \(5.4\cdot 10^6 \mathrm{eV}\), whereas the experimental value is \(8\cdot 10^6 \mathrm{eV}\).

It is interesting to note that the best results were obtained with the aid of the function (16, 4), which, for the positive value of \(\gamma\) obtained from the calculation, gives the greatest probability \(|\psi|^2\) at \(\theta=0\), i.e. the maximum binding energy is obtained for the function that gives as most probable the closest possible position of the neutrons to one another.

The conclusion of Massey and Mohr, as well as of Feinberg, that there must exist appreciable attractive forces between like particles is premature. In principle, nothing excludes the possibility of attractive forces between like particles, but these forces must be much smaller than the attractive forces between unlike particles, since otherwise the possibility disappears of explaining the conditions for the occurrence of \(\beta^+\)- and \(\beta^-\)-decay, as well as the experimental fact that, for stable light nuclei, \(A \approx 2Z\).

Recently a report appeared on a calculation of the nucleus \(\mathrm{H}_1^3\) by Present \(^{16}\). He used eight variational parameters, representing the eigenfunction of the nucleus as the product of an exponential by a power polynomial with a variable coefficient. Despite the high approximation, he again obtained too small a binding energy. For \(\rho'' = 0.7 \cdot 10^{-13}\ \mathrm{cm}\) (the equivalent width of the rectangular well) he has \(E_T = -5.3 \cdot 10^6\ \mathrm{eV}\). However, this result also does not yet give grounds for introducing appreciable attractive forces between like particles. The rapid convergence of the series could have been only apparent.

Up to the present time agreement with Thomas’s theorem has been obtained only in the calculation of Tamm and Golovin \(^{17}\). The course of this variational calculation is the same as in Massey and Mohr, except that in the triplon problem a different form of the variational function is taken. For convenience of calculation a rectangular potential well of width \(\rho''\) and depth \(a''\) is used, i.e. it is assumed that

\[ \begin{aligned} J(r) &= a'' \quad \text{for } r < \rho'',\\ J(r) &= 0 \quad \text{for } r > \rho'' \end{aligned} \tag{16, 5} \]

and the forces are assumed to be Wigner forces. The interaction of like particles is neglected. The triplon problem is solved by the Ritz method in the first approximation, with the eigenfunction taken in the form

\[ \Psi = \psi(r_1)\psi(r_2)e^{-\alpha r_3}. \]

Here \(\psi\) is the exact solution of the Schrödinger equation for the deuteron with a rectangular well, and \(\alpha\) is a variational parameter. According to the usual variational relation, in this case we have:

\[ E_T = \min \frac{\int \Psi^{*} H \Psi\, dv} {\int |\Psi|^2\, dv} = 2E_D + \min \frac{\int \Psi^{*} \Omega \Psi\, dv} {\int |\Psi|^2\, dv}, \]

where

\[ \Omega = -\frac{\hbar^2}{m} \left\{ \frac{\partial^2}{\partial r_3^2} + \frac{2}{r_3}\frac{\partial}{\partial r_3} + S\frac{r_1^2 + r_2^2 - r_3^2}{2r_1r_2} \frac{\partial^2}{\partial r_1 \partial r_2} \right\}. \]

\(S\) here denotes the sum over cyclic permutations of the indices \(1, 2, 3\).

The calculation shows that \(E_T\) reaches a minimum for a certain positive value of the parameter \(\alpha\), i.e., as in the calculation of Massey and Mohr, small distances between neutrons are the most probable.

If one uses the relation between \(a''\) and \(\rho''\) given by the exact solution of the deuteron and presented in Fig. 4, then, in accordance with Thomas’s theorem, \(E_T\) in this calculation not only reaches the experimental value \(E_T = 8 \cdot 10^6\ \mathrm{eV}\), but also exceeds it for a sufficiently narrow well (Fig. 11).

Fig. 11.

The values of the parameters found in this calculation are:

\[ a'' = -600 \cdot 10^6\ \mathrm{eV}; \qquad \rho'' = 0.44 \cdot 10^{-13}\ \mathrm{cm}. \]

These values are unsatisfactory. The most probable values for \(a''\) are \(100\)—\(150 \cdot 10^6\ \mathrm{eV}\), and, correspondingly, for \(\rho''\): \(0.8\)—\(1.0 \cdot 10^{-13}\ \mathrm{cm}\). Undoubtedly, increasing the accuracy of the triton calculation will greatly decrease the depth and increase the width of the well. At a depth of \(600 \cdot 10^6\ \mathrm{eV}\), relativistic effects become substantial; for example, the correction to the mass reaches \(8\%\). Therefore the nonrelativistic calculation ceases to be applicable.

It was noted above that in all calculations of the triton the greatest probability is obtained for the close location of the neutrons to one another. This, however, is not the result of an implicit introduction of attractive forces between them. The observed effect is of purely kinetic origin and is caused by the presence of the term \(\nabla_1\nabla_2\) in equation \((16,2)\). What has been said becomes especially clear when considering the “oscillator model” of the nucleus \(\mathrm{H}_3^1\). Indeed, suppose that two particles 1 and 2 of mass \(m\) are connected by quasielastic forces with a third particle 3 of the same mass, and do not interact with each other; then, passing to normal coordinates, we,

we can integrate equation (16.2) completely exactly. The proper function again gives close positions of particles 1 and 2 as more probable than distant ones. If, however, the term \(\nabla_1\nabla_2\), which corresponds to passing to the description of the oscillation of two independent pairs of particles, each of mass \(m\), is omitted, then the energy of the system increases, and the proper function becomes independent of the distance between particles 1 and 2. Thus the kinetic term \(\nabla_1\nabla_2\) plays a role analogous to the forces of attraction between the “like-named” particles 1 and 2. Therefore, in the triplon problem the variational function must necessarily depend on \(r_3\) and increase as \(r_3 \to 0\).

Fig. 12.

The last work on the question under discussion is Dolch’s calculation\(^{18}\) of the nuclei \(\mathrm{H}_1^3\), \(\mathrm{He}_2^3\), and \(\mathrm{He}_2^4\). However, the results obtained by him are fictitious. Dolch carried out calculations for exchange forces, using a potential function in the same form as Finberg:

\[ J(r)=a_0 e^{-\frac{r^2}{\rho_0^2}}. \]

With this form of \(J\), even the deuteron equation is not analytically integrable. Therefore Dolch computes the binding energies of the nuclei \(\mathrm{H}_1^3\), \(\mathrm{He}_2^3\), and \(\mathrm{He}_2^4\) by means of a variational method and obtains, respectively, three curves giving the dependence of \(a_0\) on \(\rho_0\). They intersect one another almost at a single point (Fig. 12). Hence

he concludes that the sought values of the parameters are:

\[ a_0=70\cdot 10^6\mathrm{eV};\qquad \rho_0=1.7\cdot 10^{-13}\ \mathrm{cm}. \]

However, the intersection of all three curves at this point is only an accident, due to the poor choice of the varied function. The function which he used for the deuteron,

\[ \psi=\alpha e^{-\beta r^2} \]

(\(\beta\) is the varied parameter) behaves well only at the point \(r=0\). Its asymptotic behavior is completely incorrect. As is easily seen from the asymptotic form of the deuteron equation (6.1), as \(r\to\infty\) the eigenfunction must have the form:

\[ \psi_{r\to\infty}=\frac{e^{-\alpha r}}{r}, \]

where

\[ \alpha=\sqrt{\frac{mE_D}{\hbar^2}}. \]

From the calculation of Tamm and Golovin it is evident that, for the binding energy, the behavior of the function in the region where \(r>\rho\) is most essential. Therefore the function

\[ \psi=\frac{e^{-\alpha r}-e^{-\beta r}}{r}, \]

where \(\alpha\) and \(\beta\) are varied parameters, should give considerably more correct results, despite its poorer behavior at the point \(r=0\). A variational calculation in which only the parameter \(\beta\) was varied, while \(\alpha\) was taken equal to \(\sqrt{\frac{mE_D}{\hbar^2}}\), gives curve \(IV\), shown in Fig. 12. This curve is considerably closer to the true deuteron curve \(V\) (obtained by Bethe\({}^{19}\) by numerical integration of the deuteron equation) than is Dolch’s curve. This once again confirms the remark made above that, in variational problems of nuclei, one must take varied functions having the correct behavior for \(r\gg \rho\). In the region \(r\lesssim \rho\) their behavior is less essential, but must satisfy the natural conditions (single-valuedness, finiteness, continuity).

As is seen from the drawing, Bethe’s exact curve can intersect the \(\mathrm{He}_2^4\) curve calculated by Dolch only near the point \(\sim 200\cdot 10^6\,\mathrm{eV}\). On the basis of this drawing it is in general impossible to say whether Dolch’s triplon curve intersects Bethe’s curve. Undoubtedly, increasing the accuracy of the calculation of the triplon and \(\mathrm{He}_2^4\) will shift curves \(II\) and \(III\) downward. But where the point of their intersection will then be found, it is as yet impossible to judge.

The nucleus \(\mathrm{He}_2^3\) differs from the nucleus \(\mathrm{H}_1^3\) only by the presence, in the former, of Coulomb repulsion. All calculations by the authors mentioned above show that the Coulomb energy

\[ E_{\mathrm{Coul}}= \frac{\int |\Psi|^2 \frac{1}{r}\,dv}{\int |\Psi|^2\,dv}, \]

where \(\Psi\) is the unperturbed wave function of the triplon, is close to the experimentally required difference in the binding energies of the triplon and \(\mathrm{He}_2^3\) (\(\sim 10^6\ \mathrm{eV}\)). Consequently, this nucleus contains no new difficulties.

Thus, the problem of the nuclei \(\mathrm{H}_1^3\) and \(\mathrm{He}_2^4\) has still not been satisfactorily solved. The available data are still insufficient to decide whether the discrepancies and failures of the various calculations are due only to the inaccuracy of the computations, or whether the hypotheses used concerning intranuclear forces are incorrect. The immediate task is to increase the accuracy of the calculation. Only this can resolve the question, and by no means the introduction of additional forces, such as, for example, the forces of attraction between like particles when using the same methods of calculation.

§ 17. Ways of Improving the Theory

Thus, the theory developed above agrees with experiment only as to the order of magnitude of the quantities obtained. The results of different calculations differ substantially among themselves. If one takes into account the Pauli principle and the grouping of protons and neutrons into \(\alpha\)-particles, then the fine structure of the mass-defect curve is also qualitatively explained.

How, then, can one explain the quantitative disagreement with experiment and among the calculations presented?

The dependence of the parameter \(a\) on \(\rho\) was obtained from the solution of the deuteron problem with the maximum accuracy that can be required of a nonrelativistic theory. In all the remaining works we have not only an approximate formulation of the problem, but also a solution in the first approximation of the variational method. For heavy nuclei one should not expect a substantial improvement from a more accurate variational calculation, since the very representation of the proper function in the form of a product of determinants is a poor approximation to reality. In fact, for the processes of collisions of neutrons with boron nuclei,\(^{20}\) it was necessary to assume that the decisive importance in the life of the nucleus is the exchange of energy between the particles. Therefore the behavior of an individual particle in the nucleus cannot be described by means of its coordinates alone. Here the properties of the collective of particles are essential. Therefore the assumption made in § 10 is incorrect, namely that the interaction of the selected particle with all the others can be approximately replaced by some conservative field. Consequently, writing the nuclear function in the form (10,1) is inadmissible.

The discrepancy between the results of the calculations presented and the exact solution for the deuteron may be regarded as confirmation of the stated point of view.

Heisenberg believes that Hartree’s method could be refined for the nucleus if the grouping of protons and neutrons into $\alpha$-particles is taken into account. In heavy nuclei it must be taken into consideration that protons are bound only with the nearest neutrons, independently of the position of particles far away. It must also be borne in mind that the most probable arrangement of particles is that in which each proton is surrounded by neutrons.

In calculations of the nuclei $\mathrm{H}_{1}^{3}$, $\mathrm{He}_{2}^{3}$, and $\mathrm{He}_{2}^{4}$, a further improvement of the variational method is possible by choosing more suitable functions to be varied. But the most exact path is the one indicated by Thomas. Unfortunately, even for the nucleus $\mathrm{H}_{1}^{3}$ it involves great mathematical difficulties. This path consists in the following. If one takes a sufficiently steep potential well, then all space proves to be divided into two regions:

1) the region where $J = 0$. In it the proper function of the triplon must satisfy the equation

\[ (\nabla_1^2 + \nabla_1\nabla_2 + \nabla_2^2)\psi = \mu\psi . \]

2) the region where $J \ne 0$. In this region $\psi$ must satisfy equations of the type

\[ (\nabla_1^2 + \nabla_1\nabla_2 + \nabla_2^2 + cJ)\psi = \mu\psi . \]

A special difficulty here is the satisfaction of the boundary conditions, since the function of one region must pass continuously and without a discontinuity in the first derivative into the function of the second region.

LITERATURE

1. See, for example, M. I. Korsunskii, Neutron, p. 153.
2. W. Heisenberg, Z. Physik, 77, 1, 1932.
3. W. Heisenberg, Z. Physik, 78, 156, 1932.
4. E. Wigner, Phys. Rev., 43, 252, 1933.
5. W. Heisenberg, Z. Physik, 80, 587, 1933.
6. E. Majorana, Z. Physik, 82, 137, 1933.
7. W. Heisenberg, “Noyaux atomiques.” Report at the Solvay Congress, 1933, p. 289.
8. P. A. M. Dirac, Proc. Cambr. Phil. Soc., 25, 376, 1930.
9. C. F. v. Weizsäcker, Z. Physik, 96, 431, 1935.
10. S. Flügge, Z. Physik, 96, 459, 1935.
11. W. Heisenberg, Z. Physik, 96, 473, 1935.
12. Eckart, Phys. Rev., 44, 109, 1933.
13. E. Feenberg, Phys. Rev., 47, 850, 857, 1935.
14. Thomas, Phys. Rev., 47, 903, 1935.
15. Massey a. Mohr, Proc. Roy. Soc., (A) 152, No. 877, 1935.
16. Present, Phys. Rev., 49, 640, 1936.
17. Golovin, Zhurn. teoret. i eksperim. fiziki 6, 508, 1936.
18. Dolch, Z. Physik, 100, 401, 1936.
19. H. A. Bethe a. R. F. Bacher, Rev. of Modern. Phys. 8, 111, 1936.
20. N. Bohr, Nature, February 29, 1936.

2. Visible reference marker in the source. ↩↩

3. ... ↩

4. Finberg. ↩