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Coulomb energy of 3H and 3He
Volume 28B. number
3
PHYSICS
COULOMB
LETTERS
ENERGY
OF
25 November
3H
AND
3He
1968
*
R. T. FOLK Lehigh
University,
Bethlehem,
Received
Pennsylvania,
15 October
USA
1968
The Coulomb energy of 3He and 3H is large enough to account for the difference in the binding energies these nuclei within the uncertainty caused by the impreciseness of the Coulomb energy operator.
Recently, several authors [1,2 have reported He which are not values of the Coulomb energy of 31 large enough to account entirely for the difference between the binding energies, AB = 0.764 MeV, of 3He and 3H. From this Okamoto and Lucas [l] conclude that we must abandon the hypothesis of the charge symmetry of nuclear interactions. We show that plausible modifications in the Coulomb energy operator lead to agreement with the experimental value of AB. In carrying out this calculation, we accept the assumption that the correct wave function that is to be used to evaluate the Coulomb energy is the ground-state wave function of a three-body Hamiltonian in which the interactions are the superposition of two-body potentials which account correctly for the two-body scattering data, the most significant of which for the present application are the lS and 35: phase shifts. We use potential II of ref. 3 and obtain an approximate solution for the wave function by an extension [4] of the independent-pair method developed in ref. 5. The magnitude of the S’ state of this wave function is 2.15 percent. Initially, the charge distribution in the nucleons is assumed to be the inverse Fourier transform of the charge form factors of the free nucleans. If we use the expressions for the form factors given by Barut et al. [6], we obtain pp(r) = g
[l + da + fb - 2b - :(a + b)/3r] exp(-@r) PY (1)
for the proton charge distribution e is the net charge, and
* Research tion.
sponsored
function,
by the National Science
where
Founda-
Pnb’) =0.382$
of
($- 1) exp(-Pr)
(2)
for the neutron, where /32 = 18.2 fmm2, a = = 0.00654 02, and b = 0.000794p4. The classical potential for the interaction of two such charge distributions at a distance Y between their centres is given by
=${I +
- exp(-pr)[l
+ 0.6892pr
+ 0.1892
0.0218 (~3r)~ - 0.00251 @r)4 + 0.000038
@r)2
+
@Y)~]}
for two protons and by similar expression for a neutron and proton, With these potential terms included in the Hamiltonians, we obtain ** (H> = - 8.79 MeV for 3H and (H) = - 8.13 MeV for 3He, and so AB = = 0.66 MeV. The n-p part of the Coulomb energy contributes 0.012 MeV to these energies. All the integrals in this calculation are evaluated numerically. In the above calculation we ignored the possibility that the charge distributions of the nucleons in nuclei are distorted by the nuclear interactions. To obtain some measure of the effect of such distortions on the Coulomb energy we consider a model in which the centre of the positive charge density of each nucleon is displaced towards the attracting nucleon and the negative charge density is distorted oppositely so that considering the absolute value of the terms in eqs. (1) and (2), the centre of the charge cloud remains at the particle centre. The polarization is assumed to vary with the distance between the nucleons by an amount that is proportional to the singlet, ** If the point charge Coulomb potential used, we obtain AS = 0.695 MeV.
Vpp = e2/r
is
159
Volume 28B. number
3
PHYSICS
LETTERS
central potential, and so this polarization is significant only when the nucleons are close. If the positive charge is displaced 0.106 fm, which is 13% of the rms radius of the free proton charge, when the nucleons are 1 fm apart, the difference between the Coulomb energy of 3H and 3He is found to be equal to the experimental value of AB. The change in the Coulomb interaction between two nucleons which is caused by the polarizations induced by the third nucleon is not included here. Let us consider another possible contribution to the Coulomb energy. One can define an equivalent charge density pex as the inverse Fourier transform of the meson exchange terms which, as suggested by Sarker [7], should be included in the expressions for the charge form factors [8] of 3H and 3He. We adjust these exchange terms so that the experimental values for the form factors [9] are obtained in an analysis in which the threebody wave functions described above and the free nucleon form factors [6] are used. Details of how the form factors are evaluated are given in ref. 4. The resulting charge density pex for 3He is given approximately by Pex = !
c$)’
(1+102d-68d2r2)
- :”
0
(1+3c-2c2r2)
exp(-&2)
here,
is given
) exp (-cr 21
References
J’
by
=
=:xi;v(q)(“2+u~+L.~)
++ [V(r,)+ (3)-i
for 3He, where the terms
V(r2)] “2’1(+5) V(r,)
uv2} dr
U, “1, and 2.~2of the wave
*****
160
function are defined in ref. 8, V is the electrical potential due to pex, and the integration is over the particle coordinates. The classical self energy of Pex was calculated and found to be very small. The net contribution to the Coulomb energy due to interactions with Pex is - 0.10 MeV for 3He and -0.03 MeV for 3H. Finally, consider another contribution to AB. The kinetic energy operator is not the same for the two three-body nuclei because the masses of the neutron and proton are not equal. We find that the effect of this is to increase AB by 0.028 MeV. We conclude that the impreciseness of the Coulomb energy operator leads to an uncertainty in the calculated Coulomb energy of at least 0.1 MeV. Thus, until a Coulomb energy operator is established by a more fundamental approach such as that of field theory, a difference of this amount between the experimental and calculated values of LB is not enough to prove that nuclear forces are not charge symmetric.
+
where e is the proton charge, c = 1.0 fmv2, and d = 0.071 fms2. A similar expression is obtained for 3H. The energy of the interaction between pex and the charges of the nucleons, assumed to be point
E
1968
(4)
= 0.061e
charges
25 November
Phys. Letters 26B (1968) 1. K.Okamoto and C.Lucas, 188; Nucl. Phys. B2 (1967) 347; Nuovo Cimento 48A (l’(67) 233. Nucl. Phys. A97 (1967) 417; 2. F.C.Khanna. Y. C. Tang and R. C. Herndon, Phys. Letters 18 (1965) 43. 3. E. W. Schmid, Y. C.Tang and R. C. Herndon, Nucl. Phys. 42 (1963) 95. 4. R. Folk, submitted to Nucl. Phys. 5. F. Folk, Nucl. Phys. 85 (1966) 449. D.Corrigan and H.Kleinert, Phys. Rev. 6. A.O.Barut. Letters 20 (1968) 167. Phys. Rev. Letters 13 (1964) 375. 7. A.Q.Sarker, Phys. Rev. 133 (1964) B802. 8. L.I.Schiff, R.Hofstadter, E.B.Hughes, A.Johans9. H.Collard, son and M.R.Yearian. Phys. Rev. 138 (1965) B57.
3
PHYSICS
COULOMB
LETTERS
ENERGY
OF
25 November
3H
AND
3He
1968
*
R. T. FOLK Lehigh
University,
Bethlehem,
Received
Pennsylvania,
15 October
USA
1968
The Coulomb energy of 3He and 3H is large enough to account for the difference in the binding energies these nuclei within the uncertainty caused by the impreciseness of the Coulomb energy operator.
Recently, several authors [1,2 have reported He which are not values of the Coulomb energy of 31 large enough to account entirely for the difference between the binding energies, AB = 0.764 MeV, of 3He and 3H. From this Okamoto and Lucas [l] conclude that we must abandon the hypothesis of the charge symmetry of nuclear interactions. We show that plausible modifications in the Coulomb energy operator lead to agreement with the experimental value of AB. In carrying out this calculation, we accept the assumption that the correct wave function that is to be used to evaluate the Coulomb energy is the ground-state wave function of a three-body Hamiltonian in which the interactions are the superposition of two-body potentials which account correctly for the two-body scattering data, the most significant of which for the present application are the lS and 35: phase shifts. We use potential II of ref. 3 and obtain an approximate solution for the wave function by an extension [4] of the independent-pair method developed in ref. 5. The magnitude of the S’ state of this wave function is 2.15 percent. Initially, the charge distribution in the nucleons is assumed to be the inverse Fourier transform of the charge form factors of the free nucleans. If we use the expressions for the form factors given by Barut et al. [6], we obtain pp(r) = g
[l + da + fb - 2b - :(a + b)/3r] exp(-@r) PY (1)
for the proton charge distribution e is the net charge, and
* Research tion.
sponsored
function,
by the National Science
where
Founda-
Pnb’) =0.382$
of
($- 1) exp(-Pr)
(2)
for the neutron, where /32 = 18.2 fmm2, a = = 0.00654 02, and b = 0.000794p4. The classical potential for the interaction of two such charge distributions at a distance Y between their centres is given by
=${I +
- exp(-pr)[l
+ 0.6892pr
+ 0.1892
0.0218 (~3r)~ - 0.00251 @r)4 + 0.000038
@r)2
+
@Y)~]}
for two protons and by similar expression for a neutron and proton, With these potential terms included in the Hamiltonians, we obtain ** (H> = - 8.79 MeV for 3H and (H) = - 8.13 MeV for 3He, and so AB = = 0.66 MeV. The n-p part of the Coulomb energy contributes 0.012 MeV to these energies. All the integrals in this calculation are evaluated numerically. In the above calculation we ignored the possibility that the charge distributions of the nucleons in nuclei are distorted by the nuclear interactions. To obtain some measure of the effect of such distortions on the Coulomb energy we consider a model in which the centre of the positive charge density of each nucleon is displaced towards the attracting nucleon and the negative charge density is distorted oppositely so that considering the absolute value of the terms in eqs. (1) and (2), the centre of the charge cloud remains at the particle centre. The polarization is assumed to vary with the distance between the nucleons by an amount that is proportional to the singlet, ** If the point charge Coulomb potential used, we obtain AS = 0.695 MeV.
Vpp = e2/r
is
159
Volume 28B. number
3
PHYSICS
LETTERS
central potential, and so this polarization is significant only when the nucleons are close. If the positive charge is displaced 0.106 fm, which is 13% of the rms radius of the free proton charge, when the nucleons are 1 fm apart, the difference between the Coulomb energy of 3H and 3He is found to be equal to the experimental value of AB. The change in the Coulomb interaction between two nucleons which is caused by the polarizations induced by the third nucleon is not included here. Let us consider another possible contribution to the Coulomb energy. One can define an equivalent charge density pex as the inverse Fourier transform of the meson exchange terms which, as suggested by Sarker [7], should be included in the expressions for the charge form factors [8] of 3H and 3He. We adjust these exchange terms so that the experimental values for the form factors [9] are obtained in an analysis in which the threebody wave functions described above and the free nucleon form factors [6] are used. Details of how the form factors are evaluated are given in ref. 4. The resulting charge density pex for 3He is given approximately by Pex = !
c$)’
(1+102d-68d2r2)
- :”
0
(1+3c-2c2r2)
exp(-&2)
here,
is given
) exp (-cr 21
References
J’
by
=
=:xi;v(q)(“2+u~+L.~)
++ [V(r,)+ (3)-i
for 3He, where the terms
V(r2)] “2’1(+5) V(r,)
uv2} dr
U, “1, and 2.~2of the wave
*****
160
function are defined in ref. 8, V is the electrical potential due to pex, and the integration is over the particle coordinates. The classical self energy of Pex was calculated and found to be very small. The net contribution to the Coulomb energy due to interactions with Pex is - 0.10 MeV for 3He and -0.03 MeV for 3H. Finally, consider another contribution to AB. The kinetic energy operator is not the same for the two three-body nuclei because the masses of the neutron and proton are not equal. We find that the effect of this is to increase AB by 0.028 MeV. We conclude that the impreciseness of the Coulomb energy operator leads to an uncertainty in the calculated Coulomb energy of at least 0.1 MeV. Thus, until a Coulomb energy operator is established by a more fundamental approach such as that of field theory, a difference of this amount between the experimental and calculated values of LB is not enough to prove that nuclear forces are not charge symmetric.
+
where e is the proton charge, c = 1.0 fmv2, and d = 0.071 fms2. A similar expression is obtained for 3H. The energy of the interaction between pex and the charges of the nucleons, assumed to be point
E
1968
(4)
= 0.061e
charges
25 November
Phys. Letters 26B (1968) 1. K.Okamoto and C.Lucas, 188; Nucl. Phys. B2 (1967) 347; Nuovo Cimento 48A (l’(67) 233. Nucl. Phys. A97 (1967) 417; 2. F.C.Khanna. Y. C. Tang and R. C. Herndon, Phys. Letters 18 (1965) 43. 3. E. W. Schmid, Y. C.Tang and R. C. Herndon, Nucl. Phys. 42 (1963) 95. 4. R. Folk, submitted to Nucl. Phys. 5. F. Folk, Nucl. Phys. 85 (1966) 449. D.Corrigan and H.Kleinert, Phys. Rev. 6. A.O.Barut. Letters 20 (1968) 167. Phys. Rev. Letters 13 (1964) 375. 7. A.Q.Sarker, Phys. Rev. 133 (1964) B802. 8. L.I.Schiff, R.Hofstadter, E.B.Hughes, A.Johans9. H.Collard, son and M.R.Yearian. Phys. Rev. 138 (1965) B57.