QUARK EFFECT ON H-ATOM SPECTRA(1S)

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1 QUARK EFFECT ON H-ATOM SPECTRA1S March 8, 18 1 A QUARK MODEL OF THE PROTON The purpose of this paper is to test a theory of how the quarks in the proton might affect the H-atom spectra. There are three quarks point particles making up the proton. There are two positively charged quarks, which are called up quarks, each one having a charge q = e/3 where e is the proton charge. There is one negatively charged quark, which is called a down quark, and has a charge q = e/3. When only the proton is present, take all three quarks to be at the origin. For a proton in the presence of an electron, it will be assumed that the electron attracts the positively charged quarks, and it repels the negatively charged quark. So a positive quark is always assumed to be displaced from the origin on the line from the origin to the electron. Label the displacement of the up quark from the origin δ. The negative quark is always assumed to be on the extension of the line from the electon through the origin. Label that displacement from the origin. The displacements δ and will be assumed to be much smaller than a, the Bohr radius. Often Taylor expansions of the exponentials will be made. For example, exp δ /a = 1 δ /a 4δ/a. This series and similar series enter the equations for the energy shifts. Only terms in the energy shifts like eq/a and eqδ/a will be kept. Terms such as eqδ n /a n1 for n are negligible in comparison, and will be discarded. 1

2 THE POTENTIAL ENERGY OF AN UP QUARK AND THE ELECTRON Let r be the distance from the origin to the electron. Consider the case where δ r <, and define to be the distance from the quark to the electron. Then r = δ. In the region defined above, the potential energy of the up quark and electron is V = eq /. Consider the case where r δ, and define to be the distance from the quark to the electron. Then r = δ. In this region, the potential energy of the up quark and the electron is V = eq / The total potential energy of the up quark and the electron is V = eq H δ eq Hδ r. 1 where the unit step function is defined by Hx = 1 for x >, and Hx = for x <. 3 THE POTENTIAL ENERGY OF A DOWN QUARK AND THE ELECTRON Let be the distance the quark is from the electron. Then = r where r <. The total potential energy of the down quark and the electron is V = eq Hr. Note that the distance between a quark and the electron has three different symbols because the relationship between r and δ is different in the three different regions. 4 THE HAMILTONIAN The unperturbed Hamiltonian for a positively charged quark at the origin and a point electron is H o = p /µ eq /r where p is the momentum operator of the up quark, and µ is the reduced mass. For mathematical simplicity, the mass of the up quark is taken to be the same as the mass of the down quark they are not the same. The Hamiltonian for an up quark and an electron is H = p µ V = p µ eq r eq r V = H o H 3

3 where the perturbed Hamiltonian for an up quark is H = eq eq H δ eq Hδ r 4 The unperturbed Hamiltonian for a negatively charged quark at the origin and a point electron is H o = p /µ eq /r. The Hamiltonian for the down quark and an electron is H = p µ V = p µ eq r eq r V = H o H 5 where the perturbed Hamiltonian for the down quark is H = eq r eq Hr. 6 Thus the perturbing Hamiltonian for the proton and the electron is H = H H. 5 ENERGY SHIFT DUE TO H The energy shift associated with H is δe = ψ rh ψr d 3 r = I 1 I I 3 7 where d 3 r = r sinθ dθ dφ dr, the unperturbed wave function for the 1S energy level is ψr = exp r/a / 4π, 4 exp r/a eq I 1 = r sinθ dθ dφ dr = eq, 8 4π r a 4 exp r/a eq I = H δ a 3 r dr, and 9 4 exp r/a eq I 3 = Hδ r r dr. 1 6 CALCULATION OF I Substitute r = δ in Eq. 9, and note H δ = H. Then I = 4eq exp δ /a exp /a δ δ d. 3 11

4 I is the sum of the following integrals: I 1 = 4eq 4eq I = 4eq 4eq exp δ /a exp /a d = r a exp δ /a exp /a a = 4 4 eq 1 δ 4 δ a a a 4 = eq eq δ, 1 a a exp δ /a δ exp /a d = a exp δ /a δ exp /a I 3 = 4eq exp δ /a δ = 4 eq a δ, and 13 exp /a d. 14 I 3 diverges. The problem is the model introduced in this paper where a quark is considered as a point charge. In a more realistic model, the quarks interact with each other, and this leads to a quark wave function and a quark cloud of charge. Think of the hydrogen atom where the potential energy of a point proton and a point electron is put into the Schrodinger equation. The solution is an electron cloud. When the point quark is replaced by a cloud of charge, the integral converges. The resulting energy contains δ which makes this contribution to the energy negligible, so set I 3 =. Add the results together, and find I = eq /a eq δ /a. 7 CALCULATION OF I 3 Substitute r = δ in Eq. 1. Note that when r = δ, then =, and when r =, then = δ. Then I 3 = 4eq exp δ a 3 /a exp /a δ δ d. δ 15 I 3 is the sum of the following integrals: I 31 = 4eq δ exp δ /a exp /a d = 4eq 4 δ d = 4eq δ. 16

5 I 3 = 4eq δ exp δ a 3 /a δ exp /a d = I 33 = 4eq exp δ /a δ 4eq δ δ a 3 δ d = 8 eq δ a exp /a d. 18 This integral diverges. For the reasons given at the end of section 6, the integral will converge. All three terms are proportional to δ, so they will be dropped. Adding all the contributions to the energy shift from an up quark yields δe = eq δ /a. 8 ENERGY SHIFT DUE TO H The energy shift associated with H is δe = ψ rh ψr d 3 r = I 1 I where 19 4 exp r/a eq I 1 = r sinθ dθ dφ dr = eq, and 4π r a 4 exp r/a eq I = r dr. 1 9 CALCULATION OF I Substitute r = in Eq. 1. Note that when r =, then = δ. Then I = 4eq I is the sum of the following integrals exp /a exp /a d. I 1 = 4eq 4eq exp /a exp /a d = exp /a exp /a a a 4 = eq a eq, a 3 5

6 I = 4eq expδ a 3 /a exp /a d = 4eq exp /a exp /a I 3 = 4eq exp /a δ a = 4eq a, and 4 exp /a d. 5 The integral is convergent, but I 3 contains δ, so I 3 is negligible and is set equal to. Adding all the contributions to the energy shift from a down quark yields δe = eq /a. 1 TOTAL ENERGY SHIFT The total energy shift of the 1S level is the sum of δe and δe, so δe1s = 4eq δ /a eq /a 6 With the center of mass at the origin and equal up and down quark masses, = δ. Substitute q = e/3 and q = e/3 into Eq. 6, and find δe1s = 8e δ 3a 4e δ 3a = 4e δ a. 7 ACKNOWLEDGMENTS I thank Ben for his many improvements to the paper. 6

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