- Measuring the Electron Magnetic Moment Anomaly. - In the framework of the Dirac equation, the gyromagnetic factor g of the electron is equal to 2. - In other words, the ratio between the magnetic moment and the spin of the electron is gq/(2m. - q/m, where q and m are the charge and the mass of the particle. - When one takes into account the interaction of the electron with the quantized electromagnetic field, one predicts a value of g slightly different from 2. - The purpose of this chapter is to study the measurement of the quantity g − 2.. - The Hamiltonian of the electron is. - where ˆ A is the vector potential ˆ A = B × r ˆ /2 and ˆ µ is the intrinsic magnetic moment operator of the electron. - In the framework of the Dirac equation, a = 0. - Using quantum electrodynamics, one predicts at first order in the fine structure constant a = α/(2π).. - 66 6 Measuring the Electron Magnetic Moment Anomaly. - One will make use of the quantity Ω = a ω.. - One neglects the interactions between the electrons of the beam. - 6.1 as a function of the time T, for a value of the magnetic field B T (data taken from D.T. - Variations of the quantity S ˆ .ˆ v , as a function of the time T. - The electron Hamiltonian is ˆ H = mˆ v 2 /2 − γB S ˆ z . - We make use of the property i¯ h(d/dt) O ˆ. - For C 2 and C 3 , we proceed in the following way:. - Hence, the general form of the evolution of S · v is. - In other words, in the absence of anomaly, the spin and the momentum of the electron would precess with the same angular velocity: the cyclotron frequency (precession of momentum) and the Larmor frequency (precession of magnetic moment) would be equal. - Measuring the difference in these two frequencies gives a direct measurement of the anomaly a, of fundamental importance in quantum electrodynamics.. - 68 6 Measuring the Electron Magnetic Moment Anomaly. - Remark: The value of the anomaly is now known with an impressive accu- racy:. - The nucleus of the tritium atom is the isotope 3 H, of charge Z = 1. - The purpose of this chapter is to study the electronic state of the 3 He + ion formed after the decay.. - h 2 /(me 2 ) for the Bohr radius and E I = mc 2 α 2 /2 13.6 eV for the ionization energy of the hydrogen atom, where α is the fine structure constant [e 2 = q 2 /(4π 0. - where q is the electron charge].. - In the ground state | ψ 0 of the tritium atom, the wave function of the electron (n = 1, l = 0, m = 0) is the same as in the normal hydrogen atom:. - (7.1) The β decay of the tritium nucleus leads to:. - ν is an antineutrino), where the emitted electron has an energy of the order of 15 keV and the helium nucleus 3 He has charge Z = 2. - the β electron is emitted with a large velocity and leaves the atomic system very rapidly. - Consequently, an ionized 3 He + atom is formed, for which, at the time t 0 of the decay, the wave function of the electron is practically the same as in tritium, and we shall assume it is still given by (7.1). - We denote by | n, l, m the states of the ionized helium atom which is a hydrogen-like system, i.e. - one electron placed in the Coulomb field of a nucleus of charge 2.. - Write the Hamiltonian ˆ H 1 of the atomic electron before the decay and the Hamiltonian ˆ H 2 of this electron after the decay (when the potential term has suddenly changed).. - What are, in terms of E I , the energy levels of the 3 He + atom? Give its Bohr radius and its ground state wave function ϕ 100 (r).. - Calculate the expectation value E of the energy of the electron after the decay. - One can for instance make use of the fact that:. - Express in terms of | ψ 0 and | n, l, m the probability amplitude c(n, l, m) and the probability p(n, l, m) of finding the electron in the state. - Calculate the probability p 1 of finding the electron in the ground state of 3 He. - Experimentally, in the β decay of the tritium atom, one observes that, in about 3% of the events, there are two outgoing electrons, one with a mean kinetic energy E k ∼ 15 keV, the other with E k ∼ 34.3 eV, thus leaving a completely ionized 3 He 2+ nucleus, as if the β decay electron had “ejected”. - In the present case, E n. - The expectation value of the electron energy in the new nuclear con- figuration is. - where R nl (r) are the radial wave functions of the 3 He + hydrogen-like atom.. - Since ψ 0 is of the form ψ 0 ( r. - The probability amplitude in the lowest energy state is (p π. - Hence the probability p and the contribution to the energy p 1 E 1. - With the numerical values given in the text, one has p 2 E 2. - 0.026 that the atomic electron is not bound in the final state.. - 41.7 eV is smaller than the total expectation value of the energy E by 0.9 eV. - an ionization of 3 He + into 3 He 2+ with emission of the atomic electron.. - There is necessarily a probability 1 − p = 0.026 for the atomic electron not to be bound around the helium nucleus, therefore that the helium atom be completely ionized in the decay. - If the mean kinetic energy of the expelled electron is E k ∼ 34.3 eV, this represents a contribution of the order of (1 − p)E k ∼ +0.89 eV to the mean energy which compensates the apparent energy deficit noted above.. - If M 1 and M 2 are the masses of the two nuclei, E β the energy. - of the β electron, E the energy of the atomic electron, and E ν ¯ the neutrino energy, energy conservation gives for each event: M 1 c 2 − E I = M 2 c 2 + E β + E ν. - For a given value of E, the determination of the maximum energy of the β electron (which covers all the spectrum up to 19 keV in the tritium atom case) provides a method for determining the minimum value m ν ¯ c 2 of E ν ¯ through this energy balance. - The positron e + is the antiparticle of the electron. - We first consider only the spatial properties of the system, neglecting all spin effects. - No proof is required, an appropriate transcription of the hydrogen atom results suffices.. - Express the reduced mass of the system µ, in terms of the electron mass m.. - Write the Hamiltonian of the relative motion of the two particles in terms of their separation r and their relative momentum p. - What are the energy levels of the system, and their degeneracies? How do they compare with those of hydrogen?. - What is the Bohr radius a 0 of the system? How do the sizes of hydrogen and positronium compare?. - 2 in terms of the fundamental constants: m, c. - h, and the fine structure constant α.. - We now study the hyperfine splitting of the ground state.. - What is the degeneracy of the orbital ground state if one takes into account spin variables (in the absence of a spin–spin interaction)?. - Explain why the (spin) gyromagnetic ratios of the positron and of the electron have opposite signs: γ 1. - One assumes that, as in hydrogen, the spin–spin Hamiltonian in the orbital ground state is:. - Recall the eigenstates and eigenvalues of ˆ H SS in the spin basis. - (b) Express the constant A in terms of the fine structure constant and the energy mc 2 . - (c) What frequency of the hyperfine transition corresponds to this calculated value of A?. - Actually, the possibility that the electron and the positron can annihi- late, leads to an additional contribution ˆ H A in the hyperfine Hamiltonian. - (a) What are the energies of the S = 1 and S = 0 states, if one takes into account the above annihilation term?. - (b) Calculate the frequency of the corresponding hyperfine transition.. - 8.3 Zeeman Effect in the Ground State
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