- H ˆ = ˆ H SS + ˆ H A + ˆ H Z (8.4) in the basis. - S, m } of the total spin of the two particles.. - Calculate the energy eigenvalues in the presence of the field B. - express the corresponding eigenstates in the basis. - S, m } of the total spin. - Draw qualitatively the variations of the energy levels in terms of B . - the probability for this system to decay during the interval [t, t+dt] if it is prepared at t = 0, is dp = λe − λt dt, where the decay rate λ is related to the lifetime τ of the system by τ = 1/λ. - with respective decay rates λ 1 and λ 2 , the total decay rate is the sum of the partial rates, and the lifetime of A is τ = 1/(λ 1 + λ 2. - In all what follows, we place ourselves in the rest frame of the positronium.. - In a two-photon decay, or annihilation, of positronium, what are the energies of the two outgoing photons, and what are their relative directions?. - In the orbital ground state ψ 100 , split by spin–spin interactions as calculated in Sect. - 2, the lifetime of the singlet state is τ s, and the. - lifetime of either of the three triplet states is τ s. - In order to determine the hyperfine constant A of positronium, it is of interest to study the energy and the lifetime of the level corresponding to the state | ψ. - defined in question 3.2, as a function of the field B.. - (a) What are, as a function of x, the probabilities p S ans p T of finding the state | ψ + in the singlet and triplet states respectively?. - (b) Use the result to calculate the decay rates λ + 2 and λ + 3 of the state | ψ + into two and three photons respectively, in terms of the parameter x, and of the rates λ 2 and λ 3 introduced in question 4.4.. - (c) What is the lifetime τ + (B ) of the state | ψ. - One measures, as a function of B, the ratio R = τ + (B)/τ + (0) of the life-. - time of the | ψ + state with and without a magnetic field. - (i) What estimate does one obtain for the hyperfine constant, A, using the value of the magnetic field for which the ratio R has decreased by a factor two?. - Variation of the ratio R defined in the text as a function of the applied magnetic field B. - The energy levels E n. - The degeneracy is n 2 for each level, as in the hydrogen atom. - The diameter of positronium is r = 3a 0 /2 = 3a H 0 , and, since the proton is fixed, the diameter of the hydrogen is 2 r H = 3a H 0 . - In the orbital ground state, the degeneracy is 4, corresponding to the number of independent spin states.. - As usual, we can express the spin–spin operator in terms of the total spin S as S 1 · S 2 = [S 2 − S 1 2 − S 2 2 ]/2. - with the energy shift:. - 2, with the energy shift:. - (a) There is a mass factor of ∼ 1/2000, a factor of ∼ 2.8 for the gyromagnetic ratio of the proton, and a factor of 8 due to the value of the wave function at the origin. - Section 8.3: Zeeman Effect in the Ground State 8.3.1.. - (b) Hence the matrix representation in the coupled basis:. - Similarly, one has the matrix representation of the full spin Hamil- tonian:. - 1 , which correspond to the same degenerate eigenvalue A of the energy. - Variation of the hyperfine energy levels with applied magnetic field. - (a) For a given value of the applied field, with the positronium pre- pared in the state | ψ. - the probabilities of finding the system in the singlet and triplet states are respectively p S = sin 2 θ ∼ x 2 /4 and p T = cos 2 θ ∼ 1 − x 2 /4.. - (b) The rate for | ψ + to decay into two photons is the product of the prob- ability of finding | ψ + in the singlet state with the singlet state decay rate:. - (c) The lifetime of the | ψ + state is. - We study the modification of the energy spectrum of a hydrogen atom placed in crossed static electric and magnetic fields in perturbation theory. - Pauli made use of the particular symmetry of the Coulomb problem. - In addition to the hydrogen spectrum, he was able to calculate the splitting of the levels in an electric field (Stark effect) or in a magnetic field (Zeeman effect). - where M and q e are the mass and charge of the electron, and where f (n) depends on n only.. - Our pur- pose, here, is to prove (9.1) in the special case n = 2, to calculate ω 0 and ω e. - We consider the n = 2 level of the hydrogen atom. - What are the energy levels and the corresponding eigenstates in the presence of B 0 only? Check that (9.1) is valid in this case and give the value of ω 0. - In the presence of E 0 only, the perturbing Hamiltonian is the electric dipole term ˆ H E. - Write the matrix representing ˆ H E in the n = 2 subspace under consideration.. - Calculate the energies of the levels originating from the n = 2 level in the presence of the crossed fields E 0 and B 0 . - All points correspond to the same energy of this level, but to different values of the static fields E 0 and B 0. - Knowing that ω e is a function of the principal quantum number n of the form: ω e = (3/2)f(n)Ω e , and that ω 0 and Ω e are the constants introduced above, answer the following questions:. - Values of the electric and magnetic fields giving rise to the same sub-level energy of the n = 34 level of a hydrogen-like atom. - Write the quantity ω 2 0 + ω e 2 in the form λ. - give the value of the constant γ, and calculate f (34).. - Section 9.1: The Hydrogen Atom in Crossed Electric and Magnetic Fields. - The orbital magnetic moment of the elec- tron is ˆ µ orb = γ 0 L ˆ , with γ 0 = q e /(2M. - Y 1 − 1 (θ, φ)) Y 1 m (θ, φ) d 2 Ω where we have incorporated the radial integral given in the text. - We want to find the eigenvalues of the matrix. - There is an obvious eigenvalue λ = 0 since the | 2p, m = 0 and | 2s states do not mix in the presence the electric field. - The shifts of the energy levels are therefore: δE = 0 twice degenerate, and δE. - If we adopt the prescription given in the text, we obtain ω e = 3Ω e
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