Question

11.17 A positronium atom is a hydrogen-like atom with a positron (m = meq = te, spin 1/2) as a nucleus and a bound electron. The hyperfine structure in the ground state of positronium is described by a perturbation Hamiltonian H' = AS, S/hwhere S, are the spins of the elec- tron and positron. a) What is the Bohr energy of the ground state of positronium (ignore hyperfine structure for now)? b) The electron and positron spins can be coupled to form the total spin S of the atom. Write down the spin states of the coupled and uncoupled bases and how they relate to each other. c) Express the hyperfine Hamiltonian in the ground state as a matrix in both the coupled and uncoupled spin bases. d) Determine the effect of the hyperfine perturbation interaction on the ground state of posi- tronium. Draw an energy level diagram to illustrate your results.

Answer 1

answer:

A POSITRONIUM ATOM IS A HYDROGEN-LIKE ATOM WITH POSITRON

IS (M = Me, q = +e, SPIN 1/2) AS A NUCLEUS AND BOUND ELECTRON.
The hypentine structure in the ground state of positionium is described by o perturbation Hamiltonian is . H = As, Sal ħ. Here, si are the spins of the election and position. - The hydrozen energy level are given by En = name .202 Нете , à is the structura constant, m is the may of election, à is relocity of light i is principle quantum number.

The hydrozen energy levels are given by En = -&mc2 212 Нетр, m' is really the reduced may. for positronium, the rieduced mas, .u- mime 'mitme Meme metme me it is approximately hall of what it is in hydrogen. - Hence the energy levely in positionium are halb of what they are in v the site in hydrozen.

. Hence the positionium ground state energo is, G.8 E = -18.6er 202 = -6.8er ... Hence, Bohen ehergo ob ground state positronium is 1-68 eV (b). The uncoupled basis staty are, Ims, ms₂> 1.17), 1+->,1-+),1--> The couple state are ISMI

1.11)=1t) 1000 = " (H=3+1++)) 11,-1) = 1 --> triplet slate 1000 = Y; (=)+=))} singled state The hyper fine Hamiltonian can be written as, =) S- Sit.se - 3² = (sitsa)² =) s² = (siets+ Sı. Szt Sz.S.) :-(sit st 25,52) 3, S2 = 1/2 (sts, "-sz)

- 1 h 15 (stil-3 - The matrix in the coupled bay is is, H' = As, SL = ln AAP [s0511-327 : A1? Loo 07 1010010 1001 0,- 2000- 300 - The matrix in the concoupled bay is is, Sio Sg = Six Sent Sly Say + 5z. 922 - all suus +5,374) +5,322) #! - Alle [89-49524) + 5.992] - ATT0001H 10 -1 2 04- 1.02- 10 0 0 0 1 - -

since the matrix is already in the coupled basis it can be used to get Hm As, S2 =lz [s(s+1) =>] <s'm'lu'/smy) = y^k' (sest) -26]. pomi S=1 I har sal Energy level . diagram is given below. S\ 3 A/4 -=D