Fermi's golden rule: ionization of hydrogen. a hydrogen atom driven by an electromagnetic wave is described by the hamiltonian (61) we are working in natural units wherein the mass of the electron m1, h1 and e2/47«o = 1. these units are called hartree units. v represents the perturbation of the atom by the electromagnetic wave. assuming that the wave has a frequency v and is linearly polarized along the z-axis and that the wavelength is long compared to the size of the atom, the perturbation has the form vaz cosvt. (65) here λ the strength of the perturbation is the amplitude of the electric field measured in hartree units. (a) (optional] hartree units. () show that the hartree unit of electric field is (66) calculate the numerical values of in v/m. the quantity λ is equal to f/f where f is the amplitude of the electric field in s. i. units. (i) show that the hartree unit of frequency is me (67) calculate the numerical value of v in hz. (b) fermi's golden rule. according to the rule the transition rate from a state |i) to a state i) under the influence of a perturbation a expl-ivt)+atexplivt) is given by (68) here we have assumed that a and a are time independent operators and ey and e are the energies of the two states with e / > e . for a hydrogen atorn driven by an electromagnetic wave it is evident from eq (65) that a -x: /2. let us take as the initial state |i) the 1s state of hydrogen denoted j1s) which has the wavefunction (69) 18 and energy e 1/2 in hartree units. as the final state i/) we take a plane wave of nomentum p which has the wave-function (70) and energy e, = p2/2. strictly if we are to calculate the ionization rate of hydrogen we should take the final state to be a positive energy cigenstate of the umperturbed hamiltonian en (61) however if the final state is a highly excited positive energy state then it is an acceptable approximation to treat it as a plane wave as described above. bxelutin of the nut (71) and hence the transition rate by (72) hint: in order to evaluate the integral to obtain eq (71) explain why /dr exp(ip . r) exp(-r). (73) the integral on the right hand side of eq (73) can be easily evaluated in polar co-ordinates (74) dr : exp(ip . r) exp(-r) dr explíp. r)expl-r) = (1+p)2 obtaining this result is a good exercise but you are not required to show it. (d) total ionization rate. the total ionization rate is defined as (75) show that (76) hint. evaluate the integral in eq (75) in polar co-ordinates. in order to perform the integral over p make the substitution