PhotoelectricEffect

PhotoelectricEffect - previous next index Photoelectric...

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previous next index Photoelectric Effect in Hydrogen Michael Fowler 3/26/10 Introduction In the photoelectric effect, incoming light causes an atom to eject an electron. We consider the simplest possible scenario: that the atom is hydrogen in its ground state. The interesting question is: for an ingoing light wave of definite frequency and amplitude, what is the probability of ionization of a hydrogen atom in a given time? In other words, assuming we can use time- dependent perturbation theory, what is the ionization rate? Formally, we know what to do. We must find the interaction Hamiltonian 1 H , then use Fermi’s Golden Rule for the transition rate with a periodic perturbation: ( ) 2 1 2 if f i R fHi E E π δω = −− But it’s not that easy! For one thing, the outgoing electron will be in some kind of plane wave state, so whatever convention we adopt for normalizing such states appears in the rate. But also the δ function is tricky for excitation into the continuum: just how many of these plane wave states satisfy fi EE ω = + ? We shall discover that with a consistent formalism, these two difficulties cancel each other. The Interaction Hamiltonian Taking the incoming wave to be an electromagnetic field having vector potential ( ) ( ) 0 , cos Art A k r t = ⋅−  The interaction Hamiltonian is given by replacing the electron kinetic energy term 2 /2 pm with ( ) 2 / p qA c m . The relevant new term is ( ) ( ) ( ) ( ) 1/2 / / m qcpA Ap em cAp ⋅+⋅ = since q = e and 0 A ∇⋅ = in our gauge. Therefore ( ) ( ) ( ) ( ) 1 0 0 cos . 2 ikr t t e H kr tA p mc e e e Ap mc ωω − ⋅−  =   = +
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2 The two different terms in this expression, having time dependences it e ω and e will give δ functions ( ) fi EE δω −− and ( ) −+ respectively in the transition rate. The e term therefore corresponds to absorption of a photon, since we are looking at a process in which the electron gains energy, > . The e term is for the process where an atom in an excited state emits a photon into the beam and drops in energy. So the relevant interaction Hamiltonian is ( ) 11 1 0 where . 2 i t ik r e H t He H e A p mc −⋅  = =   . Plane Waves: Density of States We make the assumption that the final state is a plane wave state f ik r f ke . The most straightforward way of handling the plane wave states is to confine the whole system to an extremely large cubical box of side L , and impose periodic boundary conditions (so that plane traveling wave states are allowed). The big box has volume 3 VL = , so the appropriately normalized plane wave states are 3/2 ik r ik r e L V ⋅⋅ = =  . As will become apparent, we need to count how thickly these states are distributed, both in momentum space (or k -space) and in energy. We’ll begin by reviewing the one-dimensional problem—the three-dimensional case is a simple generalization.
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