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by: Quinn Larkin

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# Quantum Mechanics I PHY 851

Quinn Larkin
MSU
GPA 3.67

Staff

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COURSE
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3
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KARMA
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## Popular in Physics 2

This 3 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 851 at Michigan State University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/207632/phy-851-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15
Lecture 24 February 7 2001 Decays The most common example where particles are destroyed and created is in the decays of particles or atoms For instance a hydrogen atom in an excited state might decay to the ground state Via the emission of a photon The photon must be created in the process Creation of a photon is surprisingly complicated due to the fact that the photon is a massless spin1 particle coupling to the current Since the Lorentz structure of such a decay is potentially confusing we will begin with a scalar form for a decay Consider hypothetical oz and 8 particles where the oz particle has a ma and the 8 particle is massless We assume both particles are spinless The a particle is in the rst excited state of a Harmonic oscillator characterized by frequency We wish to calculate the rate for decaying to the ground state Via the emission of a 8 particle which we will assume to be massless We assume the interaction term is expressed Hint g rmgm xigm rum ram Using Fermi s golden rule the decay rate is 27r 177 7 Km my n 0 k3 Hgnt n Hy 0 n 1 25E0 E Egg 193 Before proceeding further note that Hint has both a 113 and a piece in order to keep the Hamiltonian Hermittian Since we are interested in the decay of the atom which creates a particle we can neglect the destruction term Also note that the initial state assumed to be polarized along the z axis This choice is arbitrary since we are summing equally over all directions of k If we were calculating a differential decay rate dF d521 the answer would depend on the initial polarization One can now calculate the matrix element 1 m Hy n 0 k Hgnt n Hy 0 n 1 g d lm Wc k m ynl r Writing the wave functions in terms of normalized ld harmonic oscillator wave functions 0 m Hy n 0 k Hint n Hy 0 n 1 9 i145 iquot iuz 27nw Wdz cquot wax quotyywoydzcl 2V7 1952 2 zgkza mu U2 2 c I a 1 W27 where a is the characteristic size of the ground state Lecture 24 February 7 2001 Putting all this together 4 2 9 9 F 564 1 26fzkc 2 v a y lst ldls cos2 6 1 cos 62 k39 quot26hkc 27mm 921371 iksas 37rmhwc where ls wc and w is determined by energy conservation ha E1 E0 Electromagnetic decays Electromagnetic decays are complicated due to the j A A nature of the coupling Remember that minimal substitution p gt p eAc results a term in the Hamiltonian which looks like 2 Him J representing the interaction with the vector potential The last term will be neglected for now but will play the de ning role when we discuss the quantum Hall effect Remember the current is given by A2 1 7C x 2m x1 xVIl W xIlx First we must de ne the electromagnetic eld operator in terms of creation and destruction operators that make real photons l27rh2c2 1 7 1 7 1 fur t 7 63 E ezkvr zEktmukES e zkvrzEkt Lak3gt I ks V 3 Here 3 refers to the polarization or spin of the photon For each Is there are two polar izations Each must be perpendicular to the direction of k Aside from the polarization the expression for A looks peculiar for all the prefactors and the lEC term inside the sum However all these terms are necessary to ensure that E2 B2 1 1me ka awam 5 where the electromagnetic elds resulting from A are 12 cat B VXA We are now in position to consider the general problem of electromagnetic decay where a particle of charge 2 changes from state 7C to state f while emitting a photon of momentum la and polarization 8 Outlining the steps to solving the problem 2 Lecture 24 February 7 2001 1 Write down Fermi s golden rule 2 1 F 7 EEm k7 3 3351 Axiigti2 Ez E fake 2 Write down the matrix element 2 ltfksd3jxAxiigt d3xezk fltxgtv iltxgt 2 ch 27r m EkxEk Jkaf a d3xe ik x xxgt Note M does not depend on the polarization 7 7C Vd dxl 4 C20 Change the sum over k to an integral V 3 V 2 gaWd k 2703p ddelk Then eliminate the delta function in Fermi s golden rule 62k 139 f2 mgdgzkggsME One can check the units of the above expression by noting that 62k has dimensions of energy it is the same 627quot and that M has units of momentum which are the same the units of mo thus the overall expression has units of energy over 7 which is an inverse time Remember that in many books and tables F often refers to energies rather than rates in which case one erases the 7 in the denominator F 4 Finally perform the integral to nd M The dipole approximation For nuclear examples typical y energies are on the order of one MeV which gives wavelenths of a few hundred fm approximately 100 times the size of a nucleus For atomic examples emitted photons usually have wavelengths of hundreds of nm several thousand times the size of a typical atom Thus one might consider approximating the phase factor 61quot z 1 This approximation is known the dipole approximation The term dipole can be explained by noting that in the dipole approximation M m ltfPi7igt ltfinor1igt imEfEi I fltf lgtt

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