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# Quantum Mechanics II PHY 662

Syracuse

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Date Created: 10/21/15

PHY662 Quantum Mechanics II HWK 11 Due Tuesda Apr 20 start of class 1 P3 9 a Reading see lecture notes More cl tromognotism and QM 3 pts Consider a single particle with position F suh jecg to a classical electromagnetic eld E B that is computed from potentials A7 154 a Show that lt15 7 31gt is gauge invariant b Show that MS 1 a q a 7 iltp 7 7A dt 5 c Show that for uniform E and E SQUIDS 2 pts Superconducting quantum interference devi s can he used to measure changes in magnetic elds In a superconductor the effective degree of freedom is not a single e tron but is a pair of bound electrons A SQUID has two weak link in a ring through which a current is sent If the ux through the ring is increased by a ux quantum where q 2e in the formula for a ux quantum the voltage across the ring ris s then falls So one can use a S U ID to count ux quanta The SQUID magnetometer based on this principle can he used to measure the magnetic elds created by currents in the heart or the brain Suppose you have a squid magnetometer with area 1mm 4 How far would you need to raise the SQUID up for the decrease in the Earth s magnetic eld to cause a single voltage oscillation in the SQUID Make an order of magnitude estimate approm mating the Earth by a magnetic dipole and using the zqiproximate strength of the Earth s eld at its surface H quot 5 v 1 n a 2 Checking on a umption 3 pts Remember that expanding n 7 2A gave terms rst and second order in fI Check that we can neglect QfMQ A2 compared with LA 1539 for an electron in an atom if the the atom is located in the Earth s atmosphere and the electromagnetic eld is due to light from the Sun Compute the order of magnitude of this ratio Some commutation relations 2 pts Just in todz lecture we learned that the matrix element n HO may he of interest and 0 are elec tronic eigenstates with their own angular momentum l and z component of the angular momentum m We will pursue this on Tuesday In prepa ration con rm the commutation L27 L27 2FL2 L2 relation 397 X Problem 911 in for the orbital angular momentum operator L Grif ths PHY662 Spring 2004 Outline for Thurs Jan 22 2004 More Spinl Cryptography 22nd January 2004 1 Administration 0 em ail addresses 0 Acknowledgements on HWK If none please write I did these problems inde pendently o Homework due Jan 27 but will be handed out see HWK for reading and prob lems 0 Next week is magnetic resonance and possibly twolevel systems We will not use Shankar for those topics next week Today 1 A couple of results for Pauli matrices 2 Symmetry and conservation laws 3 Quantum cryptography 4 Spins and their magnetic moment magnetic elds 2 Following Shankar by using 327 Sj theiJkSk and assuming 2 states Establish the expression exercise Aai aABiix a for vector operators A E that commute with 639 Reproduce effect of 5 and S on spinors check with Pauli matrices Note interesting that one can go back and forth between and ft Not true for other spins Why not 3 31 L an 0 32 Introduction to quantum cryptography Classical cryptography Not everyone wants their mail easily readable Encryption of a plainlexl to cypherlexl is the conversion of the message to a form that is presumed to be readable only by the intended recipient Historically there is a sequence of encryption schemes simple substitution for letters transpositions polyalphabetic substitution Vignere cipher etc that are based on a shared key that is not too long Patterns lead to ability to read the message Interesting consequences of that Publickey cryptography has revolutionized classical cryptography but assumes that some problems like factoring very large integers are dif cult to solve They are based on trap door functions you hand out padlocks publicly then someone who wants to write a message locks their message Even the sender can t unlock the message But with your private key you can One code that is theoretically unbreakable code X if you will onetime pads This code is used only for very secure communications that are infrequent as key distribution can be cumbersome and insecure Quantum cryptography key distribution One use for quantum cryptography securely distributing a onetime key with no fear of eavesdropping Relies on noncommutativity of the spin operators and the collapse of the wave function upon observation Other schemes also depend on quantum en tanglem ent Here is the classic example Bennett and Brassard 1984 adapted to spinl2 usually is described with photons mp Alice sends a sequence of states randomly chosen from H iivil Bob randomly chooses to measure in the 2 or i direction for each electron Afterwards Alice and Bob publicly announce which axes they sent or measured the electron along 4 Where they used the same basis they have a sequence of up and down translated into a sequence of l s and 0 s that they agree on This can be used as a onetime pad for a subsequent message sent classically 5 No one else can know these states For if Claire measured the electrons en route from Alice to Bob Claire would have disturbed the wave function Example Alice chooses the random sequence Hi hi igt 72 igt 72 The particles are sent to Bob Bob makes measurements randomly chosen along the axes I I z z I I At this point Alice publicly announces 2 I z z I z and Bob publicly announces I z z z z I They agree on the 2nd 4th and 5th spins So they now jointly know the sequence which they can use as a bit sequence 101 key Alice decides to send the message plaintext001 which gets converted to plaintext 69 key 001 GB 101 100 Only Bob who knows that 101 is the key can properly convert the ciphertext to plaintext 100 GB 101 001 Now Alice and Bob can also check for snoops who might be peeking at their bits lf Claire had been using an apparatus in between that was making measurements it would change the amplitudes This all gets more complicated with errors in the signal etc Errors in alignment of apparatus or patterns in random number generators might provide an attacker with clues Photons work better than electrons or spinl2 atoms here one sets polarizers in 4 directions leftright updown and 2 7r4 rotations to prepare photons polarized in 4 different directions but the scheme is the same 4 Conservation and Symmetry Symmetries and Conservation Laws a reminder NOTE the rotation operators in Feynman are for a change of the frame passive trans formation So all angles are inverted to get the rotation operators in Shankar 1 For spin what is the conservation law and what is the symmetry 2 How are rotations and spin related by logarithms 3 How unitary symmetry operators are related to conserved Hermitian operators See Ch 11 in Shankar for translationmomentum 5 Spins in a constant magnetic eld This is very simple really in quantum mechanics To explain magnetic moments in a constant magnetic eld one needs all these ideas of torque etc The result is precession similar to precession in classical mechanics but easier The Hamiltonian for a magnetic moment in a magnetic eld is H i E Where the magnetic moment operator is related to the spin operator by E 75 5 Where the proportionality constant 7 has been rede ned in terms of a unitless 9 factor For orbital angular momentum g 1 For spin it depends gelectron m 2 Qproton W 5 6 and Yneutron 3 8 2Mnefmonc The time evolution operator Ut is now given by Ut exp7thh exp i7t What other operator is this equal to Example consider a spinl2 particle in the state H i in a eld oriented in the 2 direction What is its state as a function of time What are 51 ltSytgt and ltSztgt7 Note the itdecay experiment and possible indications of new particles supersymme try PHY662 Spring 2004 Feb 26 2004 4th March 2004 1 Miscellaneous 1 HWK 8 due Tuesday Mar 16 2 Continue to read Ch 17 Shankar or Gri lths Ch 6 especially for 2ndorder perturbation theory Start Ch 18 for timedependent perturbation theory for Tuesday 3 Today Transmission coef cients in 1D 2ndorder perturbation theory degener ate perturbation theory 2 Transmission coef cients Last time we reviewed the problem of a plane wave Aeik incident upon a step in the 2 2 potential AV lt h T Remember that we found the amplitude for the transmitted wave Gem115 Hg 7 2 AV with C A What is the probability of transmission through the step 3 2nd order The second order corrections to the energy require the rst order corrections to the wavefunction Note that the wavefuncl ians calculated using perturbation theory are generally less reliable than the energy estimates In any case the rst order corrections to the wavefunction are found by computing the inner product of the rst order terms in the Schrodinger equation for eigenstate n with the unperturbed eigenstate m lt 1H 1 igtlt 1H 1 2gt lt 1E21 igtlt 1Ei1 2gt Emma lt 1H 1 3gt E2lt 2 lwigt mum lt 1H l 3gt 133 E31 lt l igt 0 1 WWW lt ml ngt W Since the set of all W2 form a complete basis 1141 can be written as a sum over 3 Clearly Elm lt lH l 3gtE3 E90 SO 0 H 0 W Z lt1 Ej gm 1 577171 m n Why did we leave W2 out of the sum Well the rst order approximation to W is W2 so that any part of that is proportional to W2 is redundant In fact this choice also keeps this rst order wave function normalized at least to rst order lt nl ngt lt 3llt ill 3gtl igt 1 0 0 terms second order in i Given this we can now look at the second order correction to the energy by taking the inner product of the second order part of the Schrodinger equation with 9 imme diately canceling the rst terms on each side due to the same trick as for rst order ngHO ngEO ltw2lH lwigt Emma E Z lt 3lH l gtlt lH l 3gt 7 E2 7 E3 0 mfn lt 3lH l gt E2 Z l EH l mfn an expression that again uses only matrix elements of H 3 1 Application Here is a sample application taken from problems 61 and 63 in Gri lths Let H 0 be the square well potential for a particle con nedto 0 lt z lt a so that the normalized unperturbedwavefmictions are 11 sinclffx with E2 f0 sinW dz If the perturbing potential is a deltafunction at z a27 H 161 7 en E Aaltw2ltzgtra6ltziggtw2ltzgtdz 2a a 2 7mm a i 39 7 5 aOsmaz 2 7 270 nodd 7 07 neven The second order correction is 2 E2 Z 4710 dz Sim s1nm 6z 7 n W 7 M mfn 2167 71ltmngt2 2 2 n2 7 m2 W h mnmnbothodd a 3900 2 167212a2 J 1 2amp2 H32 2139 1 7 lt2k 1gt2gt 01 4 Degenerate perturbation theory The expressions for 11 and E72 are problematic when there is an energy degeneracy ie two unperturbed states with the same energy The energy denominators E2 7 E2 can then be zero and we don t like to divide by zero Even E71 will be affected b degeneracies this is not obvious from the expression E71 lt11ng Wig but degen eracies do affect the derivation of this result The main problem is that the perturbation selects a preferred basis among the degen erate states Consider eg a twolevel system with H0 0 0 and H0 2 3 What are the unperturbed and perturbed eigenvalues What would you get if you naively applied rstorder perturbation theory So what do we do in the face of degeneracy The trick is to choose combinations of unperturbed degenerate energy eigenstates that are eigenstates 0f the perturbation As any linear combination of unperturbed energy eigenstates has the same energy this maintains the unperturbed energies but keeps the perturbation from mixing the en ergy eigenstates Roughly the relevant matrix elements in the numerator are zero so a zero energy denominator is OK More precisely we are trying to diagonalize the total Hamiltonian H H 0 H over the degenerate subspace Let S 1 be a set of states degenerate under H 0 with identical values of E0 E0 If we transform those states to a new basis where H is diagonal then it is possible to treat at least the rst order part of the expansion HOwi H ibil 192 End by rearranging to get ltH 7 EM H wz 7 EM Contracting with 11 gives lt 3lltH 7 E l igt H E71 which is troublefree as Will H0 7 E0 0 and H is just the eigenvalue of H for 2 if 11 were not an eigenvector of H we could contract with it with n y n and H hbg not orthogonal to Mg to get 0 lt 3lH l 2gt 0 a contradiction what happens is that becomes illde ned PHY662 Spring 2004 Feb 10 2004 19th February 2004 1 Miscellaneous 1 HWK 6 not quite complete 2 Read Ch 8 Grif ths 2 WKB semiclassical approximation r A method mostly for 1D problems It is very useful for understanding tunnel ing 19 When E gt the Schrodinger equation was wave type solutions aim with Mr 4 7 V When E lt the solutions are of the form aim with H 4 V 7 This is strictly true for uniform potentials Vz const L Now apply a type of perturbation expansion Formulated either as an expansion in h or that the classical momentum changes slowly compared to the particle s wavelength almost everywhere 4 The trickiest part of this approximation is where the classical momentum van ishes This is where the oscillating solution needs to be connected to the expo nential solution Leads to connection functions a These are simple if the potential changes very rapidly over a distance much less than the particle wavelength Apply continuity or vanishing of the wave function b Example quantization condition in a well with sharp sides gives 2 12 dx x 2mE 7 V nh come back to after Merzbacher 0 Otherwise need to be careful use connection formula These give for example the quantization condition 2 2 dx 2mE 7 V n 7 h for a bound state with classical turning points x1 and 12 Now for a sketch of the details initial part follows Merzbacher Take 1M1 expli Ihl Then 7 kz2 0 dz2 dz where Mr gm 7 vltzgti Write dab 2 2 w E r k Wig then we can solve for 151 by integrating k2 iqb One can expand this root when liel ltlt W What is the physical meaning of this In any case one gets the result Watm 1 eiifzdxmmr k I 21 Airy functions These functions Az and solve 2y This function has two nice uses 1 It allows us to patch together oscillatory and damped solutions at classical turn ing points 2 We can explicitly solve problems with linear potentials Note the form of the Airy function with a sketch Also note the asymptotic forms Am 2 2141ezm 2 gtgt 0 Aiz7T7z14 1sinaids2 2 ltlt or PHY662 Spiing 2004 Outline for Tues Feb 3 2004 NlVIR lVIRI CG 3rd February 2004 1 Miscellaneous 1 Homework 4 handed out 2 Note Exam on Feb 12 I will hand out review information on Thursday Feb 5 3 Questions about HWK 3 4 Colloquium today quantum computing Very relevant Hand out abstract dis cuss qubit superposition entanglement If interested see also httparxivorgabscond mat0305461 by Mooij group 2 NMR MRI NMR Consider the Hatom which is essentially a proton immersed in a relatively strong few Tesla magnetic eld Expose the atom to a gpulse Then rotates at frequency 7B0 This rotating magnetic dipole radiates energy or creates an oscillating electric eld due to the rotating magnetic moment in a coil With enough rotating protons you can pick up a signal Note that this signal is safe as it is low energy radio frequency RF 10 s of MHZ This radiationemf provides the basis of a method to detect atoms and the effects of their chemical environment homework problem 3 set 3 People can even reconstruct the structure of proteins MRI One gets chemical information in a somewhat different fashion but more im portantly one can get the spatial location of different types of tissue where the charac teristics of the relaxation of the radiation varies The tricky part is how to encode spatial information at the submm scale in the long wavelength meters radio signal that the rotating protons generate Figure 1 NMR spectrum showing the chemical shifts of the protons sit tin in distinct parts o a molecule This ngerprint is a lot of in ten it of sig al Vs frequency for C7H7Br from httpwwwchemuclaeducgi binwebspectracgi7Problembp15ampTypeC 21 MRI One dimension 1D frequency encoding The signal picked up in a detection coil from the excited protons has a frequency t at 39 B0 depend on say the mposition a zr applying a gpulse one gets a signal composed of different frequencies with the magnitude of given location The spatial information is encoded in the frequency Something like Stdwsw a wnm nqnn r quot ulnit where s 39 39 39 39 in the plane de ned by 3 Here is an earl MRI image from 1973 that helped win aNobel prize in 2003 based upon frequency encoding Paul Lauterbur used an NMR machine with controllable linear eld gradients to measure the projection of the density ofwater along four di rections These four projections each using frequency encoding were combined to estimate the location ofthe mter tubes 22 MIer Slice selectinn Ezfme mg m when m the mum yau can 591m a mu m excxte Ths 2 can um ms us edan neuaselectedz magmas amyexcmsaZDlnya ufpzmansthz shcequot A emmBu canthenbe spphed mman slang an axxs um 23 Mer 3D imaging phase munding Ta ablamz 3D mfarmmm an cambmes 0 5m 59mm mg a iguana t m a gyms unwed hm phase encadmg39 and um m frequency encadmg Ta unplement phase mending 5 gm 3 slang mm d112cuan jxs Wham a mu m af ant There are many mhexmcks m gym sum as am gamma clever but quiz mm 51 Theresu ts are maxedlblyxmpxesswe thugh andeven allwwfmxe rhmexmagng af such mugs beahnghzans and awmg charmed acuvny m the bum as we mm usual mm ID planes Fxgure 2 MRI mug a a bum rm Th2 Wm Exam Anus 1 wwwmedharmdedu The mug 5 a 127mg whmh plus m W m mm m phase immense decays me a gyms 3 Addition of angllar momentum and Cleb schr Gordon mcienls d ngtagnhzr Anguhr mammms a fpecl c use Ufa gem pmblem Ths gznenl gven symm pmblem xs understan ng cambxmuans xepxesmtahans uf a my Rem smtahanthzaxy a 51w ya apexauanscanbeexpus a exam mm 25 lIexe 1 2 252 m quotm unba Lhaxepxesenlahanfmthe symme aw pentms atahan mam mu gemntm39s emuth 1 1M 1 1 An irreducible xepresmlahan af 5 symmetxy is an um cannalbe m k m 12 52mm Fax Imam mamas Lhs mmsum m aperamrs ammth transfarmedta bxackmgm ram Genznl g eman Nhenyautake m Mammy aftwamenhcxble representmms haw canqu decampase Lhs pramctmta dmmm afm39eduuble mummy Sgecx c examgle 1r yau Canada Wu ypxnlZ puma whatvalues can mm mm spm hm mame m Amplitude m be m nah af qese tamrspxn anus Clebscthdan caef cxents are usednatjmtfm cambmmg anguhr m am ants but 515a fax mpm numb1 srz gaup sra m gum phyacs m Clebscherdan caef cxems g a as far example betweennus ufpassxble pmcesses When adding angular mamenta gumqu afSIZ we are 5 ng mm are m mm mm angu ar mam m mates and haw are 39heyxehled m m campamnt sumo Seep 331mshmkurmmscmamafmducxmupusemuans 31 Why add together angular momenta Sample motivations 0 An electron in an atom has total angular momentum given by the sum of spin and orbital angular momentum It can be that this total is conserved though spin and orbital are not separately So the state of the electron may be given by j not 1 and 5 But you might measure ll or 52 To nd out the probability of measuring a particular value of 5 you need to be able to relate the total to the parts 0 When you combine two particles together you add their angular momenta Or when a particle decays into two other particles you split the angular momentum between the two particles and their relative angular momentum So you need to know the possible combinations and their probabilities 0 Composite particles like Cooper pairs in superconductors nuclei or particles composed of quarks have total angular momentum grven by summing the parts and their orbital motion 32 Combining representations Let a representation of a symmetry be given by a set of matrices 1f the matrices can be simultaneously written in blockdiagonal form the representation is decomposed into subrepresentations That is the vector space operated on by the subrepresentations are not mixed together by the symmetry operations 33 Review of effects of raising and lowering operators This material is also presented in Sec 125 in Shankar p 321 and following A sample question for one spin is how is J1j7 m related to j m 7 1 Here Mm is the state with J2 jj 1f r2 and an eigenstate of J with eigenvalue mh Note that JZJ JZJE 7 My JEJZ thy 7 inJ2 7 thE JJz 7 it so J Jjmgt m 71hJjmgt so J does lower the eigenvalue of J by h To nd the proportionality coef cient between J1j7 m and j m 7 1 note J J and JJ ch inJ 7 My J J 7 239J Jy 7 JngE J2 7 J th So ifW Jjmgt WW ltj7m1J7J1J2mgt 92721102 7 J mum Tangle 10 1 7m2 m1h2 This gives J1j7 m i9j mjj 1 7 mm 711jm 71gt By convention set 9jm 0 so J71j7mgt J39J391 Mm Dillm 1gt Hmw 7m11jm7 1gt and similarly Jjmgt jj1 mm1jm1gt jimjm1jm1gt Note there are two standard ways to write the coef cient 34 Combining particles Take particles with spin Let f1 and f2 be the operators that rotate particles 1 and 2 respectively De ne total spin j f1 f2 and total J J1 J2 operators More completely this notation is understood to mean f J11 I 11 eg where 11 are the identity matrices for particles 1 and 2 Note that JM and J commute for all 239 j I y 2 as the operators work on the particles separately Hence Ji7 Jj ieijk Jk so the total angular momentum components obeys the same algebra as any rotation operator So what representation does the total angular momentum together follow Phrased another way this question says how can we relate states of given individual angular momenta and zcomponents jl m1 jg mg to total angular momenta states Mm inlay Firstnote that in general Jzij1m1j2m2gt m1 m2ij1m1j2m2gt so J the total zcomponent is diagonal in the jhml jg m2 basis 35 Example Two spin 12 particles This is the most important case to know and is also the simplest Answer 1 0 351 Approach 1 as in Shankar Each spinlZ particle can have spin up or down So let s write the states for two particles as i M 7M 7 Jr 7 7 Then COO Sim 51 52 h COCO GOOD 1 0 0 0 and 2 f Sf5 2 1 2 Sf5 51gt325132 h DOOM OHHO OHHO COCO Note that 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 S Slgt121152gt h 0 0 0 0 EL 0 0 0 1 0 0 0 0 0 0 0 0 1 1 2 0 Now the central part of that matrix can be diagonalized from 1 1 to 0 0 The eigenvectors are 71 1 7 1 and 71717 This means there is a transformation so that 0 0 0 5212 52h2 COCO COCO COMO COCO GOOD 2 0 0 0 OOOH 71 This shows there are 3 spinl states and l spin0 state here Rearranging the rows and columns to re ect the matrix elements between 1 1 71 1 7 11 7 7gt and 71717 1 gives 1000 2000 7000027 0200 Sl h00710 5 h0020 0000 0000 and carrying out the details of the transformation 5 7 USLFUJr with 1 0 0 0 1 0 0 0 L L U 0 1 0 0 0 V1 W 0 0 0 0 1 0 w 7w 0 0 0 1 0 0 0 0 1 gives 0 0 0 7 7 0 0 0 5 7 511712 1152 7 h 0 0 0 0 7 0 0 0 0 577 10001990 0000 0100 07701000 1 USrU h0001 0701000 0010 00010110 10001000 11 o lo 0100 07770 0001 0653510010 0000 ih ooo 0 00 0000 Now 51 and Sy are combinations of 5 and 5 so all Si components of 57 are of the blockdiagonal form t t t 0 t t t 0 t t t 0 7 0 0 0 s where t are the triplet spinl matrix elements for spin and s is always 0 singlet representation This shows in detail how the 2 spinl2 representations combine to form 2 subrepresentations 352 Approach 2 preview of the general procedure With two spinl2 particles there are 4 states in the 817m1827 721 basis given just by 7nth ii i There is only one state with S 1 ll This implies that we must have an integer subrepresentation buried in 5 As there is no higher 5 the J for this representation must be 1 so OlH l l l i171gt l By the same argument 1 1 1gt 2 27 2 i To nd the other state that makes up the J l representation apply J3quot to ll 1 J111gt 12 7 0hl10 l 171 ii gt l2 Tangle l l l l 7 1 2 7 7 7 7 JJl22272gt l l l l l l l 7773977 hii39iiir 27 2 272 272 27 2gt The state 0 a gt 15 5 lt1 7gt17 gtgt isthem estateof the J l representation This triplet of states Jr 7 l 7 gtl 7 7 makes up a representation of J l What is remaining One more state This must be a J 0 singlet orthogonal to ll 0 This gives 7 l 7 This combination of a triplet and a singlet resulting from two spinl2 particles and the symmetry of these states is a primary example in quantum mechanics 36 General case Clebsch Gordon coef cients We wish to write the state Mm j1j2gt as a sum over states ljl 72117 m2 There is a well de ned procedure for doing this Write the maximal jg state as lj1j1j2j2gt This is ljjj1j2gt 19 Find all other states with total angular momentum j by repeatedly applying the Jim operator L Find the states with total angular momentum j 71 by nding the state orthogonal to ljj 7 l j1j2gt and then again repeatedly applying Jim r e Repeat step 3 for the j 7 2 states etc where lj 7 k j 7 k is the state orthogonal toaHOfljyji kgt7lj17j kgt7lji k17j M PHY662 Spring 2004 Apr 1 2004 1st April 2004 1 Miscellaneous 1 Reading Continue Shankar Ch 18 for tim edependent perturbation theory elec tromagnetism also Grif ths Ch 9 2 Timedependent perturbation theory especially periodic perturbations 3 Homework will be handed out by Friday and will be due next Thursday the next two weeks homework will be due on Thursday O lce hours will start at 400 on Wednesday 2 Fermi s Golden rule From the derivation in last class s notes we obtained Fermi s golden rule For H VG f dw pw coswt the transition rate from 239 A f is 7139 n r Flt vltw gtizgt pltw gt waftgt1 For H Ve i the transition rate from 239 A f is 27f a 2 T W ngW WW 5Ef Ez FM In this form the 6function has been pulled out for convolving with pw and the per turbation is taken to be of the form VG we i A coswt perturbation is the sum of two exponential perturbations with opposite frequencies The latter more common form expresses the rule in a way that allows one to integrate over either Ef or over m It can be used to obtain the other form by writing coswt eiwt ion The result is that the transition rate is proportional to the square of the matrix element between the initial and nal states of the spatial component of the perturbation This is a rst order calculation in the amplitudes with V being the part of the total Hamiltonian H H0 H that causes a transition between and Thisjust comes from the rate of change of the wave function including a part from H as per the Schrodinger equation The 6function is a result of energy conservation with one quantum of the perturbing potential having an energy hm This energy conservation comes from an integral over time of an exponential with an imaginary argument There are three parts to this argument the frequencies of the initial and nal state and the frequency from the perturbation When the sum of these is zero integrating over long times gives the 6function 3 Examples Fermi s golden rule allows one to compute real transition rates for real atoms To do so we will need to build up some background with electromagnetism as applied in quantum mechanics Before we do that let us try a more arti cial example Suppose you have a charged particle in a 1D harmonic oscillator with frequency me You modify the potential by applying a classical electric eld 8 t that is uniform in space and that behaves like white noise over time At any given instant t the change in potential is 748 tX White noise has the property that the fourier component at each frequency is uniform Writing 8 t 0 C dw E0 coswtp0 gives AVt qftX dw qXEopo coswt 0 where it is understood that there is a randomness between the frequencies that makes the perturbation incoherent one could add a random phase to each wt Fermi s golden rule then gives the transition rate from the ground state to the rst excited state as 7T Teal Thgl quEolOHQ o 2 2 12 TrqE 2h 2h2 llt1llt7m0gt aal0gtl2po W moh Let us check the units to help us believe that the calculation was properly done 4 Electromagnetism See Shankar for Maxwell s equation and this discussion The highlights are rewriting the electromagnetic eld using the potentials A and 15 E in l a 16 E ZE W the choice of Coulomb gauge for the free electromagnetic eld where sources den sity p 0 and current j 0 V A 0 lt15 0 and the equations of motion for A in the Coulumb gauge a 1 62A 2 7 if V A C2 672 0 These equations give that waves in A travel at speed c and that plane wave solutions for A are of the form a A A0 cosUc 3977 wt with the important resulting condition transverse waves 13 A0 0 The resulting E and E elds have equal magnitude and the energy density is u lam in 87r 41 Potentials in quantum theory Shankar works using path integrals This is important and I suggest you read it but we are not focusing on path integrals this term Rem ember that the Hamiltonian for a charged particle in only an electromagnetic potential is h2 a q a 2 7 lt10 7 7A 67 m C Let s rederive the conservation of current using this Hamiltonian PHY662 Spring 2004 Apr 13 2004 13th April 2004 1 Miscellaneous 1 Reading Continue Shankar Ch 18 for electromagnetism also Grif ths Ch 9 though it is a bit simpli ed especially as it sticks to twolevel systems at rst Skip higher order perturbation theory in Shankar and read up to Field Quantiza tion p 506 in the second edition by Tuesday April 13 Much of what follows in today s notes is derived from Shankar and Ch 13 of G Bayms Lectures on Quantum Mechanics 19 Of ce hours are at 400 on Wednesday this week This week s homework is due Tuesday April 20 2 Electromagnetism See Shankar for Maxwell s equation and this discussion The highlights are rewriting the electromagnetic eld using the potentials A and 15 E VXA a 18A E ZE W Choosing the Coulomb gauge for the free electromagnetic eld where the sources for the EM eld density p and current j satisfy p 0 and j 0 gives V A 0 15 0 The equations of motion for A in the Coulumb gauge are then a 1 82A 2 7 i 7 V A C2 872 0 These equations give that waves in A travel at speed c and that plane wave solutions for A are of the form A A0 cosk 3977 wt with W w2 and with the important resulting condition transverse waves 13 1410 0 The resulting E and E elds have equal magnitude and the energy density is 1 a a i E 2 B 2 u 8 lt1 1 l l gt Note on units has the units of energy so e has the units of xE L For 7 mi 00529 nm the Bohr radius has the value 136 eV as a Rydberg is 136 eV and V 72H So e2 136 X 00529 eV nm 072 eV nm mixing up CGS and SI units in a controlled way 21 Potentials in quantum theory Last time we reviewed the homework question that showed how the Schrodinger equa tion was preserved under a gauge transformation for 141 and ab 141 A 141 VA F t A qb 7 35 for an arbitrary scalar function AG t if the wave function was changed by 1 A EMAF Note that this changes the phase of 1b of course but not z tJ so we still interpret 1V as the probability of nding a particle at a particular location Shankar works using path integrals This is important and I suggest you read it but we are not focusing on path integrals this term Rem ember that the Hamiltonian for a charged particle in only an electromagnetic potential is a 2 2i pt EA qab m C Let s rederive the conservation of current using this Hamiltonian in the Schrodinger equation and multiplying on the left by 11 1w 2373934411 1mg ih71 igfl2q UAWP2ig igl gi42q 1p MAW PFLQVngN1ih1vgAQqugl1p Note that 5 fl 141 1539 2141 57 ihV Taking the complex conjugate of the above equation gives 81V 97 a 2 flmill m v 7 mgw A 7 275ng V gAQ ab 1V w 2 Adding these two equations gives 1L Zmi 2 2 4 d d d d d 7 a 71 v w1W zp HQ w AW AW ltVAgtw w 7 W w a h i i 4 a 7 0 i V wvw 7w WH ib w 7 aw w giving the current 7 h a j wvw 7 1W1V 7 wwx that satis es the continuity equation 77 a Vj 0 22 Aharonov Bohm effect Classically charged particles respond to electric and magnetic elds via the classical orce F7ME3xEy C This is a local effect the path of charged particles does not depend on distant E and E elds The phase of 1 does depend on not For example one can set A 0 along a single given path from Fore F1 parameErized by distance 3 by carrying out a gauge transformation with VA d 7A d i This gauge change modi es the wave function at 1 by a phase change a 7 a Jugnofo Ads 1103 A Mfr So A along a path gives a change in the phase of the wave function This time independent change in phase along a single path leads to no observable effects But relative changes in phases along two paths can lead to changes in interference a quantum particle can explore more than one path simultaneously so changes in A can lead to observable effects If we compare the phase change due to A for two paths of a particle about a solenoid we can see this interference effect whenever the ux is not a multiple of the ux quantum This interference effect has applications in imaging Tonomui39a s group at Hitachi lab has imaged vortices of magnetic ux in superconductors using this effect INSERT PICTURE This sensitivity to magnetic elds is used in SQUID Superconducting Quantum Intef erence Detector CHECK ACRONYM or even for electrons in mesoscopic devices 23 Electromagnetic modes We can simply express solutions to the wave equation for A using plane waves in a box of volume V as r a 1 r i rw w r 7i Haw Am 2W Akwmeuc EMA We k z 7 M where the second term is the cc of the rst to ensure that A is real which it must be in order for particle conservation to hold and for E to be real and the V l2 factor is a convenient normalization The Coulomb gauge condition 6 iii 0 is satis ed iff I X 0 The polarization vectors A can be complex but one often chooses two plane polarizations with AI L X1 L X2 The scalar Am gives the amplitude and phase of the wave The total electromagnetic energy in the volume V is W2 2 E Z 27rc2 Ami 15X 3 Calculating electromagnetic transition rates So let s imagine a singleelectron atom in its ground state and bathed in incoherent radiation such as from a typical black body or vapor lamp How long does the atom spend in the ground state 0 before it is excited to a given state of higher energy Applying Fermi s golden rule gives 27f 7e a PM gm A 0gti26ltEn 7 E0 7 rm where the interaction term that is proportional to A2 in the interaction Hamiltonian has been dropped taking the intensity of the incident radiation small compared to the electric eld due to the nucleus of the atom and we are working in the Coulumb gauge A 5 1539 A For light atoms transitions involve photons with wavelengths of the order of 100 s of nm much larger than the atomic size of order 01 nm so it is fair to write Al e just as Am in a moment we will see why this is referred to as the electric dipole approximation Writing this as a sum over incoherent positivefrequency modes since absorption gives 2 71 2 5 r 2 PM gzv Am imamw6ltEniEoihwgtA 1m Now consider the matrix element Using m ihfl H0 ltnm0gt gwm 7 H0Rgtlogt m 7 E0 7 Using En 7 E0 hm we can now write 27reQ F077 7 W ZwQV llAEXFM lt74 0gtl26En 7 E0 7 hw 11X To carry out the calculation further we need to work with the sum 2 X to convert it into an integral over energy As the dispersion relation for light is rather simple w c L 3 3 WW dewdQ dk v 2703 V ems weget F 727762 d d9 4M W lt lRlOMQME E h 07w h622m3 w w M n n7 0 w 27re2w4 h2 02 27m 3 0M dQlA E wCXl2l lt74 Consider radiation incident upon the atom from an incoherent polarized source with an intensity measured in energy per unit solid angle per frequency interval I It turns out that d w AMlQ I i w 27rc4 Substituting this in gives 27re2w4 4 74 7 2 Penn 7 mlt2 cgt w Iwlltanl0gtl i PHY662 Spring 2004 Apr 13 2004 15th April 2004 1 Miscellaneous 1 Reading Continue Shankar Ch 18 for electromagnetism also Grif ths Ch 9 though it is a bit simpli ed especially as it sticks to twolevel systems at rst Skip higher order perturbation theory in Shankar and read up to Field Quantiza tion p 506 in the second edition by Tuesday April 13 Much of what follows in today s notes is derived from Shankar and Ch 13 of G Bayms Lectures on Quantum Mechanics Of ce hours are back to 330500 on Monday This week s homework will be in your mailboxes later this afternoon and is due Tuesday April 20 N 2 Electromagnetic waves Rem ember the wave equation for the vector potential a 1 an 2 A 7 if V c2 872 These equations give that waves in A travel at speed c and that plane wave solutions for A are of the form a a a A A0 cosk Fiwt with 523 w2 and with the important resulting condition from the Coulomb gauge V A 0 transverse waves 1300 21 Aharonov Bohm effect Clari cation from last lecture one can set A 0 over any region where E 0 b carrying out a gauge transformation with VA 7A This gauge change modi es the wave function at 1 by a wellde ned phase change fagho Ad Mil A e M There are no observable effects within this region from magnetic elds But relative changes in phases along two paths can lead to changes in interference a quantum particle can explore more than one path simultaneously so changes in A can lead to observable effects If we compare the phase change due to ff for two paths of a particle about a solenoid we can see this interference effect whenever the ux is not a multiple of the ux quantum This interference effect has applications in imaging Tonomura s group at Hitachi lab has imaged vortices of magnetic ux in superconductors using this effect Here is an example electrons that pass through a superconductor niobium subject to a magnetic eld interfere with electrons that pass by the superconductor This gives an interference pattern hologram that can be used to infer the magnetic eld in the superconductor Note that the eld is not uniform V zgg it t ta5w i fit anquot i t its K 39 V 39 7 I FIG 2 Interference micrograph of a vortex lattice phase ampli ed I6X Projected magnetic lines ol force are directly observed as contour tringes They are concentrated locally at the circled regions becoming narrowly spaced These regions spatially coincide with the spots observed in the Lorentz micrograph inset and are identi ed to he quantized vortices A bend contour runs diagonal ly through the Nb foil This sensitivity to magnetic elds is used in SQUID Superconducting Quantum lntef erence Device or even for electrons in mesoscopic devices 22 Electromagnetic modes We can simply express solutions to the wave equation for ff using plane waves in a box of volume V as a 1 e e 4 4 t Z V AkXAk61k Tiwt AZAA1 671 riwt 7 M where the second term is the cc of the rst to ensure that ff is real which it must be in order for particle conservation to hold and for g to be real and the V l2 factor is a convenient normalization The Coulomb gauge condition ff 0 is satis ed iff I X 0 The polarization vectors X can be complex but one often chooses two plane polarizations with A32 J X1 J X2 The scalar Am gives the amplitude and phase of the wave The total electromagnetic energy in the volume V is w2 2 E 2wc2 AEA EX 3 Calculating electromagnetic transition rates So let s imagine a singleelectron atom in its ground state and bathed in incoherent radiation such as from a typical black body or vapor lamp How long does the atom spend in the ground state 0 before it is excited to a given state of higher energy Applying Fermi s golden rule gives 27f 7e a PM 7 gm A l0gtl26En 7 E0 7 m where the interaction term that is proportional to A2 in the interaction Hamiltonian has been dropped taking the intensity of the incident radiation small compared to the electric eld due to the nucleus of the atom and we are working in the Coulumb gauge A 5 1539 A For light atoms transitions involve photons with wavelengths of the order of 100 s of nm much larger than the atomic size of order 01 nm so it is fair to write Ame ikf just as Am in a moment we will see why this is referred to as the electric dipole approximation Writing this as a sum over incoherent positivefrequency modes since absorption gives 7 27V 71 2 5 2 PM 7 EV lAgxl lltnl ax2510M 6ltEn7Eo7rwl 1m Now consider the matrix element Using m z f t 17 H0 m a WWW EWWHO HOTl0gt m E0 7 Using En 7 E0 hm we can now write 27reQ 7 PM 7 he 2W HAM ltnlrlogtl2altEn7Eo7hwl 15X To carry out the calculation further we need to work with the sum 2 X to convert it into an integral over energy As the dispersion relation for light is rather simple wc L 3 3 Wm dedn dk v 2703 V ems we get 27re2 4 2 2 F047 i CQQWC dwd w lAEXl M 6En 7 E0 7 hm 27re2w4 MW05 39 We Consider radiation incident upon the atom from an incoherent polarized source with an intensity measured in energy per unit solid angle per frequency interval I It turns out t dnw4lAkAl2 1 w 27rc4 Substituting this in gives 27re2w4 4 4 2 F047 7 mlt2 cgt w Iwl What can we infer am Ihisfarm 0f the rate Can work with total position operator ET R Note that Ll7 R2 0 Ll7 Rx ihRy to get selection rules for m m and can also get 1 l i 1 Note this type of calculation is from Baym Shankar considers the photoelectric effect Where the nal state is a plane wave state There you also have to be careful in the initial state it is easiest for an sstate In this case the matrix element is 32 mgrlZ dSFe iif39FhgoiihVe T NED 43f d3ei r39T Tao mc 7139 87mgNflb 15f 1ao2 Pfflle and inserting into the golden rule gives 64w2ag 1 Prueh2l4 and the transition rate into the angle d9 of r 2N2M0 pf 6Ef 7 El 7 hm 4a362pfl o le h421 hum m cl pfao l Fiad This is the transition rate for the photoelectric effect

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