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# QUANTUMMECHANICS PHYS610A

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This 138 page Class Notes was uploaded by Citlalli Sauer on Tuesday October 20, 2015. The Class Notes belongs to PHYS610A at San Diego State University taught by M.Bromley in Fall. Since its upload, it has received 34 views. For similar materials see /class/225319/phys610a-san-diego-state-university in Physics 2 at San Diego State University.

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Date Created: 10/20/15

Lecture 15 Dirac s SHO o Mid term 1 discussion 0 SHO discussion section 73 o SHO Via Dirac section 74 Midterm Discussion Hermite POlynomials W5 f 8 52 NH 8 82 Where the polynomial de nes inner behaviour so demanded that recursion terminates ie results only in physical Hilbert space solns ie terminate series eg tail solns n 001 001 2nj a for E 049051th enoa9 l main SHO dot points 1 Energy is quantised E n hw the classical oscillator has continuous spectra macroscopic particle E gtgt hw Bohr s correspondence principle for large n 2 uniform energy level spacing square well En n2E1 useful creation and annihilation of FM quanta solid state phonons E M oscillator eld photons 3 non zero point energy cannot have a 019 0 best we can do is rho With AacAp 712 OlHlO 0T0 0V0 gt 0 4 solutions alternate odd and even main SHO dot points contd 5 Wavefunctions extend beyond Classical turning points 6 Probability of measuring classical particle varies as p QC L 4 Cl 1130 w xx2 ie classical particle fast in middle slows at turns Quantum 135 iba2da tends as n gt 00 to PC A 1 a 2hmw Analytic Dirac Method Worked SHO out in position basis ie can also do the same in momentum space in p2 1 2 2 A En 2m 2771a 13 ever Clever Dirac worked in energy basis ie no ibEa H i p2 mwdz only assuming that we know GCCR 51913 i711 271 so factorise operators by de ning an as if they were numbers U2 02 v z39u v 23913 mam 113 mum and CL 1 2hmw ZeGllClL mw2 32 0 So working with 6 W 213 mam M 1 lt AM A aa 2th 2p mom 2 mom 1 WUWmmW mMm mn 1 A2 A 2 i A A 2hmw p mamas 2hazp 1 A 1 A 1 CAL CAM EH 5 Mld H ha amp L A A 1 A A 1 aa EH and H hw aa 5 o and thus the commutator a cur 1 Ladder operators 0 re writing the Schredinger eqn With SHO Hlngt hw i W Enlngt o Prove the rst of the following on the board Hamplngt Enhwamplngt and Hamp lngt En hwamp ngt Bottom of the Ladder but we want H but eigenfunctions 1amplngt En hwamplngt and H5L lngt En hw5L lngt We call 6 the ladder operators since given is a eigensoln nds a new one dim Each solution dim with energy En i flu but gt 0 demands LOgt 1 0 H 1 A 1 A A h H 0 aa 0gt hw 2Ogt t us 0gt 2hw0gt which then gives us En n hw Algebraic Normalisation Need factors Lngt Cnn 1gt a111ampngt dnln 1 again the Schr dinger equation H 1 1 hwlt ti5L3F i hwn A A 1 1 ma i gtngt ltn gtngt L Lngt and L Lngt n given n 1n 1gt 71 1n 1 1 ldnl2 ldnl2ltn 1ln 1gt ltn5L ampngt n 1 Lngt xn 1n 1gt1a11d Lngt 1gt1 ngt ampnl0gt thus 1 M Coolish stuff A a Can also de ne Number operator N ampamp a now Check out amp i 6L L L z ermwaEJri erwa 2hmw 1 A 27mg 27mm 27mm h as amp amp i mw 3 ip mw 3 2hmw 1 A 2 A 2hmw 2m Z mwhp as 2mwaa and 192 772 a Matrix elements 0 Consider the matrix elements of 5 in ltm5lngt WWI 1gt ltmamp ngt WWI 1gt Using Matrix elements a calculate in Energy basis Zeroeth wavefuncion a use 1 LOgt 0 to nd 0gt o by nally jumping into x land wow LOgt 2hmw i mw 3 l0 0 W mums 1amp0 x and x ada a which has solution In tho 7 L 512 some constant wows x2 g explt x2 0 And Hm hwampamp W0 Eowo thus E0 hw General xspace soln Given ground state sohi and energy use ladder operators in x spaoe ai E 1 d 2hmw 2mg mums 0 I exp 21 932 and E0 gm to generate the of stationary states was Ammonium mowm and we also get En n hw Although it really is easier to write the wavefunotions i 1 2 Excited state calculation 0 Given the ground state solution and its energy 1 WMquot 4 GXp 73 5132 and E0 hw MW A1 amp1 150 L 413 mm W lt x2 mx hmw d9 W71 magi 2mw x x2 7TH h o remember also that H Lw E hwampw Lecture 3 Outline Matrices a projection operators Sections 16 0 start of eigenproblems Sections 18 a matrix elements and algebra Sections 16 Matrix elements 0 So ltji gt ltj 2igt E Q and gt Qagt 1 b z or bi lti gt lti9Oltgt bi ilQ ajljgtgt Za il l 2979217 0 so a linear transform has a matrix representation b1 lt1 21gt lt1 22gt a1 b2 lt2 21gt lt2 22gt a2 bn an 0 ie the rst column is the components of 1 gt Q1gt Projection operators important third example the projection operator lagt Emile Zltilagtligt ligtltilgt lagt is a linear operator Where IZzgtltzzia sometimes called the outer product lt2 gt so each ie it projects the amount of from a vector WOO igtlti06gt a igt while MIR ltO igtlti a il consecutive Projection operators 0 consider two projection operators with basis vectors 1Pin igtltiljgtltj 6ijljgtltjl 51gtij 0 insert example of electric eld polarisers Projection example Projection operator as matrix 0 matrix elements Pk j ik6kj o the outer product Pk also shows the matrix 0 0 0 0 0 0 1 0 0 1 0 0sz 1 0 0 0 0 0 0 o Finally completeness 2k Pk I Stating the Eigenproblem General operator has QM gt BUT each linear operator has eigenbits Qagt wagt eigenkets agt and eigenvalues w are speci c to each 9 Easy example is Iagt agt every agt is an I eigenvector every agt has associated eigenvalues of l A goodbad example 183 is rotation R i agt wagt has easy soln Ri algt algt other two are not so easy Recall Characteristic Polynomials a b 5131 131 A A AI10 C d 5132 5132 o For a 2 X 2 matrix the determinant is a b A 0 a A b a Ad A bc C d 0 A C d A 0 Thus characteristic eqn A2 a dA ad be 0 1 1 A 5m d i a d2 4ad be a Eigenvalues A are roots of characteristic polynomial 0 Since TrA a d and detA ad be for real matrix elements a b C d A may be complex 0 An n x n matrix has polynomial equation of order n Characteristic equation Each linear operator has eigenbits Qagt wagt In general to nd the eigenbits solve the linear equation 9 wIagt 0gt thus agt Q wI1Ogt ignoring boringtrivial agt 0gt soln only muy bueno iff det 2 wI 0 10 0 l w 0 0 R i 0 0 1 s 0 w 1 0 01 0 0 1 w Chareq 1 ww2 l 0 thus eigenvalues w 1 ii Extracting eigenvectors 0 Once we have the eigenvalues w 1 ii c Find three sets of eigenvectors using 9 wIagt 0gt 1 w 0 0 a1 0 0 w 1 a2 0 O 0 1 w a3 Extracting eigenvectors example Matrices rule ok 0 Consider the product of matrices 9A as elements a kind of obvious ie the product Via matrix product n mm ltiIQAljgt ltiIQIAljgt Zltil lkgtltklAljgt 2mm k1 T 0 Next given a vector Qagt E Qagt 0 de ne the QT operator 204 049T ie in a basis WM ltimljgt WW ltj9igt ltj9igt 9 a matrix representing W is transpose conjugate of Q a recall vector representing 04 is transpose conjugate agt transposing conjugates Previously seen that flU4L A 1SZ1 and similarly transpose conjugate of product is flUT Moi lt9Ao lt9AO WKQAT QAOzI ltQAoz ltAoz 2T edUSN proven as reqd Rule for taking transpose conjugates is summarised as reverse the product orders take transpose conjugates jogthe memory de nitions 0 Matrix is Hermitian iff 0 Matrix is antiHermitian iff HH H H aa a a 2 2 comparevvlth a 2 2 H 0 Matrix is unitary iff UT U1 ie UUTI o Unitary operators preserve inner products 0 eg consider unitary transformed vectors ngt U ozngt lt 2 lgt ltU062U061gt ltU062U061gt ltOK2IUTU061gt lt062061gt 0 Columns of unitary matrices are orthonormal vectors Hermitian matrix fundamentals 1 Eigenvalues of a Hermitian matrix operator are real 2 Eigenvectors of Hermitian matrices operators are orthogonal if they correspond to distinct eigenvalues 3 Eigenvectors of Hermitian matrices operators span the space ie matrix can be diagonalised o Firstly let Qwgt wwgt Where Q QT and dot away w 2w www and taking ajoint comma www o subtracting 0 www wwlw w wwlw 0 thus 0 w w so w of ie real asreqd Lecture 21 Outline a variation Mid term 2 review Spin Angular momentum coupling Section 15 Addition of J L S end of Section 15 Variational Principle Section 161 Yet to come skip WKB Principle Section 162 Perturbation theory Section 17 Zeeman and Stark effects Section 145 Quantum Statistics Bosons vs Fermions Section 103 Time dep Perturbation theory Section 18 Review 1 Bound States Eigenstate solns eigenfunctions eigenvalues Principle quantum numbers energy spectrum ground and excited wavefunctions nodal structure of In nite square well in 1 D Simple harmonic oscillator in 1D Va mw2x2 including annihilation creation ladder operator forms Finite square well and graphical solutions Review 2 Bound Scattering Timedep o Delta function how to deal with discontinuity in iii i Scattering states calculation of transmission and 1er dxixR reflection coef c1ents 1e match ibL rug and W W Orthogonality of states superposition of states Computation and use of Matrix elements space Review 3 3D Central potential 3D angular momentum Hatom Treatment of SE in 3 D spherical coordinates d9 Legendre functions and Spherical Harmonics range of allowed 6 and me quantum numbers etc r2dr gt Veff oo spherical well spherical Bessel Neumann Review 4 3D Hatom H atom radialequation With Coulomb potential Laguerre functions principle quantum number n relation to 6 and jmax Can you sketch the radial wavefunctions R ur Bound state H atom energies dependence on n degeneracy With 6 and me And the spectrum Evaluate expectation values in 3 D cartesian spherical eg lt7 gt ltsin9gt where do the n 6 and me quantum numbers really come from Simultaneous eigenfunctions of H W E W and mm Wm 1wgt and Lzlwgt 22mm space Triplet VS Singlet states proton vs electron S2s m Sp2S 22SpS s m h2ss1s m Singlet state 820 0 00 0 E o m m Triplet state S210 2h210 E 2h21i H Triplet state S21 1 2h21 1 E 2712 TT Triplet state S21 1 2h21 1 E 2712 ii All of these are eigenstates of S2 but also eigenstates 0f 82 ie SZsms hmp m sms Triplet VS Singlet states II c0ntd o S2s m Sp28622SpS s m h2ssl5 m o The fundamental operations are snsem 2 iT w 81986 12W H 0 Consider the singlet 0 0 E mi H SPS 0 0gt hf p H 3 m 10gt while SP2O 0gt h20 0gt and S 2O 0gt h20 0gt thus 820 0gt 3 3 2 30 0gt 00 0gt Sim 3211gt 82 M in 3712 3712 M 2h211gt Spin of composite particles 0 2 particle spin sa 51 52 51 52 17 51 52 o 3 particle spin Sb sa 53 sa 53 1sa s3 Hydrogen Atom One particle state includes spinor n mg msgt Whose total orbitalspin angular momentum range is j s s l s 2 2 H a 2 Consider a H atom electron quantum numbers n mgsmsz 1 for electron only j 6 orj lg 5 electron proton J 61 or J 6 or J 1 the J 6 has two distinct combos with i IF important later when H includes eg LS Clebsch Gordon coef cients two particle Hilbert space spins 51 and 52 possible spin 5 51 52 51 82 17 51 52 Now the tricky bit is to construct a combo state smsgt Z smsgt CSZ39SJ39S mimjms 5 m gt5jmjgt W4er ms Unitary transform between complete orthonormal sets 11gt 1 1 0 10gt 0 1 1gt 0 0 00gt 0 eg for singlet state l 0 0Hl gt E1 0 Igt I gt 0 1 gt Igt 42 0 gt gt 0gtTi H Helium atom and symmetry 0 Two electrons E wr1r2xs1s2 o For the He atom xed nucleus Z 2 with 2 electrons 712 1 Z82 712 1 Z82 H V2 V2 2m 1 4713960 7 1 2m 2 4713960 7 2 1 e2 47T 0 r1 r2 soln ibr1r2 war1wbr2 wmelmel rimmeng 1392 But QM says identical particles are indistinguishable singlet ibr1r2 AWaU39iWMI Waltr2Wbltr1gtl antisymmetry 111172 111271gt lies in spin spatial 0k triplet wr1 r2 Awar1wbr2 war2Wbr1l Lecture 21 Outline a variation Exam post mortem Spin Angular momentum coupling Section 15 Addition of J L S end of Section 15 Variational Principle Section 161 Yet to come skip WKB Principle Section 162 Perturbation theory Section 17 Zeeman and Stark effects Section 145 Quantum Statistics Bosons vs Fermions Section 103 T ime dep Perturbation theory Section 18 Triplet vs Singlet states proton vs electron S2s m Sp2 S 2ZSPS S m hss1s m Singlet state S2O 0 00 0 E o m m Triplet state S21O 2h210 E 232Ti H Triplet state S21 1 2h21 1 E 2732 TT Triplet state S21 1 2h21 1 E 2732 ii All of these are eigenstates of S2 but also eigenstates of 32 ie Szsm8 Mmp me8m8gt l Triplet VS Singlet states II again 0 S2s m Sp2 S 2ZSPS S m hss ls m o The fundamental Operations are SPS TL 2 iT w 81986 12W W 0 Consider the singlet O O E i H sps 0 0gt g s m 3 m 10gt while 319W 0gt gf o 0gt and S 2O 0gt 3W 0gt thus 820 0gt 3 3 2 30 0gt 00 0gt Clebsch Gordon coef cients 0 two particle Hilbert space spins 81 and 82 0 possible spin 8 81 82 81 82 131 32 0 Now the tricky bit is to construct a combo state sm8gt Ismsgt Z 0535ij m rmg ms Simigt3jmjgt 0 ie Unitary transform between orthonormal sets up 10 10gt 0 1 1 1gt 0 00gt 0 EIH 0 El 0 0 eg for singlet state 1 O imi H Helium atom and symmetry 0 two electrons r1r2xslsg o For the He atom xed nucleus Z 2 With 2 electrons E2 1 Z62 E2 1 Z62 H V2 V2 2m 1 47T 0gt T1 2m 2 47T 0gt T2 1 652 47T60 r1 r2 SOll l 017 1 2 T ar1 br2 wm lmgl r1 n2 2me2 r2 But QM says identical particles are indistinguishable Singlet 0171 AWaU inU ar2 bril antisymmetry Him 1121gt lies in spin spatial ok triplet r1r2 A ar1 br2 T a1 2 br1l Helium atom and symmetry 0 anti symmetry has real effects Hydrogen Atom One particle state includes spinor n mg msgt whose total orbitalspin angular momentum range is j s s lE s E2 1 e2 H V2 2m 47T 0gt 7 Consider a H atom electron quantum numbers n mgsmsz for electron only j E orj electron proton J 51 or J E or J 1 the J E has two distinct combos with i 2 important later when H includes eg LS H The periodic table Coulornbic many body Hamiltonian xed nucleus requires anti symmetric electron wavefunction Zh2v2 1 gt2621 1 62 Fl 2m 9 4713960 2 47reo jkrj rk 73 l6 I391I392 rz X XSl82 sZ One electron states wn me have n2 degeneracy n8 shells have 713 07715 0 thus 2 electrons np shells have 713 1 Mg i1 0 thus 6 electrons ll up shells according to the Pauli exclusion principle Hund s rules give con guration 2S1LJ eg Carbon 3P0 1822822192 has possible L 2 S 01 and J 3 2 1 O Variational Principle 0 Variational principle gives an upper bound on E98 0 pick any normalised trial function Wt then Egs S 39 o The proof since there exists eigenfunctions w of H W chwn suchthat HM En ngt thus 1mw2ammwzap ltmmmw2amwmw2ammmw 0 so Zn cn2En 2 E98 Zn cn2 E98 as reqd 1D Harmonic oscillator dumb guess 0 To nd an upper bound on E98 hw choose 0 trial function wt to be a Gaussian tx A bm2 o The parameter b is arbitrary While normalisation gives 00 2 all 1 A2 6 25 dxA2121b so A next calculate T V h 2 gt mw2x2gt 2m dx2 E2 A2 00 d2 bac2 E21 ltTgt 2 l 2 d m 00 93 2A2 00 2 ltVgt mw2 I OO 26 2bx2d1 0 So Si wgg2 and thus Hgt 73332 0 t with b W gives a H mm lhw minima 2h 2 Helium ground state guess 1 0 Want an upper bound on E98 7897 5 6V considering o the Coulornbic Hamiltonian H H1 H2 V66 E2 e2 2 2 1 H V2 V2 2m1 2gt47T 0lt 7 1 7 2 IF1 r2gt o ignoring the V66 interaction has hydrogenic soln 8 T141171 1OOI 1W100I 2 we 2 T2 O 0 o 1oor energy H1 Z2E1 4 X 136 544 6V if non interacting H1 H2 8E1 109 6V 0 ie wt is eigenfunction of most of H wt 8E1 Veewt 0 but this is for two non interacting electrons next add V66 Helium ground state guess 1 contd 0 Want an upper bound on E98 78975 6V given trial 71 171 2 want ltveegt iltgt 47T 0 rl rgl e2 8 2 6 4T1T2a0 V d3 d3 lt 66gt 47T 0 7mg r1 r2 r1 r2 such a running Coulomb integral is a pain but do able 5 e2 5 Vee E w 34 V 47T 0 2 so our trial function gives quite a good upper bound 5 Et ltH1H2l egt 75 6V but we can do better with a bit of thought Helium ground state guess 2 guess 1 tr1r2 e r m O gave Et 75 6V 0 o guess 2 consider shielding of nucleus ie Z lt Z 2 trial tr1r2 Lie Ztmml O exible Zt rewriting 7Ta0 E2 62 Z15 Z15 H v2 v2 1 2gt47T60 T1 742 e2 Zt Zt 1 47reo 7 1 7 2 II 1 1 2l Where 11gt 223E1 2Zt m m ltV gt 0 can show 5 5 While V66 ZtE1 s0 H 2th2 4ZtZt 2 324m 223 217mm d 27 0 when Zt 1 6 m 169 so 775 V Helium ground state research 0 guess con guration interaction wt Ci wmlimi 1 ng imam Ndim Energy eV 0 990 7834239169 1 1656 7892709569 2 2286 7898833750 3 2916 7900341868 4 3546 7900879312 5 4176 7901116327 6 4806 7901236510 7 5436 7901303773 8 6066 7901344293 9 6696 7901370133 10 7326 7901387377 11 7956 7901399320 12 8586 7901407849 gt0 7901439039 Exact 7901439390 EXP 7900519920 Lecture 14 1D SHO things about test hW 5 hW 6 two little important theorems section 56 Ehrenfest s theorem section 6 simple harmonic oscillator Via power series section 7 3 some notes on notation little theorems 1D bound states 0 Two little theorems that we ve just whipped past a There is no degeneracy in 1 D bound states b The eigenfunctions of H can always be Chosen to be real for real H while we are working in the sis basis eg if id 2 Vxgt an nun then 2m dx2 ZmW 0 so and have same ei envalue n n 412 d2 w w m 0 but 1 1 W 2 also has same eigenvalue 0 Finally 1 wn wig2239 67quot due to a and T W 27 1iCWr AW big theorem Ehrenfest s 0 Quantum mechanical expectations obey classical laws wow 0 derived in Shankar section 6 eg momentum and force man and ltgtdltpgtmdltxgt l W l 2090M b dt dt2 VXt o stationary states have 0 so p O 0 can be a shortcut to calculate pgtt but be careful Simple Harmonic Oscillator the most pervasive ubiquitous around consider srnall fluctuations about some potential rninirna 1 d2V d V Wx0x CUQ2H Vx V90 dx 9603 x0 effective potential Vr 16562 mw2x2 at rninirna Classical a mass oscillates on a spring Hooke s law d F k as a m dt2 has a solution 9315 A sinwt B coswt SHO eigenfunctions 0 Quantum problem we do the usual 9 gt 3 and p gt 13 H mcui 2 A p2 1 HEgt 57mg Egt EEgt 0 easy to work with sis space eigenfunctions Ex E2 d2 n 1 Emw2x2 MACE En nlt gt o the rst three allowed solutions 11 En are Power Series Method h 2 dwm mw2x2 E x 2m dac2 o rewrite with Change of length and energy scales mw 2E 6 Wm and K a 12 6 0 looking out towards a gt 00 ie 6 gtgt K PMS d8 0 so B O to not have explosion at a gt too we f erg2 2 soln mg gAe 22Beg22 just try and try again c we have our rescaled SE dig 2 K 0 so working with 1M6 f 67522 We want 5 d d 2 lt07 2 12 d2 df 2 d gg d gg KaTgJWg Uf i 52 c which makes the Schredinger equation look like d2f df 2 g22 2 g22 d gg 2gd 1fe 5 Kfe d2 df d gg 2gd K 1fo Power series hunt d2f df 0762 WE o hunt for a power series solution 313 Frobenius method K 1f0 fa un j0 d 00 d Jg a1 mg 3 27003534 j0 d2 00 1 6 2a2 2 x 3a3 Zj1j 2aj253 j0 Z J1j 2mm 27 K 1mg o 90 Power series formula 0 uniqueness of E linear independence means that Vj a 2 j 1 K a 2 I 3 J 1J 2 3 o A recursion formula that is the Schrodinger equation 0 Starting With a0 and a1 you get the rest 0 Split the complete solution f 6 feven fodd fevenlt gt a0 262 l and fodd G16 363 Truncate Power series a 2 j 1 K 2 j1j 2 Where to go with this Consider for large j C 72 1 1 2 ROE ijEQJC Which results in badness 1M6 f e 22 06522 Obj 2 aj2 N Eaj and aj N so we must truncate the power series gt quantises K E 1 an20 gt K2n1 gt Enn hw Meanwhile the other odd or even series an O Power series coef cients feven a0a2 239 39 39 and fodd a1 a3 3 39 39 o in terms of the principle quantum number n a Ma 3 j1j2 j o For n 0 one term a0 with a2 O and a1 O fo a0 thus 0g fog 6 82 a0 6 522 o For n 1 one term a1 with a3 O and a2 0 7M6 116 e 522 alg 6 82 0 For n 2 two terms with a2 a0 ie a4 O we 126 erg2 a0 age 682 a01 28 6 82 Hermite Polys 0 nally we also need to normalised and also given 0 x xmwhx the Hermite polynomials are Hn 14 l 2 101 H12 H24 2 2 H5 32g5 160g3 120g 0 some useful relations are Hn1 2 Hn 2an 1 Hn 1 e 2 dig n 6 52 Lecture 7 Outline QM Postulates o The basic QM postulates Section 41 and 42 0 Examples discrete and continuous Stern Gerlach and Gaussians Expectation values Homework 1 Postulation 0 Quantum Mechanics is founded on the principles of a the generalised statistical interpretation and b time evolution through the Schrodinger Eqn Postulate 1 Particle State 0 Classical particle described by 2 variable phase space eg at and pt 0 Quantum particle represented by state vector rbtgt in a Hilbert space Postulate 2 Operators i For classical particle 513t and pt are indep variables i For quantum these variables are represented by Hermitian matrices 51365 513 H a E a d A A a a zh 6a a H E zh lt m gt d9 lt gt p dx ii Classical a dynamical variable wt is a function of a and p eg Force Angular momentum ii For quantum dependent variables wa p have Hermitian operators constructed by taking classical replacing with 1 and 13 ie Q p E wa gt 719 gtp Postulate 3 Gen Stat Interp i Classical a on measurement of particle in state a and p will yield wa p i Quantum particle is in a state W measurement of the variable corresponding to Q will yield one of its eigenvalues w with probablitity 13c1ltgtltltwlwgt2 ii Classical measurement does not affect the state ii Quantum The state immediately collapses from state W to state wgt due to the measurement Postulate 4 Schrodinger Eqn Finally need to know how particle changes with time Classical state variables change according to Hamilton s equations da 8H and 8H cit 8p dtE Quantum the state vector wtgt obeys Schrodinger Eqn d A mamas Hwlttgtgt Where Hx gt 3p gtpEHx gt 3p gtp Completeness Axiom a Before showing two examples a reiteration Eigenfunctions of an observable operator1 are complete any function2 can be expressed as a linear combination of them Footnotes 1 read Herrnitian 2 function that is in the Hilbert space Iwgt Z Iwigtltwilwgt Z R Iwgt Discrete Hilbert Space S G device a An example With no classical counterpart Stern Gerlach o Completeness principle of superposition says that Iwgt awgt blw gt S G measurement In alwgt bwgt o A spin 82 measurement gives eigenvalues with prob PW lal2 ltWwgt 2 WIWgtltW gt WWW PW lbl2 Hod W 2 WIW HW IW WlP IW 0 Device useful for pure later mixed state preparation Continuous Hilbert Space The eigenvalue spectrum of i is continuous W f 93gtlt93wgtd93 where Was is also continuous smooth ie an in nite dimensional ket W in basis Wave function Mac is probability amplitude but 135 is not the probability since in nitesemally small Instead 135 is the probability density so Pada is the probability of locating object between a gt a dx Continuous Hilbert Space contd o wavefunetion 011apsesquot to a range i many identical systems before measurement 11 20a M2 ii many independent measurements a range depends on measuring device precision Expectation Values 0 Prepare N particles in identical state W o The mean value of performing 9 measurements average over ensemble lt9gt ZPltw gtw Z Iltwilwgtl2w Zwlwamlw 0 But out is an eigenvalue w w gt M ZWIMIWMIW ZWIQIMHMIW WIQIW L L 0 That s for a discrete Hilbert space Discrete Space Expectation o Stern Gerlach measuring spin particles in z dir a Each spin 82 measurement gets eigenvals wi ih 2 1 1 With wgt and wgt 0 1 O 1 11gt awgt blw gt Pltwgt a2 ltwwgt2 and Pw lbl2 ltwwgt2 82 NIDI 0 Prepare N identical initial states o On average ZiPwiwi E WWW Continuous Space Gaussian o Shankar eg 424 initial state W f ltxwgtdas 2202 0 Assume a Gaussian Was fie 5 and normalise 1ltwlwgt ltwxgtltxlwgtdx wltxgtl2dx W2 gt0 mm 0 Probability particle is measured between a and da is Pada rba2da m2eltwagt202dx 0 thus a a is the most likely NOT always the average Continuous Space Expectation a To calculate the position expectation value rst recall lt9393fgt flt9393 gtlt93 fgtd95 9355 93 f93 d93 93f95 o expectation average of many position measurements ltxgt 2 ltwaelwgt 00 ltwlxgtltxlaewgtdx 00 we mama 1 00 2 2 1 00 2 2 a 0 ZJ 0 ace das yae dy a V702 00 7T0quot2 oo 0 thus a a is the average and the most likely o for statistics interested in the spread about the average Copenhagen Interpretation 0 ideas developed by Born Heisenberg and Bohr etc 1 2 Each Observer knows a system wavefunction which is probabilistic but completely describes a system Probability of a measurement event is Pw oc lltwwgt2 Measurement is a classical process using classical devices Matter can behave as either wave or particle depending on the experiment 0 Alternate views are a breeding ground for fruit loops Lecture 8 Outline More on the postulates o QM postulates Section 41 and 42 o Expectation values 0 Gaussians o Generalised Uncertainty principle Postulations 1 Quantum particle represented by state vector tgt in a physical Hilbert space 2i Represent variables by Hermitian operators Li ExandpE ih ii Construct dependent variables via 95L gt 3 p gt 13 3i 9 measurement upon W yields one of M eigenvalues with probability Pwi oc ltwi gt2 ii State immediately collapses to eigenvector WW lwigtltwil gt w W gt ltP IP gt MaeMew 390 4 the state vector evolves according to Schrodinger Eqn Maw We Expectation Values 0 Prepare N particles in identical state W o The mean value of performing 9 measurements average over ensemble lt9gt ZPMWi Z Iltwilwgtl2wi Zltwlwigtltwilwgtwi 0 But M is an eigenvalue wiwigt 52gt ZWIMIWMW ZWIQIWMW WWW 0 That s for a discrete Hilbert space Discrete Space Expectation o Stern Gerlach measuring spin particles in z dir 0 Each spin SZ measurement gets eigenvals wi ih 2 l with wgt and wgt 1 32 O 1 O 1 MID 0 Prepare N identical initial states W awgt bwgt 1304 IOLI2 ltW gt2 and 1301 lbl2 Hui WV o On average ZiPwlwi E WWW Continuous Space Gaussian o Shankar eg 424 initial state W f x gtdx 0 Assume a Gaussian x Ari 96 U220392 and normalise 1lt gt lt Ixgtltxlwgtdx xl2dx W2 gt0 WW 0 Probability particle is measured between a and dx is Pam x2dx m2eltwagt202dx 0 thus a a is the most likely NOT always the average Continuous Space Expectation c To calculate the position expectation value rst recall lt9393fgt flt959393 gtlt93 fgtd f93593 56 f93 d93 xf95 o expectation average of many position measurements ltxgt 2 ltwatwgt mltwlxgtltxlfclwgtdx 00 wltxgtxwltxgtdx 1 OO 2 2 1 OO 2 2 90 a 0 y 0 are dx yae dy a V7702 oo 739l390392 oo 0 thus a a is the average and the most likely o for statistics interested in the spread about the average Uncertainty 0 Standard deviation uncertainty of a measurement about the average is 09 I 9 lt9gt2gt WWW lt9gt2 1 gt 0 Note Hermitian means that is real 0 cases when spectrum either discrete or continuous 03 mexwi mf and 03 madam mm o Shortcut if you already know 0le lt9gt2 1 gt 01492 2mm lt9gt2 1 gt lt 1 92 1 gt 2lt9gtltW9Wgt lt9gt2lt 1 1 gt 92gt W Z 0 Uncertainty Discrete 0 SO we have 09 92gt 92 0 Eg S G SZ measurement 82gt MSZW L0L2 b2 Uncertainty Continuous 0 SO we have 096 92 93gt2 o For Gaussian a so we need 932 Incompatible Observables 0 Going brie y back to Shankar section 110 A A dww xwawaWmdMWr hdx 0 consider the action of one measurement then the other d d 53pm am 7293 and mm ihx x 0 allows for defn of the commutation rule between 32 and 13 dww we W waLzhz x 273W amwwp wv Mx HH 0 ie the grand canonical commutation rule 3 13 iii Generalised Uncertainty Principle The problem is one measurement 9 then A keep in mind each measurement collapses wavefunction 9 lt9gt 1 l9 lt9gt Pgt A ltgt 1 A ltAgt 1 gt a 03 can show using Schwartz inequality and more that aaai 2 mm so 32313 iii thus 09260129 2 732 covered in detail in Shankar Chapter 9 i won t cover see The Heisenberg microscope experiment Chapter 4 Commuting operators Discrete o for matrices if 9 A 0 cannot nd a set of common eigenvectors that simultaneously diagonalises both 0 The S G Sm Sy 32 measurements are the archetype MID O Sm 1 1 1 O O 1 0 ie all are Hermitian measure spin in 9 y z dirs 0 But they do not commute with each other Determinate States 0 if wigt ie eigenfunction of Q 0 These are states Where every 9 measurement wi 032 Q Wi2gt lt l9 wi2 gt Q W l9 wi gt 0 0 Where wz Note 9 is Herrnitian so is Q w o The only vector that has zero inner product is O OOgt I WW 0gt SO W cow 0 Time independent Schrodinger eqn HM En ngt 0 eg energy of stationary states nH ngt Lecture 26 Outline end of the road 0 time dep PT 181182 0 Fermi s golden rule 181182 0 Symmetry again 10 o Solids free electron gas fermions in a box TimeDep Perturbation Theory Time indep H0 ie ihtgt Howe has solns Wt Zn Cntln0gt Zn Cn0e iEgmln0gt for H t 7A 0 consider solns we Zn dnteiE3thn0gt given initially only in a pure eigenstate i0gt then Z t dim cm g ltf0H lttgtIz390gte Wdt the probability of measuring particle in nal state is PM 0lf52 note that Pf0 6fl but this is the rst order term TimeDep PT 1D SHO example t Z iw0 dflttgt6fr ltf0HlttgtIz0gte Molt Time9p PT 1D SHO contd Periodic Perturbation Example i t W cm g ltf0H lttgtIi0gte Wdt Periodic Perturbation Example contd Spatial Symmetry bit more on axiom identical particles are indistinguishable 0171 AWaU iWMU i ar2 b1 1l For two bosons in same state a b which is 1 and 2 W11 1 2 Al a1 1 ar2 ar2 a1 1l a1 1 a1 2 If you put two fermions in the same state then anti symm says W11 1 2 Al a1 1 ar2 ar2 a1 1l 0 consider bosons in 1 D box they will try to ll N E1 with spin fermions in box they will ll 2E12E2Ef Freeelectron gas 0 Consider electron stuck inside in nite walls with ngglxandOgyglyandO z lZ 0 Where inside Vx y z O and outside Vx y z oo o S V2 Ew where x y z h2d2X h2052Y h2d2ZEZ 2m dx2 x 2m dy2 y 2m dz2 2 o with E E96 Ey E2 and km Wh etc Xx Ax sincxr with kxlx 71967 for n96 1 2 3 Yy Ag sinlcyy with kyly nyw for my 1 2 3 Zz AZ sin Zz with lele n27 for n2 1 2 3 o soln T nmmymz mljlz sin l7rrsinnly ywysinz Physical space VS kSpace l 4 s W33 wavevectors and kspace more soln T nm nyanz M sinquotl7rx sinnly y y sinquotl7rz 712 n3 Wavevector de ned as k km Icy k2 2 2 2 2 39 ny nz h k With Enmjnyjnz 73 7 Tm each block in k space has volume V m y 239 For fermions Pauli exclusion principle gt two per k Consider N atoms each With q free electrons N q electrons occupy an octant in k space of radius 16F N 7T3 N Tq thus 5F 37g7l392gt 3p7T2 OOIH Fermi surface and Energy 0 Waveveotor de ned as k km kg kg 7 lg 7T 712181 R2 2 2 o Ferm1 energy EF 2m 3p7r 3 o The total energy of a thin dis shell at distance 16 Energy of 16 states 2Number of states in shell 2 Energy of 16 states Volume of shell 1 state volume 2 2 47rWalk 2 dE 2 M hV 40lC W E Ot Quantum pressure on the walls 5 g thus solid object wont collapse Lecture 5 Outline Get Cracking o The 6 function Section 110 o ooD operators Section 110 a Hilbert space Section 110 Function of Operators repeat a In analogy with a power series expansion of functions f 93 Zn amquot a Can de ne equivalent functions of operators to be 00 00 2 29 o noquot Q 1 o f get eg 8 g n 2 0 makes sense if the sum sensibly converges 0 eg choosing the operator in it s eigenbasis or 0 2 0 m 9m 89 Finite dimensions a Represent the state fngt with knowledge of n positions Finite basis a Represent the state fngt with knowledge of n positions 0 Introduce the basis with 1 at the ith position 0 fnltx1 as gt 1 thus lfngt W memx 0 fn93n a basis vectors satisfy orthogonality and completeness 617 andalso Z I i1 In nite dim inner product a For the state Ifngt 2 a Want to de ne an inner product between vectors n ltfn9ngt an93 9n93 and ltfnfngt Zltfn93rz2 i1 a This defn has a problem as so does ltfngngt 0 Introduce new defn with interval spacing A51 L n 1 n L ltfngngt anltxigtgnltxigtm 2 fltxgtgltxgtdas i1 0 o for complex functions In nite dim outer product So given inner product defn lt f g fagada need to normalise at point a with require ltxx gt 0 Va 3e 13 but normalisation is NOT 1 use completeness b I a gtlta da CLb lt93Ifgt ltxx gtltx39fgtda39 ltxfgt M introduce function ltxx gt 65 513 fxvx 6xx dx x l e 6asa fa da In nite dim Dirac Delta 0 De ne the 6 function as 61 513 65 513 a Note that this function is a real number 6as x 0 Vasy x b 6a a da 1 Va 513 0 can think of this as a thin function of unit area Dirac Delta Examples Dirac Delta Derivative a Real 6 means evenfume 65 513 633 as o Think of 85 513 E 5765 513 two bumps at 95 i 6 xfxdx i xfxdx da d9 d5 7 7 7 d d I d o odd fume x 13 5 517 597 x dx Derivatives of Operators 0 Consider operator 60x with dependence on A a Can de ne derivative as d6 1 6M AA 6M 1m gt d M 0 AA 0 For matrix elements just differentiate the elements 2 AA Q 4 911912 Mi 11 2 d d Q21 922 dill31 13 0 ie is also an operator 0 Of course easiest to work in eigenbasis In nite Dimensional Operators a An example of operator is differentiation D f gt Igt 0 Where gt is the ket corresponding to function a want Matrix elements in position basis ltxlDfgt lta g da ltxDx gtltx39Ifgtdx39 few MW d2 37 fxdx xvi LUde da das da 36236 0 even though D is a matrix 513 can be hidden D is antiHermitian 0 So the matrix elements in position basis nyx d6 x 0 Now consider the hermitian conjugate 6 function is real d6a a d6a a Z w Dml I x 5 dx das da 0 As per homework if QT Q then i 2T if a so introduce new operator K z39D d6 Z ZM Ziw Z M Z Km das das da a which on the face of it is Hermitian When is K z39D Hermitian o If K is Hermitian then lt9Kfgt lt9Kfgt ltKf9gt ltfKT9gt ltfK9gt ltgxgtltxKaz gtltaz fgtdadas H atltxKaz gtlta ggtdadaz o Collapsing the inner integral of the right hand side ltmgt x dgdfldasdas iab dajxfazdaz o and similarly the left hand side 2 gamma z mdax z gltxgtfltxgt da 1 Lecture 7 Outline The postulates 0 Finish lecture 6 notes Section 110 0 Position X vs momentum spaces Section 110 o The basic QM postulates Section 41 and 42 0 Examples Discrete and continuous spectra K z39D Matrix elements Showed that D x 93 was anti Hermitian Matrix elements Kacij i6x 93 Kim lt9Kfgt glegt Kf9gt flKllggt ltfK9gt ie not Hermitian if surface term 7 0 So two classes of functions give a zero surface term Firstly those With functions that are zero at too Periodic functions With 9 f 3 927Tf27T 90f0 0 Vector spaces of such functions are Physical Hilbert spaces the spaces that satisfy 1 are also an L2 space K z39D Eigenproblem 0 vector space gt oo oroler characteristic polynomial 0 consider the eigenproblern K ltKkgt l lt95l gt xKxlgtxl gtdxl kwkcr i6x x kx dxl kwkcr dx kwkw o with solution kx A617 z39 0 Physical Hilbert space real is eigenvalues Aeikxeigenvectors 0 note the shortcut ie implied use of f dx K Eigenbasis expansion 0 Normalise using E 617 such that klk f eikklxdx 6k 16 0 So far expanded fgt in a basis afgt 0 can also expand fgt in is basis 00 1 00 ikx fkltkfgtooltklxgtltxlfgtdx m fltxgtdx 00 1 00 ikx fxltxfgtooltxkgtltkfgtdk mooe fkdk 0 Matrix elements in basis are diagonal mam mam was k X Position eigenfunctions 0 Want to solve eigenproblem X given that 0 we already know the matrix elements Xm 6x 93 ltxXfgt ltxlxlx gtltx fgtdx was x gtfltx gtdx mm 2 ltxleltxgtgt 0 s0 eigenvalues are all 9 eigenvectors 0 Consider X matrix elements in basis 1 1 I lelkgt elk xe lk xdx e lkk molar d 39 6 k k W o In X basis X acts as a K acts as i on funcs o In quantum mechanics P hK Postulation 0 Quantum Mechanics is founded on the main postulates a the generalised statistical interpretation and b time evolution through the Schrodinger Eqn Postulate 1 Particle State 0 Classical particle described by 2 variable phase space eg 9315 and pt 0 Quantum particle represented by state vector tgt in a Hilbert space Postulate 2 Operators i For classical particle 9315 and pt are indep variables i For quantum these variables are represented by Hermitian matrices 936x 93 H 3 E a d a13x gt ih6x 93 H 13 E ih ii Classical a dynamical variable wt is a function of a and p eg Force Angular momentum ii For quantum dependent variables wx p have Hermitian operators constructed by taking classical replacing with 3 and 13 ie 262315 Ewx gt 3 gtp Postulate 3 Gen Stat Interp i Classical a w measurement of particle in state a and p will yield wx p i Quantum particle is in a state W measurement of the variable corresponding to Q Will yield one of its eigenvalues w with probablitity 13wltgtltltw gt2 ii Classical measurement does not affect the state ii Quantum The state immediately collapses from state W to state wgt due to the measurement Postulate 4 Schrodinger Eqn Finally need to know how particle changes With time Classical state variables change according to Harnilton s equations dx 3H and 6H alt 3p dt E Quantum the state vector tgt obeys Schrodinger Eqn d A mama Hwlttgtgt Where Hx gtip gt3EHx gtip gtp Completeness Axiom 0 Before showing two examples a reiteration Eigenfunctions of an observable operator1 are complete any function2 can be expressed as a linear combination of them Footnotes 1 read Hermitian 2 function that is in the Hilbert space I gt Z Iwigtltwilwgt Z 1P I gt Discrete Hilbert Space SG device 0 An example with no Classical counterpart Stern Gerlach o Completeness principle of superposition says that gt alum bIW gt o A spin SZ measurement gives eigenvalues with prob 1304 W2 ltWWgtl2 WIWgtltWWgt WIPHW PW lbl2 Haj W 2 wlw HW IW WP W o This device is most useful for pure state preparation Continuous Hilbert Space The eigenvalue spectrum of 3 is continuous W f 93gtlt56 gtd93 Where x is also continuous smooth ie an in nite dimensional ket W in basis Wave function z x is probability amplitude but Px is not the probability since in nitesemally small Instead probability density Padx is probability between a gt a dx Continuous Hilbert Space contd o wavefunction Collapses to a range i many identical systems before measurement LP 20 r WHO ii many independent measurements 0 range depends on measuring device precision

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