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# INTRO QUANTUM MECH PHYS 4100

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This 256 page Class Notes was uploaded by Diamond Kirlin MD on Monday October 19, 2015. The Class Notes belongs to PHYS 4100 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 51 views. For similar materials see /class/224886/phys-4100-rensselaer-polytechnic-institute in Physics 2 at Rensselaer Polytechnic Institute.

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Quantum Physics 2005 Notes8 Threedimensional Schrodinger Equation Notes 8 Quantum Physics F2005 The Schrodinger Equation Here s what we have been working with F A r hr Notes 8 Quantum Physics F2005 A quick review of coordinate systems Cartesian coordinates form an orthogonal rectangular system with each axis at right angles to the other two mpg 231 Z x 97 6z 62 62 62 2I V 39ax2 ay2a2 y Notes 8 Quantum Physics F2005 3 Cylindrical x pcos y psin pltx2y212 Converting To Cartesian coordinates gt Line element gt Length d5 d82 if2 I392d62 dz2 gt Volume element ZZ tan6yx gt A 6 A 16 A 6 V p u6 Z ap p619 Bz Vail i L 92 5 2 pap 60 02 6192 6z2 Notes 8 Quantum Physics F2005 FIGURE 44 Spherical x rsin cos y rsin sin z r0056 2 2 Xy Z 2 b rd rsin 0d sin2 9 d 2 9 d6 l 7 sin Bd A a A 1 9 A 1 9 FIGURE 45 u Var r66 rsm66 2 V2 r ar ar r sm666 66 r sm 66 Notes 8 Quantum Physics F2005 5 The Rectangular 3D Box Quantum well Quantum Physics F2005 511 The 3D Box Vxyz lt 0 m 2 2 in b b y 2 2 00 otherwise xin CC 2 2 The 3D Box problem is a straightforward extension of the 1D infinite well or 1D box problem where 0 x in 2 2 Vx Notes 8 a 00 lxlgti 2 Quantum Physics F2005 LD Vx 9 ha Px 167 Eeih3 at 2 Pquot Vx E 2m 2 h a iax P 2 2m Vx P that lI 7 3912lt9211Vx11I 47391in 2m 9392 9t Notes 8 Quan um Physics F2005 RHVQ Vx y z 71 6 h 6 h 6 Pxeia Pyeiay PZQTE Eeih3 at Px2 P2 PZ2 VxyzE 2m 2 2 2 2 1 a71P671P671P Vxyz Piha P 2 6x2 632 6Z2 at 7amp2 2 a N0tes8 1P 1P 1D 0 Volume element d7 dx 0 Probability of finding a particle in dr at time t Pom foolwm Normalization condition 2 fool MxJ dx 1 Stationary State wave function Pxt xe h Time independent Schrodinger equation h2 62 Vx E Eigenfunction quot Notes8 Qggm ics F2005 10 2 dx 3D 0 Volume d7 dx dy dz 0 Probability of finding a particle in air at time t PxyztJ Pxyzt2dx dy dz Normalization conditon J 6196 J W J dZJ POCa y zt Stationary state wave function iEt Px y at 1006 y ze Eigenfunction nx 21 Time independent Schrodinger equation g 1V2 x yzVx my E Eigenfunction 20 x 20 y ZJn Z 1 Notes 8 Quantu Physics F2065 11 nxg f Z x M gt g 2 lJnx 0 Subsitute 90 back into time independent Schr6dinger eq to obtain allowed energies for the particle Notes 8 digit 363 5 12 3D 2 xsysbz SE 2 2 2 cos quot1 w n x y z asinax Notes 8 Quantum Physics F2005 Mpg 32gt mpg 717 7 xvyvz0 1 2 3 Notes 8 Quantum Physics F2005 14 Degeneracy amp symmetry A degeneracy always reflects the existence of some symmetry in a given problem Eg13DbOX Withab c 2 2 2 1 1243 a2 c2 h2 2 E 2m n n n 1 2 3 Exchange n1ampn2En1n2n3 stays the SAME But VJ n n 85 n n are DISTINCT 1 2 3 2 1 3 because w n n xayazal amp n n xayazal 1 2 3 2 1 3 Notes 8differ in their depe tWWf 58 15 Symmetry here is interchange coordinates x amp y Degeneracy here is two DISTINCT eigenfunctions having the SAME energy value See Fig 527 112 a b c box case 411121 Eg 2 Fig 527 a b c box case 211121112 See Figure 527 a b c box 112 211 Notes 8 Quantum Physics F2005 16 Figure 527 Rectangular boxes for the con nement of a particle The energy levels are indicated below by the quantum numbers nln2n3 The states pass through different stages of degeneracy as the box assumes higher degrees of symmetry c i c T 39 a T 3 x x b 2 a z a z a a a 222 222 222 212 212 122 122 221 221 212 122 221 112 112 211 211121 211121112 121 111 111 111 Notes 8 Quantum Physics F2005 17 Solutions for Central Potentials The 3D timeindependent Schrodinger Eq 2 H11 h V2 I V P E P 2m 2 2 h 1ir2 j 1 a sin6aw jV E PO r 2m y8r 8r r2s1n68 6 86 2sin26 8 2 2 2 2 2 2 hmh Liphcotg H 12 Lip VErlp0 2m 8r2 2mr2 86 86 s1n 6 8 Multiply both sides by r2 letting V Vr and separating variables 2 2 1P 2 a2 1P a 1P a2 1P 4211 8 r2 r2V Er Ph r2 cot6 r 12 r2 2m 8r 2m 86 86 s1n 6 8 r I urF6 2 2 2 2 F F 2 F 4211 8 Mgrr2V E 1 11 LZcot6 alt 2m ur 8r F6 2m 86 86 s1n 6 8 tWStant 1 1 Quantum Physics F2005 18 The equation Quantum Physics F2005 4 solution for central potential 2 ivzw116r2 j1 1 asm6g1 1 6 1 2mE 2 2 W 2 2 1 Pr 6r 6r Pr s1n6 66 66 Pr s1n 6 6 h 1PR ltIgt 1 R 1 8 2 E V 1 21 s1n26i rza s1n6i s1n6a r2s1n26M a 2 ml2 R 6r 6r 1 66 66 h I 6 qeiml Because the wavefunction must be single valued ml must be an integer Notes 8 Quantum Physics F2005 20 The 6 equation Quantum Physics F2005 21 9 solution for central potential 1 2 ii rz r22mE2Vr m 1 1 a sin6 ll1 Rar 6r h sm 6 s1n666 66 The angular part of Laplace39s equation is called the Legendre Equation 2 ml 1 1 as11145i111 06 sin2 6 sin6 66 Making the substitution xcos 9 d i sin6i l1X2di30 x d6 dx dx d 2 d lm E1 x dx zz1 2 m 2Glm0 x 1 Notes 8 Quantum Physics F2005 22 9 solution for central potential 2 We will start by solving the special case when mO d 2amp5 gin xc aan go To find the solution we use the power series method assume a solution of the form yx ioanxquot and substitution into gives quot iannm 1c 392 2annn 1x 2 1annx ll 1 banx 0 from which we get the recursion relation n lnl1a 2 nnm2 Notes 8 Quantum Physics F2005 23 9 solution for central potential 3 We can deduce the solution for n0 and use with the recursion relation N j 2n 2jxquot392j P 1 quotm if 2 ln 2jn 1 where Nn2 for n even and Nn12 for n odd A second set of possible solutions yields unphysical results The first few Legendre polynomials PM are POx1P1xx 132x3x2 12 or PO 1 P1 9cos6 P263cos26 12 Notes 8 Quantum Physics F2005 24 0 solution for central potential 4 Example plots of the first few Legendre funCtionS Legendreplotsmws in Notes 8 Quantum Physics F2005 25 9 solution for central potential 5 Orthogonality and normalization of Legendre Polynomials 1 2 j1PnxPmxdx 6m m Orthogonality means that we can express the angular part of any wavefunotion using a sum of Legendre polynomials Notes 8 Quantum Physics F2005 26 9 solution for central potential 6 We will not solve the m 0 case now but we will state the relation between the Legendre functions mO and the full solutions the associated Legendre functions lml 8mm 1 x2gt39m392 lml from which we can find 900 1 910 x 911 1 x212 93 1 3x2 811 x212x Notes 8 Quantum Physics F2005 27 6 and together Sperical harmonics Quantum Physics F2005 28 Associated Legendre functions x 1120 x FEW d9 2 x 130 ale x Pglel we x P30 x P m x PZ lW x P3 0l x P2w Ax 23126 Fig 123 Shapes of the associated Legendre polynomials as a function of 0 the angle between the zaxis and the equaton Notes 8 al plane denoted here by the xaxis Quantum Physics F2005 from Gasiorowicz 29 Angular solutions put together The 6 and q solutions can now be combined Flm639 lm6clgtm and when normalized yields the Spherical Harmonics Ylm 211jl m 70036eim 471 l m The orthonormality relation is YlequotVd d 6116mm39 Notes 8 Quantum Physics F2005 30 Spherical Harmonic Functions 39 m Ylmlt gt o o 43 112 1 0 34 12 0036 1 1 3 8n 12 sin 66 2 O 516Jr123COSZ 6 1 Notes 8 Quantum Physics F2005 The radial equation Quantum Physics F2005 32 The r part of the 3D TISE 6r2 1 is finite continuous and smooth so ur must go to zero at rO a ur ll1ur 2 21V EMU 0 First let39s look at ur solutions for l 0 h2 62 ur 6r2 V Eur 0 2m Notes 8 Quantum Physics F2005 33 ur example for IO particle in a box 0 rlta assumeV 00 rza hz 62ur 2m 6r2 ur AsinkrBcoskr Asin kr because u must 2 O at rO Eur 0 inside Boundary condition at a gives kn a Asin m 2 2 2 2 2 2 2 2 1Pr a Ehknhnyr thr 7 2m 2ma2 8171612 n Notes 8 Quantum Physics F2005 34 The radial equation for the Coulomb potential 2 6 471807 2 MW EMU 6W rgt lltl1gt 2m ur 6r r2 h2 Let p L where do 47180 a0 2 me 4 me Let g 3 where ER 2 ER 42180 2h2 5 u 22zl1up3upgup 6p 0 0 and so Notes 8 Quantum Physics F2005 35 0 Coulomb potential Let39s look at forms of the radial equation for l 0 in the large r limit 2 8u0 Dude J OO 109 2 Remember that e is negative because these are bound states and the potential is negative We now search for solutions to the full equation using the polynomial expansion approach The lowest order polynomial with the right behavior as p a O iS up re39 Think about the number of roots for a polynomial Notes 8 Quantum Physics F2005 36 IO Coulomb solutions Substituting back into the original diff eq and solving for g gt 81 1 me 471802 2W 136eV Rydberg Notes 8 kt3 08 Quantum Physics F2005 37 IO Coulomb solutions We look for solution of increasing polynomial order in similar manner and can find 39 L12 0C Zp pzle39pz M3 oc 270 18p2 2p3ep3 and we find that the energies of g are I um E1 2 j l 30 0 1 r Notes 8 Quantum Physws FLUU 38 V0 as E n Energy levels for solutions so far 1 1 E n n2 n 3 rnodes n 2 rnodes n 2 1 rnode n 1 0 rnodes l 0 Notes 8 1 Ry Quantum Physics F2005 39 The O wavefunctions 2 Notes 8 Probability of finding electron at r radial probability 1P1P is the probability of finding the electron in a particular position If we want the probability of finding the electron at a distance between rand rdr from the nucleus then we have to integrate 11 1 around the sphere 2 mm l grz sin 6R2ltpgtYlf 1 69 Km 6 d6d For O wp 4Jrr2R2 47m2 Notes 8 Quantum Physics F2005 41 Radial probability for O solutions n12 M 12 ll 1 ma w m EIEIE 1 m3 W2 mm m un2 M U 2 A E E in i2 14 U 2 A E H8 in i2 14 w m u ADD 3mm W3 2mm lEIEI Notess U 2 A E E in i2 14 42 Radial probability At what distance is the electron most likely to be found For the n 1 1 0 state p1 r a0 05 Angstroms Notes 8 Quantum Physics F2005 43 Solutions to Coulomb potential with IgtO 62 MW ll 1 3up woo 602 102 10 This is equivalent to an effective potential of V6 1113 p p On the next page we plot effective potential for various 1 Notes 8 Quantum Physics F2005 44 Notes 8 Effective potential with IgtO 80 higher I will push the wavefunctions out and push the energies up Quantum Physics F2005 45 Radial solutions for gt0 using r limits For large p L z 8M gt u cc 6 2 For small 0 d ZD 11 gt u oc 0l1 or 0 l p p Look for solutions of the form u oc pl1e If we plug this form as it stands back into the full equation we get a 2 11 The solution we have found corresponds to the 0 solutions for l n 1 Notes 8 Quantum Physics F2005 46 Radial solutions with gt0 1 Note that this for l gt 0 has the same energy as the 0 solutions for l n 1 Remember these are solutions for ur with no nodes Radially they actually look like the function for l 0 n 1 ln 6 they have increasing number of nodes For a given I we have a Notes 8 Quantum Physics F2005 47 Energy levels for solutions so far 1 E n quot2 n 4 l 3 3 rnodes n 3 2 rnodes l 2 2 l 1 1 n 1 rnode 4 Ry O rnodes Because the energies line up n 1 1 Ry these states are labeled 0 d 3 es by the corresponding n number Notes 8 Quantum Physics F2005 48 What about solutions with more r nodes Let39s look at a solution with one node Guess up CO Clppz1ep4 1 We find that this can be a solution if g l 22 This is the same energy as our nodeless solution with l n 2 Notes 8 Quantum Physics F2005 49 Energy levels for solutions so far N NN gt N99 1 E n n2 n 3 rnodes n 2 rnodes n 2 1 rnode Olnodes n 1 0 rnodes 0 Notes 8 Quantum Physics F2005 N N II gt D 1 hnode 50 A table of the radial wavefunctions 390 1 2 1 639 no solution no soution 2 mg 106 no solution 3 27 18p2p2eP3 p6 pequot 3 pze p3 Notes 8 Quantum Physics F2005 51 Atomic energy structure This should begin to look familiar o The energy levels correspond to the energies of the Bohr model For n1 energy level only 0 is found For the n2 energy level only 0 and 1 solutions are found For the n3 level only O12 are found In chemistry we designate the 0 case as s 1 as p 2 as d and 3 as 2 Note the mdoes not affect the energy of a state because it does not appear in the radial equation Notes 8 Quantum Physics F2005 52 A big summary of energy levels and quantum numbers S P D F 15 400 41 1 410 411 42 2 42 1 420 421 422 43i4 g4 14393314i2 300 31 1 310 311 32 2 32 1 320 321 322 200 21 1 21 11 10 gt Q s g D E m 5 0 100 Figure 89 Energylevel diagram for hydrogen showing degenerate substates Each state is denoted by the values of nlm from Semat 1972 Notes 8 Quantum Physics F2005 53 A big summary of hydrogen wavefunctions TABLE 83 SOME EIGENFUNCTIONS OF HYDROGEN w ltZgt3 2ex lt Zr 100 V7 a0 p a0 w1 awaw lt Zrgt 200 4 2 a0 10 P 200 2 1 23 an lt Zr 6 m 4V27T a0 a0 p 2a0 39 1 z 32 Zr Zr y i39 214 SW 00 a0 explt zao sm 6 exp 11 1 z 32 Z 222 Zr 27 18 2 3 81m a0 a0 L102 exp 3a0 d1 W 32lt6 2gtZ r ex Zrgt cos 0 310 81V a0 a0 a0 p 3 1 Z 32 Zr Zr Zr 39 8 81V 00 6 a0 a exPlt 340 sm 6 mm nb 1 Z 32 Zr 3 2 1 l11320 81mlta0 02 explt 3 10 COS 0 1 Z 32 er2 Zr I 32 81V a oz explt3 ao sm 0 cos 0exp z 1 Z 32 Z2r2 Zr 2 I 82A2 a0 10 explt 3 10 Sln 0 exp hZ 00 W from Semat 1972 Notes 8 Quantum Physics F2005 A big summary of pictures of wavefunctions from Semat 1972 Notes 8 Quantum Physics F2005 55 emission f What we observe 1020 10 4 DO 6 5 IJ 1 LLL see Q93 3 ltr 1 39Tquot J 0 obeVl moxom oo 3mg m wwvr00m tggg g 39 o coxrind 0 Szvsewmme e awi 9 Paschen HiHHiim HaHBH7H5 HGHE HnHle 121568 102583 97254 Balmer series Lyman series 10000 20000 30000 40000 50000 60000 170000 80000 90000 100000 110000 4 rom transitions between states E photon lRyx igure 85 Energylevel diagram for hydrogen Wavelengths of the lines of the Lyman Balmer and Paschen series are in angstroms Notes 8 Quantum Physics F2005 56 Calculating transition probabilities The probability for an transition between two atomic levels with the emission of a photon polarized along the z axis is proportional to the matrix element M alt f lav wf rcos wigt It can be shown that M is nonzero only when If ll 1 and Aml 0 or 1 Notes 8 Quantum Physics F2005 57 Angular momentum and magnetic quantum numbers land m Notes 8 Quantum Physics F2005 58 Another look at the 3D Schrodinger equa on Another way to construct the 3D Schrodinger equation is to compare it to the classical equation 2 iLVE 2m Breaking p up into components p2pip p p3pf L and In terms of angular momentum pl r 2 2 pr VE 2nz ZHWZ Notes 8 Quantum Physics F2005 59 Angular momentum If we now compare this expression to the Laplacian version we see that L2 should be an operator and we can test if it has eigenvalues and functions We already know from our prior math that the angular part of the Laplacian has the same eigenfunctions as the Hamiltonian Let39s look more closely at the L operator We think 1220 my Classically i 7x 15 so let39s do the math Notes 8 Quantum Physics F2005 60 Notes 8 Angular momentum operators In Cartesian coordinates i j k 7x x y Z Px Py Pz h a a LxYPZ ZPy7g h a a L z x z x y px pz i 6x azj Quantum Physics F2005 61 Angular momentum in spherical coordinates Lx E sin i Cot6 cos i i 66 a Ly 2 COS i CO 9 Simji l 619 6 Notes 8 Quantum Physics F2005 62 What can we tell about L We can learn quite a bit about angular momentum without doing integrals We can show that Ll and Lx cannot have a complete set of common eigenstates by considering the commutator Lva L1 xpy ypx Lx ypz zpy Ly sz xPz ggg LWLZ m zpyxpy ypx ypvxpy ypzypxzpyxpy zpwypx ypzxpy00zpyypx xypzpy ypzxpy zpwypx xpyypz xypzpy zpwypx xypypzxpyypz yxpzpy xypzr7y zpwypx 0xpyypz 00zpyypx xypypz zpwypx xypypz yzpypxzpyypx xypypz ypxzpyyzpypx xypypz yzPxpyy19xzpy zypyPx yzpypx xypypz 0 0 zypypx 0 ihxpz sz i 1Ly Similarly LXJJyi 1Lz and LWLz 1 th Notes 8 Quantum Physics F2005 63 What can we tell about L We can also show homework that L2LZO This tells us that there exists a complete set of eigenfunctions for both operators or that we can measure both quantities simultaneously with infinite accruracy We already knew that because we did the work to find the eigenfunctions and values but we could have avoided that work if all we wanted was to know whether it could be done Notes 8 Quantum Physics F2005 64 Useful commutatoroperator relations ABCABAC ABCABCBAC ABC BCA CAB 0 Quantum Physics F2005 65 Angular Momentum in Spherical coordinates 2 2 2 2 h2 a Cot61a 662 66 sin2 6 a 2 We have already found the eigenfunctions of this operator They are the spherical harmonics o We have previously found that the eigenvalues of L2 are l1 with lintegers O 1 2 3 if the potential is central o Angular momentum manifests itself as a magnetic dipole moment when the particle with L has charge It is most useful to know the projection of the dipole onto an applied magnetic field Let s say in the z direction Notes 8 Quantum Physics F2005 66 Angular momentum projection We want to know now whether the projection of the angular momentum onto the zaxis can be an eigenvalue A h a L50 L50 In the previous section we found 20 R cIgt with cp elm thus zzJ hmlw 20 is simultaneously an eigenstate of 2 and Q For L2 hll1 ml can take on integer values from Z to 1 Notes 8 Quantum Physics F2005 67 Pictorial representation of L and m Notes 8 Quantum Physics F2005 Angular momentum and magnetic moment An electron moving in a plane orbit of area A is equivalent to a current given by I LT C where T is the period of the orbit Such a plane current has a magnetic moment given by M IA 80 the magnetic moment of a classical orbiting 6A electron Is M CT We can relate the area and period to the angular momentum Notes 8 Quantum Physics F2005 69 Angular momentum and magnetic moment 275 Area can be expressed as A j r2d 0 Ar gUIar momentum is L mr2 SoA Z 2m Therefore it ii i 1 1 2mc 2mc Notes 8 Quantum Physics F2005 70 Angular momentum and magnetic moment If an atom has a magnetic moment then it will experience a force when it is placed in a magnetic field E If the magnetic field is uniform then the atom will experience a torque A2 x and precess about the direction of the field The shift in energy due to precession is AE MB cos 6 Bml 2mc Notes 8 Quantum Physics F2005 71 Notes 8 Electron spin It is observed that a free electron exhibits a magnetic moment which is quantized We have associated this moment with a new quantum number which we call spin 8 so that us 23 With s 1 2mc 2 When we compute the total magnetic moment of an atom we must include both electron spin and orbital angular momentum as a vector sum Quantum Physics F2005 72 Appendix general radial wavefunctions for V1r using series solutions from Eisberg and Resnick 2 d 2gdFiy 1IltIIgtiF0 dp p dp 0 0 Assume FppsZakpk a0 0 andSZO k0 This form is used because it assures that F will be finite at p0 i s ks k 1 10 1akpsk 2 S k 1 ygtakpsk1 0 0 It can be shown that low 1 zz 1a0p 2 iSj1Sj2 ll1aj1 s j1yajpsj1 0 j0 Notes 8 Quantum Physics F2005 73 General radial solutions Collecting terms in powers of p we see that the sum above will only be zero for all 0 if ss 1 111 0 This relation requires that 5 1 or s l 1 Only the 3 1 solution can be physically allowed and lj1 a391 aj J lj1l2 ll1 Notes 8 Quantum Physics F2005 74 Quantum Physics 2005 Notes7 Operators Observables Understanding QM Notes 6 Quantum Physics F2005 A summary of this section o This section of notes is a brief overview of the ideas in chapters 1012 of Morrison These chapters are the intellectual core of quantum mechanics as opposed to quantum physics You will address the material in chapters 1012 with greater care in Intro to Quantum Mechanics The purpose of these notes is only to introduce these ideas especially for those students who will not take Intro to Quantum Mechanics I will therefore skip lightly through these chapters picking only the most important concepts Notes 6 Quantum Physics F2005 2 Operators Dirac notation Dirac It llt ilir r ml quot W39 39 gtgt 1 Take the complex conjugate of the function inside the lt 2 Act with the operator on the function to its right 3 Integrate the integrand over all space Notes 6 Quantum Physics F2005 3 Hermitian Operators Q is Hermitian if for anytwo physically admissible state functions ltwi w2gt lea tel 1111 1112 a Qwlw2dx If an operator is Hermitian the expectation value of that operator will be real All operators for physical quantities are Hermitian Notes 6 Quantum Physics F2005 Is p Hermitian We will check explicitly whether 11141731112 ll A 6 p 4 Start by integrating IP1 15115 by parts 1P1 11 2 Oily 47angde ihlpf112 01P2ailpfdx 00 x 00 00 x i150 1P2 iq dx 0 1P2 47ain dx 1511a 1P2 6x 00 6x 00 Therefore 3 is Hermitian Notes 6 Quantum Physics F2005 5 Exercise Show by writing out the real space integral representation that the operator 2 is Hermitian Notes 6 Quantum Physics F2005 Operators and Eigenvalues In quantum physics the form of the eigenvalue equation for an observable A with eigenstate ZJa and eigenvalue a is 2isz ClZJ The collection of all eigenvalues is called the spectrum of this operator The most familiar eigenstate is the stationary state 111E ZJE iElh which satisfies H111 E111 In an eigenstate of an observable the uncertainty in that observable is zero Notes 6 Quantum Physics F2005 7 The eigenstate of one observable might not be an eigenstate of another observable but it can happen Example An energy eigenstate of the square well is 2 1Pxt Acos em With col E rm 2 L h 2mL Is this an energy eigenstate of momentum f9 47161 x 1311 ihiAcos e ia ihA sin e W 9111 NO 6x L L L Another example An energy eigenstate of the free wave is 1P Adm 13111 47161 Aeikquotquot t m YESgt X Notes 6 Quantum Physics F2005 8 Hermiticity Real eigenvalues and orthonormal eigenfunctions The eigenvalues of a Hermitian operator are real The eigenvalues of a Hermitian operator are the only values that we can observe in a measurement of that observable AND The eigenfunctions of a Hermitian operator are orthogonal constitute a complete set and satisfy closure p 465 Morrison Notes 6 Quantum Physics F2005 Proof of orthogonality Take two different eigenstates of the same observable 1 qu 61 and 2 quv 6139qu We want to demonstrate that gag 9 0 Multiply both sides of 1 by complex conjugate of 2th ltqu QM q M q Use Hermiticity of C to simplfy the LHS ltqu Q qgt ltQ qvlwqgt 6139qule q39quwqgt q39thvlwq So we have qq39ltzJq zJqgt 0 but we assumed q q39 so lt q qgt 0 Notes 6 Quantum Physics F2005 10 Hermiticity Completeness Completeness of a set of eigenfunctions means that any wellbehaved function of the variables on which the eigenfunctions depend can be expanded upon the set 1 f Zcql q We can choose to represent any function in a given range of variables by an expansion in a set of eigenfunctions of any operator It39s easy to find the coefficients cq Multiply 1 by wq fgt wq 64 qgt cq wq cqwqgt cq The coefficient cg is the quotprojectionquot off onto the eigenfunction wq Notes 6 Quantum Physics F2005 11 Summarizing Eigenfunction Expansions TABLE 109 PROPERTIES OF THE EIGENFUNCTIONS AND EIGENVALUES OF A HERMITIAN OPERATOR Q Discrete Spectrum Continuous Spectrum eigenvalues orthonormality completeness projection of II closure qn real ltqu l Xagt 6m 1 3 Cqu all q Cq Xq l 1 z XqXq 69 I all 11 q real ltqu qugt 6q a II f cqxq dq all q cq ltxq i wgt f xqz xqw dq 61 at all q Notes 6 Quantum Physics F2005 The meaning and use of expansions o The collection of coefficients in the expansion of a state function in any complete set is merely an alternate way to represent the state function These coefficients and the eigenfunctions contain the same information as the state function Expressing a state function in terms of eigenfunctions can make apparent some properties of the state func on Expressing the state function in terms of eigenfunctions can allow us to get information about other variables Expressing the wavefunction position variable in terms of momentum eigenfunctions allows us to determine the momentum properties of the state function Notes 6 Quantum Physics F2005 13 Using expansions to calculate expectation values If we have a wavefunction that is not an eigenfunction of a given operator the expectation value may be hard to calculate in the old way Expanding the wavefunction in terms of the eigenfunctions of that operator may make things easier Notes 6 Quantum Physics F2005 14 Calculation of the mean ltQgtt I PQA de llZ cnvtwnvx Q2 Cnt nxdx llzcnvltrgtwnvltxgtlzcnmwnomdx ZZCn Cn wn ngt Z n 2 qn C If we know the projection coefficients we know the expectation value Notes 6 Quantum Physics F2005 15 Example Calculate the expected value of the energy for the following nonstationary state of the SHO 1 i l iwl 1Pct le 1 7 Ze 2 1 611 639W C2 t e39i 2t all othersO 0 0 w 13 495 62 E I C 2E ha ha ha 3 quot502 0502 050 0 Note that any individual measurement of E will yield an eigenenergy Notes 6 Quantum Physics F2005 16 The Commutator A simultaneous eigenstate is one whose state function is an eigenstate of two operators 2 qty and Iqu rguw Because they are eigenstates Qq AQO and ltRgtr AR0 Q1 Q1 21 2Q0 1 Operators that commute define a complete set of simultaneous eigenfunctions 2 Two operators that share a complete set of eigenfunctions commute Notes 6 Quantum Physics F2005 17 Commutators and uncertainty principles It operators F and Q commute then they share simultaneous eigenfunctions Measurement of each observable for an eigenfunction yields a precise value The only uncertainty relation that can apply is AQAR 2 0 twill A Generalized Uncertainty Principle AQAR 2 For momentum and position 2 ihi If Q 0 then there exist no eigenstates of either observable Think about 5x and e39ikx39 Notes 6 Quantum Physics F2005 18 Commutators of x and p TABLE 113 COMMUTATOR RELATIONS INVOLVING THE POSITION AND MOMENTUM OPERATORS 53213 2iha c 1 22132 2m 11 2132 2ih25313 mi III for arbitrary operator functions f 13 and 9amp2 A d A mm 171ng IV p d 1mm and 5m v for a Hamiltonian 7 T 173 13921 VI m A w 9X A Hp m ax 1 VII Notes 6 Quantum Physics F2005 Notes 6 An aside Charles Hermite was a who did research on and and are named in his honor He was the first to prove that the base of is a His methods were later used by for the proof of his celebrated theorem that is transcendental Upon 39discovery in 1861 of continuous curves that are nowhere differentiable they possess no tangent at any point Hermite famously remarked I turn aside with a shudder of horror from this lamentable plague of functions which have no derivatives httpenwikipediaorgwikiCharlesHermite Quantum Physics F2005 20 Quantum Physics 2005 Notes2 The State Function and its Interpretation Notes 2 Quantum Physics F2005 In a particledetection scheme the wavelike interference pattern builds up particlebyparticle lC Notes 2 Figure 28 The genesis ol an electron interference pattern in the doublevslit experiment Each gure shows the pattern formed at the detector tit successive limes Note the seeming randomness ol the pattern at short time intervals From P G Mcrli G F MissirolL and G Pozzi Amer Jmu39 Flint 44 306 1976 Used with permission from Morrison Quantum Physics F2005 Note that the arrival of each particle appears to be random un lenough particles have arrived for a statistically useful sample This suggests that we should use a probability distribution approach for quantum physics calculations Discussion of probability The word event refers to obtaining a making a measurement of an observable To define the probability of an event result we consider the fraction of times that result will occur for measurements on a very large number of identical systems Such a collection of systems is called an ensemble To perform an ensemble measurement of an observable we perform precisely the same measurement on the members of an ensemble of identical systems Notes 2 Quantum Physics F2005 Normalized Probability The probability of a specific outcome is the number of times that it occurs divided by the total number of experiments quotat Pa W The probability of a set of outcomes is the sum of the probabilities of each outcome Pa orb or i Pjiinj ja N ja By this definition all possible outcomes PU 1 ja Notes 2 Quantum Physics F2005 4 Ensemble Averages Given the probabilities for a set of outcomes we can find useful quantities like the average or expectation value of a result q through a simple sum M Z quqJ quotmean of qquotltq a J391 Pltqjgt Another statistic that is useful in describing distributions is the dispersion or variance variancediSperSi0nltq 39 ltqgt2gt Notes 2 Quantum Physics F2005 Continuous distributions In many physical cases the result of a measurement can be any value within a physical range for example the position of a particle We can define the probability of finding the particle within a certain infinitesimal range as PXdX and we call PX the probability density o Given a probability density we can compute the expectation value of any parameter qx ltxgt 0 xPxdx m fl qltxgtPltxgtdx Notes 2 Quantum Physics F2005 Matter waves Just adding probability distributions does not give interference In order to get an interference pattern we must be able to superpose two waves A working hypothesis Observations occur in proportion to the probability that a particle will be in a state rt pt Et when it is measured The probability is proportional to the square of a wave amplitude Notes 2 Quantum Physics F2005 7 The State Function Every physically realizable state of a system is described in quantum mechanics by a state function P that contains all accessible information about the system in that state The phrase wave function is usually used to mean a state function that is specified as a function of position and time 1PXt Notes 2 Quantum Physics F2005 Born s Postulate If at time t a measurement is made to locate the particle associated with the wave function 11 t then the probability Ptdv that the particle will be found in the volume dv around position 7 is equal to PFtdv 1P t1Pt Where 1Pf39t denotes the complex conjugate of 1Pf39t Notes 2 Quantum Physics F2005 9 Principle of Superposition a m ing wimp a pinii Notes 2 Quantum Physics F2005 10 Normalization Sometimes you will be given wave functions that lead to an integrated probability that is not already 1 In fact lazy professors do this a lot Before you try to do any calculation with a wavefunction you should make sure it is properly normalized that is llPt1PFtdv 1 00 one naveliuncnon Din that MD ringing a M just mulli ply quotEff min steam to a new nuncftion v H MM til g t tnaii null be nomn tnj A x st H M Notes 2 Quantum Physics F2005 11 Physical constraints on wavefunctions A physical wavefunction must be normalizable The particle must be somewhere A physical wavefunction must be single valued It can t have two probabilities of being at the same point A physical wavefunction and its spatial derivatives must be continuous Notes 2 Quantum Physics F2005 12 Examples of allowed and disallowed wavefunctions Not ok Ok bl Continuous l lx 0 WM 0 r X X from Morrison 0k Not ok 0 ooooooooooo ying Notes 2 13 Expectation value In quantum mechanics the ensemble average of an observable for a particular state of the system is called the expectation value of that observable 0 1PQ11Idv Q if 1 Pdv o The uncertainty in the observable is the square root of the dispersion AQ ltQ Q2gtm lltQZgtltQgt2l Notes 2 Quantum Physics F2005 14 12 Einsteinde Broglie relations As we continue to analyze behavior of wavefunctions it is useful for us to keep in mind two basic relations for matter waves relating particle properites to wave properties Ehfhw and h hk p 1 Notes 2 Quantum Physics F2005 15 A quick peek at the Schrodinger Equation Later in this course we will deduce and solve the wave equation for matter waves The Schrodinger Equation Here I will show it to you make some simplifying assumptions and discuss the nature of the solutions Here it is in one dimension 2amp2 92 6 Px x V x 1P xt ih 2m 6x2 at Notes 2 Quantum Physics F2005 16 Free space solution for 1 For VxO we can illustrate some physical points by working through the wave function logic of this physical situation without fully solving the differential equa on A If 1P is to be normalizable it must go to zero as reinfinity B 11 must be complex because the right side of the SE has an iin it C 11 must not violate the homogeneity of free space The probability must be the same after we translate 1P by an arbitrary distance Notes 2 Quantum Physics F2005 17 Three trial functions 1 1Pxt Acoskx wt g a real harmonic wave 2 1Pxt Ae kquot a complex harmonic wave 3 1Pxt a wavepacket made up of complex harmonic waves A 1Pnormalizable B 1Pcomplex C 1 must not violate the homogeneity of free space Notes 2 Quantum Physics F2005 18 Test the harmonic function 111xt Acoskx wt e a real harmonic wave A Normalizable No unless A goes to zero B Complex No Unlikely to satisfythe wave equation C Homogeneous P No Pxt Pxt l11x t A2 cos2 kx wt 8 Px a t 111x a t 111x a t Px a t Acoskx at cos ka sinkx wt sin ka Px at Pxt cos2 ka A2 sin2 kx at sin2 ka 200s ka sin kacoskx at sinkx 11 Notes 2 Quantum Physics F2005 19 Test the complex harmonic function Pxt A610 390 A Normalizable Only if A goes to zero lPxtdx Ae iW mAeW mdx 0 Adx A zoo B Complex Yes It is indeed a solution to the SE C Homogeneous P Yes Pxt 1Pxt 111x t A2 independent of position We seem to be on the right track Notes 2 Quantum Physics F2005 20 Notes on the complex harmonic function 1Pct Aem Acoskx cut isinkx m ACospx Ethisinpx Eth Represents a state with a single value of momentum and a single value of energy therefore it is called a pure momentum state It extends over all space The phase fronts move in the x direction at a speed vp of 3 or E k p 2 For a classical particle v E M L p p 2m Notes 2 Quantum Physics F2005 21 The complex wave packet We can construct a localizednormalizable wavepacket out of complex harmonic waves using the Fourier transform 1I xl L Akeiltkx gtdk m 00 gt Because this wavefunction can be localized it is likely to be normalizable We will need to test specific cases gt Because each term in it is a solution to the wave equation the wave function must also be a solution 2 Is the shape of the probability of the sum of complex harmonics translation invariant Pxl deiklx 136 deiklx 56 ye I39m 5e39ik2xdeik1x 136 2 2 39k k 39k k 2 2 39k39 39k39 a b abe 1 2xbael1 2xa b abelxbael x Px a a2 b2 ampe ik1x ik1ageik2xik2a 5e ik2x ikzadeik1xnkla ik39xa ik39xa b a6 2 2 a b abe This function is just the old one moved over Notes 2 Quantum Physics F2005 22 06 IIX 0 04 02 Px 0 m l lnclass exercise 1 Illx o Itquot2 it 10 08 I l 4L 1 l l 30 20 10 10 20 30 40 Hi Morrison 31 p 89 This is a normalized state function at t0 The magnitude of 1Px0 continues to decay smoothly to zero as X9100 1 Is this function physically admissible Why or why not 2 Where is the particle most likely to be found 3 Estimate the expectation value of the position Is it the same as for part 2 4 Is the uncertainty in momentum of this state equal to zero Justify Notes 2 Quantum Physics F2005 23 Morrison 33 p 90 InClass exercise 2 33 Measurement in Quantum Mechanics ShortAnswer Questions Consider a system whose state function III t at a xed time to is shown in Fig 331 Notes 2 a V b c d e 3 x to 0 X Figure 331 Describe how you would calculate the expectation value of position for this state from the functional form of Ila t What is the value of ensemble average of position for this state at time to Describe brie y how you would calculate the position uncertainty for this state from Ila t Is As zero positivebut nite or in nite Why Is the momentum uncertainty Ap zero positive but nite or in nite Why Can this state function be normalized If not why not If so describe brie y how you would normalize IIc t Quantum Physics F2005 24 In which I try to clarify what we are doing Notes 2 From evidence of interference in particle experiments we introduced the idea that matter must be described by some sort of wave De BroglieEinstein relations ph A Ehf The wave is not directly observable but the resulting probability of various outcomes is hence we introduced a probability interpretation of quantum physics experiments We then assumed that there existed a wave that could yield the correct observables with classical relations between time energy momentum and position The ramifications were then explored The probability of an outcome is proportional to the square of the wavefunction for that outcome The probability interpretation leads to the ability to compute the expectation average value for appropriate observables once the wave function is known The superposition of waves to form a packet leads to a mathematical statement of the Heisenberg Uncertainty Principle We could deduce the form of a useful wave equation for matter waves We will now explore some examples of wave functions and their observable ramifications Wavepackets momentumposition transforms More general state functions the momentum state Quantum Physics F2005 25 Important results so far Einstein deBroglie relations p hk T ha hf The probability interpretation ltq lqxPxdx Pxdx 11 de The general form for a wave packet using waves 1 m with welldefined momentum 1Pxt Ake dk Notes 2 Quantum Physics F2005 26 In class exercise 3 Developing the matterwave equation from Morrison Problem 27 Write down an expression in exponential form for a harmonic traveling wave because our matter wave packet is composed of simple traveling waves Rewrite your wave function in terms of pk and Ea Deduce the simplest differential equation that might give rise to such a traveling wave consistent with Ep22m 62 6x2 Lab 6260 b1 2 atz a a0 a1 a2 6x Notes 2 Quantum Physics F2005 27 lnclass exercise notes You should now have something like 4 i 1 2 i 2 P a a a 19 19 0 1hW 2h2p 12mhp 24m2 which can only be true if 100 110 1920 and amp12m 91 h so a plausible differential equation for the motion of a free particle is 2 2 h a y 1756 90 2m 6x at Notes 2 Quantum Physics F2005 28 A solution to the matter wave equation the infinite square well 2 2 Starting with h a1 1716 1 2m 6x Bl go to zero at xL2 leads to some interesting results Only certain quantized values of wavelength are allowed and specifying that the wavefunction must 11xl Ncosj e i t is one solution For further work we want this function to be normalized L2 L2 1 ll VJCIYJWN2 J0082 jdxN2 2 NZ L2 L2 L 2 L 2 x iat xt cos e W gt L L Notes 2 Quantum Physics F2005 29 The square well continued Now find the expectation value of position L2 L2 x I xPxtdx 2 I ccos2 dx 0 L2 L L2 L You can do this integral directly or argue its value based on symmetry Notes 2 Quantum Physics F2005 30 Notes 2 The square well continued Now find the variance of position so we can assess the range of positions over which we might expect to find the particle LZ LZ x2gt I x2Px tdx E I 62 cos2 dx L2 L L2 L x2gt El j f u2 cosz udu L 77 7r2 2L2 U3 L12 1 ucos2u 7m 3 s1n2u Jr 6 4 8 4 rm 2 3 we ali 2 Jr 24 4 12 277 Quantum Physics F2005 31 Particles and waves the Gaussian wavepacket We will simplify calculations by assuming that the spatial part of the wavefunction can be described at t 0 by a Normal Gaussian function centered at x0 and moving with average wavenumber k0 ikoxe x x02 Ma 1PX e We have already solved a similar problem in Notes 1 12 80 1 1 J eik0 kxe x x024a dx 0x 55 Notes 2 Quantum Physics F2005 32 CX Gaussian wavepacket continued to simplify further let39s take x0 0 12 1 1 0 2 2 so Ak J felk039kxe39x 2 de V27 axxZ oo Then letting k39 k k0 we solve by completing the square 2 1 zk39x x202k392 02k392 x zak39 022 202 x x x x X X 1 1 12 Ak 63901ka2 6 427 awE 00 12 1 1 2 2 2 letx39 x i0xk39 then Ak 0ka le x dx39 xEax J27 axxZ 1 1 Inf 2 2J 1 i quot k k 2 Ak Zaxe axk39 yr 0 e 427 0x5 It Notes 2 Quantum Physics F2005 33 Gaussian wavepacket continued 1 7 a k k02 Aki04e 2 7139 This is a Normal Gaussian k distribution centered at k0 with width 0k i U x Although we specified a wave momentum k0 we find that the wavefunction 1 actually contains a spread of wave momenta which increases as the 0 6 spatial Gaussian narrows Remember that Ak A ff Waves of differing k move at different speeds so the original wavefunction will spread with time Notes 2 Quantum Physics F2005 34 lnversely correlated widths Note that if 1P is properly normalized DIWde 1 then the waveveotor amplitude function is also normalized to 1 lAk2 dk 1 Notes 2 Quantum Physics F2005 35 An aside on the definition of the width of a Gaussian 1 The width of a Gaussian can A be defined in various ways Two conventional ways are A by the Full Width at Half A j 1 Maximum FWHM and twice 1 i3 1 2 3 4X 394 3 2 1 3 1 2 3 4 the standard deviation Morrison uses ALstandard deviation of the position C Figure 48 A cornucopia of de nitions of the width mm of a function f1 hand is fc 6 12 a The fullwidthat halfmaximum wm 1 665 b function at the special point where f z f mmax e The function e z2 18 al to 1 e of its maximum value at x i1 so this de nition ives w 1736 39 39 39on mac A1 0560 Notes 2 Quantum Physics F2005 36 Time evolution of the wavefunction The general form of the wavefunction is 1 E This means that if you are given 1Px0 you can find 1Pxt 1 E Note that to do the top integral you must know wk 1mm 0 Ake39 quotquotquot dk Ak 0 1Pc0e39ikxabc Notes 2 Quantum Physics F2005 37 In class exercise 4 Constructing a wave packet from momentum amplitudes Consider the amplitude function 1 0 kltkO Ak 2 Ak lt k lAkltkltk lAk Mk 0 2 0 2 0 kgtk0Ak a Find 1Pc0 b Find Px0 c Sketch Px0 and find the positions of the first zeros d Estimate AxAk Notes 2 Quantum Physics F2005 38 Formal derivation of group velocity for a wavepacket wow jAke39 kquot39 k dk Assuming that Ak has a fairly narrow range of important k39s we will allow ourselves to expand ak da wk a wk0k k0 jko Letting k39 k k0 w 361 5 eik0x wk0t Jdk39eik xli t0 Notes 2 Quantum Physics F2005 39 Momentum Probability Amplitude Note that Ak clearly contains information about momentum because k ph It is useful later to be able to describe the state function directly in terms of momentum states 13p 5Aj the momentum probability amplitude Then we can rewrite the wavefunction as Px0 0 lt1gt peipxmdp and lt1gt p o Px0e39ip C739dx Note that the factor of fl is required for proper normailzation of 1P and 1 Notes 2 Quantum Physics F2005 40 Interpretation of the space and momentum state functions 1Pxz the wavefunction and directly gives information about the probability of finding a particle at a specific position but it is difficult to directly calculate properties like the expected momentum because momentum is not written as a function of position c1gtp the momentum probability amplitude and directly gives information about the probability of finding a particle with a specific momentum They both describe the same quantum state If the members of an ensemble are in the quantum state 1Pxt with Fourier transform c1gtp F1Px 0 then Ppdp lt1gtplt1gtpdp is the probability that in a momentum measurement at time t a particle39s momentum will be found to be between p and p dp Notes 2 Quantum Physics F2005 41 Expectation values Given the momentum amplitude function and the associated probability distribution we can compute expectation values p 0 ltIgt p pltIgt pdp expectation value average of p Ap momentum uncertaintyp 119 2 192 119 2 p2 ltIgtltp pzclxmdp TABLE 41 POSITION AND MOMENTUM INFORMATION FOR A GAUSSIAN STATE FUNCTION WITH L 10 m0 0 AND po 0 Position Momentum 1 14 2 2 2 L2 14 2 2 2 II 0 1 4L cl p L h z 27d 6 p 1rh2 e w 0 p 01 m 7 2L Notes 2 42 Quantum Physics 2005 Notes3 Observables Chapter 5 Quantum Physics F2005 Introduction Observables are physical attributes of a system that can be measured in the laboratory In quantum physics in the absence of a measurement a microscopic system does not necessarily have values of its physical properties A particle does not have a position until we measure it It has a set of possible positions We want to find out how to calculate observables from wavefunctions The mathematical approach is through the introduction of operators Notes 3 Quantum Physics F2005 In quantum scale measurements the measurement itself affects the state of the particle so we cannot use classical definitions of observables Example position and momentum The momentum is measured by measuring velocity This involves the measurement of position at two times The first measurement affects the momentum and hence the later position and consequent velocity Notes 3 Quantum Physics F2005 Two ways to compute momentum 1 Given 1Pxt use the wave function to compute CIgtpp Compute p from chgtpcIgtdp 2 Find a way to express p as a function of x Compute p from 1Ppx1de Let s compute each way for the lowest state of the square well Notes 3 Quantum Physics F2005 4 Momentum in the square well 1I x Z E cos em 1 oo 1 L2 if q 1P x0 e39lPXhdx cos P 42m Jami2 L 1 mm m Md cos cos x zs1n x x min L p p he NW Mb III I S NEI Notes 3 Quantum Physics F2005 Momentum in square well Sinai2H mama pL 1 L h 2 L h 2 2 W j LBJL L h 2 L h 2 sin sin 19 ltIgtltpgt i 2 2h 43 L j LL j 2 2hp 2 2hp for plotting let39s set 2 in some units Notes 3 Quantum Physics F2005 Momentum in square well One part of the momentum distribution looks like this The Other OOkS like thIS peaked atp hn L The sum see hi for suare wemws Notes 3 Quantum Physics F2005 7 Momentum in square well Now to calculate expectation value of momentum lei ll r Li 7 7 s1n iip L 2 2h 2 272 I Dd d I0 p p 4m pp 2 272p 2 272p We don39t really want to calculate this awful thing so let39s think about symmetry We note that I is symmetric aboutp 0 This means that pd is an odd function and therefore the integral is zero The situation would be much worse if I asked you to calculate p2gt because you39d really have to do the integral Notes 3 Quantum Physics F2005 The Momentum Operator a plausibility argument We want to be able to do something like this 1 fl 11 xtpxwltxtgtdx but we don39t yet know how to write px Consider a pure momentum function which is used to make up a wavepacket IPP 1 eipx Eth 23th to extract p from this we might take the derivative wrt x 6 1 ip 6 P III gt ih 1P xt 11 6x h p 6x p p p Notes 3 Quantum Physics F2005 9 The Momentum Operator From a plausibility argument for a pure momentum state we found a momentum operator A 6 E lh PX ax We generalize this idea and will apply this operator to any wavefunction to extract information on the momentum pxgt OE Pxt X Pxtdx 0 1 Py xt ihaijlpxtdx 00 00 x Notes 3 Quantum Physics F2005 10 Momentum in the square well A 0 1Py ct x Pxtabc OE PxI ihaij11 xtdx 00 00 x ih2 in a in Icos cos dx L 00 L 636 L lh22 0 cos sin dx lh22 0 udu 0 L 00 L L L CX Notes 3 Quantum Physics F2005 11 Where are we and where are we going We are in the middle of Chapter 5 learning the concept of the operator for a quantum observable o We will review a few of the most frequently used operators and some obvious ramifications of this approach o By the end of this class we will formally introduce the Schrodinger equation and begin to address some wellknown solutions stationary states and the timeindependent Schrodinger equation the time dependent SE Notes 3 Quantum Physics F2005 12 A more general momentum operator derivation following Morrison prob 51 A reasonable starting point dltxgt m m dz We will also use Schrodinger39s equation 2 2 E af 1wand 6111316 1 is or 2m 6x 1h 6t 2m 6x 1h 80 6211 allyquot J 2 2 ax ugntum Ph Notes 3 sics F2005 13 A more general momentum operator derivation ih 6211 6211f 1P x x P x ltpgt 2 6x2 6x2 1 11 LIP x P 2 dx 2 6x 6x 6x 6x Px x P P P ihj P dx 2 6x 6x 6x CX CX ltpgt 1P ih 1Pdx O 111 de oo 6 00 A 611 E lh p 6x Notes 3 Quantum Physics F2005 14 Notes 3 Momentum Operator h L r 357 U Quantum Physics F2005 Notes 3 The Operator Postulate irulm l 1mm m Quantum Physics F2005 Operator mathematics You already know about operators you ve used them in Math Theo Phys Optics They are essentially a set of instructions on what to do to a function take the derivative multiply it by Some rules Operators must act on a function Operators act on every function to their right unless their action is constrained by brackets The product of operators implies successive operation starting with the rightmost The sequence of operations does not commute addition or subtration of operators means to distribute the operations Notes 3 Quantum Physics F2005 17 How general operators differ from the specialized ones you learned in elementary school Noncommuting operators Let39s look at operators Q1 and Q2 Take Q 3 47161 and Q 22 x 6 Is Ql zfo Qz lfm A A 6 a gem lh xf mf mi 6x 6x A A a Q2Q1fx thL 6x A measure of whether operators commute is the difference due to order Q2Q1Q1Q2f ihf Exercise 5 Check this expression explicitly for fe2X Notes 3 Quantum Physics F2005 18 Exercise 5 solution Q1 i 471 and Q 2 x and fx e Is Ql zfo Qz lfo Q1Q2fx 471xezx ihe2x 2xe2x Q2Q1fx ihxezx 4mm A measure of whether operators commute is the difference due to order QZQI Q1Q262x e ih2xe2x ihezx 2mm ihe Notes 3 Quantum Physics F2005 19 Some other operators To find the expectation value of a general observable Q First construct an operator to extract Q For wavefunctions this can usually be accomplished by rewriting the physical quantity in terms of x and p Then compute the expectation integral as before Q PQx de Notes 3 Quantum Physics F2005 20 Some other operators 2 Classical kinetic energy T L 2m When an operator is squared that means to perform the operation twice A 1 1 1 6 6 T A2 M 39h ih w 2m p w 2m p 2m ax 6x Position fc x Constant C C Notes 3 Quantum Physics F2005 21 Some other operators Potential energy V V Usually the potential energy V is a function of x and t so we just contruct it from Vxt Example a spring V 082 gtV 10222 If it has time dependence we need to include that too The total energy operator in x F f V iS called the Hamiltonian The energy operator in terms of t 13 mg Notes 3 Quantum Physics F2005 22 The Schrodinger Equation The Schrodinger equation is a relationship between the space and time representations of the energy operators This equation allows you to calculate the timedevelopment of the state functions 6 2m 2 8x2 1921 Ew 911 V h i x 011 l at Notes 3 Quantum Physics F2005 23 Quantum Physics 2005 Notes4 The Schrodinger Equation Chapters 6 7 Notes 4 Quantum Physics F2005 The Schrodinger Equation Flt151p r22 92 92 V xt I ih 2m 6x21 W at We can use the relationship between time and space derivatives to explore the behavior of the wavefunctions Notes 4 Quantum Physics F2005 Moving particles and probability flow aPxtazuzp a tazp h 6220 162 at at at at 2im 6x2 2im 6x2 h 62 6 a h 620 6 2im w 6x2 w 6x2 6x 2im w 6x 6x w Definingjxt i a aiwj we have 21m 6x 6x 6Pxt 6jxt at 6x where jxt is called the probability current density Notes 4 Quantum Physics F2005 3 Interpreting probability flow We can understand this equation if we think about probability in a finite region and current flow into and out of this region a 6 b aPabt 5Pxtdx Jat JUN This is a conservation law that states that the only way the probability of finding a particle in a region can change is if the probability of finding it outside the region increases accordingly Notes 4 Quantum Physics F2005 Probability current for a pure momentum state Taking IPPO xi AeiPox Eolh h t 62 62 at x 2im w 6x 6x w h u u I I 6 iltpox Eorgth 1P0 eiltp0x Eorgth 1Poj e iltpox Eorgtheiltpox Eorgthj 39 2im h h hA2 Zipo IAIZ 2 39W739 O39IAI V0 probability current is constant in space therefore what flows into a volume must flow out Notes 4 Quantum Physics F2005 5 Probability flow in three dimensions We are interested in the probability of finding the particle in a finite volume V PV j Ptdr We define jm g mhwm lmw Amax w an so m E j 0 at 3 mvPtdr if dEi 0 at A Notes 4 Quantum Physics F2005 6 Time evolution of a wave packet 1 A wave packet is a linear combination of an infinite number of extended single wavelength with infinitesimally differing wavenumbers 1mm 0 Akequotltquotquotquot quotdk 1 M This definition allows us to predict the time evolution of the wave packet where Ak 0 1Pc0eik xa k Notes 4 Quantum Physics F2005 Time evolution of a wave packet 2 Gaussian example Assuming the spatial part of the wavefunction at t 0 is a Normal Gaussian function centered at 0 and moving with average wavenumber k0 12 1 ikx X2402 1P x0 e 0 e X q Zn 12 1 1 0 2 2 SO elk0 kxe x 40x dx 221 q Zn Jo We have already solved for Ak in Notes 2 page 33 1 Ak 2034 e a k ko2 7139 Notes 4 Quantum Physics F2005 8 Time evolution of a wave packet 3 Gaussian example Now since we know Ak we can solve for the time evolution of 1Pxt 1 00 l 2 m 2 Z w 2 2 i 2 m 1110C L IAke kx hk tZ dk 23 Ie axk k0 6PM hk t2 dk 1 2 1ix 2 mm 0g e My Jexp c b11 uz iEEQZ Lx u du 27 w 2m m which is an integral of a standard form lexp 0m2 udu xhkotj m 40 iht 2n2 Notes 4 Quantum Physics F2005 9 OO eXp 1 2 0 4 ikx thZm wow x3 e 0 J 277 Time evolution of a wave packet 4 Gaussian example An animation of a travelling quantum wave packet can be found at Notes Look at the probability distribution You can see the motion of the center of the wavepacket is the same for whatever initial width you choose The speed only depends on the initial choice of k0 When the initial packet is narrower the packet spreads more quickly You can see why this is by looking at the real part of the wave packet where you can see the waves Short wave components travel faster than longer wave components as you can see in the display of the real part of psi This effect can be observed in the transport of light pulses down fiber optics Notes 4 Quantum Physics F2005 10 Thus far in our story We have shown some properties of wave functions and state functions Some rules governing the form of the wave functions and some plausible wave functions Given a wave function to describe a particle how do we deduce observable properties What is a plausible wave equation We will now start finding solutions to the wave equation under various wellstudied conditions Notes 4 Quantum Physics F2005 11 Things you should understand and be able to do Understand and do simple calculations related to the important experiments listed in Notes 1 Photoelectric Compton Diffraction Understand and use the Einstein and DeBroglie equations relating particle properties kinetic energy and momentum to wave properties frequency and wavelength p 39 in Morrison Understand and perform simple calculations using the Heidenberg Uncertainty Principle p 9 Morrison Do calculations with probability distributions normalization expectation value variance and standard deviation You should also be able to look at a graph of probability and make a reasonable guess of average position and standard deviation on a quiz for example Notes 4 Quantum Physics F2005 12 Things you should understand and be able to do 2 Understand the concept of the complex state function and its relation to probability especially the wave function and the probability of finding the particle somewhere Know the physical limitations on the form of wave functions p 78 Recognize a pure momentum state single wavelength travelling wave and pick out the wavelength wave number frequency period and phase velocity pp 107 ff Transform back and forth between the wave function real space representation and the momentum amplitude function momentum space representation of a quantum state Set up a transform in each direction Perform the math for some simpler cases Use symmetry to simplify or avoid calculation Have a rudimentary understanding of how the real space wave function is related to the amplitude function especially the Heglsenbepgygcrsigegginty Principle Notes 4 uantum 13 Things you should understand and be able to do 3 Understand the concepts of the wave packet and group velocity p 113 Be able to perform simple operator arithmaticp 159 Know and love useful operators for important observables in real space p 173 and momentum space position momentum kinetic energy total energy Know the Schrodinger equation and how one simplifies it into the time independent Schrodinger equa on Notes 4 Quantum Physics F2005 14 Another motivational interlude Many of the really amazing apparent paradoxes of physical understanding have roots in quantum physics A particle interfering with itself is odd A particle being everywhere at once Quantum cryptography o The understanding of many of the most common things is based in quantum physics Light emission by a fluorescent bulb Lasers Semiconductors The folding of proteins Photosynthesis Chemical processes Developing new nanotechnology requires an understanding of quantum physics Notes 4 Quantum Physics F2005 15 A special set of solutions Constant V The full TDSE if 92 911m t P xt V xt P xt ih 2Inaxz at Let39s assume that V does not Change with time 2 2 h a 2 Pxt Vx Pxt ih 61110 2m 6x Now we39ll search for solutions using the separation of variables method 1PM t 1JXTf Notes 4 Quantum Physics F2005 16 ConstantV solutions h2 azmxgw WWW m ammo 2m 6x at 1706w 2x Vxw xrt xihaT t 2m 6x at Divide both sides by Pxt iiazw gmm 2m xrt 6x xrt Bi Now we see that the left is only a in of x and the right only t The only way this equation can be true for all x and t is if each side is equal to the same constant Notes 4 Quantum Physics F2005 17 ConstantV solutions 80 now we have two separated equations 722 1 6210x wx 6x2 LiiLz 670 a 1t at Looking at the t equation 610 a Vx a ih aw 0 yields 1t 10e39iath and by association with the EinsteinDeBroine relations a 7260 E Notes 4 Quantum Physics F2005 18 ConstantV solutions 80 now we have Wow wow wxe iE h Where the total energy is a constant of the particle The new equation in x 31 2 62 Vltxgt ltx we 2m 6x is called the timeindependent Schrodinger Equation The wave function solutions to the time independent SE are called stationary states because the energy is constant Notes 4 Quantum Physics F2005 19 Special solutions for the one dimensional Schrodinger equation with constant potential Stationary States es 4 Quantum Physics F2005 20 An unconfined particle Constant potential VO everywhere Notes 4 Quantum Physics F2005 21 Free Space V same everywhere Stationary states of the free particle wave 1 62 x 2mE V 6x2 7amp2 should look familiar from your diff eq course wltxgt0 2 or Schaum39s Outline S kzy 0 X The solutions are sinkx coskx emquot or e39ikquot wherek w 75 By convention we set V 0 for free space Notes 4 Quantum Physics F2005 22 The general solution vs the specific case The free particle wave 2 o There are an infinite number of possible solutions to the free space Schrodinger equation All we have found is the relation between the possible time solutions and the possible space solutions o We need to give more information about the state for you to limit the set of possible solutions If we specify the energy E then the set of possible k s is limited to two possibilities and but this still leaves us with sine and cosine or k k solutions We need to be given other limitations such as the value of the wavefunction at certain points certain times o We will spend a significant portion of this course solving special cases given interesting Vx and boundary conditions andor initial conditions Notes 4 Quantum Physics F2005 23 Wavefunction for a free particle wave 3 The spatial part of the wave function is a linear combination of a complete set of solutions Choosing the exponential function representation Ux are lye 7 SO 1Pxt Uxrt aeikx be jmeiElh aeikx l lye mm If we specific the direction of propagation to be x then mom Uxz t be ilo eiElh be kxw Looking at the probability density 111111 b2 constant Let39s try normalizing this 1011J de b2 0 dx This means we have to let b gt O as x gt 00 Notes 4 Quantum Physics F2005 24 Wavefunction for a free particle wave 4 momentum Since the position is completely indeterminate let39s see what the momentum is 0 1113266de b2 0 ei39 ih ike ikquot dx x 00 p w 00 hk I 1 de 2 I dx 0 2 62 2 0 ikx wl 2 92 ikx wl 1 h ijdx 9 e h 726 dx ltp2gt 00 6x 00 00 6x h2k2 I 1 de 2 I dx oo 00 SO p2gt ltpgt2 fisz hk2 0 This state has a perfectly defined momentum Notes 4 Quantum Physics F2005 25 Wavefunction for a free particle 5 Probability current jxt 5 i qquot 31 Apiltf 2m 6x 6x hb2 eikxwlikeikxwt e ikx wlike ikX wl m b2 E b2 E Pv m m where P is the probability density Notes 4 Quantum Physics F2005 26 A particle confined to a region of space The Particle in a Box with Infinitely High Walls Notes 4 Quantum Physics F2005 27 Stationary states of a particle in a box 1 Outside the box 1Pxt0 Inside the box VO so 1Pxt a coskx b sinkxeiElh Voo or QKXJCJMkmdBm we are free to choose which one to use V200 xO 5 xL Boundary conditions x s OO x 2 LO Note that l have chosen to place the boundaries at O and L rather than L2 as we did previously I have done this to provide a slightly different approach and solutions but the basic result is the same Notes 4 Quantum Physics F2005 28 Particle in a box 2 Starting with the xO boundary condition 0 asinObcosO 0 The only way this can be true is if bO Now for the xL boundary condition L asinkL 0 This can be true if aO no wave function at all or if CL mt nx asinknx and normalizing gives w x sinknx Notes 4 Quantum Physics F2005 29 Particle in a box 3 The imposition of the two boundary conditions that confine the wavefunction to a region of space leads to quantization of the stationa state particle energies Notes 4 Quantum Physics F2005 30 Particle in a box 4 If we chose to place the boundaries of the box at L2 we would have found spatial wave functions like those in the book 2 rm 6 for nodd w x L cos L x ZWWW 3sin x for neven The corresponding energy level for a given n is the same as we found for the OL box Notes 4 Quantum Physics F2005 31 Particle in a box 5 Wave functions for stationary states Note that however we choose to specify the box the states have definite even or odd parity about the center of symmetry of the potential The existence of this symmetry in the potential led to simple solutions for the wavetunctions A u39X 4w 11 1 as n5 15 ms 1 n m n 14 X ms na 12 14 X ms ma 1 n2 14 ms as M ins ins ins Dr M 4 V 4 4 DE 1 14 X as P P P 1 1 1 us us us 14 E14 14 n2 n2 n2 2 14 X DE DE 1 Equot 12 14 X DE DE 1 Equot 12 14 X DE DE 1 Equot 12 14 X as Notes 4 Quantum Physics F2005 32 Particle in a box 6 Momentum distribution for stationary states In a previous homework and in Notes 3 we deduced the momentum amplitude functions for these wavefunctions mt kL mt kL 1 111 Sln Ak for even n r w ea L Notes 4 Quantum Physics F2005 33 Particle in a box 7 Expectation values for stationary states By observing symmetries and from our previous work on particle in a box wave functions we know Expectation value of position xn the center of the box Uncertainty in position Jltx gt Ali r 1 2 n 7139 0 Expectation value of momentum p mm Uncertainty in momentum p5 L Notes 4 Quantum Physics F2005 34 Particle in a box 8 Expectation value and uncertainty in energy A2 The kinetic energy operator is A m Since we previously solved for ltp gt we can easily solve for ltTgtf n The lowest allowed state has n1 so expectation value of the kinetic energy is always gt zero Another remarkable observation is that T2gt Tgt2 0 There is no uncertainty in the energy Notes 4 Quantum Physics F2005 35 Particle in a box 9 Probability current 1Pnx sin x e W jxtE lh1PELP Pilp 2m 6x 6x sin knx sin km k sin knxsin km 0 2m L everywhere in the box Notes 4 Quantum Physics F2005 36 A problem with this example A small mon st efr in the closet here is that these wavefunctions do not satisfy all of the physical limitations on wavefunctions that we specified earlier continuous yes normalizable yes continuous derivative no there is a kink at the boundanes This situation has arisen because we have actually specified an infinite potential energy discontinuity at the boundaries Notes 4 Quantum Physics F2005 37 A general observation about energy levels We have found for the particle in the box that the energy is quantized it can only take on certain discrete values that are related to the dimensions of the box and the mass of the particle We found no such quantization for the free particle plabe wave It is a general observation that situations in which particles are bound give rise to quantized energies whereas unbound systems lead to an energy continuum Notes 4 Quantum Physics F2005 38 Another observation about energy The Time Independent Schrodinger Equation for a free particle also leads to another useful observation 2 2 lnspectin h a 21J EZJ we see that the left hand side 2m 6x is a measure of the curvature of the wavefunction whereas the right hand side is the energy Basically the curvier the wave function the more energy is has This is useful of you want to get an idea of what the wavefunction might look like for a special potential shape Notes 4 Quantum Physics F2005 39 An addition to the language Eigenvalue equations Eigenfunctions Eigenvalues The Time Independent Schrodinger equation is a secondorder ordinary differential equation in X It is linear and homogeneous and has 2 independent solutions 19w EzJ is known in mathematics as an eigenvalue equation because the operator returns a constant times the original function 20 is called the eigenfunction of the operator F E is the corresponding eigenvalue for Hzp Notes 4 Quantum Physics F2005 40 The next set of notes will deal with several specific cases Step Potential Barrier Potential Harmonic Well Potential Particle in a 3D Box Quantum Dot Notes 4 Quantum Physics F2005 41 Nates 4 QUantum Physics F2005 4 2 Quantum Physics 2005 Notes1 Course Information Overview The Need for Quantum Mechanics Notes 1 Quantum Physics F2005 Contact Information Peter Persans SCfoO x2934 email persap rpiedu Physics Office SC 1025 x6310 mailbox here Office hours Thurs 24 M12 2 and by email appointment Please visit me Home contact 7861524 710 pm Notes 1 Quantum Physics F2005 Notes 1 Topic Overview Why and when do we use quantum mechanics Probability waves and the Quantum Mechanical State function Wave packets and particles Observables and Operators The Schrodinger Equation and some special problems square well step barrier harmonic oscillator tunneling More formal use of operators The single electron atom angular momentum Quantum Physics F2005 Intellectual overview The study of quantum physics includes several key parts Learning about experimental observations of quantum phenomena Understanding the meaning and consequences of a probabilistic description of physical systems Understanding the consequences of uncertainty Learning about the behavior of waves and applying these ideas to state functions Solving the Schrodinger equation andor carrying out the appropriate mathematical manipulations to solve a problem Notes 1 Quantum Physics F2005 The course 2 lecturestudios per week TF 122 Reading quiz and exercises every Class 15 of grade Homework35 o 2 regular exams 30 Final 20 Notes 1 Quantum Physics F2005 Textbook Required Understandinq Quantum Physics by Michael Morrison Recommended Fundamentals of Physics by Halliday Resnick Handbook of Mathematical Formulas by Spiegel Schaums Outline Series References Introduction to the Structure of Matter Brehm and Mullen Quantum Physics Eisberg and Resnick Notes 1 Quantum Physics F2005 Academic Integrity Collaboration on homework and inclass exercises is encouraged Copying is discouraged Collaboration on quizzes and examinations is forbidden and will result in zero for that test and a letter to the Dean of Students Formula sheets will be supplied for exams Use of any other materials results in zero for the exam and a letter to the DoS Notes 1 Quantum Physics F2005 Notes 1 Quantum Physics F2005 8 Classical Mechanics A particle is an indivisible point of mass A system is a collection of particles with defined forces acting on them A trajectory is the position and momentum rt ptof a particle as a function of time If we know the trajectory and forces on a particle at a given time we can calculate the trajectory at a later time By integrating through time we can determine the trajectory of a particle at all times 2 min VVFI 131 mg Notes 1 Quantum Physics F2005 Some compelling experiments The particlelike behavior of electromagnetic radiation Simple experiments that tell us that light has both wavelike and particlelike behavior The photoelectric effect particle Double slit interference wave Xray diffraction wave The Compton effect particle Photon counting experiments particle Notes 1 Quantum Physics F2005 10 The photoelectric effect Light Collector M Vext FIGURE 310 Apparatus for ob serving the photoelectric effect The ow of electrons from the emitter to the collector is measured by the am meter A as a current i in the external circuit A variable voltage source Vext establishes a potential difference between the emitter and collec tor which is measured by the volt meter V Krane Current i 2 211 Krane 11 0 ultj Potential Difference V FIGURE 311 The photoelectric current i as a function of the potential difference V for two different values of the intensity of the light When the intensity is doubled the current is doubled twice as many photoelec trons are emitted but the stopping potential VS remains the same A N 39srope41x1o15vs Stopping potential VS volts 0 60 80 100 120 Radiation frequency 1013 Hz FIGURE 312 Millikan s results for the photoelectric effect in sodium The slope of the line is he the exper imental determination of the slope gives a way of determining Planck s constant The intercept should give the cutoff frequency however in Mil Krane In a photoelectric experiment we measure the voltage necessary to stop an electron ejected from a surface by incident light of known wavelength Notes 1 Quantum Physics F2005 11 Notes 1 Interpretation of the photoelectric effect experiment Einstein introduced the idea that light carries energy in quantized bundles photons The energy in a quantum of light is related to the frequency of the electromagnetic wave that characterizes the light The scaling constant can be found from the slope of eVSwp vs wave frequency v It is found that eV hv c1 Ephm q where h 6626x103934 joulesec Stop material For light waves in vacuum 6 2w 3x108 ms so we can also write hc eVStop A A convenient nonSI substitution is hc 1240 eVnm Quantum Physics F2005 I material from Millikan s 1916 paper 1 That there eXists for each exciting frequency V above a certain critical value a de nitely determinable maximum velocity of emission of corpuscles 2 That there is a linear relation between V and V dV 3 That 2 or the slope of the V v line is numerically equal to he 4 That at the critical frequency ya at which V a p hvo i e that the intercept of the Vv line on the V axis is the lowest frequency at which the metal in question can be photoelectrically active 5 That the contact EMF between any two conductors is given by the equation Contact EMF hevu 39 vo39 Va Vo Notes 1 Quantum Physics F2005 13 ant5 0 Volts 9 Eu Notes 1 Millikan Phys Rev 7 355 1916 12 a mo ii W f m h L rL 4115 l x g 5539 30 139 1561 Quantum Physics F2005 Compton scattering 2t Target X ray source FIGURE 319 Schematic diagram of Compton scattering appara tus The wavelength A of the scattered X rays is measured by the detector which can be moved to different positions 0 The wave length difference A A varies with 6 from Krane The Compton effect involves scattering of electromagnetic radiation from electrons The scattered xray has a shifted wavelength energy that depends on emitting results for X ray scattering scattered direction Notes 1 Quantum Physics F2005 15 Compton effect We use energy and momentum conservation laws Energy hv39 hv K electron K kinetic energy Momentum x component i cos ymvcos A A39 h y component 0 F31n ymvsm6 A39 AAAL1 Cos mc m mass of electron yreativistic correction to electron momentum mv Notes 1 Quantum Physics F2005 16 Conclusions from the Compton Effect Xray quanta of wavelength A have Kinetic energy K Momentum p Notes 1 Quantum Physics F2005 17 Double slit interference Plane wavefronts Double slit Maxima b Screen FIGURE 510 Results from the twoslit experiment using photons Minima 1 Measured photon intensity as a function of position From M Cagnet Atlas of Optical Phenomena SpringerVerlag 1962 b Calculated photon intensity The intensity maxima in a double slit wave interference experiment occur at d sin 6 1M where d is the distance between the slits The width of the overall pattern depends on the width of the slits Notes 1 Quantum Physics F2005 18 Xray diffraction In xray diffraction xray waves diffracted from electrons on one atom interfere with waves diffracted from nearby atoms Such interference is most pronounced when atoms are arranged in a crystalline lattice X rays Bragg diffraction maxima L111 Rt39 f s are observed when HERE 3XQeLJQLLQQi 2d Sin 639 1M ected from a set of crystal planes of spacing d The beam re ected from thesecondplalrlletrivelasadistagcezj d sm Ogreatert ant e earn re ecte planes of atoms in the crystal from the rst plane Notes 1 Quantum Physics F2005 19 Laue Xray Diffraction Pattern A Laue diffraction pattern is observed when xrays of many wavelengths are incident on a crystal and diffraction can therefore occur from m a ny p a n es satisfy the Bragg condition for a particular wavelength Bottom Laue pattern of NaCl crystal S 39 m n I u u y pretty rom Krane FIGURE 38 Laue pattern of a quartz crystal The differ NOtGS 1 ence in crystal structure and spacing between quartz and NaCl m makes this pattern look different from Figure 37 The bottom line on light In many experiments light behaves like a wave cphase velocity vfrequency Azwavelength In many other experiments light behaves like a quantum particle photon with properties h olon energy Ep hv h momentum p Z and thus E pC Notes 1 Quantum Physics F2005 21 Some compelling experiments The wavelike behavior of particles Experiments that tell us that electrons have wavelike properties electron diffraction from crystals waves Other particles proton diffraction from nuclei neutron diffraction from crystals Notes 1 Quantum Physics F2005 22 Electron diffraction from crystals RIC 43 Comparison of X rztv diffraction and electron gtion The upper half of the gure shows the result of rz39ing of 0071 nm X rays by an aluminum foil and the hull shows the result ol scattering of 600 eV electrons minum The wavelengths are different so the scales of 90 halves have been adjusted electron diffraction patterns from single crystal above and polycrystals left from Krane Notes 1 Quantum Physics F2005 23 Proton diffraction from 101 O O o L 39H Ii Ily nl 1114 u i In NH HI prulnn gt p 1 c C O N M O L o c39 105 106 o 4 8 12 16 2o 24 28 Scattering angle degrees FIGURE 48 Diffraction of 1 GeV protons by oxygen nuclei The pattern of maxima and minima is similar to that of singleslit diffrac tion of light waves from Krane Notes 1 Quantum Physics F2005 nuclei 24 Neutron diffraction Region of lquotumbra dZA l I A Width of central l maximum Shadowdiffraction DMd pattern at very large distance D gtgt dZA 20 80 wave of fast neutrons gt Advancing plane FIGURE 47 Diffraction ofquot neu frons by a sodium chloride crystal C from Krane Differential cross section barns per steradian U 8 A 0 30 0 O o O 24 O D O N a O on a s 03 N h 0 a Neutron scatterinq andle decrees Diffraction of fast neutrons from Al Cu and Pb nuclei from French after A Bratenahl Phys Rev 77 597 1950 Notes 1 Quantum Physics F2005 25 Electron double slit interference Electron interference from passing through a double slit I x b x from Rohlf FIGURE 511 Results from the twoslit experiment using electrons Ia Measured electron intensity as a function of position From C Jonsson Zeit Phys 161 454 r 1961 2 Calculated electron intensity Notes 1 Quantum Physics F2005 26 Helium diffraction from LiF crystal gtlt Ei LiF crystal Collimating Detector Intensity l Detector setting ab a b Fig 216 a Experimental arrangement used by Stern et al to investigate crystal di rraction ofneutral helium atoms b Experimental results showing central re ec tion peak 4 0 plus rstorder di raction peaks d 11 In the experiment 0 185 from French after Estermann and Stern Z Phys 61 95 1930 Notes 1 Quantum Physics F2005 27 Alpha scattering from niobium nuclei 0001 20 l l l l 39 4o 60 80 Scattering angle degrees Angular distribution of 40 MeV alpha particles scattered from niobium nuclei from French after G lgo et al Phys Rev 101 1508 1956 Notes 1 Quantum Physics F2005 28 The bottom line on particles In many experiments electrons protons neutrons and heavier things act like particles with mass kinetic energy and momentum 2 p mv and K mv2 g for nonrelativistic particles m In many other experiments electrons protons neutrons and heavier things act like waves with h A ZmK elw Notes 1 Quantum Physics F2005 29 The De Broglie hypothesis for everything Quantum Physics F2005 30 Some other wellknown experiments and observations optical emission spectra of atoms are quan zed the emission spectrum of a hot object blackbody radiation cannot be explained with classical theories Notes 1 Quantum Physics F2005 31 Notes 1 Emission spectrum of atoms comiuuous specrum Emission Specmun Absorption Specimm stron omy website Quantum Physics F2005 32 Blackbody radiation di 3 I i I I l x f 8nhc prii d1 a 2 15 a E Visible 2 region a QE 1 4000 K i 0 05 I 10 I 15 x 10 3 Figure 112 Planck s energy density of blackbody radiation at various temperatures as a function of wavelength Note that the wavelength at which the curve is a maximum de creases as the temperature increases from Eisberg and Resnick Notes 1 Quantum Physics F2005 ekeAkaquot 1 33 A review of wave superposition and interference Many of the neat observations of quantum physics can be understood in terms of the addition superposition of harmonic waves of different frequency and phase In the next several pages I review some of the basic relations and phenomena that are useful in understanding wave phenomena Notes 1 Quantum Physics F2005 34 Harmonic waves yA sinkxvtqp or yASinkXIatp sinx A wavelength k 27 T period w27 2 v Notes 1 Quantum Physics F2005 35 Phase and phase velocity yAsinkXwt kxiwt phase when change in phase2n repeat Phase velocity speed with which point of constant phase moves in space Traveling wave distance m Notes 1 Quantum Physics F2005 36 Complex Representation of Travelling VVaves Doing arithmetic for waves is frequently easier using the complex representation using a el cos zs1n6 So that a harmonic wave is represented as 20xt Acoskx out 8 11006 ReAeikx at Notes 1 Quantum Physics F2005 37 Adding waves same wavelength and direction different phase 111 1101 sinkx wt Q 102 1002 sinkx wt 2 taking the form 10R 1p1 102 100 sinkx cut a we find 1151 1152 211011102 COSlt 1 2gt and 1101 Sin 1 44102 Sin 2 1101 COS 1 44102 COS 2 Notes 1 Quantum Physics F2005 38 tana Adding likewaves in words Amplitude When two waves are in phase the resultant amplitude is just the sum of the amplitudes When two waves are 1800 out of phase the resultant is the difference between the two Phase The resultant phase is always between the two component phases Halfway when they are equal closer to the larger wave when they are not see Notes 1 Quantum Physics F2005 39 Another useful example of added waves diffraction from a slit 3 path length difference Let s take the field at the view screen from an element of length on the slit ds as 5st Ignore effects of distance except in the path length l ST l Notes 1 Quantum Physics F2005 40 Single slit diffraction dw Re11xeik ds zs zo ssinB m w6wxelkzrm elmmads raZ w eiazrma sin9 where sin 6 4n 75 m m 2 27 A 2 First zero at so S1116 7 i 2 2 5111 ak a 1 1461 7 Notes 1 Quantum Physics F2005 Adding waves traveling in opposite directions Equal amplitudes ZJR zJ0sinkx cot 81 sinkx cot 82 ZJR 214 sinkx L282 cos cot The resultant wave does not appear to travel it oscillates in place on a harmonic pattern both in time and space separately Standing wave Notes 1 Quantum Physics F2005 42 Adding waves different wavelengths 3 do lb quot o N no A 5k diff BEATS 901 001 cosk1x cult 2 001 cosk2x cuzt 1 w 2 01COSEUk1k2xw1w2t xcosk1 k2x w1 w2t Ak A0 cos kx th cos x t a t i 2 2 l The resultant wave has a quickly varying part that waves at the average wavelength of the two components It also has an envelope part that varies at the difference between the component wavelengths Notes 1 Quantum Physics F2005 43 Adding waves group velocity Note that when we add two waves of differing co and k to one another the envelope travels with a different speed monochromatic wave 1 v wl phasel k1 502 monochromatic wave 2 vphm2 k 2 A0 beat envelope VWP E see Notes 1 Quantum Physics F2005 Adding many waves to make a pulse In order to make a wave pulse of finite width we have to add many waves of differing wavelengths in different amounts o The mathematical approach to finding out how much of each wavelength we need is the Fourier transform 00 00 fx l Ak cos kxdk Bk sin kxdk 7 0 0 Where Ak Offbr cos kxdx Bk Offbr sin kxdx Notes 1 Quantum Physics F2005 45 The Fourier transform of a Gaussian pulse We can think of a Gaussian pulse as a localized pulse whose position we know to a certain accuracy Ax20x U3 1 77320 fX e 076 Notes 1 Quantum Physics F2005 46 Notes 1 Finding the transform I will drop overall multiplicative constants because I am interested in the shape of Ak Ak 06 j e xZZage ikxdx j e axze ikxdx where a 1 20 solving by completing the square ik 2 xJE T k2 4a Akoc Maizem Je J5 dx 39k lettln xxla l g 3 2 a 0C ie kZm 0 e32d3Eek24a Zekza 2 v m a a Quantum Physics F2005 47 Transform of a Gaussian pulse The Heisenberg Uncertainty Principle We can rewrite this in the standard form of a Gaussian in k Ak oc equot 22quot393 where 02 1a2 k x The result then is that oxok1 for a Gaussian pulse You will find that the product of spatial and wavenumber widths is always equal to or greater than one Since the deBroglie hypothesis relates wavelength to momentum phgt we thus conclude that oxopgthl2m This is a statement of the Heisenberg Uncertainty Principle Notes 1 Quantum Physics F2005 48 The Heisenberg Uncertainty Principle o This principle states that you cannot know both the position and momentum of a particle simultaneously to arbitrary accuracy There are many approaches to this idea Here are two Notes 1 The act of measuring position requires that the particle intact with a probe which imparts momentum to the particle Representing the position of localized wave requires that many wavelengths momenta be added together The act of measuring position by forcing a particle to pass through an aperture causes the particle wave to diffract Quantum Physics F2005 49 The Heisenberg Uncertainty Principle Position and momentum are called conjugate variables and specify the trajectory of a classical particle We have found that if one wants to specify the position of a Gaussian wave packet then AxAp h o Similarly angular frequency and time are conjugate variables in wave analysis They appear with one another in the phase of a harmonic wave AwAt 1 Since energy and frequency are related Planck constant we have for a Gaussian packet AEAt 71 Notes 1 Quantum Physics F2005 50 The next stages We have seen through experiment that particles behave like waves with wavelength relationship phA The next stage is to figure out the relationship between whatever waves and observable quantities like position momentum energy mass The stage after that is to come up with a differential equation that describes the wavy thing and predicts its behavior There is still a lot more we can do before actually addressing the wave equation Notes 1 Quantum Physics F2005 51 References Krane Modern Physics Wiley 1996 Eisberg and Resnick Quantum Physics of Atoms Wiley 1985 French and Taylor An Introduction to Quantum Physics MIT 1978 Brehm and Mullin Introduction to the Structure of Matter Wiley 1989 Rohlf Modern Physics from ato Z Wiley 1994 Notes 1 Quantum Physics F2005 52 Addendum Energy and momentum The notation for energy momentum and wavelength in Morrison is somewhat confusing because he does not always clearly distinguish between total relativistic energy which includes mass energy and kinetic energy which does not Here goes my version 2 P 2m0 1 Classical kinetic energymomentum relation T m0v2 2 Relativistic total energymomentum relationship E2 17262 m c4 With E T mocz When you want to find the wavelength from classical kinetic energy use 1 andp p2 h2 11 he T gt 2m0 2mOt2 lszT lszCZT When you want to find the wavelength for a relativistic particle use 2 and p A hCE hcTmocz mcz 2 2 2 1 0 J 1 moc E Tmocz Notes 1 Quantum Physics F2005 53 Addendum comparing classical and relativistic formulas for wavelength h he I zmoT lZmOCZT hc T m c2 2 Relativistic A 0 hc hc To compare the two expressions let39s Taylor expand eq 2 in T 1 Classical 1 CZ A hc g hc he 2 T 2 m 62 2 T l2moczT moc 2 1 1 0 m CZ moc 0 They become the same at small T Notes 1 Quantum Physics F2005 54 Addendum comparing classical and relativistic Notes 1 formulas for wavelength h he 1 CIaSSIcaI A lzmoT ZmoczT 2 Relativistic A ho 2 moc2 T 2 1 1 moc To find the difference between the two forms let39s keep all the terms in T 0 2 A hc 2 E M he 2 T T moo2 72H 1 mocz T2 2 T21 1 VZTmocz ZJHJ moC moc moc 2moc 1 T ARelativistic E AClassical E AClassical T 4moc 2 J 2moc Quantum Physics F2005 55

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