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University Physics III

by: Sonny Breitenberg

University Physics III PHYS 262

Marketplace > George Mason University > Physics 2 > PHYS 262 > University Physics III
Sonny Breitenberg
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Paul So

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Paul So
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This 172 page Class Notes was uploaded by Sonny Breitenberg on Monday September 28, 2015. The Class Notes belongs to PHYS 262 at George Mason University taught by Paul So in Fall. Since its upload, it has received 62 views. For similar materials see /class/215186/phys-262-george-mason-university in Physics 2 at George Mason University.


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Date Created: 09/28/15
ii More Realistic Particle Wave Packets Notice that since Vx for a free particle extends into all space it can t be normalized Under more practical circumstance a particle will have a relatively well defined position and momentum so that both Ax and Ap will be nite with limited spatial extents A more localized quantum particle can not be a pure sine wave and it can be described by a wave packet with a combination of many sine waves Re 1m Vx IAke quot dk a wave packet linear combination of 0 x many sine waves Wave Packets Re I m a 0 u it Re W b 0 0 M vn n lvnl Al I39v39V39v I k0 it I M MA A ll 1 I A A V VV I quotquot1quot W W VV W W A nan ow l function Ak k H M X Vquot Comb1nat10n of two s1ne waves 9 more localized than a pure sine wave t l l l l l l l l l l l l l l l l l D gives a wave function 41100 with a broad spatial extent Re x Ap smaller Wave Packets gives a wave function lJX with a narrow spatial extent c Ak d R6 lIX A broad function Ak AP bigger Ax smaller Ea Chapter 40 Quantum Mechanics II Particle inaBox Wave function 3 Energy levels I Potential Wells II Potential Barriers and Tunneling I The Harmonic Oscillator E Particle in a Box A 1D box with hard walls m U 0 U L non enetrable lt Ij 00 p U m A free particle inside the box Ux 0 inside box L x No forces acting on the particle except at hard walls 0 L P in x is conserved between bounces gt P is fixed but P switches sign between bounces E Particle in a Box The situation can be described by the following potential energy Ux The potential energy U is zero in the interval 0 lt x lt L and is infinite everywhere outside this interval U00 2 0 O S x S L CC 39 DC 00 elsewhere The limeindependent Schrodinger equation is quot3 395 2 2 a h d Vx L105 UX gW l39UOCWx EVx U O I I x Recall this is bas1ca11y 9 M L KE PE Total E Problem statement For a given Ux what are the possible wave functions Vx and their corresponding energies E Wave functions for a Particle in a Box Inside the box 0 g x g L Ux O and the particle is free From before we know that the wave function for a free particle has the following form linear combination of the two Winside x Alelkx A26 lkx possible solutions where A1 and A2 are constants that will be determined later Outside the box U x oo and the particle cannot exist outside the box and woumde x 0 outside the box At the boundary x 0 and x L the wavefunction has to be continuous f ID Wavefunctions for a Particle in a BOX Let see how this boundary condition imposes restrictions on the two constants A1 and A2 for the wave function Using the Euler s formula we can rewrite the interior wave function in terms of sine and cosine winsidex A1coslocisinkxA2coskx isinkx A1A2coskxiA1 A2sinkx Imposing the boundary condition at x O winside0A1A2cos0W A1 A2 0 WWW x 21A1 sin kx C sin kx where C2iA1 ID Wavefunctions for a Particle in a Box Now consider the boundary condition at x L Illinside C Sin Z O For a nontrivial solution C 70 only certain sine waves with a particular choice of wave numbers can satisfy this condition 5 lilgifwf C9 379 1735113939 This implies that the wavelengths within the box is quantized 3 W L2g3g ua Allowed wavelengths are the ones which can t within the box Wavefunction for a Particle in a BOX 6 lx zz 722 HZLZDSDW n5 512 kn 1 n4 21 L must be an integral number of m wmyebng mLLnAJ2 n3 312 Walgggm similar to standing waves on a cramped string 0 Quantized Energies for a Particle in a BOX Since the wave number kn is quantized the energy for the particle in the box is also quantized 521 3952 1 39 39 3 3 3 IE s mg LIE f E a ma r M n IS called the 1222 3214 Li 7 g quantum number 255 Note the lowest energy is not zero h2 E gt 1 8mL2 n 0 gives yx 0 and it means no particle 16151 0 95 Ex E Probability and Normalization Recall that 1x2 and not the wavefunction itself Vx is the probability density function In particular lyx2 dx C2 sinz quotdex gives the probability in finding the particle in an interval x xdx Within the box a W b Illml2 3 4 quot3 n l ii Probability and Normalization b 2 N0t 53 The positions for the particles are lW probabilistic We just know that it has to be in the box but the exact location within the box is uncertain Not all positions between x 0 and L are equally likely In CM all positions are equally likely for the particle in the box There are positions where the 0 a L particle has zero probability to be found Probability and Normalization Although we don t know exactly where the particle might be inside the box we know that it has to be in the box This means that 3336 A u tda cit l normal1zat1on condltlon ILCZ sin2 xjdsz2 l 0 L 2 So the normalization condition xes the nal free constant C in the wavefunction C JZL This then gives particle in a box Time Dependence Note that With lX found we can write down the full wavefunction for the timedependent Schrodinger equation as W x7 W xeiEth 395 I 16 Note that the absolute value for eiEth is unity ie e e el so that Pnx t2 lx2 is independent of time and probability density in nding the particle in the box is also independent of time lX is called the stationary state of the system eiEthlz iEth iiEth o i Finite SquareWell Potential 0 L Squarewell with nite height U0 d osst elsewhere Ux In Newton s mechanics a particle will be trapped inside the well if E lt U0 In QM such a trapped state is called a bound state If E gt U0 then the particle is not bound For the in nitely deep well as in the particle in a box problem all states are bound states For a nite squarewell there will typically be only a nite number of bound states ID Finite SquareWell Potential Similar to the particle in a box problem Ux 0 inside the well we have 2 d WEX k217x where k 2mE 01quot 1 inside the well wx Aleikx A2641 0r yimde x A cos kx B sin kx and A B are constants to be determined by boundary conditions and normalization But for a nite squarewell potential the wavefunction is not identically zero outside the well The Schrodinger equation is given by 2 2 U E K 21496 Where K outside the well Finite SquareWell Potential Since U0 gt E Kis positive and the wavefunction outside the well is given in terms of exponentials instead of harmonic functions 1x Ce De where C and D are constants to be determined by BC and normalization again For this problem there is a new type of BC at large distances from the origin wavefunction must remain nite not blowing up at large X 1pm x 2 Ce and KM x 2 De ID liq Finite SquareWell Potential As in the particle in a box problem both vx and dVx dx must also be continuous at x 0 and x L Matching yxlt0x yWdex and WM x at x 0 and x L will enforce a certain set of allowed functions to be fitted within the well and the bound state energy is correspondingly quantized 3 Mn b U Continuum U0 2 6E1 3ri V 7 i quot3 39E3509EZ Ur 257 mtg quot239 ElZr l3El 040qu E7 l 5 n l i E 0625151 X r 0104 L Example 6 in a SquareWellQuantum Dot An electron trapped in a SquareWell potential with width L 05nm size of an atom U x Continuum a What is Eco1 if this is a infinitely deep well U0 00 zhz 721055x10 34Js2 1 2mL2 2911Xmelkg050x10 9m2 224x10 19J150eV b For a finite well with U0 613mJ 90eV The energy levels for the finite well are given as shown on the next slide DQdot Example 6 in a SquareWellQuanturn Dot U x Conlinuum U0 652 1 F 39E 76eV n 2 0 r 52 36eV E1 391 l E 094eV What is the wavelength of light released if the electron was originally at the 1st excited state n2 and relaxed back to the ground state nl gt A hc 1240eVs 460nm E2 E1 36eV 094eV I Refraction at a Spherical Surface At B Snell s law gives r Ilt 8 7 393 i 1 na sm 6a 2 nb sm 6b 1 39 s l s I From trigonometry we also have From APBC 9a a tanaz tan z 39 FromAP BC 6b S s 6 gt 9b tan L R 5 Refraction at a Spherical Surface Again consider only paraxial rays so that the incident angles are small we can use the small angle approximations sin 0 tan 0 0 With this Snell s law becomes 71119 11b9 Substituting Ba and 6 om previous slide we have na06 HIM naa na quot13 nb naanb quotb na With the small angle approximations the trig relations reduce to h h aztanaz ztan z tan s s39 R ID Refraction at a Spherical Surface Substituting these expressions for a 8 and into Eq and eliminating the common factor h we then have obj ectimage relationship spherical refracting surface I Refraction at a Spherical Surface To calculate the lateral magni cation m we consider the following rays V 0 y M I I S I S l Ray 1 from Q going toward C along the normal to the interface will not suffer any de ection Ray 2 from Q going toward Vwill be re acted into nb according to Snell s law E Refraction at a Spherical Surface V 0 7 y i Q 1 l s 39 s l 1 From geometry we have the following relations tan a ys tan bz yys39 From Snell s law we have na sin 6a 2 nb sin 6b Refraction at a Spherical Surface ID Using the small angle approximation again sin 6 tan 6 the Snell s law can be rewritten as v 1116 sin6tan6 X 1 quotas a nbs 3 na quot3 s s39 Substituting these into the definition for lateral magni cation we have My 0 V m 2 39 a lateral magni cation spherical refracting surface liq Refraction at a Flat Surface For a at surface we have R 00 Then the Obj ectImage relation can be reduced simply as n n n n a b b a 39 0 s s 00 I n 11 ns ab a gt 39 gt 1 s s nbs Itl33 water Combing this with our result for lateral magni cation we have m 1 so that the image is unmagnifzed and upright i g Example 345 amp 346 I Images formed by a spherical surface can be real or virtual depending on ma nh S and R l 39 r 396 S CHI S gt I 2L DB Pearion Edubalmn Inc pubhsmng as P eavson Admson Wesley legs 800 cm Copyrigh1 2008 Pearson Educalson incquot publishing as Pearson AddisonWesley IE Example 345 amp 346 EX345 z 13900 gt s39113cm s s39 R 8006m s39 200cm EX346 ni z 133 E 019 s21 3cm s s39 R 8006m s39 200cm E Thin Lenses Consider a thin lens as two closely spaced spherical surfaces S2 RSI a aquot i I CopyrlgnKQ 2m Pearson Educabun Km publlsmng as Pearson Addisonrwesley means that t ltlt other lengths For images produced by these two refracting surfaces we will use the image Q from the rst surface as the object for the second surface E Thin Lens 11quot 11 m Q P PM P Wig C cz mk klN 2 Q39 91 K l Q lt quot gt S i c I canyngm 2mm Pearsun Educalxzm Inc puDHshmg as Pearsun Aumsummsxey For the situation indicated here Q is on the side of the outgoing light By the Sign convention we know s1 39 image distl gt 0 and s2 object dist 2 lt 0 But since they represent the same physical distance to Q for consistency we need to have I 52 51 E Thin Lens Now we apply the objectimage relation to the two spherical refracting surfaces nanm ncnc nb S1 S1 39 R1 52 52 39 R2 left surface right surface NOTES Since both C1 and C2 are on the outgoing side of light R1 and R2 are by convention Since the material outside of the lens is typically air or vacuum we take na and nc l o For simplicity we will call nb for the lens itself 11 We also apply the imagetoobject consistency relation 32 sl 1 n n l 1 S1 S1 R1 Thin Lens To eliminating sl by adding these two equations 1 l l l n l S1 52 39 R1 R2 Calling our original object distance s1 simply as s and our nal image distance sz simply as s we have Where g Converging amp Diverging lens Depending on the values of n n R1 and R2 fcan be positive or negative outside l l l nnmtside f l 2 f 9 positive f 9 negative Converging lenses Diverging lenses Meniscusquot V Rlanoconvex Double convex MeniscusV H lanoconcave Double concave E Focal Points of a Converging Lens Consider a far away object s 2 00 so that the incoming rays are parallel to the optical axis these rays will all at one point the right focal point F 2 at a distance f the focal length to the right of the lens 1 l 00 s39 gt S39Zf This gives the focusing capability of a thin lens liq Focal Points of a Thin Lens Now consider the reverse situation if the outgoing rays are all parallel to the optical axis so that the image is at infinity s 39 00 where is the object originally located 1 l l s 00 f 3 This simple calculation indicates that all F1 F2 rays must originate from a single point the left focal point F1 at a distance f the focal len h to the left of the lens gt lef gtlefl Locating Mins amp Maxs Minimum 2 39 2 7r requlres that I 10 M 07 Where Ez asma 32 2 A 7 7r 3Eas1n6mm mi1i2 39 dition as same con T quot prev1ous1y derlved Note that 8 2 0 is not a solution for a minimum In fact g a sin 6 or 6 0 is the central maximum with 11m 5 aO in 2 32 1 Locating Mins amp Maxs Maximum For singleslit Fraunhofer diffraction patterns maximum occur near E 37 57 77 7 7 7 1 2 2 2 2 but not exact y To nd the maximum exactly we need to find the extremum for the intensity function ie asin d1x isinx dx dx Nlm Z x 71 d sinx 2 sinx cosx sinx 2 0 dx x j x x x2 j 2 20 Where x Locating Mins amp Maxs There are two solutions sinx 1 0 xWT gt x cosx sinx 2 2 0 gt xtanx x x tanx y yx 7r 27r 37r X 2 x asin6 2 A 37r2 Sir2 77r2 7m sin 6 m7r gtasin6m same minimum condition 0 intercepts are solutions to tan x x They gives the locations of the maximum of the intensity function 9H Width of the SingleSlit Pattern a t If the slit width is equal to or narrower than the wavelength only one broad maximum forms locations of 1St min on both I 1 left and ght of central max 0 A a l l l L 0 10 20 6 The wider the slit or the shorter the wavelength the narrower and sharper I is the central peak I 9 10 20 9 9 20 10 0 10 20 ID Intensity of TwoSlits Diffraction Patterns With two slits we have diffraction from the individual slits and interference form the two slits The combined intensity is the superposition of the two effects interference diffraction factor factor where JI S 7r 8 7r d 9 separation bet slits 7 51 and 3 7 51 a a width ofboth slits Intensity of TwoSlits Diffraction Patterns md 2md 10mdl md2 Copyrighl 2m Pearson Educahan Inc publlsmng as Pearsnn Addisnnrwesley mi 8mi 4 O mi4mi8 Copyrigmg 2mm mmquot m punishing as Addmwmy d4a Ddouble slit Envelope of intensity function Calculated intensity 0 0 For I 4a Every fouth i terference maximum at the sides mi 2 4 8 N is missing E Diffraction Patterns from Several Slits Let consider an example with N8 slits On the screen at P maximum will occur at dsin6 m t m 0ili2 when waves from adjacent slits have a path difference which is exactly mi This condition for maximums is the same 1 21 9 v for the twoslits patterns Maxima occur where he path difference for adjacent slits is a whole number of wavelengths rlsin 6 771A Diffraction Patterns from Several Slits Now let look at the condition for the minima Phasor diagram for qb 7139 Phasor diagram for 1quot I h q gt lt25 2 w 2 180 Phasor diagram for qb g d5 3 90 ForN 8 there are a total of 7 nummaw1th m 113 7r5 3 7 m2 m1N 1 4 2 4 4 2 4 N 1 1E Diffraction Patterns from Several Slits a N 2 lwo slits produce une minimum between adjacent mnxinm c N I6 with 16 slilsi the maxim are even taller and nun ower Willi more intervening minima mZim 1110 mi 6 b N eight slits produce inllcr nurmwer mnximzl in the same luculinns saptrailed by seven minimal I llljv l 1710 ml 11 H m0 ml ii The Diffraction Grating 6 an array of a large number of parallel slits all with the same width a and separation d slits in a grating is typically called rulings or lines typically a grating will have 1000s of lines per mm As in the discussion with our previous example the diffraction pattern from this grating will have the condition for maxima similar to the twoslits patterns with the same 0 separation d Ka m m ka l m l The Refraction Diffraction Grating 9 Grading Spectrographs Diffraction Grating can be use to measure the spectrum of light emitted by a source spectroscopy or spectrometry 25610 I l H m l m O m 1 Recall that the maxima will get sharper with a larger number of slits lines Locations of these peaks can be measured precisely and one can then measure the xi accurately Microscopic Interpretation of Entropy Microstate vs Marcrostate Macroscopic State I 00 0 Four heads Three heads one tails Recall Corresponding microscopic states Microstate a specific description of the properties of the individual constituent of the system Macrostate a bulk description of a system in terms of its marcoscopic variables Macroscopic stulc Two heads two rails One head three tails Four tails Corresponding microscopic states OOQQ OQQQ 0amp9 00 0Q 00 aaao acne aoaa ea 7 1 n inc publishing as Pearson Addznanrw capyngm 2006 Pearson Educaiia esiay Microscopic Interpretation of Entropy ID Observations 1 For a given rnacrostate typically there are many possible microstates If the of coins or molecules is large N A the of microstates corresponding to a particular macrostate can be astronomically large All individual microstates are equally likely Each coin has exactly 50 being head or tail and each toss is independent Microscopic Interpretation of Entropy ID Observations 3 However for a given rnacrostate the of possible microstates are different Since all microstates are equally likely the probability of the different macrostate is different 4 Some macrostate are much more probable than others When NNA this disparity is much much larger Microscopic Interpretation of Entropy ID Observations 5 The less probable macrostates all heads or all tails correspond to more ordered microstates 6 The more probable macrostates 50 50 heads and tails correspond to the most disordered microstates E Microscopic Interpretation of Entropy These observations motivate the following microscopic de nition of entropy r 1 1 4rquoti w TT39IC IIIIF39LL 1 nwvmwy x where Wis the of possible microstates for a given macrostate and k is the Boltzmann constant Example 4 coins macrostate all heads macrostate 3 H amp l T of microstates l of microstates 4 O Skln10 52km O CO All matched coins ordered unmatched coins less order 2nd Law Quantitative Forrn ID AS 7 v ital 3 reversible gt 0 iri rl fevarsflfblzej The total entropy disorder of an isolated system in any processes can never decrease Nature always tends toward the most probable macrostates states with the highest S disorder in any processes ID Entropy Changes for Different Processes 1 General Reversible Processes Note S is a state variable AS is the same for all processes including irreversible ones with the same initial and nal states NOTE in most applications it is the change in entropy AS which one typically needs to calculate and not S itself ID Entropy Changes for Different Processes 2 Reversible Cycles 3 Any Reversible Processes for an Idea Gas 1St Law gives Note Thru the Ideal Gas Law P is xed for a given pair of T amp V Tl9V1 gt Tfan dU er dW er dUdW nCVdTPdV 019 nCVdT quotif dV Entropy Changes for Different Processes Dividing T on both sides and integrating f f er nCVdT dV AS 7 I lR i i We have Entropy Changes for Different Processes 4 Calorimetric Changes dQmch fdQ fmch AS iT I T If c is constant Within temperature range A3 E ml m 2r 7 I 39 Entropy Changes for Different Processes 5 During Phase Changes or other isothermal Processes dQ 1 T stays constant during a AS J dQ phase change 139 quot WM r x l a r39 g at Entropy Changes for Different Processes 6 Irreversible Processes Although dS dQT is not valid for an irreversible process AS between a well definite initial state a and final state b can still be calculated using a surrogate reversible process connecting a and b S is a state variable Example 208 free expansion of an ideal gas Initial State a VJ Final State b 2K7 Since QW0 AU0 For an ideal gas this means that AT0 also Y Although Q0 but AS is not zero AS in a Free Expansion Important point Since S is a state variable AS is the same for any processes connecting the same initial a and nal 7 states In this case since Tdoes not change we can use an surrogate isothermal process to take the idea gas from state 61 KT to state b 2K7 to calculate AS Applying our general formula Tf Vf ASnCVln nR11 T V we have Surrogate Isothennal Expansion AS nCV ln jannj ann2 576JK 141 3 2nd LaW AS gt O amp Clausius Statement Clausius Statement Heat can t spontaneously transfer from T C to T H We will prove this by contradiction using AStot gt 0 Assume the contrary Q ASH Q heat absorbed from T H Astor ASC heat released into T C Astor Q C T H T C with the explicit signs Q Not Possible 1s taken to be violated ASM gt 0 Einstein s Photon Explanation In 1905 Einstein published his theory on photoelectric effect which resulted in his Nobel price in 1921 Built upon Max Planck s hypothesis of quantized light photon later energy of a photon Where h 6626 gtlt1034J S is a universal constant called Planck s Constant T Note the smallness of this number Einstein s Photon Explanation NV Incident light as a collection of photons Each photon has energy hf Intensity of light of photons The interaction is an allornone process Electrons bounded to the surface of the metal can absorb all of a single photon energy or none at all If hf is large enough to overcome 45 a electron Will be ejected with kinetic energy Kmax By energy conservat1on we have rug 1 depends on the metal surface Einstein s Photon Explanation ID Unexpected Results Explained Since Kmax has to be positive if hf lt 45 no electrons Will gain enough kinetic energy to leave Therefore there is a threshold frequency and is given by hf0 15 Since intensity I is proportional to the of photons Kmax or V0 should not depend on 1 Increasing intensity will only increase the of electrons being ejected and thus the photocurrent but it Will not affect the stopping potential V0 Kmax linear dependence on f It is obvious with the equation E Notes Convenient Energy Units leV energy required to move one unit of charge across an electric potential of 1 V leV 1602gtlt10 19C1V 1602gtlt10 19J 16V 1602x10 19J hc 4136x10 15eVs300x108ms 1241x10 6eVm 124leVnm h6626gtlt1034Js j4136gtlt10156Vs Photon Momentum momentum of a photon Atomic Line Spectra amp Energy Levels An Electric Discharge Tube with Diluted Gas Diluted gas containing trace elements such as H He Na Hg Observation Energetic electrons from cathode excite gaseous atoms in the tube light can be emitted Gas is diluted so that the emission process is by individual atoms The spectrum of light emitted are sew of unique lines characteristic of the specific type of atoms in the gas liq Photo Emission by Atoms In order to explain the observed discrete spectra lines from atomic emissions Niels Bohr combined the following two central ideas in 1913 in his model Light as packet of energy photon amp energy levels in atoms Visualization of Energy Levels in a typical atom i m Ei An atom drops from an initial level i to a E lower energy final level E 3 f by emitting a photon with E2 energy equal to E Ef f E1 E Each atom has a specific set of possible internal energy states An atom can possess any one ofthese levels but cannot take on any intermediate values E The Hydrogen Spectrum n 00 n 6 n 5 n 4 3646 nm 4Ol mu 434d nm 486 nm 6563 um um 1 15 1717 H5 H H H HT and Hans Ill he visith region of Hi specu39um All Balmer lines beyond H39s are m lhc uluuwolci spectrum Hydrogen is the simplest atom and it also has the easiest spectrum to analyze In 1885 Johann Balmer derives an empirical relation to accurately describe the wavelengths for these lines R 1097 x 107 m lis called the Rydberg constant which was experimentally found to match with observed data The Hydrogen Spectrum The formula for the Balmer series can be interpreted in terms of Bohr s hypothesis Multiplying Balmer s empirical equation by hc we have hc l l l 22 n2 Comparing this with Bohr s equation for the energy of an emitted photon om an excited atom we can identify the two terms as the energy values for the energy levels 1 and f Ez hCR and Ef I 2 I n th 22 ID The Hydrogen Spectrum This suggests that the hydrogen atom has a series of discrete energy levels 39 5L J39L Then according to Bohr s explanation the various lines om the Balmer series corresponds to the transition of an excited atom from n 3 or above to the n 2 level The constant th has the following numerical value th 219 x1018 136eV Thus for a hydrogen atom the lowest possible energy level is given by E1 l36 eV n 1 called the ground level and all the higher levels 11 gt1 are called the excited levels The Hydrogen Spectrum Other special series corresponding to different transitions have also been described Lyman Series R Paschen Series Brackeii Series Pfund Series R i2 g XRay Diffraction XRays were discovered by Wilhelm Rontgen in 1895 There were indication that Xrays are EM waves with a very short wavelength A 103910 In 12 Luuc dn39rmcnun pulh m or H mm ccuun ofquun crywl a Basic sump for xrruy um mcxinn Thin Luad crystal Clccn r eru my 0 Xm hem Film in holder Max Von Laue in 1912 proposed that scattered Xrays from the crystalline solid might produce a diffraction pattern when they interfere with each others The successful experiment verify that 1 Xrays are waves 2 crystalline solids are arranged in regular repeating patterns l XRay Diffraction ii 103 75quot Nr i 4 Sntlium simpli cation v b Scattering from adjacent atoms in a row Interference from adjacent atoms in a row is constructive when the path lengths u cos 9 and t cos JY are equal so that the angle of incidence 0 equals the angle of reflection scattering 6 1 cos 6 n cos 9r a Scattering of waves from a rectangular array Incident plane waves Scuttcrers eg titnms l 9r l l611 91 C Scattering from atoms in adjacent rows Interference from atoms in adjacent rows is constructive when the path difference 2d sin 9 is an integral number of wavelengths as in Eq 36 Conds for constructive interference 6576 2dsingmlml2m Bragg cond liq XRay Diffraction XRay Diffraction has been very useful in exploring the crystalline structure of solids and structures of organic molecules Rosalind Franklin was the first to use Xray diffraction to image a DNA molecule in 1953 The dark crossed bands was the first evident of the helical structure of the DNA molecule 399 W Chapter 3 7 Relativity El Events and Inertial Reference Frames Principles of Einstein s Special Relativity Relativity of Simultaneity Time Intervals Length Lorentz Transformation Relativistic Momentum amp Energy General Relativity Albert Einstein and the Special Theory of Relativity The Special Theory of Relativity was published by Albert Einstein in 1905 when he was still working in a Swiss patent office lt stands as one of the greatest intellectual achievement of the 20th century In the same year two groundbreaking works on the theory of Brownian motion and the Photoelectric Effect were also published Special Relativity led to fundamental rethinking of the concepts of space and time 0 The concept of simultaneity The measurement of a time interval The measurement of length Causality Summary of Changes Before Special Relativity space and time are thought to be as absolute properties of the background or stage Where objects act upon 9 Different observers om different inertial reference frames Will agree on all the measurements on the previous slide After Special Relativity These space and time measurements are relative to the observer s frame of reference A stationary observer With respect to a moving observer Will see a moving clock runs slower a moving meter stick gets shorter two events being simultaneous for one inertial observer Will not necessary be simultaneous for the other ID ID Reality Check One should note that these relativistic kinematic effects are only dramatic for objects moving at large relative speeds near 6 For everyday objects moving with ordinary slower speeds these effects are small In fact one can show that Einstein s theory reduces to Newtonian mechanics in the limit u ltlt 6 There are numerous experiments as well as practical applications that have demonstrated Einstein s predictions Muon s life time Symmetry between magnetic Atomic clocks on ight amp electric forces Trajectories of subatomic GPS system particles in accelerators NAVSTAR Nuclear reactors Relativistic Momentum amp Energy ID As we have seen the coordinates for an event in spacetime xyz l changes depending on the observer s frame of reference xt lt gt x39t39 Lorentz Transformation As we will see below other dynamical quantities such as momentum energy etc must also be adjusted so that the laws of physics satisfy the following conditions Laws of physics e g conservation of momentum conservation of energy Newton s laws apply equally to all inertial observers Laws of physics apply equally in both relativistic and nonrelativistic regime The modi ed relativistic dynamical quantities should reduce to the classical ones for u ltlt 6 Relativistic Momentum amp Energy Relativistic Momentum Momentum of a particle moving with velocity v as measured in the lab frame S frame Relativistic Energy Total energy of a particle moving with velocity 7 as measured in lab frame S frame Relativistic Momentum amp Energy ID All laws of physics remain valid in all inertial reference ames Relativistic l3 amp E must conserve 9 Experimentally it has been shown repeatedly that it is 7m and 717102 rather than their classical counter parts that are conserved in high energy collisions l a a a 1 so P ymv gt mv l V262 Ill 9Forvltltc 7 So that for nonrelativistic speeds relativistic 13 reduces to classical 13 Relativistic Force Newton s 2nd Law same form as in the classical case but with relativistic momentum Note If F 0 no external force momentum I3 as expected Will be conserved in both relativistic amp classical regimes Let use the relativistic force to consider the workenergy theorem Work done by F to f f dP accelerate a relativistic W I F dx 2 dx t particle from v to vf for simplicity F is in xdirection only Relativistic Work amp Energy f f dP dP f dx gt idt I tV f W IvdP Note dPv vdP Pdv gt vdP dPv Pdv Substituting this into the above equation for W f f f WjvdpjdPv jpdv 1St term 2nd term Relativistic Work amp Energy lSl term PVIZr 7mvf Vf 7mvivi For s1mpl1c1ty choose v1 0 and vf v mv2 l vZc2 f Then the 1st term becomes IdPv gmv2 Now let consider the 2nd term f f IP dv I mv dv 139 139 V 1 VZC2 We can integrate this by a simple change of variable E Relativistic Work amp Energy Let s 1 v26212 a s1 vZ62U2 2vczdv vc2 dv f f mv 2f 2 So JP dv dv mc Ids mc sf sl mcz 1 vZc2 mcz 1 02c2 mcz 1 vZc2 mc2 d5 E Relativistic Work amp Energy Putting everything together 2 W m0221 v202 mc2 l v c 2 2 2 mv 2 2 2 II v c 2 mc II v c mc 1 V262 21 v202 2 mv mc2 1 vzc2 1 22c2 Mmcz7rr f 2 mc2 2 mc mc l vzc2 l vzc2 1 v202 m02 g Relativistic Work amp Energy Applying the workenergy theorem this amount of work done to accelerate the particle from 0 to v should equal to the change in KE Laws of physics should be unchanged in relativity Rcluliristic kinetic energy heroines infinite as r approaches L Since vi 0 KEI 0 Vf V KEf HAPPENSEP i i i i 1 i i l l l i DOESN39T i i l l i i i HAPPEN imr39z y 1 Relativistic Kinetic Energy of a particle moving with 7 as measured in the lab frame 0 i Newlnniun mechanics incnrl eully predicts ihnl kinCIiD energy becomes infiniic only it39zi becomes infinite Relativistic KE 9 Classical KE Slow moving particle regime v ltlt 6 2 712 2 2 2 2 Using binomial theorem 1V 21 1 V 0 V E 1 11 62 2 c2 c2 2 62 Substituting this into the equation for Relativistic KE 2 712 V 2 2 KE l 2 mc mc c 2 E 11v 2 ljmc2 lmv2 2 2c This is the classical result for v ltlt 6 Total Relativistic Energy Let look at the equation for the relativistic kinetic energy of a moving particle again 2 me KB l V262 2 I lZC 9 It separates into two terms 1St depends on speed of the particle 2nd is a constant term independent of motion 9 KB can be interpreted as the difference between a total energy term and constant energy term 2 KE mczzE mcz l v c z total energy constant energy Total Relativistic Energy E is the total relativistic energy of a relativistic particle Note 1 for v O KE O BUT total rel energy E 2 me2 72 0 There is a residual energy even for a particle at rest The quantity mc2 is called the rest energy Rest Energy mcz 9 Independent of velocity Proportional to the mass of the particle Mass is a form of energy Total Relativistic Energy ID Note 2 Since E 7mc2 is the total relativistic energy of the system E is expected to be conserved in all processes This statement on the conservation of Total Relativistic Energy is more general than the classical conservation of energy since it contains the rest mass energy Note 3 9The mass m Which we have been using is a constant in our analysis It is called the rest mass and is the mass of an object measured by an observer stationary With the object 9The quantity mrel 7m is called the relativistic mass and is not a constant for a moving object and is measured by an observer not at rest With the object ii Conservation Laws 2 g g with W RPf szmv These set of equations form the generalized conservation laws in Special Relativity AND these conservation laws apply to all processes equally in all inertial reference frames Example 3712 A Relativistic Collision 167 X 1017 kg BEFORE U Initially both protons move in opposite directions net linear momentum is zero The three particles after collision are at rest again with net linear momentum equals to zero Thus relativistic momentum is by design AFTER QQ conserved Pm 240 X T18 kg We now need to consider the conservation of total relativistic energy Proton Proton Find initial velocity of proton IE Example 3712 Before After 2 2 2 2 yMprotonc szroton c mpionc 2y 1Mpr0t0n m i pan m 28 714 quot0quot 1M1072 2Mpm 2167 X 10 27 kg 2 32 C 2 gt yZCZ y2v2202 gt VC1 172 vc 1 110722 203600 IE EnergyMomentum Relation E ymc2 P 77quotquot E2 y2m264 2p2 yZmZVZCZ E2 mzc4 l vzc2 Numerical Notes 1 Keep u in units of c e g u 08c since we always have the ratio uc Then y 11 086262 1 1082 can be calculated simply 2 Since the ratio u c occurs so often it has a special symbol uc and 721 ll 2 3 Approximation When u ltlt 6 low speed nonrelativistic Recall Binominal Expansion 1 5 51 quot5 for 8 small Numerical Notes 3 Approximation when u ltlt 6 low speed nonrelativistic 7 l ll uZc2 l u26212 Relativistic u03c LIZ202 0322 2 0045 y 2124 2 Ll 262 quite diff from 1 If u c is small uZZczwill be really small Problem calculator might treat 1 Liz202 simply as 1 and you Will loose all signi cant gures if you combine 1 and Liz202 NonRelativistic u003c LIZ202 2 00322 000045 y 1 00045 not much diff from 1 Typically if u 2 016 one should consider speed to be relativistic Lengths Perpendicular to the Relative Motion Mav1s 61 2 7 i i i if 7 Vlfr J i l p Zi l S u l K A l y r i Stanley 1 m P i Claim Length measurements in directions perpendicular to the direction of relative motion will not be contracted Lengths Perpendicular to the Relative Motion fr 3 39l i 4 it i x ri i A Stanley 1 5 Ma We can prove this assertion by contradiction Let assume that in Stanley s viewpoint S frame Mavis meter stick is shortened Then when they pass by each others the shorter stick will make a mark on the longer one For this case a mark will be made on Stenley stick since Mavis one is shorter by assumption The actual mark making is a physical occurrence that no observers will disagree Lengths Perpendicular to the Relative Motion wk 3 H ii K lm M e Pyrii 1 r7i Stanley i ii Now consider the experiment in Mavis viewpoint she will see Stanley s moving stick shortened and a mark will be made on her meter stick instead This ends in an asymmetric physical events but the experiment is set up symmetrically paying no favorites between Mavis amp Stanley ID Lengths Perpendicular to the Relative Motion According to Einstein s I 5 postulate While different observers can differ on their interpretation of a physical event all physical laws and physical occurrence must be the same for all observers Therefore the asymmetric result cannot be correct and the only symmetric result that can be consistent with Einstein s 1St postulate is the assumption that lengths in the perpendicular direction 610 not change for different observers How an Object Looks Moving Near C Di 3 Array at rest g i zAvQ39 AF Jllu Illll Win El 39393939 Kr 2 sJ b Array moving to the right at 02c C Array moving to the right at 0 um 77 ll NIH Example Muon Life Time cosmic raYs 14 30m h muon Mu011 is an unstable elementary particle with 3007 56 Produced by cosmic rays mostly high energy protons collide With atoms high up in the atmosphere gt4 km In a the rest frame of the muon it has an average life time of T 22us muon Without Einstein s SR if umm E 09 avg dist travelled 0993 gtlt108 m s22 gtlt10 6 s 2 600m Example Muon Life Time cosmic rays 14 atom 0 Muon is typically produced high up in the mu0n atmosphere 4 to 13 km So according to Galilean relativity muon can t reach the surface of the earth amp we should not be able to detect them However we do detect cosmic rays muons on the ground Einstein s Relativity resolves this paradox Example Muon Life Time ID Earth s frame T u099c Earth is stationary amp muon move downward toward the ground The two events creation and decay that de ne the lifelime of a muon occurs in the rest frame of the muon so that the 22 us life time is also the proper time for an observer in muon s rest frame Ato 22m In contrast in Earth s frame these two events do not occur at the same place and the interval Al de ned by these two events will not be a proper time interval as observer by someone in the Earth s frame In fact AI will be time dilated At ArzyAroz O 22m 16us ll uZcz 1 0992 Example Muon Life Time Earth s frame So according to an earth observer avg dist travelled by these cosmic rays muons Will be avg dist travelled 16 us0993 gtlt 108m s 4800m So some muons Will reach the surface of the earth Muon s frame Muon is stationary amp the Earth rushes upward at 099c O In the Earth s rest frame the atmosphere is at rest with the Earth and its height 04000m is the I 20990 proper length measured by an earth observer Example Muon Life Time Muon s frame But 1n muon s Vlewpomt the Earth 1s rush1ng toward it at 099c and the height of the atmosphere will be length contracted in muon s frame O I 0 1 0992 4000m 560m lt 600m 7 i u099c So that on average some muons will be able to reach the surface of the Earth before it decays I Another Example The Twin Paradox A lt 20 light years Eartha A First consider Time of Travel according to Astrid This is proper according to Eartha lst leg distance from earth to star appears to be contracted according to Astrid dist to starAsmd 201y1 71262 201y1 0952 6251 7 Another Example The Twin Paradox ID Time oftravel 6251y 095c 66 yrs 2nd leg return trip distance also appeared to be contracted according to Astrid dist to starAstrid 6251y And time of travel 66 yrs So if Astrid started her journey at 20 years of age when she comes back to earth she will be 20266 33 years old Now we consider Time of Travel according to Eartha dist to travelEaltha 201y x 2 4OZy And time of travel 40ly 095c 42 years So if Eartha said good bye to Astrid at 20 years of age when they meet back she will be 2042 62 years old Another Example The Twin Paradox So according to this description Astrid Will be almost half her twin sister s age after she comes back from her trip to the star Here is the apparent paradox If all inertial reference frames are equivalent can Astrid make exactly the same arguments with the space ship being stationary and the earth moving so as to conclude that Eartha Will be younger at the end of her journey Both of these descriptions cannot be true but SR seems to suggest that they should be paradox Another Example The Twin Paradox To resolve this apparent paradox one need to realize that the following two situations are not symmetric decelerate Astr1d has to switch 20 light years inert ref frame here accelerate A 9 i i 06960 Earth did 4 20 light years W not 2100 UZO95C 6 nor decc O 4 00 Another Example The Twin Paradox ID Notes Conclusion Astrid was not in a single inertial reference frame She switched inertial reference frame at midpoint But her journey can be approximated by two separate inertial reference ames Eartha was in a single inertial reference frame thoughout the journey The seemingly symmetric descriptions between Astrid and Eartha are not symmetric at all and their actual physical experience needs not be the same Actually since Eartha was on a single inertial reference frame her description should be correct SR remains correct but one needs to be careful on making claims which are consistent with SR s assumptions 9H Lorentz Transformation Transforming the spacetime coordinates from S to S correctly according to SR y y39 The Lorentz coordinate S S transformation relates Frame 539 moves relative m i the spacetinie coordinates x frame S with constant of an event as measured 7 velocity it along the T 13 4 III the two irames common xivaxis x y 3 f in frame 5 and r iquot z t39 in frame S Origins 0 and 0 0 coincide at time 0 i m ConyngmtaQDDE Pearmii Education inc V Dubiishitig as Pearmii Addisnaneiiey u in Xdir only xyzt lt gt x39y39z39t39 Chapter 3 5 Interference II Interference and Coherent Sources II TwoSource Interference of Light I Intensity of Interference Patterns II Interference in Thin Films I The Michelson Interferometer Wave Nature of Light III Previous Chapters Geometric Optics 9 ltlt L Ill Rays Model is an approximation of EM waves with rays pointing in the direction of propagation II Next Couple of Chapters WavePhysical Optics 7 L Ill Like water waves light spreads and interferes with each other Observed phenomena cannot be accounted for by rays Diffraction Interference constructive destructive interference patterns caoyngme 2003 Pearson Education Inc publishing as Pesrsnn Addrsnn Wesley Interference and Superposition ID II Interference refers to any situation in which two or more waves overlap in space 9 The resultant displacement at any point is governed by the principle of superposition the resultant disturbance at any point and at any instant is found by adding the instantaneous disturbance that would be produced at the point by the individual waves as if each waves was present alone Interference and Superposition Constructive Interference peaks lt gt peaks wave 1 wave 2 5 Destructive Interference peaks lt gt peaks 22 apart wave 1 wave 2 3 ID Conditions for Observable Sustained Interference 1 The sources have to be coherent a The individual waves must maintain a constant phase relationship oscillate in unison with each other e g two speakers driven by the same ampli er two regular light bulbs don t interfere since they are not coherent Emission from a light bulb is from a thermal process of random motions of charged particles in the filament Conditions for Observable Sustained Interference 2 3 The waves should have the same polarization Two or more interfering waves must have the same wavelength monochromatic You can have white light interference pattern if the source is coherent but the effect will appear for different colors corresponding to the different wavelengths in white light lll 9H TWOSource Interference of Light a Point a is symmetric with respect to the two coherent tho sheets lengths differ by an integral number of wavelengths r2 Iquot mA 39l r1 distance to S1 Constructive Inter rz r1 2m m0i1i2 r2 distance to 82 r2 r1 path difference b Conditions for constructive interference Waves interfere constructively if their path sources Waves will arrive in phase constructively r2 r1 0 C Conditions for destructive interference Waves interfere destructively if their puth lengths differ by a haltlintegrnl number of wavelengths r i r i I71 7 V sI Destructive Inter 1 rz rlz m5 m0i1i2 Young s Double Slit Experiment m m 12 constructive destructive interference interference bright regions dark regions Coherent wave Cylindrical fronts from two slits wave fronts e Monochromatic light 3 l Bright bands Where 1 wave fronts arrive in r m y r 3 phase and interfere 1 a Dark bands where wave fronts arrive out 1 gt of h d 39nterfere 2 gt n quot39 Spreading of light behind slits Recall Huygen s Principle D2 slits Copyright 2008 Pearson Education Inc publishing as Pearson Addisoanesley l Young s Double Slit Experiment b Actual geometry seen from the side c Approximate geometry 5d dsin 0 Screen 52 U dsm 0 r 2 1 dl lee H 6 51 51 y I P F To screen quotV R aquot so we can reu the ra s as In real situatlons the drstancc R to the y y 7 7 7 parallel 1n whlch case the path length screen ls usually very much greater than d 1 l 9 1 erence lS srm r r r sm the d1stance 1 between the sins p y 3 1 mm m BMW publishing as Mmmwgsw cauyngme mos Pearson Education the nubllshlng as Pearson Addzsonrwesley If screen is far away so that R gtgt 0 we can assume rays from S1 and S2 to be approximately parallel Then from the simpli ed geometry right panel we have an explicit expression for the path difference 6is the angular location of observation point P on the screen ConstructiveDestructive TwoSlit Interference m m 12 t39 d t39 Applying the conditions for constructive gigging intiiiiiiii bright regions dark regions destructive interference we have the following conditions Constructive Interference Two Slit Interference Destructive Interference Two Slit Interference a n39 ileum a 33 Kl 2m 2 g go w The brightdark bands in the pattern are called fringes m is the order of the fringes Locating Fringes In normal situations P ym Rlm d quotquot quot am I d S tenths of mm xi S tenths of um 400700 nm So typically we have the condition that d ltlt L so that 6 is small Thus we can approximate sin 9 E tanH E 9 The linear distance to a particular ordered fringe ym is given by ym Rtan 6m With the small angles approximation we have Ema isi fi39ieim E Example A Thermodynamical Cycle El Example 194 P 80 x 104Pa b 30 x104Pa 1 I I I C I 0 20 x 10 31113 50 X 10 3m3 Given Qab 150 J de 600 J Find n AUab AUabd Qacd 5 Special Processes 3 Adiabatic no heat exchange Q O AUQ W A Maximum Initial volume compression Note The compression stroke in an internal V1 combustion engine is quick and it can be well approximated by an adiabatic process will come back to this example later E Adiabatic Processes 2 examples El Quasistatic adiabatic expansion m 5 2 Expanding gas push piston up 9 work is done by gas 9 W gt 0 9 AUlt 0 energy ows out of gas l l 385 m For an Ideal Gas U is a function of T only 20L So AU lt 0 also implies AT lt 0 temperature drops insulation El Adiabatic free expansion nonquasistatic 5 1 7 Inwlmiun Gas expands into vacuum 9 no work done W0 H l 19 Adiabatic 9 Q 0 1st law gives AU 0 20L U remains unchanged and T is a constant l Vacuum 51 L lircukalhlc his il 30 K nurlnmu g Special Processes 4 Isochoric constant V NO volume change 9 NO work WO AUQ W 5 Isobaric constant P Since P is a constant dW PdV can be integrated easily and gives IE Special Processes II Isothermal constant T For an Ideal Gas UT depends only on T 9 AU 0 AU Q W gtAU0 Recall work done by an isothermal process in an Ideal Gas nRT V dV 2 nRTln V 1 V2 V2 W Pelsz V1 V1 Summary 1 Adiabatic Q20 3 ITSSOEIriT 2 Isochroic AVO 3 Isobaric APO 4 Isothermal AT 0 Isothermal T4 Ta I V 0 V1 Isochoric Adiabatic T2 lt Tu T1 lt T 1st Law for in nitesimal changes W diggm ggg E Cp and CV for an Ideal Gas 9 T1 U1 T2 U2 Constant volume process gas does 2 no work Qa AU Constant pressure process gas does Pr I b I I I I W I I I I I 0 V1 V2 Couynghl 2003 Pearson Educahnn nn publIsl39IIng as Pearson AddIsoanesley Two different ways to change dTT2 T1 Process a Constant V work Qb AU W Container of fixed volume an nCVdT H T gtT IT 11 moles of ideal gas Heat added an E Cp and CV for an Ideal Gas I9 For the same dT T1 U 1 T2 U2 Process b Constant P Constantvolume de quotCPdT processl gas dogs M T T IT p2 no wor 39 Q U Piston motion Constant pressure 7JI process gas does i1r p1 work Qb AU W Container with 1 f I b I movable piston m0 es 0 l I that applies Ideal gas W constant pressure I l 7 n 0 V1 V2 V Heat added dQ b Which is bigger an or de Cp and CV for an Ideal Gas ID First let consider the constant volume process a gt No work done byon gas 9 W O gt Then 1St law gives all z dQ nCvdT Now for the other constant pressure process b With the same dT gt We have dW Pd V and de nCpa T by de nition gt Substitute these into the 1St law gives all dQ dW ndeT PdV Cp and CV for an Ideal Gas With P constant we can consider the following differential using the Ideal Gas Law dP V PdV anT Substitute this into the previous equation we have dUb ndeT anT nCp RdT Cp and CV for an Ideal Gas Important point Since U for an ideal gas is a function of T only and process a and process b have the same dT T2 T dU dU Note This fact is very useful Basically one can use dU nCva T to calculate the internal energy change for a given dT whether V is constant or not gt nCvdT nCp RdT Finally this gives True for all Ideal Gases ID ig Quantum Wavefunction In QM the matter wave postulated by de Broglie is described by a complex valued wavefunction Pxt Which is the fundamental descriptor for a quantum particle 1 Its absolute value squared xt2 dx gives the probability of finding the particle in an infinitesimal volume dx at time t 2 For any Q problem A l A 9 Thejob is to find f xt for the V U ReIm Pxt particle for all time 9 Physical interactions involves operations A on this wavefunCtiom AWO t I xt is a complexvalued function of 9 Experimental measurements Will space and time involve their products f xtA xt Ea Probability and Uncertainty The outcome from the singleslit and doubleslit electron diffraction experiments An electron must be considered as a matter wave passing thru both slits This also implies that the location of the electron cannot be determined precisely This idea is formally stated as the Heisenberg Uncertainty Principle Alix Allowed Ax is the uncertainty in posrtion x M A 2 Apx is the uncertainty in the xdir momentum This principle states that the product of these two uncertainties taken together can t never be smaller than 7 1 a iszffgfgfm We can t obtain the exact location and the exact momentum MAM lt f of a particle at the same time AX Apx Probability and Uncertainty The Heisenberg Uncertainty Principle is not a statement on the accuracy of the experimental instruments It is a statement on the fundamental limitations in making measurements To understand this there are two important factors 1 The waveparticle duality 2 The unavoidable interactions between the observer and the object being observed Uncertainty Principle Conceptual Example Suppose we want to measure the position of a particle originally at rest by a laser light The measurement is accomplished by the scattering of photons off this particle photon xi 1k lsycattered phone 0 particle Uncertainty Principle Conceptual Example But photons carry momentum and the particle after collision will recoil ls7cattered particle phone recoil While the scattered photon gives us a precision information on the position of the particle the photon Will also inevitably impart momentum onto the particle Uncertainty Principle Conceptual Example If the measurement is made by a photon with wavelength ft then at best we expect our position measurement to be accurate up to an uncertainty Ax xi And the photon will transfer parts of its momentum to the particle along the measurement direction x A le Combining these two expressions gives AxApx l l Note By improving Ax using a smaller ft does not help because Apx hxl gets larger in inverse proportion to ft Uncertainty Relations It is the component of the momentum px in the direction of the measurement xdir being affected by the measurement process Thus uncertainty relations only hold for conjugate pairs of variables AxApx 2 h AyApy 2 h AzApZ 2 h There are no restrictions for unconjugated variables AxApy 0r AxAy etc 9H Uncertainty Relations There is another important conjugate pair A metastable state At small will have high This can be understood from the fact that E uncenaimy Medium AE At 1 uncertainty in frequency is inversely E Medium A gt A f proportional to uncertainty in time 3 gt AE hAf Planck s hypothesis 1139 Short Ar Small AE MN 2 h E1 Long AI So precise measurements of energy and time can t be made at the same time Wavefunction and Probability 2 ILPOC f is called the probability distribution function or probability density In other words LPx t2 dx shaded area I l is the probability in nding the particle in the interval x x clx at time t dx X Also since pxdx E I Px t2 dx is a probability it has to be normalized At any instance of time t the particle must be somewhere in space ID The Schrodinger Equation In Classical Mechanics we have the Newton s equation which describes the trajectory xt of a particle F 2 mi In EM we have the wave equation for the propagation of the E B elds 62E B 1 62E B ax 0 2 at derlved om Maxwell s eqns In QM Schrodinger equation prescribes the evolution of the wavefunction for a particle at x I under the in uence of a potential energy Ux I quot quot r725 quot note th1s eq 1s complex timedependent Schrodinger equation The Schrodinger Equation For most problems we can factor out the time dependence by assuming the following harmonic form for the time dependence Pxt 1xe 6 6LPX 4m z39at lh lh xe 21quot x 10 e at at m gt m gt hwyAx Ewxei t note E hf ZL27rf ha is the energy ofthe particle also 02 1 xt dam 2 2 6x dx g The Schrodinger Equation Substituting these into the timedependent Schrodinger equation we have Jim meow Evx7 f 2m a x2 I t time dependence can 2amp3 be cancelled out glittfm timeindependent Schrodinger equation lx does not depend on time and it is called a stationary state ID More on Wavefunction In general the probability in nding the particle in the interval a b is given by MK 2 pub I wxl dx Note yx is not the probability density Iqx2 is the probability density Other physical observables can be obtained from lX by the following operation example position x x xpxdx 2 x 1x2 dx X is called the expectation value of x it is the experimental value that one should expect to measure in real experiments More on Wavefunction In general any experimental observable position momentum energy etc 0x Will have the expectation value given by lt0 2 OOC llx2 dx Note Expectation values of physically measurable functions are the only experimentally accessible quantities in QM Wavefunction VX itself is not a physically measureable quantity Standard Procedure in Solving QM Problems with Schrodinger Equation with the timeindependent Schrodinger Equation Equotff 39 mug l39ix mm m IIL micrm Ild i lulluilc ccllcrc ommlc Given A particle is moving under the in uence of a potential Ux Examples Free particle Ux 0 lt lt V A M 39 Particle in a box U06 2 0 0 x L E 00 elsewhere U39 U0 0 S x S L UUJ k l Barrier Ux O elsewhere E HMO Uxk x2 ID Standard Procedure in Solving QM Problems with Schrodinger Equation Solve timeindependent Schrodinger equation for lX as a function of energy E with the restrictions 0 lX and are continuous everywhere x 60 2 lX is normalized ie Low90 dx 1 Bounded solution lX gt 0 as x gt ioo Then expectation values of physical measurable quantities can be calculated Wave Equation for a Free Particle ID A free particle means that Ux0 and there is no force acting on the particle The timeindependent Schrodinger equation gives h2d2wltxgt 2m a x2 where k Jng is a constant The boxed ordinary differential equation is in a standard form which has the following solution one can verify this 39 A 39 by direct substitution dam 46 E WOO 07 dx where A is a constant Note there are no restriction on the allowed values for k that is not case for nonzero Ux as we will see later Wave Characteristics of a Free Particle For a free particle with Ux O the total energy E of a free particle is given by its kinetic energy term only 2m Z m 2 E 2 k 222E h2 2 Rewr1t1ng L217 gt k 2 h2 h 1 h 1 And k is the wave number a is the angular freq A ee particle with energy E and momentum p lX is basically a harmonic sinusoidal function with a wavelength ft and frequency f This is consistent with de Broglie s matter wave hypothesis ID The Kinetic Energy Term in the Schrodinger Equation Substitute Aeikx into the LHS of the Schrodinger equation we have 2m 6ch2 2m 6ch2 2m 2m 2 ikx hz armorial d A6 21k2Aei le wltx Thus the rst term in the Schrodinger equation is associated with the Kinetic Energy of the particle Together with the Potential Energy term Ux lX the Schrodinger equation is basically a statement on the conservation of energy 11 M dxg UxVxEVx KE PE TotalE Blackbody Radiation Max Planck in 1900 proposed a solution to this problem A different way to calculate E E is not a continuous variable and E can only take on discrete values En OAE2AE3AEnAE 1101 2 3 7 7 7 7 And the discrete energy increment for the EM waves is given by EnPEn Then E I M oo ekeM 1 ZIPEH Blackbody Radiation Then the spectral emittance in this quantum calculation is 27m hc A A4 ekeMT 1 21 Planck s Blackbody Radiation Law Notes As it gt 0 ekeMT E1 hc xlkT so E hc l W kT Planck s result reduces to the classical result for if h AE O ehclkT1 h 1 1 711 9 O O O Planck s Result 1n CM amp QM L1m1ts 1 gt 0 short wavelengths high energy ekeMT gt oo exponentially faster than 15 gt 0 as l gt 0 1 so 15ehcka 1 gt O as l gt 0 There is no ultraviolet catastrophe in QM limit xi gt 00 long wavelengths low energy Classical R J Law ekeMT 12 kc xlkT 2 1 Then MAT 27rhc E 27r c JCT 27w kT is ekeMT 1 154 1 l4 Correspondence Principle QM agrees well with CM in the range where AE is small Energy is more like a continuum than a discrete set For light photons the basic energy unit is the energy of a photon hf EM 9 QM if f is large 1 is short EM 9 CM iffis small 1 is long Radio waves low freq behaves like classical waves with dif action and interference easily observable Correspondence Principle Visible lights mid freq have both wave classical amp particle quantum behaviors T T dif action photoelectric effect XRays high freq mostly particle quantum behaviors XRays productions and Compton Scattering liq WaveParticle Duality The Principle of Complementarity First stated by Niels Bohr in 1928 The wave and the particle descriptions are complementary We need both descriptions to complete our model of nature but we never need to use both descriptions at the same time to describe a single part of an occurrence Movable photomultiolier detector C Modify our light diffraction 03 experiment shghtly use a gtsm m photomultiplier tube to measure 3 1m en 5 ity the number of photons arnvmg at z a given location Monochromatic light A Screen lg WaveParticle Duality After 21 photons reach the screen Photon detections by photomultiplier tube must be done over time LOW intensi result After 1000 phpons retah the screen In principle the one or a few photons passing the slits at one time arrival of a photon at a particular location will be probabilistic And results are statistical in nature Cugytlgm e 2003 Pearson Educanan m publmmng as Pearson Admsawwgaey g WaveParticle Duality After 21 photons reach the screen The distribution pattern of photons arriving at the screen requires a wave description After IOOO photons reach the screen But the detection of a single photon in the photomultiplier is a particle measurement capyngm 3 ma Pearson Enunahuu m Dublbmng as Pearson Addisonrwesley Chapter 39 The Wave Nature of Particles II De Broglie Waves III Electron Diffraction II Probability and Uncertainty II Wave Function and Schrodinger Equation l E De Broglie Waves As we have seen light has a duality of being a wave and a particle By a symmetry argument in nature de Broglie in 1924 proposed that all form of matter should process this duality also Implying that a regular particle such as an electron or a baseball should have wave characteristic also For a particle with momentum p mv or 7mv and energy E de Broglie proposed Bopynng 2003 Pearson Education lnc punnsnmg as Pearson Addisonrwesley de Broglie wavelength De Broglie Waves and the Bohr s Model In the Bohr s model angular momentum of the electron in a particular Bohr s orbit is quantized Using the de Broglie wave hypothesis one can imagine an electron as a standing wave in a given energy state 11 In order for the standing wave to t around a given Bohr s orbit it must satisfy the periodic boundary condition the wave must match back onto itself 2727quotquot nt nl23 From de Broglie s relation we have A h mv Putting these together we have h h 272739quot n gt mvrn n mv 27 Bohr s angular momentum quantization rule Example de Broglie wavelength of an electron and a baseball 9 an electron mv 911x10 kg100x107ms moving at 728 X 107mm v 100X107ms experimentally assessable li 663x10 34Js mv 145x10 3kg400ms a baseball 114 x10 34m moving at v 400m s immeasurably small Since h is so small typical daily object s wave characteristic is insignificant in p tase Ineidentwaves dsinsoo 1 Electron Diffraction Clinton DaVISson and Lester Genner 1927 A heated l ilttment The electrons The detector can are aceelerttled by he moved to detect electrodes and scattered electrons threcteil at a crystal at ttny tingle H Atoms on surlacc 01 crystal emits electrons eVap22m hphl2meVZ El tt mts trike I Va 54V peak 39tl nickel cry stttl Electron beam in vacuum Voltage source Nickel crystal acts as a diffraction grading Constructive interference is expected at dsinezmi m12 0 igg Diffraction of 6 off Aluminum Foil In 1928 GP Thomson son of J J Thomson performed another demonstrative experiment in showing electrons can act as a wave and diffract from a polycrystalline aluminum foil Top Xray diffraction Debye and Sherrer XRay diffraction experiment done a few years earlier using a similar setup GP Thomson s electron diffraction experiment produced a qualitatively similar result Bottom electron diffraction publishing as Pearson Auutsonrwesley Cagyngm Eons Peavsmn Eaucstmn Inc Electron TwoSlit Interference a Photographic Electron interference Fl 1 n 1 quot p u r b Afrer28 AfterlOOO After10000 electrons electrons electrons 0 db 39 quot o Slit Electron beam vacuum CD 0 O J 39 7 0 a h Slll o 399 o 399 Copyrlgm o 2003 Pearson Educalmn Inc Dubllsmng as Paavsml Aduisoanesley The probability of arrival vs y can only be l explained by two interfering matter waves may Similar to the twoslit experiment with photons electrons demonstrate both particle detection of single e s on lm and wave interference pattern characteristics Sequence of Events according to Mavis S frame At l O lightning strikes on B and B Note lplatform is length A 0 contracted and the box car s rest length is longer 0 2Speed of light is the same u in both frame according v to O A Some time later lightning strikes on A andA A O f 0


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