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Blackbody RadiationPhotoelectric EffectWave-Particle DualitySPH4UEverything comes ungluedThe predictions of ―classical physics‖ (Newton‘slaws and Maxwell‘s equations) are sometimescompletely, utterly WRONG.• classical physics says that an atom‘s electrons should fallinto the nucleus and STAY THERE. No chemistry, nobiology can happen.• classical physics says that toaster coils radiate an infiniteamount of energy: radio waves, visible light, X-rays,gamma rays,…The source of the problemIt‘s not possible, even ―in theory‖ to knoweverything about a physical system.• knowing the approximate position of a particle corruptsour ability to know its precise velocity (―Heisenberguncertainty principle‖)Particles exhibit wave-like properties.• interference effects!The scale of the problemLet‘s say we know an object‘s position to an accuracy Dx.How much does this mess up our ability to know its speed?Here‘s the connection between Dx and Dv (Dp = mDv):DpDxh4That‘s the ―Heisenberg uncertainty principle.‖ h 6.610 -34 J·s―It is physically impossible to predict simultaneously the exactposition and exact momentum of a particle.‖1

Atomic scale effectsSmall Dx means large Dv sincehDv4mDxExample: an electron (m = 9.110 -31 kg) in an atom isconfined to a region of size x ~ 510 -11 m.How is the minimum uncertainty in its velocity?Plug in, using h = 6.610 -34 to find v > 1.110 6 m/secExampleThe speed of an electron (m = 9.110 -31 kg) is measured tohave a value of 5 x 10 3 m/s to an accuracy of 0.003 percent.Determine the uncertainty in determining its position.p mv31 3 m 9.1110 kg 5.0010 s 27kg m4.5610sDp 0. 00003p27kg m0.000034.5610s 31kg m1.3710shDxDp434h 6.6310JsDx 4Dp 31kg m41.3710s 43.58 10m 0.385mmExampleThe speed of an bullet (m = 0.020 kg) is measured to have avalue of 300 m/s to an accuracy of 0.003 percent. Determinethe uncertainty in determining its position.p mv m 0.020kg300 s kgmh 6DxDps434Dp 0. 00003ph 6.6310JsDx 4Dp 4kg m kgm41.8 10 0.000036 s s 31 2.9310m4kgm1.810sExampleA proton has a mass of 1.67 x 10 -27 kg and is close tomotionless as possible. What minimum uncertainty in itsmomentum and in its kinetic energy must it have if it isconfined to a region :(a) 1.0 mm(b) An atom length 5.0 x 10 -10 m(c) About the nucleus of length 5.0 x 10 -15 m2

ExampleA proton has a mass of 1.67 x 10 -27 kg and is close tomotionless as possible. What minimum uncertainty in itsmomentum and in its kinetic energy must it have if it isconfined to a region :(a) 1.0 mmhDxDp432s34h 6.6310J Dp 34Dx41.010m 5.2810kg ms1KE mv22DpmDv56.6310Js34 27 3hDv 4mDx 41.67 10 kg 1.0 10 m 3.1610ms1 2 1.67 10 7 5 3.16 m kg 102 s 37 8.3310J2ExampleA proton has a mass of 1.67 x 10 -27 kg and is close tomotionless as possible. What minimum uncertainty in itsmomentum and in its kinetic energy must it have if it isconfined to a region :(b) An atom length 5.0 x 10 -10 mhDxDp434h 6.6310JsDp 104Dx45.010m1.061025kg ms1 2KE mv2DpmDv6.6310Js34 kg 0 m27 10hDv 4mDx41.6710 5.01m 63. 2s1 1.67 1027kg 63 . 224 3.3310J2m 2 s ExampleA proton has a mass of 1.67 x 10 -27 kg and is close tomotionless as possible. What minimum uncertainty in itsmomentumNoticeand inthatitswhenkineticweenergyconsidermustaitparticlehave if it isconfined to(saya regiona proton),:that is confined to a small(c) Aboutregion,the nucleusthe Quantumof lengthMechanics5.0 x 10 -15 mrequiresthat such a particle cannot have a precisehDxDp momentum (or even Dpmomentum mDvof zero).434This means 34that even at absoluteh 6.6310Jsh zero,6.63this10JsDv Dp proton must have kinetic27 20154menergy. Dx 44Dx45.010m1.This 6710 kg 1.0610menergy is called the ―zero point6 m energy‖,20kg m 6.32101.061and 0 there is no way to avoid this. ss1 2KE mv21 2 1.67 10 7 6 m kg 6.32 102 s 14 3.3310mJ2Quantum Mechanics!• At very small sizes the world is VERYdifferent!• Energy can come in discrete packets• Everything is probability; very little is absolutelycertain.• Particles can seem to be in two places at sametime.• Looking at something changes how it behaves.3

Maxwell‘s Classical TheoryThe Ultraviolet CatastropheRayleigh-Jeans Law‣ This formula also had aproblem. The problem wasthe term in the denominator.‣ For large wavelengths it fittedthe experimental data but ithad major problems atshorter wavelengths.2ckTI(,T)4Planck LawBlackbody Radiation:First evidence for Q.M.2hc21I( , T)5 hcekT1Max Planck found he could explain these curves if heassumed that electromagnetic energy was radiated in discretechunks, rather than continuously.The ―quanta‖ of electromagnetic energy is called the photon.Energy carried by a single photon isE = hf = hc/Planck‘s constant: h = 6.626 X 10 -34 Joule secHigher temperature: peak intensity at shorter E = nhf, n=1, 2, 3, 45

Blackbody Radiation:First evidence for Q.M.It was more difficult for atoms to absorb very high energyphotons (short wave lengths thus high frequency).E = nhf, n=1, 2, 3, 4Planck himself matched mathematics to the data. Heused mathematics as a device to obtain the correctanswer which he initially believed was still in classicalNewtonian physics.QuestionsA series of light bulbs are glowing red, yellow, and blue.Which bulb emits photons with the most energy?Blue! Lowest wavelength is highest energy.Red!The least energy?E = hf = hc/Highest wavelength is lowest energy.Which is hotter?(1) stove burner glowing red(2) stove burner glowing orangeHotter stove emits higher-energy photons(shorter wavelength = orange)Colored LightWhich coloured bulb‘s filament is hottest?1) Red2) Green3) Blue4) SameVisible LightColoured bulbs are identical on the inside – the glass is tintedto absorb all of the light, except the color you see. maxPhotonA red and green laser are each rated at2.5mW. Which one produces morephotons/second?1) Red 2) Green 3) Same# photons Energy/secondsecond Energy/photonPowerEnergy/photonPowerhfRed light has less energy/photon so if they both havethe same total energy, red has to have more photons!6

Wein‗s Lawb maxTWein Displacement Law- It tells us as we heat an object up, itscolor changes from red to orange towhite hot.- You can use this to calculate thetemperature of stars.The surface temperature of the Sun is5778 K, this temperaturecorresponds to a peak emission =502 nm = about 5000 Å.Wien‘s Displacement Law(nice to know)• To calculate the peak wavelengthproduced at any particular temperature,use Wien‘s Displacement Law:T · peak = 0.2898*10 -2 m·Ktemperature in Kelvin!The Wave – Particle DualityORLight WavesUntil about 1900, the classical wave theory of light describedmost observed phenomenon.Light waves:Characterized by:‣ Amplitude (A)‣ Frequency (n)‣ Wavelength ()Energy of wave is a A 27

Waves or Particles ?Physical Objects:Ball, Car, cow, or point like objects called particles.They can be located at a location at a given time.They can be at rest, moving or accelerating.Falling BallWaves or Particles ?Common types of waves:Ripples, surf, ocean waves, sound waves, radio waves.Need to see crests and troughs to define them.Waves are oscillations in space and time.Direction of travel, velocityUp-downoscillationsGround levelWavelength ,frequency, velocity and amplitude defines wavesParticles and Waves: Basic difference in behaviourWhen particles collide they cannot pass through each other !They can bounce or they can shatterWaves and Particles Basic difference:Waves can pass through each other !As they pass through each other they can enhance or canceleach otherLater they regain their original form !8

And then there was aproblem…In the early 20 th century, several effects were observedwhich could not be understood using the wave theory oflight.Two of the more influential observations were:1) The Photo-Electric Effect2) The Compton EffectPhotoelectric Effect Electrons are attracted to the (positively charged) nucleus by theelectrical force In metals, the outermost electrons are not tightly bound, and canbe easily ―liberated‖ from the shackles of its atom. It just takes sufficient energy…Classically, we increase the energyof an EM wave by increasing theintensity (e.g. brightness)Energy a A 2But this doesn‘t work ??PhotoElectric EffectNobel Trivia An alternate view is that light is acting like a particle The light particle must have sufficient energy to ―free‖ theelectron from the atom. Increasing the Intensity is simply increasing the numberof light particles, but its NOT increasing the energy of eachone! Increasing the Intensity does diddly-squat!For which work did Einstein receive the Nobel Prize?1) Special Relativity E = mc 22) General Relativity Gravity bends Light3) Photoelectric Effect Photons4) Einstein didn‘t receive a Nobel prize. However, if the energy of these ―light particle‖ is related to theirfrequency, this would explain why higher frequency light canknock the electrons out of their atoms, but low frequency lightcannot…9

Photoelectric EffectThe Apparatus• Light shining on a metal can ―knock‖electrons out of atoms.• Light must provide energy to overcomeCoulomb attraction of electron to nucleus• When the emission of photoelectrons from the cathode occurs, theytravel across the vacuum tube toward the anode, due to the appliedpotential. Even when the variable potential is dropped to zero, thecurrent does not drop to zero, because the kinetic energy of theelectrons is still adequate enough to allow some to cross the gas (thuscreating a current).• If we make the variable source of electrical potential negative then thishas the effect of reducing the electron flow. If the anode is made morenegative, relative to the cathode, a potential difference, the cutoffpotential, V 0 , is reached when the electrons are all turned back.• The cutoff potential corresponds to the maximum kinetic energy of thephotoelectrons. They do not have the KE to make it across the gap.Classical physics prediction• Electrons can be emitted regardless of the incident frequency, though itwill take longer time for smaller incident wave amplitude.• There should be a time delay between the wave illumination and theemission of electrons.Modern physics explanation• The electromagnetic wave consists of many lumped energy particlescalled photons.• The energy of each individual photon is given by the Joule• The higher the wave intensity, the higher electron energy, and thus thehigher the stopping voltage. 1fE hf10

Modern physics explanation• If N is the total number of photons incidentduring time interval T, then the total incidentoptical energy in Joules is: E Nhf• The incident energy per second (power) is givenby: NP hf Watt = J/Sec.TModern physics explanation• Interaction (absorption / emission) between theelectromagnetic wave and matter occurs throughannihilation/creation of a quantized energy (photon).• In the photoelectric effect, each single absorbed photongives its total energy (hf) to one single electron.• This energy is used by the electron to:• Overcome the attraction force of the material.• Gain kinetic energy when freed from the material.• n=N/T is the number of incident photons persecond.Modern physics explanation• Work function (): It is the minimum required energyrequired by an electron to be free from the attraction forceof the metal ions.• Some of the electrons may need more energy than thework function to be freed.+ve-veZeroTotal EnergyThe mostenergeticelectrons in thematerialModern physics explanation+ve-veTotal EnergyZerohfhfThe mostenergeticelectrons in thematerial11

Modern physics explanationTotal Energyhf+ve-veZero12The most energetic electronoutside the material2mv maxhfModern physics explanation• The electrons that need only the work function tobe freed, will have the greatest kinetic energyoutside the metal.1 2hf mvmax2• The electrons requiring higher energy to befreed, will have lower kinetic energy.Modern physics explanation• Thus, there is a minimum required photonenergy (hf o ) to overcome the work function of thematerial (note f 0 is called the cutoff frequency).hf o• If the incident photon energy is less than thework function, the electron will not be freed fromthe surface, and no photoelectric effect will beobserved.If hf

Modern physics explanation1hf 22mv maxhf eV o hf oeV h f of ohVo f feo Slope = h/e• The stopping potential doesn’t depend on the incidentlight intensity.• The stopping potential depends on the incidentfrequency.Photoelectric Equation• Since the cutoff potential is related to the maximumkinetic energy with which the photoelectrons areemitted: for a photoelectron of charge e and kineticenergy E k , and retarding potential V 0 . Then we have(loss is KE = gain in PE) : E k =eV 0 .• E photon (hf)=Φ+E k (Φ, the work function, is energywith which the electron is bound to the surface, E k isthe kinetic energy of the ejected photoelectron)• E k =hf-Φ : This tells us that if f is small such thathf=Φ, no electrons will be ejected.Threshold Frequency• Photoelectrons are emitted from thephotoelectric surface when the incident lightis above a certain frequency f 0 , called thethreshold frequency. Above the thresholdfrequency, the more intense the light, thegreater the current of photoelectronsThreshold frequency• The intensity (brightness) of the lighthas no effect on the thresholdfrequency. No matter how intense theincident light, if it is below the thresholdfrequency, not a single photoelectron isemitted.13

Photoelectric Effect Summary• Each metal has ―Work Function‖ (Φ) which isthe minimum energy needed to free electronfrom atom.• Light comes in packets called Photons• E = h fh=6.626 X 10 -34 Joule sec• Maximum kinetic energy of releasedelectrons• K.E. = hf – Φ• Photoelectrons are emitted from thephotoelectric surface when the incident lightis above a certain frequency f 0 , called thethreshold frequency.Photoelectric Effect (Summary)“Classical” MethodIncrease energy byincreasing amplitudeelectronsemitted ?NoNoNoNoWhat if we try this ?Vary wavelength, fixed amplitudeelectronsemitted ?No electrons were emitted until the frequency of the light Another exceededa critical frequency, at which point electrons were emitted symbol from forthe surface! (Recall: small large n) frequencyNoYes, withlow KEYes, withhigh KEPhoto-Electric Effect (Summary) In this ―quantum-mechanical‖ picture, the energy of the light particle(photon) must overcome the binding energy (work function, Φ) of theelectron to the nucleus. If the energy of the photon does exceed the binding energy, theelectron is emitted with a KE = E photon – E binding . The energy of the photon is given by E=hn, where the constant h =6.6x10 -34 [J s] is Planck‘s constant.SummaryIf light is under your control: You can set the frequency (wavelength, colour)and intensity. Your apparatus can count any ejected electrons. You create ahigher potential relative to the metal plate, then the ejected electrons will bepulled into the collector and forced into the ammeter circuit. If you areinterested in the energy of the ejected electrons, you would make thepotential of the collector for and more negative with respect to the surfaceand eventually you will reach a voltage level where the ejected electronscan no longer reach the collector. This potential is called the Stoppingpotential, V o .The maximum kinetic energy of the ejected electrons will then be:KEelectron qV0―Light particle‖By the definition of the eV, the Stopping Potential expressed in volts willhave the same numerical value as the electron energy expressed in eV.That is a Stopping Potential of 2.7 V implies a maximum electron energy of2.7 eVBefore CollisionAfter Collision14

SummaryHow does this explain the photoelectric effect? For our metal with 2.7 eVwork function, then a single photon would need an energy of 2.7 eV to ejectan electron. If you used red light (650 nm), then the photons in the beamwould have energyhc6.6310 34 310819Ephoton hf 3.0610 1.91eV9650101eV=1.60x10 -19 JThese photons will be absorbed, but they do not have enough energy toeject electrons.Often the photoelectric equation is illustrated on a graph of KE vs frequency. On this graph, theslope ALWAYS equals Planck's constant, 6.63 x 10 -34 J sec. It NEVER changes. All lines on thistype of graph will be parallel, only differing in their y-axis intercept (-f) and their x-axis intercept(the threshold frequency).The threshold frequency is the lowest frequency, or longest wavelength, that permitsphotoelectrons to be ejected from the surface. At this frequency the photoelectrons have noextra KE (KE = 0) resulting in0 = hf – Φhf =ΦE photon =ΦEnergy (eV)f o (material 1)Φ (material 1)Curve for material 1Slope= Planck‘s constant, hf o (material 2)Curve for material 2Frequency (Hz)Φ (material 2)Note that red light has such a low frequency (energy) that it will never eject photoelectrons -that is, the energy of a red photon is less than the work function of the metal.ReviewQuestion• What happens to the rate electrons are emittedwhen increase the brightness?• more photons/sec so more electrons are emitted.Rate goes up.The If The suitable negative Minimum light potential is amount allowed of of the to energy fall plate on which 'C' plate which is 'P', necessary it will the give photo out start electric photo photo current electrons electric becomes as emission shownin zero is the called is figure. called Work The Stopping Function. photo Potential electrons If the amount or are cut-off attracted potential. energy by the of Stopping incident collector radiation potential 'C' connected is is less that to than value the theThreshold +ve of work retarding terminal function frequency potential of of a battery. metal, is difference defined no The photo glass between the electrons tube minimum is two evacuated. are plates frequency emitted. which When of is incident just the sufficient collector light which to 'C' halt is kept can thecause at most It +ve is energetic photo denoted potential, electric by photo the Φ. emission photo Work electrons function electrons i.e. emitted. this of are frequency a material attracted is just by given it able and by to a Φ=hf current eject 0 . electrons flows in with theout circuit It It is giving is denoted a which property them by is additional indicated "Vo" of material. by energy. the Different galvanometer.It is denoted materials by have f 0 . different values of work function.• What happens to max kinetic energy whenincrease brightness?• no change: each photon carries the same energy aslong as we don‘t change the color of the light15

Photoelectric Effect: Light Frequency• What happens to rate electrons are emittedwhen increase the frequency of the light?• as long the number of photons/sec doesn‘t change,the rate won‘t change.QuestionWhich drawing of the atom is more correct?• What happens to max kinetic energy whenincrease the frequency of the light?• each photon carries more energy, so each electronreceives more energy.This is a drawing of an electron‘s p-orbital probabilitydistribution. At which location is the electron most likely toexist?123QuestionYou observe that for a certain metal surface illuminated withdecreasing wavelengths of light, electrons are first ejectedwhen the light has a wavelength of 550 nm.a) Determine the work function for the material.b) Determine the Threshold Potential when light ofwavelength 400 nm is incident on the surfaceQuestionYou observe that for a certain metal surface illuminated withdecreasing wavelengths of light, electrons are first ejectedwhen the light has a wavelength of 550 nm.a) Determine the work function for the material.hc 34 8 m 6.6310 J s310s 955010m193.6210J 2.25eVIt is quicker is weuse hc=1240eV nmhc 1240eV nm550nm 2.25eV16

QuestionYou observe that for a certain metal surface illuminated withdecreasing wavelengths of light, electrons are first ejectedwhen the light has a wavelength of 550 nm.b) Determine the Threshold Potential when light ofwavelength 400 nm is incident on the surfaceKE Ephotonshc 1240eV nm2.25eV400nm 0.85eVQuestionSuppose you find that the electric potential needed to shutdown a photoelectric current is 3 volts. What is the maximumkinetic energy of the photoelectrons.The given potential is the stopping potential V 0U qVo1.610 19 C3V194.810J 3eVThis is the maximum kinetic energy of the photoelectronQuestionIf the work function of the material is known to be 2eV, what isthe cut-off frequency of the photons for this material.The cutt-off frequency is the frequency above which electronscan be freed from the material. That is, the frequency ofradiation whose energy is equal to the work functionSo is light awave or aparticle ?E hf cfch2eV154.1410eV s144.8310HzorE hfcfch1921.610J346.6310Js144.8310HzOn macroscopic scales, we can treat a large number of photonsas a wave.When dealing with subatomic phenomenon, we are often dealingwith a single photon, or a few. In this case, you cannot usethe wave description of light. It doesn‘t work !17

Is Light a Wave or a Particle?• Wave• Electric and Magnetic fields act like waves• Superposition, Interference and Diffraction• Particle• Photons• Collision with electrons in photo-electric effectBoth Particle and Wave !Are Electrons Particles or Waves?• Particles, definitely particles.• You can ―see them‖.• You can ―bounce‖ things off them.• You can put them on an electroscope.• How would know if electron was a wave?Look for interference!Young‘s Double Slit w/ electronElectrons are Waves?Source ofmonoenergeticelectronsd2 slitsseparatedby dLScreen a distanceL from slits• Electrons produce interference pattern justlike light waves.• Need electrons to go through both slits.• What if we send 1 electron at a time?• Does a single electron go through both slits?18

Electrons are Particles• If we shine a bright light, we can ‗see‘which hole the electron goes through.(1) Both Slits (2) Only 1 SlitElectrons are Particles and Waves!• Depending on the experiment electron canbehave like• wave (interference)• particle (localized mass and charge)• If we don‘t look, electron goes through bothslits. If we do look it chooses 1.But now the interference is gone!Electrons are Particles and Waves!• Depending on the experiment electron canbehave like• wave (interference)• particle (localized mass and charge)• If we don‘t look, electron goes through bothslits. If we do look it chooses 1.I‘m not kidding it‘s true!Schroedinger‘s Cat• Place cat in box with some poison. If wedon‘t look at the cat it will be both deadand alive!PoisonHereKitty, Kitty!19

Momentum of a PhotonCompton found that theconservation ofmomentum did hold forX-ray scattering collisionsat an angle (Comptoneffect)p mv E p v2 cEchfchff hThe Compton EffectIn 1924, A. H. Compton performed an experimentwhere X-rays impinged on matter, and he measuredthe scattered radiation.Incident X-raywavelengthMATTER 1Scattered X-raywavelength 2 > 1 2eLouis de BroglieElectron comes flying outProblem: According to the wave picture of light, the incident X-ray gives upenergy to the electron, and emerges with a lower energy (ie., the amplitudeis lower), but must have 2 1 .Quantum Picture to the RescueIf we treat the X-ray as a particle with zero mass, and momentum p = E / c,everything works !Incident X-rayp 1 = h / 1Electroninitially atresteep eScattered X-rayp 2 = h / 2 2 > 1Compton found that if the photon was treated like a particle withmometum p=E/c, he could fully account for the energy & momentum(direction also) of the scattered electron and photon! Just as if 2 billiardballs colliding!Compton Scattering (nice to know)• Compton assumed thephotons acted like otherparticles in collisions• Energy and momentum wereconserved• The shift in wavelength ishD o (1 cos )mcCompton wavelengthe20

DeBroglie‘s Relationp = h / The smaller the wavelength the larger the photon‘s momentum! The energy of a photon is simply related to the momentum by:E = pc (or, p = E / c ) The wavelength is related to the momentum by: = h/p The photon has momentum, and its momentum is given by simply p = h / .Quantum Summary• Particles act as waves and waves act asparticles• Physics is NOT deterministic• Observations affect the experimentParadox 1 (non-locality):Einstein’s BubbleFour QuantumParadoxesSituation: A photon is emitted from an isotropic source.21

Paradox 1 (non-locality):Einstein’s BubbleSituation: A photon is emitted from an isotropic source.Its spherical wave function Y expands like an inflatingbubble.Paradox 1 (non-locality):Einstein’s BubbleSituation: A photon is emitted from an isotropic source.Its spherical wave function Y expands like an inflatingbubble.Question (Albert Einstein):If a photon is detected at Detector A, how does thephoton’s wave function Y at the location of DetectorsB & C know that it should vanish?Paradox 1 (non-locality):Einstein’s BubbleIt is as if one throws a beer bottle intoLake Ontario. It disappears, and itsquantum ripples spread all over theAtlantic.Then in Copenhagen, the beer bottlesuddenly jumps onto the dock, and theripples disappear everywhere else.That’s what quantum mechanics sayshappens to electrons and photons whenthey move from place to place.Paradox 2 (Y collapse):Schrödinger’s CatExperiment: A cat is placed in a sealed box containing a device that has a 50%chance of killing the cat.Question 1: What is the wave function of the cat just before the box is opened?When does the wave function collapse?1 1( Y dead + alive ?)2 2Question 2: If we observe Schrödinger, what is his wave function during theexperiment? When does it collapse?The question is, whenand how does thewave functioncollapse.•What event collapsesit?•How does thecollapse spread toremote locations?22

Paradox 3 (wave vs. particle):Wheeler’s Delayed ChoiceA source emits one photon.Its wave function passesthrough slits 1 and 2, makinginterference beyond the slits.The observer can choose to either:(a) measure the interference pattern at**plane s 1 , requiring that the photon travelsthrough both slits.or(b) measure at plane s 2 which slit image itappears in, indicating thatit has passed only through slit 2.The observer waits untilafter the photon haspassed the slits to decidewhich measurement todo.Paradox 3 (wave vs. particle):Wheeler’s Delayed ChoiceThus, the photon does notdecide if it is a particle or awave until after it passesthe slits, even though a particlemust pass through only one slit and a wave must passthrough both slits.Apparently the measurement choice determineswhether the photon is a particle or a wave retroactively!Paradox 4 (non-locality):EPR ExperimentsMalus and FurryAn EPR (einstein Poldalsky Rosen)Experiment measures the correlatedpolarizations of a pairof entangled photons, obeyingMalus’ Law [P( rel ) = Cos 2 rel ]Paradox 4 (non-locality):EPR ExperimentsMalus and FurryAn EPR Experiment measures thecorrelated polarizations of a pairof entangled photons, obeyingMalus’ Law [P( rel ) = Cos 2 rel ]The measurement gives the same resultas if both filters were in the same arm.23

Paradox 4 (non-locality):EPR ExperimentsMalus and FurryAn EPR Experiment measures thecorrelated polarizations of a pairof entangled photons, obeyingMalus’ Law [P( rel ) = Cos 2 rel ]The measurement gives the same resultas if both filters were in the same arm.Furry proposed to place both photons inthe same random polarization state.This gives a different and weakercorrelation.Paradox 4 (non-locality):EPR ExperimentsMalus and FurryApparently, the measurement on the rightside of the apparatus causes (in somesense of the word cause) the photon onthe left side to be in the same quantummechanical state, and this does nothappen until well after they have leftthe source.This EPR “influence across space time”works even if the measurements arelight years apart.Could that be used for FTL signaling?Sorry, SF fans, the answer is No!FourInterpretationsof QuantumMechanicsThe CollapseInterpretationScientists who subscribe to the Collapse interpretation make achoice. They believe that when you accept the electron‘s wavenature, you must give up on the electron‘s particle nature.In this interpretation, the electron leaves the source as a particle thatis governed by one set of laws, but then ―expands‖ into a spread-outwave as it passes through the slits. The electron is now governed bynew laws. However, before we can measure this wavy, spread-outquantum electron it ―collapses‖ back into a particle and arrives atonly one of the many possible places on the screen.The consequence of choosing the Collapse interpretation line ofthinking is that you must accept that an electron physically changesfrom particle to wave and back again. These two realities, includingthe laws that describe them, alternate uncontrollably24

The Pilot Wave InterpretationThe Pilot Wave interpretation avoids this unexplained collapse altogether. Scientistswho subscribe to this interpretation choose to believe that the electron always existsas a classical particle and is only ever governed by one kind of physical law, for boththe familiar classical as well as quantum phenomena. However, to account for theelectron‘s wave behaviour this description requires the introduction of an invisibleguiding wave.In this interpretation, wave-particle duality is explained by assuming that electronsare real particles all of the time, and are guided by an invisible wave. The electron‘swave nature is attributed to this abstract wave, called a Pilot Wave, which tells theelectron how to move. To obtain the interference pattern in the double-slit experiment,this wave must be everywhere and know about everything in the universe, includingwhat conditions will exist in the future. For example, it knows if one or two slits areopen, or if a detector is hiding behind the slits.The Pilot Wave interpretation embodies all of the quantum behaviour, including all theinteractions between classical objects like the electron, the two-slit barrier, and themeasuring devices. In contrast to the Collapse interpretation where the collapsingelectron wave was considered real, in the Pilot Wave interpretation the wave is anabstract mathematical tool. This interpretation has a consequence. The Pilot Waveinterpretation, which was invented to deal with an electron as a real physical object,suffers the fate of being permanently beyond detectionThe Many-Worlds InterpretationSupporters of the Many Worlds interpretation, similar to the Pilot Wave idea,choose to accept that electrons are classical particles. Then they go even further,demanding that all elements of the theory must correspond to real objects—unlikethe collapsing electron or the Pilot Wave. Supporters insist on only measurable,physical objects within the world. This world is constantly splitting into manycopies of itself.When electrons demonstrate wave behaviour they exist in a superposition of manydifferent states. To Many Worlds supporters, who maintain the idea of an electronas a classical particle, a parallel universe must exist for each of the electron‘spossible states. When the electron reaches the slits, it has to choose which slit togo through. At that moment, the entire universe splits into two. In one universe, theelectron passes through the left slit as a real particle. In the other universe itpasses through the right slit as a real particle. The consequence of accepting theMany Worlds interpretation, with many quantum particles constantly facing similarchoices, is the requirement that our universe must be constantly splitting into analmost infinite number of parallel universes, each having its own copy of every oneof usThe Copenhagen InterpretationAdvocates of the Copenhagen interpretation choose to limit their discussion directly to theexperiment and to the measurements on physical objects. Questions are restricted to whatcan be seen and to what we actually do. They try to think about experiments in a veryhonest way, without invoking extra theoretical ideas like the on-off switching of the Collapseidea, or the guidance supplied by the invisible Pilot Wave, or the proposed splitting intoMany Worlds.Let’s CompareIt is tempting to come up with mental pictures about what is happening that go beyond theresults of an experiment, and to try to interpret what is happening by means of those hiddentheoretical mechanisms. The previous interpretations attributed the mysterious wave–particle duality to imaginative mathematics. In the Copenhagen interpretation much of thismystery is attributed to what happens when an experimenter enters the lab and interactswith the quantum mechanical system. With the Copenhagen perspective, the mathematicsonly deals with the experimenter‘s information about measurement interactions with thequantum mechanical system.The consequence of accepting the Copenhagen interpretation is a fundamental restrictionon how much you can read into experimental results. We know that electrons are particleswhen they are fired from the source, and we know that they are particles when they hit thescreen. What happens to electrons in the middle, what they are ―doing‖, or what they really―are‖ is not possible to know. In the Copenhagen interpretation these are unfoundedquestions. We may call an electron a wave or a particle, but ultimately those names are nomore than suitable models.25

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