1Lecture 23 (Friday)Reminders:Reading for this week TZD Chapter 6 and through 7.5.Homework #8 due next WednesdayToday’s topic: More on Matter Waves e e eModern evidence of matter waves. Double-slit experiment with matter: Electrons, Protons, Neutrons, Buckyballs (60 carbons)! Send electrons one at a time through double slit…. 1973: λ=0.004nm & D=3μm & screen 16 cm away DSimilar to case with photons…Interference pattern builds up one electron at a timeEach electron must be a wave, going through both slits, interfering with itself.At screen, electron localizes to a point where detected.200nmDouble-slit Experiment with LightWaves from slits constructively interfereAfter slit, wave spreads in all directionsInterference patternThe interference pattern is proportional to:A) B) C) D) E) None of the above.ErErBr2ErIntensity!Actually see single photons!Clicker questionDouble-slit Experiment with Electrons! Waves from slits constructively interfereAfter slit, wave spreads in all directionsActually see single electrons!Probability of observing single electrons is well described by a simple interference pattern.WHAT IS THE WAVEFUNCTION, Ψ(r, t) THAT PREDICTS THIS PATTERN??Analogy with LightInterference pattern1) E-field (a vector field) is calculated via Maxwell’s Eqs. Partial differential equations must be solved for E.2) Interference pattern from the intensity,Single photons2IE∝rConnection to the quanta:2Probability of photon arrivalE∝rGoal for matter wavesInterference pattern1) Want partial differential equations to solve for Ψ.2) Interference pattern from the squared magnitude,Single particles()2,rt∝ΨrConnection to the quanta:2Probability of particle arrival ∝Ψ2OK, so we have an analogy. How does it look to you?I don’t get what you’re saying, and can’t even think of a question.I sort of get the analogy, but am troubled by some questions.Some questions, but it’s probably ok to continue.No problems!A:B:C:D:Clicker questionGo back a step….In an Electromagnetic wave, we know what an electric field E(x,t) is. We also know that |E|2gives the intensity of the light, and thus the probability for where a photon might hit.For matter waves, we postulate that there is some wavefunction ψ(x,t), even though we do not know what it physically means.However, we know that we want |ψ(x,t)|2to give the probability for where an electron (for example) might hit.is the probability density for finding the electron at point x.To get probability, multiply the probability density by a region δx.2)(xψThe probability of finding the electron in the region δx shown isxxδψ2)(xxδ2)(xψSimilar to the relationship between linear mass density and mass.Probability versus Probability DensityProbability and probability density are similar to mass and linear mass density.2)(xψ2)(xψClicker questionThis is the wave function for a neutron. Around which point is the neutron most likely to be found?A. x = 0B. x = xAC. x = xBD. x = xCE. There is no most likely placeWave functions exist for any object, including electrons and neutrons.The wave functions we have been looking at came from two-slit interference and so were simple sine waves but different functions are also possible.Remember the probability is related to so need to square the wave function. 2)(xψThe Wave Functionψ(x)xL−LThe wave function does not tell you the path of an object like a normal traveling wave function.The wave function (squared) gives the probability of finding the object at a given x.Some details about the wave function:Wave function is really a function of the 3D position, ψ(x,y,z) but many of the problems are 1D so we will start with that: ψ(x).Wave functions are complex valued (have real and imaginary parts) and are never directly observable.The probability density is observable and is)( )( )( )( )(2imag2real2xxxxxψψψψψ+==∗ψ*(x) is complex conjugate: replace i by −i in ψ(x).3Getting Probabilities from Wave Functionsψ(x)xL−LWave function ψ(x):Probability density 2)(xψ2)(xψProbability of electron being between a and b is∫badxx2)(ψFor an infinitesimal distance δx the probability isxxδψ2)(To find the probability over a finite distance we need to integrate. xL−LabProbability of finding electron between a and b is the area under the curve.NormalizationProbability of electron being between a and b isProbability density = 2)(xψ2)(xψ∫badxx2)(ψWhat if this calculation gives a number larger than 1?A properly normalized wave function obeys the normalization constraint:xL−LabDoesn’t make any sense!1)(2=∫∞∞−dxxψThis is simply a statement that the electron is located somewhere!That is, the probability of finding the electron somewhere is 100%.If ψ(x) is a properly normalized wave function, what is the value of A as shown in the figure?2)(xψA. A = 1B. A = 1 / LC. A = 1 / 2LD.E.xL−L1)(2=∫∞∞−dxxψClicker questionA1=∫−LLAdx1=−LLAx1)( =−− LAAL12=ALLA 2/1=24/1 LA =The probability is 0 for |x|>L so we only need to consider the region between –L and L. The electron must be somewhere in that region. In that region we know so applying the normalization condition we get:Ax =2)(ψ⇒⇒⇒so A = 1 / 2LNotice it has units of 1/LengthElectron Wave Function of C60Is this really the wave function?A) Yes, it could really be Ψ(x,y,z)B) No, but it could really be | Ψ(x,y,z)