THE WAVE PROPERTIES OF MATTER (CHAPTER 5)

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1 1 / 29 THE WAVE PROPERTIES OF MATTER (CHAPTER 5) Quantum Mechanics I (October 19-26, 2018)

2 Next Homework (#6) 2 Will be assigned this coming Monday, and due in a week Come and see me about the midterm, in office hours

3 3 Introduction Along with relativity, quantum mechanics was a stunning achievement of the early 20 th century Although, even though ~100 years have passed, it is safe to say that we don t really have a good understanding of why quantum mechanics is the way it is Some physicists here at Albany are trying to derive quantum mechanics from principles of entropy, information theory In this and the next chapter, we will cover some of the concepts in QM Be aware that at times, concepts can be very counterintuitive!

4 Wave-Particle Duality Sections 5.1, 5.2, 5.3, and Is a photon a wave or a particle? If we do an experiment where we probe the wavenature, then that s what we will see Double-slit experiment, x-ray scattering If the experiment probes the particle nature, then that is what we will see Photo-electric effect Well, it appears to be both!! This also applies to matter entities like electrons Experiments show both aspects, i.e., electron as a wave or as a particle! (double-slit experiment)

5 X-Ray Scattering 5 Wavelength of x-rays is a ~spacing between atoms in crystals; from optics, leads to effects like interference You ll come across details of this in 250. Points to notice: There is a difference in path lengths travelled by the upper line vs. lower line (and is = 2 d sin θ). If this path difference = n λ, then we get constructive interference; n λ/2 => destructive interference n λ = 2d sin θ -> Bragg s Law Experiments revealed pattern of alternating dark and light bands (let us look at Figure 5.7) n = 1 has the max intensity; let s do a group problem now

6 First Example 6 X-rays scattered from crystal have a first-order diffraction peak at 12.5 degrees Bragg s Law is the only formula you will need Determine the angles for the 2 nd and 3 rd -order peaks Angle between incident beam and crystal is theta Note that in Example 5.1 of textbook problem is phrased differently and angle there is 2*theta X-ray scattering demonstrated the wave nature The derivation (in PHY250) of the Bragg equation is based solely on details of the properties of waves.

7 De Broglie Waves (Section 5.2) 7 In 1924, De Broglie proposed the idea that matter, e.g., e- s would also exhibit wave and particle-like behavior Where lambda = h / p h is Planck s constant, p is the electron's momentum, and lambda is the resulting wavelength of the electron One can see how this comes about For a photon E = hf (frequency) and is also =pc (relativity) This implies that hf = pc. hf/c = p. Now λf = c or f/c = 1/λ Means h / λ = p or λ = h / p De Broglie simply extended this idea to matter, like e- s

8 8 Second Example Calculate what the de Broglie wavelength is of (a.) a tennis ball of mass m = 57 g and v = 25 m/s? (b.) an electron of kinetic energy = 50 ev? Formulae which you shall require to solve this problem p = γmv, lambda = h / p, and h = x 10^-34 J-s To do part (b.) relativistically, you need E = KE + mc^2 or KE = γmc^2 mc^2 = (γ-1)mc^2, and γ = 1/ (1-β^2) β = v/c, c = 299,792,458 m/s, hc = 1240 ev-nm (handy) m = 9.11x10^-31 kg or 511 kev (1 ev=1.602x10^-19 J) Recall wavelength of visible light is ~( )x10^-9 m, so answer to (a.) is very small! (10^-9 is nm)

9 9 Continued If you could pass these objects through slits or crystals with spacing ~λ, you will see diffraction & interference effects In the last chapter, we came across the Bohr atom. In it, one postulate was that angular momentum = n * h-bar De Broglie waves show how this can come about If the electron can be represented by a wave => it should fit in the orbit, i.e., n λ = 2 π r r = nλ/(2π) = nh/(2πp) so L = mvr = rp = nh/(2πp)*p = nh/ (2π) = n * h-bar, bingo! Now -- Section 5.3, electron scattering Similar to x-ray scattering, experiments were done with e s, and similar results were obtained: e s wave-like too The geometry was a little different, but concept was the same

10 Double Slit Experiment (Section 5.5) 10 screen On Screen Scattering experiments, double-slit experiment show that γ and electrons are waves. So, what are they? Is it that MWF -> waves; Tu, Th, Saturday ->Particles; Sun -> both? This is still a mystery and it is accepted as fact. Bohr s complimentary principle states that it is not possible to describe physical observables simultaneously in terms of particles and waves Extra path difference=d sin θ Both e s and γ s showed this effect -> Figs and 5.20 Even if you send in 1 at a time In Chapter 3, we studied photoelectric effect & Compton effect which only worked if γ s and electrons are particles undergoing (elastic) collisions.

11 Before we go into describing particle as waves, Section 5.4: A Mathematical Description of Waves The section has no physics. Deals with a mathematical description of waves.

11 11 Before we go into describing particle as waves, Section 5.4: A Mathematical Description of Waves The section has no physics. Deals with a mathematical description of waves. Here, we will simply go over the basics PHY250 -> Waves and Oscillations goes into details. You ll get ripples (waves) in these scenarios Rope attached to a wall. You jerk it up+down Rock in pond, or small wooden block in water n It will bob up and down, due to water waves Sounds waves, electromagnetic waves, et al. New Zealand Institute of Physics Most of these examples are called transverse waves: wave travels in one direction, but disturbance is transverse to this Sound is longitudinal wave instead: both directions are the same

12 The Wave Equation 12 All these phenomena can be represented by the wave eqn. Assuming wave (psi Ψ) travels along x-axis Can be physical amplitude or E\B field val The general equation of a travelling wave speed of wave A is the amplitude, i.e., max value of displacement ψ ( x, t ) is the displacement at any x & t Writing k = 2π/λ and ω = 2πf = 2πv/λ ψ = A sin (kx ωt) Where k is the angular wave# Wave traveling in +x: A sin(ωt - kx) and x: A sin(ωt + kx) v 2

13 13 Phase ψ = A sin (kx-ωt) : known as the phase in parentheses sin(θ) goes between -1 and 1 as θ goes from 0 -> 360 At kx ωt = π/2 ψ=+a =0, π, 2π ψ=0 =+3π/2 ψ=-a We talk of phase velocity The velocity at which the phase moves v_phase = ω/ k The most general wave equation: ψ = A sin (kx ωt+φ) Where φ is the phase at t=0 & x=0. It can have any value

14 14 Superposition of Waves If two waves are in the same region, their displacements can get added up (throw two rocks in a pond). For ex.: ψ1 = A sin(kx ωt) φ = 0 but positive x direction ψ2 = A sin(kx + ωt) φ = 0 but negative x direction ψ = ψ1+ ψ2 = A [ sin ( kx ωt ) + sin ( kx + ωt ) ] = A[sin(kx)cos(ωt)-cos(kx)sin(ωt)+sin(kx)cos(ωt)+cos(kx)sin(ωt)] ψ = [2Asin(kx)]cos(ωt) Notice that ψ looks different -> x and t have separated If phases, frequencies, amplitudes are different, then the resulting ψ can be anything (for instance, Figure 5.14) In general, ψ ( x, y, z, t ) -> we ll use 1-D, for simplicity

15 15 Wave Packet When we add many waves together, that is the name for it We can think of a particle as a sum of many waves [in right example of square waves => almost continuous spectrum of frequencies]. To simplify our calculations, we can assume that a particle can be represented as a Gaussian wavepacket. See below For instance, if I have the following, we end up with There is a technique called Fourier Analysis which decomposes the square wave into a sum of sine-waves. At time t=0 Consult Figure 5.16

16 Classical Uncertainty 16 The probability of finding the particle between ±Δx is high, where Δx (or Δk) is such that e -Δk^2Δx^2/4 ~ 0.6 Taking log of both sides => Δk^2 Δx^2 / 4 ~ 0.5, i.e., ΔkΔx ~ 1.4 [actually it is ~0.5]. Formula has implications... To get something localized in space, i.e., Δx must be small, => Δk must be large, i.e., superposition of large # of waves Similarly, we can get ΔωΔt ~ 1 (Δt is localization in time)

17 17 Group Velocity When we have wavepackets, we have another variable Called group velocity, this tells you how the envelope moves v_group = dω / dk [evaluated at k = k 0 ]. Let s see what we get. Take de Broglie waves: this is Problem #28 For a relativistic particle p=γmv=mv/ (1-v^2/c^2) E=γmc^2= = h-bar times k We shall see that the group velocity of a wave packet is simply the velocity of a relativistic particle!

18 18 For a Relativistic Particle Putting equations and together we get that v = pc^2/e (is = to the v_group!) Exercise #1: (a.) Show that the phase velocity of a relativistic particle is c/β, where β = v/c 1. (b.) What does it mean that phase velocity can be >1? Equations you will need: v_phase=λ*f, E=γmc^2=hf (photonlike), p=h/λ (de Broglie), & p=γmv, where v is v_group here Exercise #2 (26). λ = 4.0 cm and v_phase = 4.2 cm/s. What is the (a. ) frequency (b.) period (c.) wavenumber (d.) angular frequency? 4 eqn s on board

19 Uncertainty Principle (Section 5.6) 19 We already got a hint of this when discussing wavepackets ΔkΔx~O(1) i.e. if we want Δx to be small => localized packet n Then Δk, the range of wave-numbers, must be large For de Broglie waves: p = h/λ and by definition λ = 2π/k. n Momentum p = h k / 2π = h-bar * k. Previous relationship evolves Heisenberg s Uncertainty Principle: n This means that we cannot determine the momentum and position simultaneously to arbitrary precision. n If Δx is small => Δp is large. Physically, let s see what this means Imagine an electron moving along: how do we know where it is? We could send a photon beam and by looking where the scatter happens, we can estimate where the electron is

20 Example Continued However, an electron with momentum p has λ = h/p, i.e., size of wave packet To see a small object, one needs a small probe, which means λ of γ must be small => E=hc/λ must be large.

in which direction it is going => we do not know its p. Read Example 5.

20 20 Example Continued However, an electron with momentum p has λ = h/p, i.e., size of wave packet To see a small object, one needs a small probe, which means λ of γ must be small => E=hc/λ must be large. So, we have collision of a γ and electron of roughly the same momentum In this collision, the electron will get scattered, but we don t know in which direction it is going => we do not know its p. Read Example 5.8 on your own, but now we will do #37: a neutron is inside a deuterium nucleus dia. ~3.1x10-15 m. What is its minimum (a.) p and (b.) KE from the Unc Prin?

21 21 The Energy-Time Uncertainty Another form of the uncertainty principle comes from ΔωΔt ~ ½ (for a Gaussian packet). E = hf =hω/2π->ω=2πe/h or (2π/h)ΔEΔt~1/2 => What this implies is that conservation of energy can be violated, but only if the time period < h-bar/(2δe)!! In QM, we have the concept of virtual particles, i.e., particles for which E^2 p^2 c^2 + m^2 c^4 A photon can spontaneously convert to a e + e - pair and then convert back to a γ. It turns out that these effects are observable -- we shall next consider an electron sitting at rest e- creates an E-field, which is transmitted by photons

22 Virtual Photons 22 These photons can spontaneously turn into cloud of e+e- pairs and since the e+ will be attracted to the central e-, we have a screening effect. If we observe the electron from a distance, we will get a measure of its charge, e As we come closer, we penetrate this cloud and the screening effect reduces. What we ve observed is that as the energy of the probe increases (comes closer due to smaller size) α changes

23 Next Example Problem (#43) 23 An atom is in an excited state of 4.7 ev and emits a γ to come to the ground state. The time for this transition is 1 x 10 ^ -13 seconds. (a.) What is the energy uncertainty of this photon? (b.) What is its wavelength uncertainty? Equations you will require include the following E = hf = hc/λ (needs to be differentiated; no need for J) ΔE Δt ~ h-bar/2 (h-bar is just h/2-pi) h = 6.6x10^-34 J-s, c~3x10^8 m/s (h*c = 1,240 ev-nm)

24 Section In the section on waves, we denoted the displacement as ψ(x,t) where this could represent A wooden cork bobbing up and down in a pond with ripples E/B fields in an EM wave Pressure in a sound wave, etc. If we super-posed many waves, we can still write the overall function as ψ(x,t,), although the functional form can get a little complicated (see Figure 5.14) Since a wave-packet is the sum of many waves, we can use the same language. However, as will discover later, the wavefunction in QM can be complex, so ψ(x,t) by itself does not possess physical meaning.

25 Probability and Wave Functions 25 Instead we say that ψ(x,t) ^2 or ψ * ψ is the probability (density) of finding a particle (=wave-packet) Between x & x + dx and t & t + dt. The total probability of finding the particle between x =- and x=+ is obviously = 1. So That is, if we write ψ(x,t) = A * f(x,t), then the above condition gives the value of A. Next we consider the Copenhagen Interpretation of QM Where Niels Bohr of the Bohr model of the atom worked

26 26 The Copenhagen Interpretation Bohr, Pauli, Born, Heisenberg, et al. had the following to say about QM: Heisenberg's uncertainty principle is in operation Bohr s complementarity principle is as well Probability of something occurring = ψ(x,t) ^2 wave-function Many other scientists Einstein, Planck, de Broglie, Schrodinger were not comfortable with this probabilistic nature However, the original interpretation seems to be OK, and it s generally accepted However, there are spooky phenomena, e.g., EPR paradox

27 EPR Paradox and Schrodinger s Cat 27 Two photons are prepared in an entangled state, e.g., if one has spin up, then other has spin down. Now, as the 2 photons travel far from each other, either photon can be spin-up/ down -> that s what wavefunc encodes. As soon as you measure spin on one γ, spin of other γ is instantaneously determined! Imagine a cat in a closed box Depending on whether some poison is released based on a random event, the cat will be either alive or dead. The wavefunction has to allow for both possibilities simultaneously, i.e., cat is dead and alive at the same time.

28 Section Consider a particle in a 1-D box The particle is somewhere in here. What can we say about its energy? Some observations: Since the particle is inside the box, then ψ(x<0,t)=ψ(x>l,t)=0 Probability of finding the particle at x=0 and x=l -> 0. n This is because the wavefunction is taken to be continuous, so if ψ(x>l)=0, then it goes to = l Assuming the particle to be represented by a sine wave, we get that we can fit many wavelengths inside the box.

29 29/29 Particle in a Box We can use this very simple idea to do reasonably sophisticated problems (in Ch. 6). The answer we get here is different from what we would get using the uncertainty principle (there KE min = h-bar^2/(2ml^2), which is larger by 1/pi^2 - This is not a serious problem, as the uncertainty principle is used for making approximate calculations As long as λ/2 = l/n where n integer, our previous conditions hold λ n = 2l/n Assuming no potential E and non-relativistic effects, E = KE = p^2/2m = h^2 / ( 2 m λ^2 ) Lowest energy state = E 1 E 0 corresponds to infinite λ

PHYS 3313 Section 001 Lecture #16

PHYS 3313 Section 001 Lecture #16 Monday, Mar. 24, 2014 De Broglie Waves Bohr s Quantization Conditions Electron Scattering Wave Packets and Packet Envelops Superposition of Waves Electron Double Slit