The particle-wave duality Lec. (3) The particle-wave duality

Photons and Matter Waves de Broglie wavelength Photons Matter

The particle-wave duality The physics of the nanometer scale tends to become dominated by quantum physics. The first laws of quantum physics dealt with energy quantization. A new universal constant, Planck’s constant h, was introduced in physics in addition to other constants like the speed of light, c, the gravitation constant, G, and the charge quantum, e.  

The particle-wave duality The photoelectric effect: small energy particles of light, the so-called light quanta, are able to knock out electrons from metals just like one billiard ball hitting another. Fundamental objects in the physical world, electrons, protons, neutrons, photons and other leptons, hadrons, and field quanta, all have the same dual nature: they are at the same time both particles and waves. the nature of light was proposed in the 17th century, it was debated whether light were particles (corpuscles), as claimend by Newton, or waves, as claimed by Huygens.

The particle-wave duality Young’s famous double-slit interference experiments that demonstrated that light were waves. In 1913 Niels Bohr published his theory of the hydrogen atom, explaining its stability in terms of stationary states. Bohr did not explain why stationary states exist. Planck’s formula for the energy distribution in black-body radiation demonstrated that light possesses some element of particle nature. This particle aspect became more evident with Einstein’s explanation of the photoelectric effect: small energy parcels of light, the so-called light quanta, are able to knock out electrons from metals just like one billiard ball hitting another.

The particle-wave duality Two strong evidences for the existence of electron waves. (a) An electron diffraction from the quasicrystal Al70Co11Ni19 (b) Electron waves on the surface of copper detected by scanning tunnel microscopy (STM). The waves are trapped inside a ring of iron atoms. The ring is created by pushing the iron atoms around on the copper surface using an atomic force microscope.

The particle-wave duality de Broglie’s famous relation between the momentum p of a particle and the wave length (or wave number ) of its associated wave Existence of diffraction patterns when electrons are shot through a thin metal film. In the beginning of 1926 Schroodinger published his wave equation providing a firm mathematical foundation for de Broglie’s ideas.

de Broglie waves First a little basics about waves. Waves are disturbances through a medium (air, water, empty vacuum), that usually transfer energy. Here is one:

de Broglie waves The distance between each bump is called a wavelength (λ), and how many bumps there are per second is called the frequency (f). The velocity at which the wave crest moves is jointly proportional to λ and f: V = λ f Now there are two velocities associated with the wave: the group velocity (v) and the phase velocity (V). When dealing with waves going in oscillations (cycles of periodic movements), we use notations of angular frequency (ω) and the wavenumber (k) – which is inversely proportional to the wavelength. The equations for both are: ω = 2πf and k = 2π/ λ

de Broglie waves The phase velocity of the wave (V) is directly proportional to the angular frequency, but inversely proportional to the wavenumber, or: V = ω / k The phase velocity is the velocity of the oscillation (phase) of the wave. The group velocity is equal to the derivative of the angular frequency with respect to the wavenumber, or: v = d ω / d k The group velocity is the velocity at which the energy of the wave propagates. Since the group velocity is the derivative of the phase velocity, it is often the case that the phase velocity will be greater than the group velocity.

Derivation for De Broglie Equation De Broglie, in his research, decided to look at Einstein’s research on photons – or particles of light – and how it was possible for light to be considered both a wave and a particle. Let us look at how there is a relationship between them. We get from Einstein (and Planck) two equations for energy: E = h f (photoelectric effect) & E = mc2 (Einstein’s Special Relativity) Now let us join the two equations: E = h f = m c2

Derivation for De Broglie Equation From there we get: h f = p c (where p = mc, for the momentum of a photon) h / p = c / f Substituting what we know for wavelengths (λ = v / f, or in this case c / f ): h / mc = λ De Broglie saw that this works perfectly for light waves, but does it work for particles other than photons, also?

Derivation for De Broglie Equation In order to explain his hypothesis, he would have to associate two wave velocities with the particle. De Broglie hypothesized that the particle itself was not a wave, but always had with it a pilot wave, or a wave that helps guide the particle through space and time. This wave always accompanies the particle. He postulated that the group velocity of the wave was equal to the actual velocity of the particle. However, the phase velocity would be very much different. He saw that the phase velocity was equal to the angular frequency divided by the wavenumber. Since he was trying to find a velocity that fit for all particles (not just photons) he associated the phase velocity with that velocity. He equated these two equations: V = ω / k = E / p (from his earlier equation c = (h f) / p )

Derivation for De Broglie Equation From this new equation from the phase velocity we can derive: V = m c2 / m v = c2 / v Applied to Einstein’s energy equation, we have: E = p V = m v (c2 / v) This is extremely helpful because if we look at a photon traveling at the velocity c: V = c2 / c = c The phase velocity is equal to the group velocity! This allows for the equation to be applied to particles, as well as photons.

Derivation for De Broglie Equation Now we can get to an actual derivation of the De Broglie equation: p = E / V p = (h f) / V p = h / λ With a little algebra, we can switch this to: λ = h / m v This is the equation De Broglie discovered in his 1924 doctoral thesis! It accounts for both waves and particles, mentioning the momentum (particle aspect) and the wavelength (wave aspect). This simple equation proves to be one of the most useful, and famous, equations in quantum mechanics!

De Broglie and Bohr De Broglie’s equation brought relief to many people, especially Niels Bohr. Niels Bohr had postulated in his quantum theory that the angular momentum of an electron in orbit around the nucleus of the atom is equal to an integer multiplied with h / 2π, or: n h / 2π = m v r We get the equation now for standing waves: n λ = 2π r Using De Broglie’s equation, we get: n h / m v = 2π r This is exactly in relation to Niels Bohr’s postulate!

De Broglie and Relativity Not only is De Broglie’s equation useful for small particles, such as electrons and protons, but can also be applied to larger particles, such as our everyday objects. Let us try using De Broglie’s equation in relation to Einstein’s equations for relativity. Einstein proposed this about Energy: E = M c2 where M = m / (1 – v2 / c2) ½ and m is rest mass. Using what we have with De Broglie: E = p V = (h V) / λ Another note, we know that mass changes as the velocity of the object goes faster, so: p = (M v) Substituting with the other wave equations, we can see: p = m v / (1 – v / V) ½ = m v / (1 – k x / ω t ) ½ One can see how wave mechanics can be applied to even Einstein’s theory of relativity. It is much bigger than we all can imagine!

Quantum Theory Schrodinger started with the idea of Conservation of Energy: KE + PE = Etotal . He noted that KE = (1/2)mv2 = p2/2m, and that =h/p, so that p = h/ = (h/2)*(2/) = k = p, so KE = 2k2/2m Etotal = hf = (h/2)*(2f) = .

Example (1): What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) 1.0 eV and (b) 1.0 keV. (a) The momentum of the electron is and

(b) The momentum of the electron is Example continued: (b) The momentum of the electron is and

Problem What is the de Broglie wavelength of a 50 kg person traveling at 15 m/s? (h = 6.6 x 10-34 J s) comparable to spacing

Emission Energetics - I Problem: A sodium vapor light street light emits bright yellow light of wavelength = 589 nm. What is the energy change for a sodium atom involved in this emission? How much energy is emitted per mole of sodium atoms? Plan: Calculate the energy of the photon from the wavelength, then calculate the energy per mole of photons. Solution: ( 6.626 x 10 -34J s)( 3.00 x 10 8m/s) h x c wavelength Ephoton = hf = = 589 x 10 -9m Ephoton = 3.37 x 10 -19J Energy per mole requires that we multiply by Avogadro’s number. Emole = 3.37 x 10 -19J/atom x 6.022 x 1023 atoms/mole = 2.03 x 105 J/mol Emole = 203 kJ / mol

Emission Energetics - II Problem: A compact disc player uses light with a frequency of 3.85 x 1014 per second. What is this light’s wavelength? What portion of the electromagnetic spectrum does this wavelength fall? What is the energy of one mole of photons of this frequency? Plan: Calculate the energy of a photon of the light using E=hf, and wavelength l C = f . Then compare the frequency with the electromagnetic spectrum to see what kind of light we have. To get the energy per mole, multiply by Avogadro’s number. Solution: 3.00 x 108m/s 3.85 x 1014/s wavelength l = c / f = = 7.78 x 10 -7 m = 778 nm 778 nm is in the Infrared region of the electromagnetic spectrum Ephoton = hf = (6.626 x 10 -34Js) x ( 3.85 x 1014 /s) = 2.55 x 10 -19 J Emole = (2.55 x 10 -19J) x (6.022 x 1023 / mole) = 1.54 x 105 J/mole