Quantum Mechanics. Watkins, Phys 365,

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1 Quantum Mechanics Objectives: quantitative description of the behavior of nature at the atomic scale Central Idea: Wave-particle duality Particles obeyed classical physics: discrete, indivisible, could be counted Waves obeyed Maxwell s equations: diffraction, interference But objects actually behave like particles and waves. Most noticeable in the smallest particles eg. Electrons. Where did this idea come from? 1905: A number of experiments that could not be explained by classical ideas: Photoelectric effect: Light behaves like particles 1

2 Quantum Mechanics Photoelectric effect: Light is incident on a metal surface in a vacuum. Electrons are then emitted from the surface of the metal (photoemission). For light with wavelength above some well defined frequency, electrons are emitted. The higher the frequency, the higher the kinetic energy of this photoelectron Fig (a): photoelectrons are captured by nearby plate with same potential resulting in a current I Fig. (b): by applying voltage V m this current can be suppressed. Initial kinetic energy of the photoelectron is therefore E m =qv m. Watkins, Phys 365,

3 Now vary the light frequency ν: Observed value of E m changes with frequency E m versus ν is linear, with slope h = 6.63x10-34 J s and intercept -qφ. Φ = work function and varies from metal to metal. Measures how tightly the electrons are bound to the solid. Conclusion: light is absorbed in discrete quanta of energy of magnitude hν and that a packet of light can only kick out an electron if its energy is greater than or equal to Φ. These quanta of light are called photons. Photomultiplier Tube: Very important in optical spectroscopy Based on the photoelectric effect. Watkins, Phys 365, 006-1

4 Light is incident on a metal coated with certain materials to reduce its work function (e.g. Cs-coated GaAs for optical detection for λ~ nm). Photoelectrons are accelerated by an electric field towards a dynode. Collision with dynode kicks up many secondary electrons. This process can be repeated over an over again with as many as 1 dynodes. Large multiplication factor of up to 10 7 is achieved. Original photoelectron is detected at the final electrode, called the anode, as a charge pulse of roughly 10 7 electrons. Good way to measure light intensities by counting photons. Advantage of digital detection, namely immunity from variations in the gain of the electronics. Very useful for low light levels. Wave-particle duality Watkins, Phys 365,

Problem: For a fixed incident wavelength, increases in the light intensity did not change the energy of the electrons emitted, just their number (curren. Changes in the wavelength changed the energy.

5 Problem: For a fixed incident wavelength, increases in the light intensity did not change the energy of the electrons emitted, just their number (curren. Changes in the wavelength changed the energy. There was a threshold wavelength for emission associated with the surface work function Φ. Compton Effect: High energy photons (X-rays) collide with electrons and loser energy (longer wavelength). Entirely explainable with light as particles. Planck suggested that molecular vibrations were quantized. Einstein first to hypothesize that light was actually many discrete quanta (photons) (1905). Proposal 1: E = hν = hω (1) Atomic emission spectra: Emission from various gases such as hydrogen gave a set of discrete lines. Orbiting electrons would radiate due to acceleration.

Compton Scattering http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html 1 of 1/14/013 1:15 PM Arthur H.

The shift of the wavelength increased with scattering angle according to the Compton formula: Index Compton explained and modeled the data by assuming a particle (photon) nature for light and

6 Compton Scattering 1 of 1/14/013 1:15 PM Arthur H. Compton observed the scattering of x-rays from electrons in a carbon target and found scattered x-rays with a longer wavelength than those incident upon the target. The shift of the wavelength increased with scattering angle according to the Compton formula: Index Compton explained and modeled the data by assuming a particle (photon) nature for light and applying conservation of energy and conservation of momentum to the collision between the photon and the electron. The scattered photon has lower energy and therefore a longer wavelength according to the Planck relationship. Great experiments of physics Reference: Compton At a time (early 190's) when the particle (photon) nature of light suggested by the photoelectric effect was still being debated, the Compton experiment gave clear and independent evidence of particle-like behavior. Compton was awarded the Nobel Prize in 197 for the "discovery of the effect named after him". Calculation Compton scattering experiment Develop formula HyperPhysics***** Quantum Physics R Nave Go Back

7 Bohr s hypothesis: Electrons in stable orbits energy E n n = 1,, 3 E E 1 = hν p = nh Expression derived for E n = K((1/n 1 ) -(1/n ) ) where K is a constant. This fit the experimental data establishing the above hypotheses as facts. De Broglie Hypothesis: If a light wave has particle properties then could particles such as electrons have wave-like properties? Proposal : p = h/λ or p = hk, where k = the electron wavevector () Later confirmed by Electron diffraction (Davisson and Germer, 197) and Neutron diffraction: 3

$Light has wave properties e.g. diffraction, and interference. Photoelectric effect=> light has particle-like properties. Wave-particle duality describes this apparent contradiction.$ de Broglie: a free particle with mass, such as an electron can be described as a matter wave with a wavelength λ = h/p where p = the particle momentum. Examples: Baseball with mass 1.

de Broglie: a free particle with mass, such as an electron can be described as a matter wave with a wavelength λ = h/p where p = the particle momentum. Examples: Baseball with mass 1.

8 Light has wave properties e.g. diffraction, and interference. Photoelectric effect=> light has particle-like properties. Wave-particle duality describes this apparent contradiction. de Broglie Wavelength Light has both wave-like and particle-like properties Do particles with mass have wave-like properties? de Broglie: a free particle with mass, such as an electron can be described as a matter wave with a wavelength λ = h/p where p = the particle momentum. Examples: Baseball with mass 1.0 kg, and velocity 10 m/s : p=10 kgm/s, λ= h/p = 6.6 x10-35 m Electron with kinetic energy 100 ev: p h KE = = m λ e m e λ = h m KE e λ= m Comparable to an atomic spacing in a lattice or molecule. Can be directly observed by aiming a beam of electrons at a crystal lattice and observing resulting diffraction pattern. Left: Electron diffraction pattern formed by passing a 00kV electron beam through a thin 50nm GaAsSb sample. Such Watkins, Phys 365,

9 Electron diffraction from GaAsSb thin film (APL 79 (001) 384). 4

10 Similar things can be done with neutrons which are roughly three orders of magnitude heavier than electrons. Classical wave mechanics Classical waves are described by a wave equation from elementary wave mechanics. Quantum mechanics gives us a wave equation to describe wave- and particle-like properties of matter and radiation. Watkins, Phys 365,

11 Review of classical wave mechanics: e.g. a travelling transverse wave on a string. Perpendicular displacement u(x,of a stretched string as a function of distance x along the string is given by the wave u 1 u equation: = x c t u(x, is the amplitude of the displacement of the wire from its equilibrium value c is the wave velocity, which is a function of the wire tension and the mass per unit length (Feynman, Vol. 1). If u(x) is the shape of a wave disturbance at t=0, then u(x-c describes a moving wave solution For infinite wire, can also find sinusoidal solutions of the form: [1] u( x, = u0 cos( kx ω where k=π/λ and ω=π/t. We usually write [1] using the more convenient notation: Watkins, Phys 365,

12 u( x, j( kx ωt ) = u0e where j = -1 (we use j instead of i to avoid later confusion with curren Real part of u(x, gives the equation for a freely propagating wave, in this case moving to the right with wave velocity: c=λ/t = ω/k [] Consider wire that is confined between two fixed ends: Can be viewed as a superposition of right-going and leftgoing waves with nodes at the ends (standing wave). Solution: u(x, = A e jωt sinkx where k is chosen to give amplitude of zero at the ends at all times. The sine function guarantees that the amplitude is zero at x=0. The constraint at x=l is therefore: kl=nπ. Watkins, Phys 365,

13 First few normal modes: kl = nπ ω k = c ω = n nπc L Quantized frequencies: cannot assume arbitrary values but are quantised These modes are stationary in the sense that the position of the peak amplitude does not vary ( stationary states ) For Quantum Mechanics: We need an analogous wave function for particles like electrons. We state below, without any justification, the prescription for finding the wave function of particles. (For more details see Eisberg and Resnick) Watkins, Phys 365,

14 Quantum wave mechanics: 1. Each particle is described by a complex wave function of the form Ψ( x, y, z, which plays the role of the wave amplitude u(x, in the classical wave equation. The function Ψ and its derivatives, -continuous -finite -single valued Ψ x etc. are:. Classical variables such as momentum, and energy are replaced by quantum mechanical operators which act on the particle wavefunction Ψ. Classical variable Quantum operator x f(x) p E x f(x) h j x h j t where h h = π. Watkins, Phys 365,

15 3. Probability of finding a particle in a volume element dv = dx dy dz is given by Ψ * Ψdxdydz where Ψ * is the complex conjugate of Ψ. (if Z = X+jY, then Z * = X-jY, or if Z = e jφ then Z * =e -jφ ) Since the probability of finding the particle somewhere is unity: Ψ * Compare: Ψdxdydz = 1 [1] Classical physics: the location of a particle is a well defined quantity which can be represented by a position vector (x,y,z). Quantum physics: the value of a given coordinate such as x, p x, energy, etc. is given by calculating the expectation value of the relevant q.m. operator: < x >= Ψ * Ψ ( x, xψ( x, dx * ( x, Ψ( x, dx [] Similar to the mean value of a probability distribution: Watkins, Phys 365,

16 x = xp( x, P( x, In the case of quantum mechanics the denominator (eq. ) is usually chosen (normalized) to be unity, so we usually write: * < x >= Ψ ( x, xψ( x, dx Implications of statistical picture of QM: We cannot specify the location of a particle exactly, but only as the mean of a probability distribution, weighted by the particular coordinate. We call this mean value the expectation value of that particular coordinate : The expectation value of any function (operator) f(x) is given by * < f ( x) >= Ψ ( x, f ( x) Ψ( x, dx Time dependent Schroedinger wave equation How do we find a particle s wavefunction? Classical equation for energy of particle in a potential well: K.E. +P.E. = total energy Watkins, Phys 365,

17 We can rewrite this as p m + V ( x, = E In quantum mechanics we replace each variable in this equation with its Q.M. operator counterpart, acting on the wave function Ψ(x,: Schroedinger Wave Equation: h m x Ψ + V ( x, Ψ = h j Ψ t [3] Strong resemblance to the classical wave equation (derivatives with respect to both x and t.) Presence of the imaginary factor on the right hand side. While all physical solutions to the classical wave equation can be expressed in terms of pure real numbers, this is not the case for the solutions of the Schroedinger equation. Ψ(x, is in general a complex function of time and space. Time independent Schroedinger Wave Equation Look for solutions that are separable: i.e. Ψ(x, = ψ(x)φ( Assuming that the particle potential V is only a function of x, one can easily show that this form of the solution gives two separable versions of the SWE: Watkins, Phys 365,

18 dφ( dt je + φ( h = 0 [4] d ψ ( x) h + V ( x) ψ ( x) = Eψ ( x) m dx [5] E = "separation constant". Eq. 5 is time independent SWE, since it depends only on x. Exercise: Show that solutions to [4] are given by φ( = e jet / h Next: To solve the complete Ψ(x, we have only to solve for ψ(x) using equation [5]. The full value of Ψ(x, is then given by Ψ( x, = ψ ( x) e jet / h Watkins, Phys 365,

19 Exercise: Show that E is equal to the expectation value of the energy operator h j x Next: Consider various solutions of SWE for different forms of potential energy V(x). Special case: Free particle, 1-d V = 0 and therefore the time independent SWE is: h m d ψ ( x) dx = Eψ ( x) [6] Identical to the simple harmonic oscillator equation Solutions of the form: Substitute ψ(x) into [6] and find: ψ ( x ) = h k me E = or k = [7] m h 1 1 p In classical mechanics: E = mv = m Ae jkx Watkins, Phys 365,

20 and so we have p = hk Recall: k = π/λ for a wave vector k Therefore: p= h/λ for a free particle (de Broglie relation) Full value for Ψ(x, is now given by: Ψ ( x, = Ae jkx e jet / h Define a frequency ω = E / h (Note: E = h ω = hν ) Solution is of the form of a travelling wave: j( kx ωt ) Ψ ( x, = Ae [8] In summary: Free particle with mass m and momentum p can be described as a travelling wave, with wavelength given by p = h k = h / λ i.e. de Broglie postulate. Particle energy and frequency related by E = hω. Or E = hν Watkins, Phys 365,

21 Note: Eq. [8] describes an infinite plane wave travelling along the x axis. This means that our particle is essentially everywhere in space. Not physical Solution: Fourier analysis. Wave packets of finite extent can be formed by the superposition of plane waves with a range of frequency components (k is a spatial frequency) Bandwidth Theorem: Well known theorem from the study of classical waves called the bandwidth theorem states that Δk Δx 1 [9] Δω Δt 1 [10] To form a wave packet (pulse) of width Δx in real space, we need a range of spatial frequency components Δk which is given by Δk 1/Δx. Figure below shows how we can build up pulse from a wave by adding different Fourier components with different amplitudes: Ψ = n A cos( k x) n n Watkins, Phys 365,

22 (1/4)cos(k 9 x ) (1/3)cos(k 10 x) Bandwidth Theorem k 9 =(π)9 k 10 =(π)10 λ=1/9 λ=1/10 (1/)cos(k 11 x) k 11 =(π)11 λ=1/11 cos(k 1 x) k 1 =(π)1 λ=1/1 (1/)cos(k 13 x) k 13 =(π)13 λ=1/13 (1/3)cos(k 14 x) (1/4)cos(k 15 x) k 14 =(π)14 k 15 =(π)15 λ=1/14 λ=1/15 1 A n 1/ 1/3 1/ k (units of π) Watkins, Phys 365,

23 For above example: Δk~4π Δx~1/6 Therefore Δ kδx ~ 4π / 6 ~ This is consistent with the bandwidth theorem Quantum Mechanical Case: Using the relations p = hk, and E = hω we obtain Heisenberg Uncertainty Relation: Δp Δx h [11] ΔE Δt h [1] Physical significance of [11]: The better we know the particle s position, the larger the range in the particle s momentum and vice versa. If Δp = 0, then Δx is infinite: this corresponds to our perfect plane wave solution above. If Δx = 0, then Δp is infinite If ΔE=0 then Δt=is infinite etc. Watkins, Phys 365,

24 SWE Solution for particle in an infinite potential well (1-d) Previous solution for free particle similar to free travelling wave solution of classical wave equation for infinite wire. Energy E and wave vector k can have any value in a continuous range. If a particle is bound in a potential well, we can get quantization of the allowed energy levels, as for standing wave on a string. Only discrete values of the energy are allowed, in contrast to the continuous range of energies for a free particle. e.g. Infinite square potential well: V(x) = 0 for 0<x<L V(x) = for all other x values. Infinite square potential well For 0<x<L we have V=0 General solution is of the form e j(ωt-kx), however there is an important difference because of the boundary conditions. Since the potential is infinite outside the range 0<x<L, the wave function must be zero in that region, since otherwise the particle energy would be infinite. Watkins, Phys 365,

25 Seek solutions which have zeros at x=0 and x=l Obvious choice is the sine function: ψ ( x ) = Asin( kx) [1] This function has zeros at x=0 and x=l This happens when: kl = nπ i.e. k = nπ/l. Examples of the first three values of n are shown on previous page. Note the similarity to the form of the standing wave modes of a string discussed previously. Left: First three stationary states of the infinite square well As before: Since E = h k m E n nπ k =, therefore L ml (obtain by subst. [1] into SWE) n π h Allowed energies of these bound states are quantized in a very similar manner to that of the normal modes of a string. We call the integer n a quantum number. = Watkins, Phys 365,

26 Note: Particle has zero point energy i.e. minimum energy is not zero! Find the normalization constant A from: * ψ ( x) ψ ( x) dx = 1 Therefore L 0 A (sin nπx ) dx = A L L = 1 i.e. A = L Example: Find expectation value of position for the particle in a square well (shown in class) Next: Finite Square Well Suppose square well is not infinite: Cannot assume wavefunction goes to zero at boundary. Solution is beyond the scope of this course (numerical) Boundary conditions: Wavefunction and its derivatives are continuous at boundary. Wave function rapidly tends to zero inside barrier region. Watkins, Phys 365,

27 Finite Potential Well With the finite well now get the similar oscillations But now wave penetrates into the barrier regions Wave dies out inside the region

28 Finite Potential Well With the finite well now get the similar oscillations But now wave penetrates into the barrier regions Wave dies out inside the region

29 Finite Potential Barrier Classical wave would bounce off barrier But with SWE wave penetrates and dies out

30 Finite Width Barrier With finite width barrier the wave now declines exponentially But does not complete disappear has significant probability Result is penetration of the barrier Called Quantum Tunnelling As barrier width decrease probability of penetration increases Hence today s chip s transistors have 40 nm structures But only 1. nm insulators (gate insulator later in course) Result is significant current leakage through these One of the limits in how small a transistor you can make

SWE solution for an electron bound to a proton: the hydrogen atom Example of a particle bound to a 3-d well, in this case a Coulombic potential well, 1 V ( r) α r where r is the separation of the

31 SWE solution for an electron bound to a proton: the hydrogen atom Example of a particle bound to a 3-d well, in this case a Coulombic potential well, 1 V ( r) α r where r is the separation of the electron from the proton. Separate the wave function into three factors of the form Ψ( r, θ, φ) = R( r) Θ( θ ) Φ( φ) For 1-d infinite well, we observed quantization of the allowed energy levels. One quantum number needed. Watkins, Phys 365,

32 For 3-d Coulomb well we also observe quantization of the allowed energy levels. Three quantum numbers needed. Here we only state the results: 3 principle integer quantum numbers labelled n, l and m. Only certain values of these integers are allowed: A quantum state with particular values of n, l, m is called an orbital n =principal quantum number associated with the radial function R(r) and takes the values 1,,3... with state 1 being the lowest state in energy, and state n= corresponding to complete ionization of the electron. l is associated with the wavefunction Θ(θ) and describes the angular variation of the wavefunction. l takes values of 0, 1,, 3, 4 etc. m is associated with Φ(φ) wavefunction. For each orbital with l>0, there are several orbitals in general with the same energy (in the absence of magnetic fields). m can take integer values in the range -l to +l. Orbitals are labelled according to their value of l : l=0: s-orbitals; l=1: p-orbitals; l=: d-orbitals, l=3: f-orbitals Label n l M Watkins, Phys 365,

33 Orbital Shapes for Quantum Numbers The orbital shapes come from spherical wave solution s have spherical solutions p have 3 solutions on x, y, z axis d have 5 solutions

34 1s s 0 0 p 1-1, 0, 1 3s p 3 1-1,0,1 4s d 4 -,-1,0,1, States with l=0 correspond to spherically symmetric wavefunctions. Energies of s-orbitals are indicated below: E n represents the energy of the ns state. For n = the particle energy is 0 and above this value is a free particle. This is the ionization limit. (Figure from Eisberg and Resnick) Energies of the various s orbitals are given by: 4 μq 13.6eV E n = = ( 4πε ) h n n 0 [1] Watkins, Phys 365,

35 Define Rydberg energy, 13.6 ev which represents the binding energy of an electron to the proton. μ = reduced mass of the electron-proton system given by 1 = μ m e m p For hydrogen atom, m p >> m e and therefore μ ~ m p Quantization of hydrogen energy levels was one of the first pieces of evidence for the new quantum theory. When hydrogen is excited e.g. by a high energy electrical discharge, it emits light due to transitions between the various discrete energy levels: (Above: From Eisberg and Resnick) As principal quantum number n increases, so does the radius of the wavefunction. s-orbital wavefunctions have Watkins, Phys 365,

36 maximum amplitude at the origin, and decrease exponentially with radius. e.g., the 1s wavefunction is given by ψ 100 ( r) 1 1 = π a0 3 / e r / a 0 Notation 100 indicates n=1, l=0, and m=0. a o = Bohr radius = effective radius of the 1s ground state 4πε 0h a o = μq which turns out to be 0.59A. Other orbitals more complicated and their energies are in general different from those of the s states with the same n. e.g. p orbital wavefunction (n=, l=1, m=0) is given by: ψ 10 3 / 1 1 r r / a0 ( r) = e 4 π a 0 a0 cosθ Has node at the origin and strong angular dependence Probability distribution has dumbbell shape. Spin Quantum Number: An additional quantum number which does not come from SWE is found to be necessary to describe properties of matter. A fourth number has been found necessary to describe the intrinsic spin or angular momentum of the electron. Electrons have an intrinsic magnetic moment Watkins, Phys 365,

37 h In units of the intrinsic angular momentum of the electron is given by the quantum number s=±1/. In the absence of external magnetic fields, the energy associated with this two values are identical ( degenerate ) Multielectron Systems (atoms with Z>1) Shell model: electron levels in a multielectron atom are similar to those of H. generate the ground states of the various elements by adding electrons to the lowest available hydrogen orbitals Pauli exclusion principle: "We cannot place two electrons in the same quantum state" Makes the periodic table possible. e.g. for He, one electron may be placed in the 1s ground state hydrogen orbital. Because of the fact that the electron can have two possible spins, however, the second electron can also be placed in the 1s state, as long as it has the opposite value of the spin quantum number. If we want to add a third electron to form Li, there are no more available empty 1s states and so the third electron must occupy a s state. By adding electrons to the next available state, we can build up the entire periodic table: Number of electrons Atomic Number 1s s p 3s Watkins, Phys 365,

38 Z 1 H 1 He 3 Li 1 4 Be 5 B 1 6 C 7 N 3 8 O 4 9 F 5 10 Ne 6 11 Na 6 1 Note that there are 6 available p-states: p corresponds to l= 1 and so there are three possible states m=-1,0, and +1 times because of spin for a total of 6. This completes the shell and so the next electron must go into a 3s orbital. Electrons successively occupy orbitals that are further and further removed from the nucleus. The bonding behaviour of atoms is dominated by the outer electrons or valence electrons. See your text for more details. Watkins, Phys 365,

Quantum Mechanics & Atomic Structure (Chapter 11)

Quantum Mechanics & Atomic Structure (Chapter 11) Quantum mechanics: Microscopic theory of light & matter at molecular scale and smaller. Atoms and radiation (light) have both wave-like and particlelike