There are a number of experimental observations that could not be explained by classical physics. For our purposes, the main one include:

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Chapter 1 Introduction 1.1 Historical Background There are a number of experimental observations that could not be explained by classical physics. For our purposes, the main one include: The blackbody radiation spectrum - Max Planck The photoelectric effect - Albert Einstein The hydrogen atom spectrum - Niels Bohr The wave-like nature of matter - Louis de Broglie The birth of quantum mechanics came about quickly with the striking resolution of these experimental inconsistencies with classical physics by the introduction of a quantum hypothesis. We will discuss each of these in more detail, ringing up some mathematical and physical tools that will be useful later. 1.1.1 Blackbody Radiation Blackbody radiation is the radiation spectrum emitted by a heated body, and is characterized by the distribution function ρ. This quantity can be expressed as a function of frequency ν or wavelength λ. The integrated radiation spectrum gives the total energy density E V : 1

2 CHAPTER 1. INTRODUCTION E V = ρ ν (ν, T ) dν = ρ λ (λ, T ) dλ (1.1) where the photon frequency is ν = c/λ (c is the speed of light), and thus ( ) ( ) dν c ρ λ (λ, T ) dλ = ρ ν (ν, T ) dλ = ρ ν (c/λ, T ) dλ (1.2) dλ λ 2 The quantities E V, ρ ν (ν, T ) dν and ρ λ (λ, T ) dλ are termed energy densities and have units of energy/volume (J m 3 ). Experiment: There were known empirical relations about the blackbody radiation spectrum before an accurate theoretical model was developed that could explain them. The most famous were the Stefan-Boltzmann and Wein displacement laws: Stefan-Boltzmann law R T = c 4 E V = c 4 ρ ν (ν, T ) dν = σt 4 (1.3) where R T is the radiancy (the rate of total radiated energy per unit area with units J s 1 m 2 ), c is the speed of light, ρ ν (ν, T ) is the blackbody radiation spectrum in terms of frequency ν and temperature T, and σ is the Stefan-Boltzmann constant with experimentally observed value σ = 5.6697 1 8 J s 1 m 2 K 4. Wein displacement law λ max T = constant = 2.9 1 3 m K (experimental value) (1.4) where λ max is the wavelength corresponding to the maximum of the radiation energy density spectrum and T is the temperature. Classical Physics: From classical physics, the equation that predicts this spectrum as a function of temperature and radiation frequency is given by the Rayleigh-

1.1. HISTORICAL BACKGROUND 3 Jeans distribution law (Eq. 1.5 below). The derivation of this equation is beyond the scope of this course - and besides it is wrong! - but it is very useful to know the form of this equation to get a feel for how classical physics differs from quantum physics. Rayleigh-Jeans distribution law ρ RJ(ν) (ν, T ) dν = 8π c 3 k BT ν 2 dν (1.5) where ρ RJ(ν) dν is the radiative energy density, ν is the frequency of the radiation, T is the temperature, c is the speed of light, and k B is the Boltzmann constant. Eq. 1.5 was derived from classical physics, i.e., it is allowable that the radiative energy can have any value, and so the energy is said to be described by a continuum of states. The Rayleigh-Jeans distribution law does not correctly predict the observed blackbody radiation spectrum, and does not obey the empirically derived Stefan-Boltzmann and Wein displacement laws. Quantum postulate: Planck s insightful hypothesis was that the energy states of the emitted radiation must be quantized, unlike the continuum of states allowed by classical physics. This lead to a new theoretical model, Planck s distribution law ρ MP(ν) (ν, T ) dν = 8π hν 3 dν (1.6) c 3 e hν/k BT 1 where ρ MP(ν) dν is the radiative energy density, ν is the frequency of the radiation, T is the temperature, c is the speed of light, k B is the Boltzmann constant, and h is Planck s constant. The fundamental assumption that Planck made was that the radiative energy must be quantized, in particular, E(ν) = nhν where n =, 1, 2...

4 CHAPTER 1. INTRODUCTION Problem 1.1 : Verify that Planck s distribution law obeys the Stefan- Boltzmann law (Eq. 1.3) and the Wein displacement law (Eq. 1.4), and give the predicted values for the Stefan-Boltzmann constant and Wein displacement constant. 1) Inserting Eq. 1.6 into the Stephan-Botzmann formula of Eq. 1.3, we have R T = c 4 E V = c 4 = c 4 8π c 3 ρ MP(ν) (ν, T ) dν (1.7) hν 3 e hν/k BT 1 dν If we let x = hν/(k B T ), dx = (h/k B T )dν, and the fact that we obtain which gives R T = 2πh c 2 2) From Eq. 1.2, we have x 3 dx e x 1 = π4 15 (1.8) ( ) 4 kb T π 4 h 15 = σt 4 (1.9) σ = 2π5 k 4 B 15h 3 c 2 (1.1) ( ) c ρ MP(λ) (λ, T ) dν = ρ MP(ν) (c/λ, T ) dλ = 8πhc 1 dλ (1.11) λ 2 λ 5 e hc/λk BT 1 To find the extremum value λ max, we solve ρ MP(λ) λ = 8πhc [ ( ) ( )] 1 e hc/λk B T hc 5 + = (1.12) λ 6 e hc/λk BT 1 e hc/λk BT 1 λk B T the term in [ ] must equal zero. Introduce the new variable x = hc/(λk B T ) we obtain 5 ex e x 1 x = (1.13) or 5/x = ex e x 1 = 1 (1.14) 1 e x

1.2. PHOTOELECTRIC EFFECT 5 inverting both sides and moving all terms involving x to one side we obtain e x + x/5 = 1 (1.15) This equation cannot be solved analytically; however, we can approximate the solution via a low-order Taylor series expansion if we expand about a point close to the solution. We note that if we ignore the nonlinear term, we obtain the solution x = 5, and if we substitute this back into the true equation with the exponential e 5 =.67, we see we this value of x is very close to the solution. Hence, expand the exponential in a Taylor series about x = 5 by introducing a new variable y = x 5, which should be close to zero. We get e (y+5) + (y + 5)/5 = 1 (1.16) or e 5 e y + y/5 = (1.17) If we take the first order expansion e y 1 y we obtain or e 5 (1 y) + y/5 = y(1/5 e 5 ) + e 5 = (1.18) y = e 5 /(1/5 e 5 ) = 1/(e 5 /5 1) (1.19) this gives to reasonable accuracy y =.35 or x = y 5 = 4.965. 1.2 Photoelectric Effect The photoelectric effect is the phenomena that metals emit electrons when irradiated with light of sufficient frequency. The kinetic energy of the electrons varies linearly with frequency above some threshold frequency ν, and is independent of the intensity of the incident radiation, in complete contradiction with classical physics. The kinetic energy of the electrons is measured by the stopping potential V S required to bring the electron to rest. Hence, we have 1 2 m ev 2 = ev S (1.2)

6 CHAPTER 1. INTRODUCTION NOTE: The kinetic energy is always positive, and with the convention that e is positive (as in the table in the inside cover of McQuarrie, and any other text), V S must be positive also. This differs by a minus sign from Eq. 1-1 in McQuarrie, and in my view is inconsistent. Einstein used a quantum hypothesis to reconcile the photoelectric effect: The Photoelectric effect 1 2 m ev 2 = ev S = hν φ = h(ν ν ) (1.21) where m e is the mass of an electron, v is the initial velocity of the ejected electron, h is Planck s constant, ν is the frequency of the incident radiation, and φ = hν is the work function of the metal (a characteristic of the material related to the ionization potential), and ν is the threshold frequency below which no electrons are ejected. 1.3 The Hydrogen Atom Spectrum The hydrogen atom spectrum is predicted by the Rydberg formula The Rydberg formula ν 1 ( 1 λ = R H 1 ) n 2 1 n 2 2 (1.22) where ν is the wavenumber (a frequently used unit in spectroscopy), R H = 19, 677.57cm 1 is the Rydberg constant, and n 1 and n 2 are integers with n 1 = 1, 2, 3 and n 2 = n 1 + 1, n 1 + 2,. Different values of n 1 give rise to different line series named after people who discovered them. The first one was by Balmer (n 1 = 2). For each n 1, the corresponding line series has the limit lim ν = R 1 H ν max (n 1 ) (1.23) n 2 n 2 1

1.4. THE BOHR MODEL OF THE ATOM 7 1.4 The Bohr Model of the Atom Classical physics does not predict that atomic spectra exist. Niels Bohr proposed an atomic model based on a quantum postulate that did correctly predict the atomic spectrum of hydrogen. Here we derive a general model, a particular case of which is Bohr s hydrogen atom model. Angular Momentum Table 1.1: Translational/rotational energy for rigid bodies. Define: M = i m i I = i m i (r i R) 2 R = 1 M i r i Translational Rotational momentum p = Mv l = Iω Kinetic energy K = 1 2 Mv2 = p 2 /(2M) K = 1 2 Iω2 = l 2 /(2I) Where r i is the distance of particle i from the center of mass. where For a 2-particle system, the inertia is simply I = µr 2 12 (1.24) ( ) m1 m 2 µ = m 1 + m 2 (1.25) is the reduced mass and r 12 is the distance between the two particles. Note when m 1 m 2, µ = m 1, and when m 1 = m 2 = m, µ = m/2. Linear Responses Consider an energy function that is a quadratic in a parameter x: E = 1 2 Ax2 bx (1.26)

8 CHAPTER 1. INTRODUCTION At the minimum of E, we have E x = Ax b = (1.27) which gives solution an energy E = 1 2 A ( b A) 2 b x = b/a (1.28) ( ) b = 1 bx (1.29) A 2 The Bohr Model Consider the energy of an electron (charge e) orbiting a nucleus Z (charge Ze) with fixed angular momentum l. The total energy is E = K + U = 1 l 2 2 µr + ( e)(ze) 2 4πɛ r = 1 ( ) ( ) l 2 1 Ze 2 1 2 µ r 2 4πɛ r (1.3) The most energetically favorable distance r for the electron to orbit the nucleus can be determined by minimizing the energy with respect to the parameter r. Alternately, if one notices that with the change of variables x = 1/r, Eq. 1.3 becomes the same form as the quadratic equation in linear ) and b = ( Ze 2 4πɛ ). From Eqs. 1.28 and 1.29, response theory with A = ( l 2 µ we immediately obtain the orbital radius and energy in terms of the angular momentum l, reduced mass µ and nuclear charge Z. The critical insight of Bohr was to introduce a quantization for the angular momentum: l = n h, n = 1, 2, (1.31) The expressions for the orbital radius and energy for a quantum state n in the Bohr atom model with nuclear charge Ze is

1.4. THE BOHR MODEL OF THE ATOM 9 The Bohr atom quantum states ( ) 4πɛ h 2 n 2 r n (Z) = Ze 2 µ = r 1(1) n2 Z (1.32) and E n (Z) = 1 2 ( Ze 2 4πɛ ) 2 µ h 2 n 2 = E 1(1) Z2 n 2 (1.33) where r 1 (1) and E 1 (1) are the orbital radius and energy of the first quantum state (n = 1) in the Bohr hydrogen model (Z = 1). If the reduced mass is replaced by the electron mass m e (the limit that the proton mass is infinite compared to the electron mass), the radius defines a length unit called the bohr: a o 4πɛ h 2 /(e 2 m e ) =.529177 Å, and the energy defines a Hartree: 1 Ha e 2 /a o = 27.212 ev. The hydrogen atom spectrum results from transitions between quantum states in the Bohr model. The Bohr atom atomic spectrum E n1 n 2 (Z) = E n2 (Z) E n1 (Z) ( ) Ze 2 2 ( µ 1 ) = 1 2 4πɛ h 2 1 n 2 1 n 2 2 = hν = hcν (1.34) Comparison with Eq. 1.22 shows that the Bohr model accurately predicts the Rydberg atomic line spectrum for hydrogen: R H = ( µe 4 8ɛ 2 ch 3 which is very close to it s experimental value. ) = 19, 681 cm 1 (1.35)

1 CHAPTER 1. INTRODUCTION 1.5 de Broglie Waves The energy of a photon according to Planck is E = hν = hc/λ, and for Einstein s relativity theory E = mc 2. Louis de Broglie combined these two theories by postulating that all masses have wave-like behavior. If we equate the Planck and Einstein energy expressions hν = mc 2 h c λ = mc2 h λ = mc = p (1.36) de Broglie s generalization was that wave-like behavior was not only applicable to photons traveling at the speed of light, but any mass m traveling at velocity v: The de Broglie wavelength or h λ = mv = p (1.37) λ = h p (1.38) de Broglie s wave theory helps explain Bohr s quantum postulate. Recall Bohr s assumption (Eq. 1.31) where the angular momentum is (for µ = m e ) l = Iω = (mr 2 )(v/r) = mvr = pr (1.39) where p is the linear momentum. Substitution of the above equation into Eq. 1.31 and using Eq. 1.37 gives or nh 2π = h λ n r n (1.4) λ n = 2πr n /n = 2πa o n (1.41) where we have used subscripts to emphasize the different quantum states. The above formula states that the radius of the Bohr atom is an integer multiple of the de Broglie wavelength.

1.6. THE HEISENBERG UNCERTAINTY PRINCIPLE 11 1.6 The Heisenberg Uncertainty Principle The Heisenberg uncertainty principle states that it is a fundamental law of Nature that it is impossible to know with certainty the position and momentum of a particle simultaneously. In particular, The Heisenberg Uncertainty principle x p x h/2 (1.42) where x and p x are the uncertainties in the x position and x component of momentum, respectively.