Notes for the Physics GRE - Harvard University Department of Physics

Notes for the Physics GRETom RudeliusHarvard University

Contents1 Introduction 22 General Advice 23 Classical Mechanics (20%) 34 Electromagnetism (18%) 65 Optics and Waves (9%) 176 Thermodynamics and Statistical Mechanics (10%) 247 Quantum Mechanics (12%) 298 Atomic Physics (10%) 369 Special Relativity (6%) 3910 Miscellaneous (15%) 401

1 IntroductionHello, potential physics Ph.D. student! If you are reading these notes, chancesare you are preparing to take the physics GRE subject test (pGRE). This exam,administered by the Educational Testing Service (ETS), consists of 100 multiplechoice problems in 170 minutes. Students score one point for every correctanswer and lose a quarter of a point for every incorrect answer, so arbitraryguessing is costly. The raw score out of 100 is converted to a scaled score out of990, which roughly corresponds to a raw score of 85 or above and a percentileof 95.The test is designed to measure the student’s physical intuition and education,but in practice it also measures a student’s calculation speed: most studentswho take the test are unable to complete it in the time allowed. Hence,you do not want to have to waste time on the exam re-deriving formulas andshould try to commit as much of your undergraduate physics education to memoryas possible. In these notes, I will present some of the most prolific formulasfound on the pGRE, along with some general tips for taking the exam. The restis in your hands. Good luck!2 General AdviceThese are the best tips I can give to someone preparing for the pGRE.1. Take the test seriously. ETS isn’t going to give you any extra pointsbecause the name of your undergraduate institution is “Harvard,” “Stanford,”“MIT,” “Princeton,” “Cornell,” etc. The test might not be as difficultor rigorous as the exams you’re accustomed to, but taking it lightlyis a big mistake (and one that some of the smartest and most talentedundergrads make every year). The pGRE is probably about as importantas the rest of the exams you’ve taken in your undergraduate career combined:treat it as such. If you want to study theoretical physics at one ofthese top schools, you should probably shoot for a score of 900 or better.2. Take every available practice exam, and do so under testing conditions.ETS should mail you a practice booklet when you register forthe pGRE, and you can find a set of four other practice tests on the OhioState Physics web page. You should take all five of these exams undertesting conditions (i.e. in 170 minute intervals). As Vince Lombardi said,“Practice does not make perfect. Only perfect practice makes perfect.”Don’t waste your time by doing a few practice problems here and there:approach the practice test in the same manner you approach the real test,and get accustomed to 170 minutes of furious thought.3. Start studying early. I started studying for the October physics GREsubject test the day that I finished the math GRE subject test in April.This is probably a tad extreme, but you should probably start studying2

at least 2 months before the exam. A good strategy would be to take thefirst practice test 2 months before the exam to get an idea of what youneed to learn, the second practice test 1 month before the exam, the third2 weeks before the exam, the fourth 1 week before the exam, and the fifth2 days before the exam.4. Set a consistent sleep schedule of at least 8-9 hours per night.Anyone who has taken Psych 101 at Cornell with Professor James Maascan tell you that the average undergrad needs 9.25 hours of sleep per nightto be well-rested. Furthermore, it is critical to develop a consistent sleepschedule. This can be a problem for many aspiring physics grad studentsbecause the pGRE is administered at 8:30am, so you will probably haveto wake up at 7:30am or so. For the week before the exam, you shouldreally try to get on an 11pm to 7:30am sleep schedule (as tough as thatmight be for a physics party animal such as yourself). Whatever you do,don’t stay up late studying for the exam.5. Take your own, handwritten notes. Actually writing formulas downimproves your ability to remember them. While the notes presented hereare nice as a guide, you should create your own notes, even if they are justparaphrases of mine.6. Use dimensional analysis. A decent fraction of the problems on thepGRE can be solved using dimensional analysis. Before starting a 5-minute calculation, pause to see if you can determine the answer basedsolely on its dimensionality.3 Classical Mechanics (20%)Newton’s laws:F = ma, (1)Newton’s gravitational law:F 12 = −F 21 . (2)⃗F = GMmr 2 ˆr (3)where r is the distance between two objects of mass M and m, respectively, andG is the gravitational constant. On earth, g = GMr= 9.8m · s −2 . Kepler’s laws2of planetary motion:1. The orbit of every planet is an ellipse with the Sun at one of the two foci.2. A line joining a planet and the Sun sweeps out equal areas during equalintervals of time.3

3. The square of the orbital period T of a planet is directly proportional tothe cube of the semi-major axis a of its orbit.In mathematical terms, the third law isThis can be seen, for circular motion, fromGMmr 2= mv2rT 2 ∝ a 3 . (4)= mω 2 r = mr(2π)2T 2 . (5)An elliptical orbit may be parametrized by the radius r and angle θ byr =where ɛ is the eccentrity of the orbit.p1 + ɛ cos θ , (6)ɛ = 0 ⇒ circleɛ < 1 ⇒ ellipseɛ = 1 ⇒ parabolaɛ > 1 ⇒ hyperbola.(7)The relation between escape velocity v e and circular orbit velocity v c :v e = √ 2v c . (8)For a spring of spring constant k,⃗F = −k⃗x, (9)√Thus, ω =kmV = 1 2 kx2 = 1 2 mω2 x 2 . (10), and the solution to the equation of motion is harmonic motion,x(t) = A sin(ωt + φ). (11)For springs of spring constant k i in parallel, the spring constants sum,For springs in series, the reciprocals sum,k = ∑ k i . (12)1k = ∑ 1. (13)k iIf two free masses m 1 and m 2 are connected by a spring k, we use the reducedmass,µ = m 1m 2m 1 + m 2, (14)4

soω =Note that if m 1 = m 2 = m, then µ = m/2.The Lagrangian L of a (classical) system is√kµ . (15)L = T − V (16)where T (q i , ˙q i is the kinetic energy (T = 1 2 mx˙i 2 , in cartesian coordinates) andV (q i ) is the potential energy. The Euler-Lagrange equations are,d ∂L= ∂L(17)dt ∂q˙i ∂q iThese give rise to the equations of motion. If our system is multidimensional,we get one equation of motion for each coordinate x i . The energy of the systemis defined as,E = ( ∑ ∂L˙q i ) − L. (18)∂ ˙qi iFurthermore, one can show that the Euler-Lagrange equation yields,dEdt = −∂L ∂t(19)Note that if L has no explicit time dependence, ∂L∂t= 0, then energy is conservedin the system. The Lagrangian is related to the Hamiltonian H of the systemby,H = ( ∑ p i q i ) − L. (20)iHere, q i is the canonical coordinate, and p i = ∂L∂q˙iis the conjugate canonicalmomentum. Note that if ∂L∂q i= 0, then p i is conserved by the Euler-Lagrangeequation. After computing the Hamiltonian in terms of q i , ˙q i , and p i , you shouldbe able to substitute p i in for ˙q i and get H = H(q i , p i ). Hamilton’s equationsare,˙q i = ∂H , (21)∂p iṗ i = − ∂H∂q i. (22)We may define the Poisson bracket between two functions f(q i , p i , t) and g(q i , p i , t)by{f, g} = ∑ ∂f ∂g− ∂g ∂f. (23)∂qi i ∂p i ∂q i ∂p iWith this, Hamilton’s equation generalizes to,d∂ff = {f, H} +dt ∂t . (24)5

GeometryI1Solid disc/cylinder, radius R2 MR2Hollow hoop/cylinder, radius R MR 22Solid sphere, radius R5 MR22Hollow sphere, radius R3 MR2Table 1: Moments of inertia for some common objects of mass M. Note that theaxes of the disc/hoop/cylinder are the ones through the center of ring/cylinder.The moment of inertia I for a rigid body around an axis isI = ∑ m i R 2 i , (25)where the sum is over all the point masses m i and R i is the distance from themass to the axis. The kinetic energy for a body rotating about its center ofmass isT = T rotation + T translation = 1 2 Iω2 + 1 2 mv2 , (26)where ω is the angular speed of rotation. The angular momentum of a rotatingbody isL = Iω. (27)The moments of inertia for some common geometries are shown in Table 1.In an elastic collision, both total energy and total momentum are conserved.If the objects have initial velocities vi 1 and vi 2 along the x-direction, then theywill have final velocities vf 1 and v2 fthat satisfy,v 1 i − v 2 i = v 2 f − v 1 f . (28)In an inelastic collision, such as when two globs collide and stick together, onlytotal momentum is conserved, energy is not.For an object sliding on a surface with friction, the force of friction if proportionalto the normal force exerted orthogonal to the surface,F fr = µF N . (29)If the object is not moving, use the coefficient of static friction, µ = µ S . If it ismoving, use the coefficient of kinetic friction, µ K . In general, µ S > µ K .4 Electromagnetism (18%)Under Franklin’s model of electricity, electric charge is always conserved. Thereare 2 kinds of charge, positive and negative, which are quantized. The forceobeys an inverse-square law, F ∝ 1r. 2In conductors, electric charges move freely. Insulators do not readily transportcharge. Semiconductors are in-between.6

Coulomb’s law:where k = 1charge:4πɛ 0F = kQqr 2 , (30)and ɛ 0 is the permittivity of free space. Electric field from pointE = F/q = kQ ˆr. (31)r2 Total E-field = vector sum of E-fields,⃗E Tot = ∑ ⃗ Efields . (32)This becomes an integral for a continuous charge distribution,∫ ∫ dq ρdV⃗E = kr 2 ˆr = k ˆr. (33)r2 Rules for drawing E-field lines:1) Begin on positive charges, end at negative–can begin/end at infinity ifthe total net charge ≠ 0.2) No two field lines can cross or touch.Electric flux through a surface S:Φ = ∫ E ⃗ · dA⃗ S= EA , if the electric field and surface are orthogonal= EA cos θ , if θ is the angle between the electric field and surface normal.(34)For a closed surface S, we have Gauss’s Law:∫Φ =where q in is the charge enclosed by the surface.A conductor in electrostatic equilibrium satisfies:1. ⃗ E = 0 inside the conductor.2. Any charge resides on the surface.S⃗E · d ⃗ A = q inɛ 0= 4πkq in , (35)3. The ⃗ E field just outside the surface is perpendicular to the surface withmagnitude q/Aɛ 0= ρ ɛ 0.4. Charge accumulates at sharp points (a.k.a. points with the smallest radiusof curvature).Change in potential energy for a particle of charge q 0 moving from point Ato point B in an electric field ⃗ E(⃗x):∆U = −q 0∫ BA⃗E · d⃗s. (36)7

Here, the d⃗s implies that this is a path integral.change in electrostatic potential:For the same process, the∆V = ∆Uq 0∫ B= − ⃗E · d⃗s. (37)AIf the E-field is uniform, ⃗ E(⃗x) = ⃗ E 0 ,∆V = ⃗ E 0 · ⃗d, (38)where ⃗ d is the displacement vector.An equipotential surface is a surface with points of constant V . The E-field lines are always perpendicular to these equipotential surfaces. For a pointcharge q moving around near another point charge Q:The potential energy:F = kQqr 2 . (39)U = kQqr . (40)The field generated by Q:E = kQr 2 . (41)The electrostatic potential due to Q:V = kQ r . (42)More generally,⃗E = − ⃗ ∇V. (43)For a spherical charge distribution, the radial component of the electric field isgiven by,E r = − ∂V∂r . (44)The electrostatic potential due to several different charges is a sum,V = k ∑ iq ir i. (45)In the continuous limit, this becomes an integral∫ dqV = kr . (46)A grounded plane induces an “image” charge, which is opposite the actual chargein both position and sign. For instance, if the yz-plane is grounded and thereis a charge Q at position (1, 0, 0), then an image charge −Q will be produced8

at (−1, 0, 0). A charged particle moving in this space will feel the electric fieldfrom both the actual charge and the image charge.The electric displacement field in a material:D = ɛ 0 E + P = ɛ 0 (1 + χ)E ≡ ɛ 0 κE, (47)Here, κ is called the “relative permittivity” or “dilectric constant,” and P is thepolarization,P = ɛ 0 χE, (48)where χ is the electric susceptibility.For a (parallel-plate) capacitor of capacitance C, with charge Q on eachplate, voltage drop V ,Q = CV. (49)Hence, the electric field inside the parallel-plate capacitor isFor capacitors C i in parallel, the capacitances sum:For capacitors in series, their reciprocals sum:E = V d = Q Cd . (50)C = ∑ C i . (51)1C = ∑ 1C i. (52)The work done in charging a capacitor = the potential energy U stored in thecapacitor, which is given by,U = 1 2 CV 2 = Q22C = 1 QV. (53)2When a dielectric material is inserted between the plates of a capacitor, thecapacitance is multiplied by a factor κ = dielectric constant,C dielectric = κC 0 . (54)When two charges of charge q, −q, respectively, are separated by a distance2a, the electric dipole moment is given by,⃗p = 2q⃗a, (55)where the vector ⃗a points from the negative charge to the positive charge. Thetorque on an electric dipole in a uniform E-field is,The potential energy of the electric dipole is,⃗τ = ⃗p × ⃗ E. (56)U = −⃗p · ⃗E. (57)9

GeometryIsolated sphere, charge Q, radius RParallel plate capacitor, area A, separation dTable 2: Geometry and Capacitance.CapacitanceC = 4πɛ 0 RAC = ɛ 0 dThe capacitance for two important geometric setups can be found in Table 2.With a dielectric, κ > 1 always, andE dielectric = 1 κ E free , (58)V dielectric = 1 κ V free . (59)The bound charge density on plates is related to the free charge density byGauss’s Law with dielectrics:The electric current I in a conductor:ρ bound = −( κ − 1κ )ρ free . (60)Φ = Qκɛ 0. (61)I = dQdt = nqv dA, (62)where n is the number density of charge carriers, q is the charge per carrier, v dis the drift velocity, and A is the cross-sectional area of the conductor (usuallya wire). The current density:J = nqv d = I A . (63)Current density is proportional to the E-field in the conductor:where σ is the conductivity.⃗J = σ ⃗ E, (64)Resistivity = ρ = 1 σ . (65)A material is said to obey Ohm’s Law if and only if its conductivity σ is independentof the applied field. The resistance R of a conductor is related to thecurrent I through it and the voltage drop V across it,V = IR. (66)10

Resistivity is further related to resistance,R = ρ l A(67)where l is the length of the conductor and A is the cross-sectional area of theconductor. Resistivity is linearly dependent upon temperature:ρ = ρ 0 [1 + α(T − T 0 )]. (68)Here, α is called the temperature coefficient of resistivity.In the classical model of electron conduction in a metal, electrons ∼ moleculesof a gas:⃗v d = e E ⃗ τ, (69)mwhere τ is the average time bectween collisions with atoms of the metal, e isthe electron charge, and m is the electron mass. The resistivity ρ obeysρ =Here, n is the number density of free electrons.The power dissapated by a resistor:mne 2 τ . (70)P = IV = I 2 R = V 2R . (71)The EMF of a battery is the voltage across terminals of the battery when thecurrent is 0. For resistors in series, resistances R i sum,For resistors in parallel, reciprocals sum,Note that this is opposite to the rule for capacitors.Kirchoff’s Rules for circuits:R = ∑ R i . (72)1R = ∑ 1R i. (73)1. The sum of the currents going into a junction = the sum of the currentscoming out of the junction.2. The sum of potential differences across any closed loop = 0.If a capacitor (capacitance C) is charged with a battery of EMF E througha resistance R, the charge q on the capacitor and the current I in the circuitvary according to,q(t) = Q[1 − e − tRC ], (74)I(t) = E R e− tRC . (75)11

Here, Q = CE is the maximum charge on the capacitor. The quantity RC isoften called the time constant of the circuit. If the capacitor with charge Q isdischarged through a resistance R,q(t) = Qe − tRC , (76)I(t) = I 0 e − tRC , (77)where I 0 = Q RCis the initial current.Force on a charged particle moving with velocity ⃗v in a magnetic field B:⃗F = q⃗v × ⃗ B. (78)Force on a wire or other conductor of length L carrying current ⃗ I in a uniformmagnetic field:⃗F = L( ⃗ I × ⃗ B), (79)or, in infinitesimal form,d ⃗ F = Id⃗s × ⃗ B. (80)Magnetic moment µ of a current loop carrying a current I:⃗µ = I ⃗ A. (81)Here, ⃗ A points through the center of the loop in a direction determined by theright-hand rule; put your right thumb along the direction of the current, andcurl your fingers through the center of the hoop. The direction they point is theright (no pun intended) one.Torque on a current loop in a magnetic field:⃗τ = ⃗µ × ⃗ B. (82)A charged particle of charge q, mass m in a uniform magnetic field B withan intial velocity v perpendicular to the field will henceforth experience circularmotion with radiusr = mvqB . (83)Hence,since ⃗v = ⃗ω × ⃗r.The Biot-Savart Law:d ⃗ B = k mId⃗s × ˆrr 2ω = qB m , (84)= k mId⃗s × ⃗rr 3 . (85)k m = µ 0 /4π = 10 −7 Tesla m A −1 . µ 0 is the permeability of free space. Inintegral form:⃗B = µ ∫0I d⃗s × ˆr4π r 2 . (86)12

Given two parallel wires separated by a distance a, each carrying a current I i ,the force per unit length f between the wires isf = µ 0I 1 I 22πa . (87)If the currents are in the same direction, the force is attractive. If they are inopposite directions, the force is repulsive.Ampere’s Law: ∮⃗B · d⃗s = µ 0 I(+µ 0 I d ). (88)Here, the term in parentheses is Maxwell’s correction term, and I d = ɛ 0dΦ Edt .The loop around which we are integrating should be taken to encircle thewire/conductor of interest. For an infinitely long solenoid, this gives an interiormagnetic field of,B = µ 0 In (89)where n = N/l is the number of turns per unit length. Outside the infinitelylong solenoid, the magnetic field vanishes.For a long, current-carrying wire of radius R,B = µ 0 I 12πr , r ≥ R= µ 0 Ir 12πR, r < R.2Magnetic flux through a surface:∫Φ B =For uniform ⃗ B, planar ⃗ A,S(90)⃗B · ⃗A. (91)Φ B = BA cos(θ). (92)For a closed surface, Gauss’s law for magnetism:∫Φ B = ⃗B · ⃗A = 0. (93)SThis is assuming that there are no magnetic monopoles in nature, which shouldbe your working assumption for the pGRE.Given a magnetic field B 0 and a material, the magnetic field in the materialis:B = B 0 + B m = B 0 + µ 0 M = µ 0 (H + M), (94)where H is the magnetic field strength and M is the magnetization of thesubstance. For para/dia-magnetic materials,M = χH, (95)13

where χ is the magnetic susceptibility. χ < 0 for diagmagnetic materials, χ >0 for paramagnetic materials, and χ ≫ 1 for ferromagnetic materials. Themagnetic permeability µ m is defined according toFaraday’s Law:∮µ m = µ 0 (1 + χ). (96)⃗E · d⃗s = E = − dΦ Bdt . (97)The minus sign in this equation is so important that it gets a name of its own:Lenz’s law. E is the EMF produced in the loop of wire by the time-varyingmagnetic flux through it. For a bar of length l moving in a magnetic field Bwith velocity v perpendicular to B, a motional EMF is produced,E = −Blv. (98)The Lorentz Force Law:⃗F = q( ⃗ E + ⃗v × ⃗ B). (99)Maxwell’s Equations:∮∮∫∫⃗E · d ⃗ A = Q ɛ 0(100)⃗B · d ⃗ A = 0 (101)⃗E · d⃗s = − dΦ BdtThe EMF through an inductor of inductance L:(102)⃗B · d⃗s = µ 0 I + ɛ 0 µ 0dΦ Edt . (103)E = −L dIdt . (104)For a coil of wire with N turns carrying current I with magnetic flux Φ B ,For a solenoid in particular,L = NΦ B. (105)IL = µ 0N 2 A, (106)lwhere A is the cross-sectional area and l is the length of the solenoid.When the battery is turned on in an L-R circuit:I(t) = E R (1 − e−t/τ ) , τ = L/R. (107)14

When the battery is turned off:The energy stored in the magnetic field of an inductor:I(t) = E R e−t/τ . (108)U = 1 2 LI2 . (109)The energy per unit volume in the B-field region of the inductor:u B = B22µ 0. (110)The current and charge of an LC-circuit oscillate in time:Q(t) = Qmax cos(ωt + φ) (111)I(t) = −ωQmax sin(ωt + φ). (112)Here, ω = √ 1LCand the phase φ depends upon the initial state of the circuit.The energy U stored in an LC-circuit:U = U C + U L = Q2 max2Ccos2 (ωt + φ) + LI2 max2sin 2 (ωt + φ). (113)Note that U is constant in time. In an RLC-circuit, we get damped harmonicmotion for I, Q.In AC-circuits,E = Vmax sin(ωt). (114)If only a resistor, I is in phase with V , andIrms = Imax/ √ 2, (115)Vrms = Vmax/ √ 2. (116)For RLC in series, inductive reactance χ L = ωL, capacitive reactance χ C = 1ωC ,impedenceZ = √ R 2 + (χ L − χ C ) 2 . (117)Voltage and current are out of phase by φ, wheretan φ = χ L − χ C. (118)RThe average power produced by the generator of RLC AC-circuit isPavg = IrmsVrms cos φ = I 2 rmsR, (119)and the energy is dissipated as heat in the resistor. For this general RLC circuit,Irms = V rmsZ . (120)15

The current in the RLC AC-circuit reaches a maximum at the resonant frequency,ω = ω 0 = 1 √LC. (121)Electromagnetic waves obey the wave equation (in 1-D),Thus, the solutions are∂ 2 E∂x 2∂ 2 B∂x 2The speed of light is related to µ 0 and ɛ 0 byAlso,And,= 1 c 2 ∂ 2 E∂t 2 , (122)= 1c 2 ∂ 2 B∂t 2 . (123)E = Emax cos(kx − ωt), (124)B = Bmax cos(kx − ωt). (125)c = 1 √µ0 ɛ 0. (126)c = |E||B| . (127)c = νλ = ωλ = ω/k. (128)2πThe rate of flow of energy corssing a unit area is given by the Poynting vector,Radiation pressure:Furthermore,At cavity walls,⃗S = 1 µ 0⃗ E × ⃗ B. (129)P = |S|c . (130)Sav = E maxBmax2µ 0. (131)E ‖ = B ⊥ = 0. (132)During reflection by a conducting surface, the E-field gets a shift of π, but theB-field does not.The characteristic impedence of a transmission line is given by,Z 0 =√LC , (133)where L is the inductance per unit length and C is the capacitance per unitlength. The speed of the wave on this transmission line is given by,c = 1 √LC. (134)16

5 Optics and Waves (9%)Waves propagate in the direction orthogonal to the wave fronts. When lighthits a surface, it may be:1. Partially or totally reflected.2. Scattered in random directions.3. Partially transmitted by refraction.4. Partially absorbed in either media.Specular reflection is the name for regular, geometric reflections. Diffuse is thename for irregular reflections from jagged surfaces. We concentrate on specularreflections.Law of reflection: angle of incidence = angle of reflection. These angles aremeasured from the normal to the surface of reflection. Index of refraction:n ≡ c v . (135)Here, c is the vacuum speed of light and v is the speed of light in the medium.Note that n ≥ 1 for all materials, and n air ≈ 1. Snell’s Law relates the angle ofincidence and the angle of refraction:n i sin θ i = n f sin θ f . (136)Here, θ i , θ f denote the initial and final angles with respect to the surface normal,respectively. n i and n f denote the indices of refraction for the initial and finalmaterials, respectively. Note that ifn i sin θ in f> 1, (137)then no value of θ f will satisfy Snell’s Law. In this case, we get evanescentwaves leaking into the final region. This phenomenon is called “total internalreflection.”The variation of refractive index n with wavelength λ is called dispersion.In general, as λ increases, n decreases slightly. This is why we see rainbows.Concave and convex mirrors have different properties, as do concave andconvex lenses. For the convex mirror, the image is virtual, upright, and smallerthan the object. Think “objects in mirror are closer than they appear.” Forconcave mirrors, the image is real, inverted, and smaller if you are behind thefocal point and virtual, upright, and enlarged if you are closer than the focalpoint (see Figure 1). For both concave and convex spherical mirrors, the focallength (distance from the mirror to the point where light rays focus) is half theradius of curvature of the mirror. The mirror equation for spherical mirrors:1f = 1 + 1 , (138)d O d i17

where f is the focal length, d 0 is the object distance, and d i is the image distance.The magnification equation:M = h ih O= − d id O, (139)where h i is the image height and h O is the object height. The sign conventionis:• d O > 0 if object is in front (real).• d O < 0 if object is in back (virtual).• d i > 0 if image is in front (real).• d i < 0 if image is in back (virtual).• f > 0 for concave.• f < 0 for convex.• M > 0 for upright image.• M < 0 for inverted image.We will soon see that the sign conventions are different for lenses.For a lens of diameter D, focal length f, the “f-number” or “relative aperture”is given by,f-number = f D . (140)Increasing f-number means increasing light-gathering power. The “numericalaperture” is given byn.a. = n sin α, (141)where n is the refractive index of the medium between the object and the lensand α is the half-angle of the limiting ray. The image brightness is inverselyproportional to both the square of the f-number and the square of the numericalaperture. For a thin lens,(11f = (n − 1) − 1 ). (142)R 1 R 2Here, f is the focal length, n is the index of refraction of the lens, R 1 is theradius of curvature for the lens nearest the object, R 2 is the radius of curvaturefor the lens farther from the object. Analogously to the mirror equation,But the sign conventions have changed:• d O > 0 if object is in front (real).1d O+ 1 d i= 1 f . (143)18

Figure 1: Convex (top left) and concave mirrors. Note the position of the imagerelative to the object for each case. Source: Wikipedia.19

Figure 2: Convex (top) and concave (bottom) lenses. Notethe position of the image relative to the object for each case.Source: http://physics20p1.blogspot.com/2011 12 01 archive.html andhttp://images.tutorvista.com/content/refraction-light/.• d O < 0 if object is in back (virtual).• d i > 0 if image is in back (real).• d i < 0 if image is in front (virtual).• R 1 > 0 if center of lens is in back.• R 1 < 0 if center of lens is in front.• R 2 > 0 if center of lens is in back.• R 2 < 0 if center of lens is in front.Rather than trying to remember all of these sign conventions, it is probablyeasier to remember the images of Figure 2. When two thin lenses are in contact,1f = 1 f 1+ 1 f 2. (144)When light is (at least approximately) perpendicularly incident on a medium,the fraction of light reflected is,R =(n1 − n 2n 1 + n 2) 2, (145)20

and the fraction of light transmitted is,T = 1 − R = 4n 1n 2(n 1 − n 2 ) 2 . (146)Here, n 1 is the index of refraction of the original medium, n 2 is the index ofrefraction of the final medium. If light incident on a sufrace is polarized in sucha way that its E-field is in the plane of the path formed by the incident andreflected ray (p-polarized), then there exists an angle of incidence θ B for whichthe reflection coefficient R p = 0 and all of the light is transmitted. θ B is calledBrewster’s angle. No such Brewster’s angle exists if the incident material is ametal or semiconductor.The wave equation:∂ 2 u∂t 2 = c2 ∇ 2 u. (147)The solution to the 1-dimensional case isu(x, t) = A sin(kx − ωt) + B cos(kx − ωt), (148)where ω/k = c. When two waves on a string meet each other, the resultingamplitude is the sum of the amplitudes for each individual wave. Two sinewaves going in the same direction give interference. For two waves sin(kx − ωt)and sin(kx − ωt + φ), we get constructive interfence for φ = 0 (in-phase) anddestructive interference for φ = π (out-of-phase). Two sine waves heading inopposite directions give standing waves,u = A sin(kx − ωt) + A sin(kx + ωt) = 2A sin(kx) cos(ωt). (149)For these standing waves, the points of 0 amplitude (kx = nπ) are called “nodes”and the points of maximum amplitude (kx = π/2 + nπ) are called “antinodes.”Two waves traveling in the same direction with different frequencies but thesame speed give “beats,”u(x, t) = A sin(k 1 x − ω 1 t) + A sin(k 2 x − ω 2 t)= 2A cos((k1 − k2)x/2 − (ω 1 − ω 2 )t/2) sin((k1 + k2)x/2 − (ω 1 + ω 2 )t/2).(150)The “beat frequency” is the difference of the individual frequencies,Characteristic impendence Z of a string:ν beat = ν 1 − ν 2 . (151)Z = µc, (152)where µ is the mass per unit length of the string and c is the wave speed. Whena wave on the string meets a hard, fixed boundary, there is a phase shift of π.When it meets a soft, free boundary, there is no phase shift (see Figure 3).When a wave moves from a place of low density to high density, the reflectedwave experiences a π phase shift. The transmitted wave feels no shift. When21

Figure 3: Fixed (left) and free (right) boundary. Note the phase shift in thefirst case. Source: http://snvphysics.blogspot.com/2011/03/waves.html.Figure 4: Waves move from less to more dense regions (left) and more toless dense regions (right). Note the phase shift in the first case. Source:http://www.physicsclassroom.com/class/waves/u10l3a.cfm.a wave moves from high density to low density, neither the reflected nor thetransmitted wave feels a π phase shift (see Figure 4). The amplitudes of thewaves are related byA incident = A transmitted − A reflected . (153)The sign convention (see Figure 4) is + for above, − for below.For single-slit diffraction, minima occur at,d sin θ n = nλ , n = 1, 2, 3, ..., (154)where d is the width of the slit, θ n is the angular location of the minima, and λis the wavelength of the light. For the double-slit experiment, maxima occur atd sin θ n = nλ , n = 0, 1, 2, ..., (155)22

where d is now the distance between slits. This equation also applies to diffractiongratings. For Bragg diffraction,For a film of thickness t,For a telescope,2d sin θ = λ. (156)2t = λ/2 ⇒ constructive interference, (157)2t = λ ⇒ destructive interference. (158)Magnetization = f Of e, (159)where f O is the focal length of the objective lens and f e is the focal length of theeyepiece. The distance between the objective lens and the eyepiece is f O + f e .The aperture formula:θ = 1.22λ/d. (160)Here, θ is the angular aperture, d is the diameter of the lens, and λ is thewavelength.The E-field for a light wave traveling in the z-direction can be written as,⃗E(⃗r, t) = (A x cos(kz − ωt), A y cos(kz − ωt + φ), 0). (161)If φ = 0, π, the light is linearly polarized, and the direction depends on theamplitudes, A x , A y . If φ = π/2, −π/2 and A x = A y , the light is circularlypolarized. Otherwise, the light is elliptically polarized. A polarizer allows onlycertain polarizations of light to pass through. The intensity I of the waveemerging from the polarizer is related to the incoming intensity I 0 and theangle θ between polarization and polarizer by,I = I 0 cos 2 θ. (162)If the incoming light is unpolarized, I = I 0 /2. The B-field is perpendicular toboth the direction of propagation and the E-field.The Doppler effect for sound is,( ) v + vrω obs = ωem . (163)v + v sHere, v is the speed of sound, v r is the speed of the receiver, and v s is the speed ofthe source. The sign convention can be determined by remembering that movingcloser corresponds to a blueshift (greater bserved frequency) and moving apartcorresponds to a redshift (lesser observed frequency). Note that a blowing winddoesn’t affect v r or v s , since the time of propagation for consecutive pulses isnot changing with time.23

Phase velocity is the speed of the wave. Group velocity is the speed of thewave packet. Phase velocity:v phase = ω k . (164)vgroup = dωdk . (165)For waves on a string, v phase = vgroup. For water waves, v phase = 2vgroup.For the wavefunction of a free particle in quantum mechanics, v phase = 1 2 v group,and vgroup is the classical velocity of the particle. For de Broglie waves, the deBroglie-Einstein wave equation gives v phase vgroup = c 2 .6 Thermodynamics and Statistical Mechanics (10%)An equilibrium state is one in which all bulk physical properties are uniformthroughout the system and don’t change with time. State functions are functionsthat take unique values at each equilibrium state. Adiathermal = adiabatic =there is no interaction between systems. Diathermal = there is a thermal interactionsbetween systems. Thermal equilibrium is transitive (i.e. A ∼ B, B ∼ Cimplies A ∼ C). Thermodynamic equilibrium means thermal equilibrium, mechanicalequibrium, and chemical equilibrium.Isotherms are surfaces of constant T . For an ideal gas,P V = nRT, (166)so isotherms are hyperbolas. A quasistatic process is one that passes througha series of equilibrium states. Reversible processes are quasistaic processes forwhich no dissipative forces like friction are present.Bulk modulus:K ≡ −V ( ∂P∂V ) T . (167)Volume thermal expansivity:Linear expansion coefficient:β ≡ 1 V (∂V ∂T ) P . (168)α ≡ 1 L ( ∂L∂T ) F . (169)Here, F is the tension. In general, β = 3α. Young’s Modulus:Compressibility:Y ≡ L A (∂F ∂L ) T . (170)κ = 1 K . (171)24

Force is related to pressure and area by F = P dA, sodW = −P dV (172)for a reversible process. Thus, the total work done on a system is,∫ V2W = − P dV. (173)V 1The adiabatic free expansion is not reversible, and there is no work done on thesystem and no change in entropy. For a wire:The first law of thermodynamics:dW = F dx. (174)∆U = W + Q, (175)where U is the internal energy of the system, W is the work done on the system,and Q is the heat added to the system. For a reversible process,The heat capacity:C =The specific heat c = C/m. C VC P is the heat capacity at constant pressure,Here, H = U + P V is the enthalpy. γ is defined by,dU = dQ − P dV. (176)lim Q/∆T = dQ/dT. (177)∆T →0is the heat capacity at constant volume,C V = dQ V /dT = ( ∂U∂T ) V . (178)C P = dQ P /dT = ( ∂H∂T ) P . (179)γ = C PC V> 1. (180)For an ideal gas,C P = C V + nR. (181)For a monotomic ideal gas,For an adiabatic free expansion,U = 3 2 nRT , C V = 3 2 nR , γ = 5 3 . (182)W = 0 , Q = 0 , ∆U = 0. (183)25

Figure 5: The Carnot cycle. Source: http://hyperphysics.phyastr.gsu.edu/hbase/thermo/carnot.html.For an ideal gas, U = U(T ) is a function of temperature only. In reality, allgases cool slightly during expansion, due to intermolecular attraction potentialenergy increasing with expansion. For an adiabatic,For a Van der Waals gas,P V γ = const. (184)(P + a )(V − nb) = nRT, (185)v2 where a and b are constants and v = V/n is the molar volume.Figure 5 depicts the Carnot cycle. The top and bottom curves are isothermsin which heat enters at temperature T H and exits and temperature T C , respectively.The side curves are adiabatics. The work done is the area bounded bythe four curves. The efficiency of the engine:η = W Q H= 1 − Q CQ H= 1 − T CT H. (186)The second law of thermodynamics says that 100% efficiency is impossible.More precisely, it is impossible to construct a device that, operating in a cycle,will produce no effect other than the extraction of heat from a single bodyat uniform temperature and perform an equivalent amount of work. Carnot’stheorem says that the Carnot engine is the most efficient engine between anytwo reservoirs. The heat entering in an isotherm is,Q = nRT ln( V endV start). (187)26

Running the Carnot cycle backward gives a refrigerator with efficiency,η R = Q CW =The efficiency of a heat pump is,The Clausius equation:Q CQ H − Q C=T CT H − T C. (188)η H = Q HW = Q H T H= . (189)Q H − Q C T H − T C∮ dQT 0≤ 0. (190)Here, T 0 is the temperature of the auxiliary reservoir, and equality holds if andonly if the cycle is reversible. Change in entropy is defined by,∆S ≡∫ fwhere the path is reversible. For any reversible process,For an isothermal free expansion,idQ RT , (191)dQ R = T dS. (192)∆S = nR ln( V fV i). (193)In general,dQ≤ dS, (194)Tand equality holds if and only if the process is reversible. For a thermallyisolated system,∆S ≥ 0, (195)with equality holding if and only if the process is reversible. The central equationof thermodynamics:T dS = dU + P dV, (196)which holds for a reversible process. For an ideal gas, of molar entropy s,In general,s = c V ln T + R ln v + s 0 . (197)S = k B ln Ω, (198)where k B is Boltzmann’s constant and Ω is the thermodynamic probability, orthe number of microstates giving rise to this macrostate. The specific heatcapacity in terms of the entropy is,C V = T ( ∂S∂T ) V , (199)27

The Helmholtz function:C P = T ( ∂S∂T ) P . (200)F = U − T S. (201)For a process in which the end point temperatures are the same as the surroundings,the maximum work obtainable equals the decrease in F . F at a minimumis the condition for thermodynamic equilibrium for a system held at constantvolume. Gibbs function:G = H − T S. (202)G at a minimum is the condition for thermodynamic equilibrium for a systemin contact with a heat and pressure reservoir. For a chemical reaction at temperatureT 0 ,∆H − T 0 ∆S ≤ 0. (203)Maxwell’s relations:( ∂V∂S ) P = ( ∂T∂P ) S, (204)( ∂S∂V ) T = ( ∂P∂T ) V , (205)−( ∂S∂P ) T = ( ∂V∂T ) P , (206)−( ∂T∂V ) S = ( ∂P∂S ) V . (207)These can be remembered by going around the wheel twice in opposite directionsfrom different starting points,−PSTV(208)The mean free path of gas molecules isl =1πd 2 n v√2, (209)where d is the diameter of the molecule and n v is the number density. Thecollision frequency is,√2πd 2 n v = ¯v l , (210)where ¯v is the average speed. The pressure exerted by N molecules of an idealgas is,P = 2 N3 V (1 2 m ¯v 2 ). (211)The temperature is,3k B T= ( 1 2 2 m ¯v 2 ). (212)28

The partition function:Z = ∑ jg j e − E jk B T, (213)where the sum runs over all microstates and g j is the degeneracy of the jthmicrostate. The probability of being found in state j is,The pressure is,The entropy is,P r(j) = g je − E jk B T. (214)ZS =P = k B T ln Z. (215)∂∂T (T k B ln Z). (216)7 Quantum Mechanics (12%)Time-dependent Schrodinger Equation:Time-independent Schrodinger Equation:i ∂Ψ∂t = − 22m ∇2 Ψ + V Ψ. (217)Eψ = − 22m ∇2 ψ + V ψ. (218)The wavefunction Ψ can be factored into a spatial and time component,For a stationary state,Ψ(⃗x, t) = ψ(⃗x)ϕ(t). (219)ϕ(t) = e −iEt/ . (220)The squared wavefunction gives a probability distribution,∫|Ψ(⃗x, t)| 2 d 3 x = 1. (221)Thus, the expectation value of ⃗x is∫〈⃗x〉 =⃗x|Ψ(⃗x, t)| 2 d 3 x. (222)The expectation value of momentum ⃗p is∫〈⃗p〉 = −iΨ ∗ ∇Ψ. (223)29

From here, we see the expressions for the operators ˆx and ˆp,ˆx = x, (224)ˆp = h ∇. (225)iFor a general operator ˆQ,∫〈Q〉 =d 3 xΨ ∗ ˆQΨ ≡ 〈Ψ| ˆQ|Ψ〉. (226)Note that this 3-dimensional formalism simplifies to the 1-dimensional case inthe obvious way–simply replace the integral measure d 3 x with dx and replace∇ with ∂∂x. I will now switch over to 1-d.The de Broglie wavelength λ of a particle with momentum p:The Heisenberg uncertainty principle:λ = h p = 2πp . (227)σ x σ p ≥ 2 . (228)Stationary states ψ n are states with definite total energy E n . The generalsolution to the time-dependent Schrodinger equation is,Ψ(x, t) =∞∑c n ψ n (x)e −iEnt/ , (229)n=1where the {ψ n } form an orthonormal basis of eigenstates of the Hamiltonian. c ncan be loosely thought of as “how much of ψ n is in Ψ.” These constants satisfyˆQ is called Hermitian if∞∑|c n | 2 = 1, (230)n=1c n = 〈ψ n |Ψ〉, (231)∞∑E n |c n | 2 = 〈Ĥ〉 = 〈E〉. (232)n=1〈 ˆQψ|ψ〉 = 〈ψ| ˆQψ〉. (233)Observables are represented by Hermitian operators because Hermitian operatorshave only real eigenvalues. The spectrum of eigenvalues for an operatormay be either discrete or continuous. If it is discrete, eigenfunctions belongingto distinct eigenvalues are orthogonal, and you may construct a complete,orthonormal basis of eigenfunctions provided it is Hermitian.30

The conserved probability current density is given by⃗j = Re(ψ ∗ ∇ψ). (234)imThe momentum space wavefunction Φ(p, t) is related to the position spacewavefunction by Fourier transform,Φ(p, t) = √ 1 ∫e −ipx/ Ψ(x, t)dx, (235)2πΨ(x, t) = √ 1 ∫2πe ipx/ Φ(p, t)dp. (236)In momentum space,ˆx = i ∂ ∂p(237)The commutator of two operators:ˆp = p. (238)[Â, ˆB] ≡ Â ˆB −ˆBÂ. (239)For momentum and position,The generalized uncertainty principle says,[ˆx, ˆp] = i. (240)σ 2 Aσ 2 B ≥( 12i 〈[Â, ˆB]〉) 2. (241)Note that the expectation value of the commutator of two Hermitian operatorsis always imaginary, so this is indeed non-negative. It is true thatddt 〈Q〉 = i 〈[Ĥ, ˆQ]〉 + 〈 ∂ ˆQ 〉. (242)∂tAssuming that ˆQ has no explicit time-dependence yields the time-energy uncertaintyprinciple:σ H σ Q ≥ 2 |d〈Q〉 |. (243)dtAs long as we are dealing with finite values of the potential V , the wavefunctionψ is C 1 continuous (i.e. ψ is continuous, ∂ψ∂xis continuous. Contrariwise,for the infinite square well,V (x) = 0 , 0 ≤ x ≤ a= ∞ , otherwise,(244)the time-independent Schrodinger equation becomesd 2 ψdx 2 = −k2 ψ , k = √ 2mE/. (245)31

This yields solutionsψ n (x) =with corresponding energy spectrum√2a sin(nπx ) , n = 1, 2, 3, ..., (246)aE n = n2 π 2 2, n = 1, 2, 3, ... (247)2ma2 These solutions satisfy the orthogonality condition,And,∫ a0√2c n =aψ ∗ mψ n dx = δ mn . (248)∫ aThe quantum harmonic oscillator has potential0sin( 2πx )Ψ(x, 0)dx. (249)aV = 1 2 kx2 = 1 2 mω2 x 2 , (250)√kwhere ω =m . To solve the Schrodinger equation with this potential, weintroduce creation and annihilation operators,â ± =The Hamiltonian may then be rewritten,The ground state is a Gaussian:1√2mω(∓ip + mωx). (251)Ĥ = ω(a − a + − 1 ). (252)2ψ 0 (x) =( mω) 1/4e− mω2 x2 (253)πThe set of eigenstates can then be built up from the ground state ψ 0 by applyingthe creation operator,ψ n (x) = 1 √n!(a + ) n ψ 0 (x). (254)The spectrum is evenly spaced,E n = (n + 1/2)ω , n = 0, 1, 2, ... (255)The annihilation operator kills the ground state,a − ψ 0 = 0. (256)32

For the finite square well, there is always at least one bound state. This is nottrue for the spherical finite square well, which need not have a bound state.To solve the 3-d Schrodinger equation with a central potential, we use separationof variables ψ(r, θ, φ) = R(r)Y (θ, φ) and arrive at the radial and angularequations,1Y1 dR dr( 1sin θ(r2dRdr ) − 2mr2 2 (V (r) − E) = l(l + 1), (257)∂ ∂Y(sin θ∂θ ∂θ ) + 1 ∂ 2 )Ysin 2 θ ∂φ 2 = −l(l + 1). (258)The solutions to the angular equation are spherical harmonics,Y ml (θ, φ). (259)For m = 0, these spherical harmonics have no φ dependence. The solution tothe radial equation depends on the potential. For the spherical square well,the solutions are Bessel functions. For the hydrogen atom potential, see §8 onatomic physics.Classically, angular momentum is given by L ⃗ = ⃗r×⃗p. In quantum mechanics,the operators of angular momentum similarly obeyThese operators obey commutation relations,ˆL = ˆr × ˆp. (260)[ˆL x , ˆL y ] = iL z , [ˆL y , ˆL z ] = iL x , [ˆL z , ˆL x ] = iL y . (261)The eigenvalues of ˆL z are m, where m = −l, ..., +l. The spherical harmonicsare eigenfunctions of both ˆL z and ˆL 2 ,In spherical coordinates,ˆL 2 Y mlˆL z Y ml= l(l + 1) 2 Y ml , (262)ˆL z = i= mY ml . (263)∂∂φ . (264)ˆL 2 commutes with each of the individual components of angular momentum,[ˆL 2 , ˆL i ] = 0. (265)Spin operators Ŝi obey the same commutation relations,[Ŝi, Ŝj] = iɛ ijk Ŝ k , (266)For eigenstates |sm〉 of Ŝ2 , Ŝz,[Ŝ2 , Ŝi] = 0. (267)Ŝ 2 |sm〉 = s(s + 1) 2 |sm〉, (268)33

Ŝ z |sm〉 = m|sm〉. (269)Here, s = 0, 1/2, 1, 3/2, ..., m = −s, −s + 1, ..., s − 1, s. Π 0 mesons have spins = 0, electrons have spin 1/2, photons have spin 1, and gravitons have spin 2.Bosons have integer spin and fermions have half-integer spin.Spin 1/2 is the most important case. We pick a basis and define spinors,χ + =( 10), χ − =( 01), (270)where χ + corresponds to spin up and χ − corresponds to spin down. ThenŜ 2 χ ± = 3 4 2 χ ± , (271)Ŝ z χ ± = ± 2 χ ±. (272)The spin operators are related to the Pauli matrices in this basis, Ŝ i = 2 σ i,where( ) ( ) ( )0 10 −i1 0σ x = , σ1 0 y = , σi 0z = . (273)0 −1The eigenspinors of Ŝx in this basis are,χ (x)± = √ 1 ( ) 1. (274)2 ±1The magnetic dipole moment for an electron in a magnetic field:⃗µ = γ ⃗ S, (275)where γ is the gyromagnetic ratio. The Hamiltonian for this system isĤ = −ˆµ · ⃗B = −γB ⃗ · Ŝ. (276)When you have two spin 1/2 particles A and B, the spins don’t sum in theobvious way. Instead,|11〉 = | 1 12 2 〉 A| 1 12 2 〉 B (277)|10〉 = | 1 12 2 〉 A| 1 2 − 1 2 〉 B + | 1 2 − 1 2 〉 A| 1 12 2 〉 B (278)|1 − 1〉 = | 1 12 2 〉 A| 1 12 2 〉 B (279)form the triplet states, and|00〉 = | 1 2is the singlet state. For two distinguishable particles,12 〉 A| 1 2 − 1 2 〉 B − | 1 2 − 1 2 〉 A| 1 12 2 〉 B (280)ψ(r 1 , r 2 ) = ψ A (r 1 )ψ B (r 2 ). (281)34

But, for indistinguishable particles,ψ(r 1 , r 2 ) = N[ψ A (r 1 )ψ B (r 2 ) ± ψ B (r 1 )ψ A (r 2 )]. (282)Here, N is a normalization constant, the + corresponds to bosons, and the− corresponds to fermions. Note that for fermions, if ψ A = ψ B , then thewhole expression vanishes. This is the Pauli Exclusion Principle: two fermionscannot occupy the same state. Identical bosons tend to be somewhat closertogether than distinguishable particles, and identical fermions tend to be fartherapart. This is called an “exchange” force. For hydrogen H 2 , the bonding stateis the antisymmetric spin-singlet state, hence the spatial part is symmetric,hence electrons are closer together in the middle, pulling the protons inwardand creating a covalent bond.Perturbation theory for a Hamiltonian H = H 0 + λH ′ is done in a differentway depending on whether the spectrum of energies is degenerate or nondegenerate.In the nondegenerate case, we expand in the small parameter λ,(H 0 +λH ′ )[ψ 0 n+λψ 1 n+λ 2 ψ 2 n+...] = (E 0 n+λE 1 n+λ 2 E 2 n+...)[ψ 0 n+λψ 1 n+λ 2 ψ 2 n+...].(283)This can be solved to give, to first order,E 1 n = 〈ψ 0 n|H ′ |ψ 0 n〉, (284)ψ 1 n = ∑ m≠n〈ψm|H 0 ′ |ψn〉0En 0 − Em0 ψm. 0 (285)To second order,E 2 n = ∑ m≠n|〈ψm|H 0 ′ |ψn〉| 0 2En 0 − Em0 . (286)In the degenerate case, in which each {ψ i } has the same energy, if we define thematrixW ij = 〈ψ 0 i |H ′ |ψ 0 j 〉, (287)then the adjusted energies are, to first order, the eigenvalues of W .Finally, we examine blackbody radiation. For light, m = ±1 but m ≠ 0.The most probable occupation number for photons isN ω =e ωd kkT − 1 , (288)where d k = Vπ 2 cω 2 dω, and V is the volume of the blackbody.3density in the frequency range dω is ρ(ω)dω, whereThe energyThe total energy density isω 3ρ(ω) =π 2 c 3 (e ωkT − 1) . (289)EV = σT 4 . (290)35

8 Atomic Physics (10%)The potential for the hydrogen atom:The effective potential is defined by,V (r) = −e24πɛ 01r . (291)V eff = V + 2 l(l + 1)2m r 2 . (292)The energy spectrum for eigenfunctions of the Hamiltonian:[ ] mE n = −2 2 ( e2 1) 24πɛ 0 n 2 ≡ E 1n 2 . (293)The Bohr radius:a 0 = 4πɛ 0 2me 2 = 0.529 × 10 −10 m. (294)Here, E 1 = −13.6 eV. In general, for hydrogen-like atoms, we replace e 2 → Ze 2 ,where Z is the central charge. The mass m appearing in these equations isactually the reduced mass µ for the system. Hence, if the proton in the centeris replaced by a positron, then m = 0.5m e . The ground state wavefunction hasl = m = 0, n = 1, andψ 100 = √ 1 e −r/a0 . (295)πa30The energy of the photon released when an electron drops from an initial energystate to a final energy state is the difference in energies,E γ = E i − E f = −13.6eV ( 1 n 2 i− 1n 2 ). (296)fIn terms of the wavelength of the photon, this becomes1λ = R( 1n 2 f− 1 n 2 ), (297)iwhere R is the Rydberg constant. The Paschen series corresponds to n f = 3and is in the IR range. The Balmer series corresponds to n f = 2 and is in thevisible range. The Lyman series corresponds to n f = 1 and is in the UV range.From the above analysis, we would expect helium to have a ground stateenergy of 8E 1 = −109 eV. Once we include for shielding of the nucleus, however,we find a true value of −79 eV. The ground state of helium is parahelium, whichmeans a spin singlet state with both electrons in the n = 1 state. Orthoheliumis the spin triplet state, which has an antisymmetric spatial part, so one electronis in the n = 1 state and one is in the n = 2 state. In general, the energiescorresponding to the same n values are lower for the spin triplet state than for36

the singlet state, because exchange forces lead the electrons to be farther apartin the triplet state.Knowing the n, l, m values of an electron tells you the orbital of the electron.There are two electrons per orbital, one for spin up and one for spin down. Thereare n 2 wavefunctions for any particular n, hence 2n 2 electrons per n. We haven = 1, 2, 3, ..., l = 0, ..., n − 1, and m = −l, −l + 1, ..., l − 1, l. l = 0 is giventhe letter “s,” l = 1 is given the letter “p”, l = 2 is “d,” then f, g, h, (skip j)i, k, l, ... As l increases, the electrons screen the nucleus more effectively. So, thestate with the lowest energy and most tightly bound electron is l = 0. As nincreases, Z increases, so there is more pull on the electrons.The electron configuration of carbon, which has 6 electrons, is (1s) 2 (2s) 2 (2p) 2 .In general, the orbitals fill up in the order of 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s,4d, 5p, 6s, 4f, 5d, 6p, 7s, and so on. The symbol people made up for the stateof an atom is,2S+1 L J , (298)where ⃗ J = ⃗ L + ⃗ S is the total angular momentum. For instance,H = 2 S 1/2 , He = 1 S 0 . (299)For two electrons, L = |l 1 − l 2 |, ..., l 1 + l 2 , S = |s 1 − s 2 |, ..., s 1 + s 2 , and J =|L − S|, ..., L + S.Hund’s Rules tell electrons how to fill up the orbitals. They are,1. The state with the highest total spin will have the lowest energy.2. If more than one state has that spin, the ground state is one with thegreatest energy.3. If a subshell is less than half-filled, the one with lowest J is the groundstate. If it is more than half-filled, the one with the highest J is the groundstate.Selection rules tell which electron transistions are allowed. For dipoles:∆J = 0, ±1 , J ≠ 0 → 0 , ∆m J = 0, ±1. (300)For electric dipole transitions,For magnetic dipole transitions,∆l = ±1. (301)∆l = 0. (302)Furthermore, if ∆S = 0, then for an electric dipole transition∆L = 0, ±1 , L ≠ 0 → 0, (303)and for a magnetic dipole,∆L = 0. (304)37

If ∆S = ±1, then ∆L = 0, ±1, ±2.The fine structure constant is given by α =e24πɛ ≈ 10c 137. The fine structuresplitting of the hydrogen atom, which is due to coupling of the spin of theelectron to the orbit of the electron as well as relativistic effects, is handled usingperturbation theory and gives an O(α 2 ) correction to the energy eigenvalues.For this correction, we must use eigenstates of total angular momentum Ĵ 2 andĴ z rather than eigenstates of ˆL z and Ŝz. Taking the fine structure into accountalways lowers the energy values, with lower j corresponding to lower energy. Tobe explicit,E nj = E 1 α2[1 +n2 n 2 ( n− 3/4)]. (305)j + 1/2Note that j = |l − 1/2|, |l + 1/2|, so each energy level for l > 0 is split in twoby the fine structure. The Lamb shift gives an O(α 3 ) correction to the energies,and the hyperfine splitting gives an O(α 4 ) correction. Hyperfine splitting is dueto coupling of the electron spin with the proton spin, and is positive for the spintriplet state, negative for the spin singlet state:Ehf 1 =g p 43m pm, 2 e c2 a 4 0triplet= − gp4m pm, 2 e c2 a 4 0singlet.(306)Here, g p ≈ 5.59 for the proton’s gyromagnetic ratio.The Stark effect describes the physics of an atom in an electric field Ĥ′ =eEẑ. For a hydrogen atom in the ground state, the first-order correction to theenergy vanishes. The second-order correction is,E 2 1 =∑n≠1,l,m|〈nlm|eEz|100〉| 2E 1 − E n. (307)The Zeeman effect describes the physics of an atom in a magnetic field,H ′ z = −(⃗µ l + ⃗µ s ) · ⃗B, wheree2m ⃗ B · 〈 ⃗ L + 2 ⃗ S〉 =⃗µ s = − e m ⃗ S , ⃗µ l = − e2m ⃗ L. (308)For the weak-field case, we want eigenstates of Ĵ 2 and Ĵz, and[Ez 1 =1 +e2m ⃗ B · 〈 ⃗ J〉j(j + 1) − l(l + 1) + 3/42j(j + 1)]. (309)This splits the ground state into two levels for m j = ±1/2. For the strong-fieldcase, we want eigenstates of ˆL z and Ŝz, andE ′ z = µ B B(m l + 2m s ), (310)where µ B ≡ e2mis the Bohr magneton.As we head left and down on the periodic table, atomic size increases, ionizationenergy (energy required to remove the outermost electron) decreases,and electron affinity (ease of gaining electrons) decreases.38

9 Special Relativity (6%)∆s 2 = −c 2 ∆t 2 + ∆x 2 + ∆y 2 + ∆z 2 = ∆s ′ 2(311)is a Lorentz-invariant distance. We may define γ = 11−(v/c) 2√moving at speed v. Time dilation:for a particleτ = γt. (312)Here τ is the proper time, which is the time measured by a clock in the particle’srest frame. Remember that “moving clocks appear to tick slowly.”Length contraction:L ′ = L γ . (313)Here, L is the proper length of the object, as measured in its rest frame. Rememberthat “moving rods appear shorter.” Note that length contraction occursonly along the direction of motion: perpendicular distances are unaltered.Lorentz transformation for motion at velocity v:t ′ = γ(t ± vx ), (314)c2 x ′ = γ(x ± vt). (315)Note that the ± is the same for both time and position. To figure out whichsign should be used, consider the Newtonian limit for position. This tells youthe sign for time.Einstein’s famous formula generalizes toE = γmc 2 = γE rest = √ p 2 c 2 + m 2 c 4 . (316)To lowest order in v, this gives the Newtonian limit,The four-vector for momentum:Here, p i = γmv i . The fource four-vector:E = mc 2 + 1 2 mv2 . (317)p ν = (E/c, p x , p y , p z ). (318)F ν = dp νdτ . (319)Simultaneity is relative to the observer. A and B are simultaneous events withrespect to observer O if light rays emitted from A and B will reach O at thesame time. If ∆x ′ is the distance between events in the simultaneous frame,then∆t = γ(v/c 2 )∆x ′ (320)39

gives the time difference between the events in a frame moving with velocity vin the x ′ direction relative to the simultaneous frame.Two parallel velocities v and u sum according tos = v ± u1 ± vu , (321)in units where c = 1. Note that this formula maintains the speed limit of 1 inthese units.The relativistic Doppler effect for motion parallel to the propagation of light:√1 ± v/cω obs = ωem1 ∓ v/c . (322)The sign can be determined in the usual way–if the observer and emitter areheading toward each other, there is a blueshift, and ω obs > ωem. If they aremoving apart, the light is redshifted, and ω obs < ωem.10 Miscellaneous (15%)This section involves both “laboratory methods” and “specialized topics” suchas nuclear and particle physics.One important rule of thumb: the weak force/interaction is just about alwaysthe culprit. This is because the weak force doesn’t conserve lepton flavor. Sowhenever the question asks, “Which force is responsible for this decay process?”you should probably answer “weak force.”Compton scattering is the process by which an incoming light ray is scatteredby an electron. The change in wavelength is,λ f − λ i = 2π (1 − cos θ), (323)m e cwhere θ = 0 corresponds to no change in direction of the light ray. The photoelectriceffect describes the release of an electron from a metal when struckby a photon. Einstein’s equation relating the maximum kinetic energy of thereleased electron Kmax to the stopping potential V 0 to the frequency ν of thephoton is,Kmax = eV 0 = hν − φ, (324)where φ is the work function. Some strange facts about the photoelectric effectthat helped lead to a better understanding of quantum mechanics were:1. Rate of ejection is proportional to intensity of light.2. There is a minimum frequency below which no photoelectron emissionoccurs.3. The maximum kinetic energy of the electron is independent of intensity.40

During solid formation, energy levels of outer shell electrons get split upinto many closely-packed energy levels and form bands. The current-carryingelectrons in the conduction band are called free electrons. In insulators, theconduction band and valence band are very far apart. In conductors, the twobands overlap, and electrons move freely in a solid like an ideal gas. In n-typesemiconductors, dopant atoms provide extra conduction electrons. In p-typesemiconductors, dopant atoms steal extra conduction electrons. The Fermi energyfor a p-type semiconductor is lower and closer to the valence energy band.The Fermi energy for an n-type semiconductor is higher and closer to the conductionenergy band. p-type and n-type semiconductors can be brought together,creating a depletion region. In p-type semiconductors, dopant atoms steal extraconduction electrons. Semiconductors involving dopants are called extrinsicsemiconductors, whereas intrinsic semiconductors are free from impurities.If there is no electric field applied, free electrons move in random directions.If an external E-field is applied, the average drift velocity is,v d = −eEtm , (325)where t is the average scattering time. The Wiedemann-Franz law relates thethermal conductivity κ, the electrical conductivity σ, the Lorenz number L, andthe temperature T ,κ= LT. (326)σThe Wiedemann-Franz law fails at low T .Superconductivity is the phenomenon of zero electrical resistance occuring insome materials below a characteristic energy. The Meissner effect is the completeexpulsion of a magnetic field from a superconductor during its transition to thesuperconducting state (χ M = −1). There are two types of superconductors,type I involve a first-order phase transition and type II involve a second-orderphase transition. All high temperature superconductors are type II, and areusually made of metal alloys or ceramics. Cooper pairs are formed when twoelectrons (or other fermions) bind together at low T to become bosons. A pairis bound if and only if the energy is lower than the Fermi energy. Cooper pairsare responsible for superconductivity. In conventional superconductors, the pairis due to electron-phonon interactions (interaction with ionic lattice).In quantum chromodynamics (QCD), momentum, energy, and charge areconserved. Lepton flavor is conserved, where the six flavors of leptons are tauon,electron, muon, and one each for their corresponding neutrinos. The numberof leptons and the number of baryons are conserved, and quarks have baryonicnumber 1/3. Antiparticles have equal magnitude but opposite sign of flavornumbers. The weak interaction does not conserve each of the six flavor numbers,but the number in each family (muon, tauon, electron) is conserved.When a charge accelerates, it radiates, with Poynting vector⃗S(R, θ) =q24πc 3 R 2 sin2 θ|¨⃗x|ˆn, (327)41

where θ is the angle with respect to the direction of propagation. The totalpower is,P = 2 q 2 |¨⃗x| 23 c 3 . (328)The Auger effect describes the process by which an electron from a higherenergy level drops into a lower energy level and kicks another electron out. Betadecay happens via the weak interaction,or,Electron capture:energy + p → n + e + + ν e , (329)n → p + e − + ¯ν e . (330)energy + p + e − → n + ν e . (331)Protons consists of two up quarks and one down quark, while neutrons consistof one up and two downs.If a U-tube is filled with water on the right and mercury on the left, thenthe height of the mercury L, the difference in heights between the left and rightarm d, the density of water ρ w , and the density of mercury ρ m are related by,ρ w=Lρ m L + d . (332)For cosmic microwave background radiation, distance is proportional to temperature,which is proportional to redshift.References[1] Wikipedia.[2] Assorted websites found by Google searches.[3] Tipler, Paul. Modern Physics.[4] Finn, C.B.P. Thermal Physics.[5] Griffiths, David. Introduction to Quantum Mechanics.42