Formula SheetConstantse = 1.6 × 10−19Coul , c = 2.998 × 108m·s−1,h = 6.626 × 10−34J·s = 4.136 × 10−15eV·s ,~ = 1.055 × 10−34J·s = 6.582 × 10−16eV·sLorentztransformationct0= γ (ct − βx) , x0= γ (−βct + x) , y0= y , z0= zβ = v/c γ =1p1 − β2Relativistic energyand momentum~p = γm~u E = γmc2=pm2c4+ p2c2Time dilation &length contractionτ = γτ0l = l0/γRelatvisitic Dopplereffectf = f0r1 + β1 − βRelavistic velocityadditionu0x=ux− v1 − uxv/c2u0y=1γuy− v1 − uxv/c2, u0z=1γuz− v1 − uxv/c2Photoelectric effects Kmax= eVS= hf − ΦCompton scattering λ0− λ =hmc(1 − cos θ)hmec= 0.0243˚ABragg diffraction nλ = 2d sin θBohr Atom rn= a0n2, En= −ke22a01n2a0=~2meke2= 0.529˚A ,ke22a0= 13.6eVde Broglie wave p =hλ= ~k , E = hf = ~ω ~ =h2π, λ =2πk, ω = 2πfUncertainty relation ∆px∆x ≥~2, ∆E∆t ≥~2Wave packet vphase=Ep, vgroup=dEdpWavefunction P = |Ψ|2,R∞−∞|Ψ(x)|2dx = 1, P (a < x < b) =Rba|Ψ(x)|2dxSchr¨odinger’sequation−~22m∂2∂x2Ψ (x, t) + V (x)Ψ (x, t) = i~∂∂tΨ (x, t), Ψ (x, t) = φ(x)θ(t)−~22md2dx2φ(x) + V (x)φ(x) = Eφ(x), i~ddtθ(t) = Eθ(t) → θ(t) = e−iE t/~Infinite square well kn=nπL, En=~2π22m L2n2S.H.O. V (x) =12mω0x2, ω0=pk/m, En= (n +12)~ω0Typical operators ˆx = x , ˆpx= −i~∂∂x,ˆH =ˆp22m+ V (ˆx) ,ˆE = i~∂∂tOperator exp. value hˆOi =R∞−∞Ψ∗(x, t)ˆOΨ(x, t)dx ∆Q =phQ2i − hQi2Barrier tunneling T (E) ≈ exph−2~√2mRforbidd.pU(x) − Edxi1Schr¨odinger’s equa-tion 3D−~22m∇2Ψ (~r) + U (~r) Ψ (~r) = EΨ (~r)Wavefunction in 3D 1 =RΨ∗ΨdV =R|Ψ (x, y, z) |2dx dy dz =R|Ψ (x, y, z) |2r2sin θ dr dθ dφExpectation value hˆOi =RΨ∗ˆOΨdV , e.g. hxi =RΨ∗(x, y, z) xΨ (x, y, z) dx dy dz3D Box Energy: Enx,ny,nz=~2π22m n2xL2x+n2yL2y+n2zL2z!Wavefunction: Ψnx,ny,nz(x, y, z) = Anx,ny,nzsinnxπ xLxsinnyπ yLysinnzπ zLz3D Quantum Oscil-latorU =12mω0r2=12mω0x2+ y2+ z2Enx,ny,nz=nx+ ny+ nz+32~ω0Central force U (~r) = U (r) → Ψ (r, θ, φ) = R (r) Θ (θ) Φ (φ)1 =R|Ψ (r, θ, φ) |2r2sin θ dr dθ dφ =R∞0|R(r)|2r2drRπ0|Θ(θ)|2sin θdθR2π0|Φ(φ)|2dφAngular part∂2Θ∂θ2+cos θsin θ∂Θ∂θ−m2sin2θΘ (θ) = l(l + 1)Θ (θ) ,∂2Φmdφ= −m2Φm(φ)Radial part∂∂ rr2∂∂rR (r)−2mr2~2[U(r) − E] R(r) = l(l + 1)R(r)Hydrogen atom U(r) = −ke2r, Rnl(r) = Ln−l−1(ra0)ra0le−r/(na0)Angular momentumoperatorsˆL2= −~2∂2∂θ2+cos θsin θ∂∂θ+1sin2θ∂2∂φ2ˆL2Yml(θ, φ) = l(l + 1)~2Yml(θ, φ)ˆLz= −i~∂∂φˆLze−imφ= m~e−imφ2Useful formulae:sin2x + cos2x = 1 ,2πZ0sin2u du = π ,πZ0u sin2u du =π24,2πZ0u sin2u du = π2,Zpa2− x2dx =x2pa2− x2+a22arctanx√a2− x2, arctan (±∞) = ±π/2 ,∞Z0rne−r/adr = n! an+1Quiz 41. Tunneling. A particle of mass m = 1 and energy E = −1 encounter an inverted pendulum-potential:U(x) = −x29(You should sketch the potential and identify the classically forbidden region first). Calculatethe approximate transmission probability for the particle to tunnel through this barrier. (4 pt.)2. Particle in a 3D box. Consider a three dimensional box of size Lx= L, Ly= 2L, andLz= 2L.(a) Find the five lowest allowed energy and identify their degeneracy. (2 pt.)(b) For the (nx, ny, nz) = (1, 2, 1) state, find the normalization constant A1,2,1(2 pt.)(c) For the same (1, 2, 1) state, find the average position of the particle: hxi, hyi, hzi. (2 pt.)3. Angular wavefunction. The angular part of a particular central-force potential wavefunc-tion is:Θl,m(θ) Φm(φ) =10532π1/2sin2θ cos θe+2iφWhat is the total angular momentum and the z-direction component of the angular momentumof this state? (3+2 pt.)4. Hydrogen radial wavefunction. The n = 2, l = 1 radial wavefunction for the hydrogenatom has the form:R21= A r exp (−r/(2a0))(a) Find the normalization constant A. (2 pt.)(b) What is the average distance of the electron from the nucleus? (3 pt.)3Quiz 4 solution1. The classical forbidden region is where the energy of the particle is less than the potentialenergy:E = −1 < −x29∴ −3 < xforbidd.< 3The approximate tunneling probability is:T (E) ≈ exp−2~√2mZforbidd.pU (x) − E dx= exp−1√2 ~3Z−3r−x29+ 1 dx= exp−13√2 ~3Z−3p−x2+ 32dxFrom this point, you can either use the provided formula with a = 3:3Z−3p−x2+ 32dx ="x√32− x22+322arctanx√32− x2#x=+3x=−3=π 322,or recognize that the integrand√32− x2is just the equation for the upper half of a circlewith radius 3, and the area under the curve is simply the area of a half circle:π322.Either way, the final result is:T ≈ exp−3π2√2 ~2. (a) This is exactly problem 1 of chapter 8. The energies of state (nx, ny, nz) is:Enx,ny,nz=~2π22m n2xL2+n2y4L2+n2z4L2!=~2π22 m L24n2x+ n2y+ n2zLet E0=~2π22mL2. The energies of the first few states are:nxnynzE0(4n2x+ n2y+ n2z)1 1 1 6E0X1 2 1 9E0X1 1 2 9E0X1 2 2 12E0X1 3 1 14E0X1 1 3 14E0XnxnynzE0(4n2x+ n2y+ n2z)1 2 3 17E0X1 3 2 17E0X1 4 1 21E01 1 4 21E02 1 1 18E02 2 1 21E02 1 2 21E0So the first five energies are E1= 6E0, E2= 9E0, E3= 12E0, E4= 14E0, and E5= 17E0with degeneracy d1= 1, d2= 2, d3= 1, d4= 2, and d5= 2.4(b) The wavefunction of the (1, 2, 1) has the form:Ψ121(x, y, z) = A121sinπxLsin2πy2Lsinπz2LThe normalization condition decides A121:1 =Z|Ψ121(x, y, z) |2dV = A2121LZ0sin2πxLdx2LZ0sin2πyLdy2LZ0sin2πz2LdzEach integral is very similar to each other:LZ0sin2πxLdx =LππZ0sin2(ux) dux=Lππ2=L2← ux=πxL2LZ0sin2πyLdy =Lπ2πZ0sin2(uy) duy=Lππ = L ← uy=πyL2LZ0sin2πz2Ldx =2LππZ0sin2(uz) duz=2Lππ2= L ← uz=πz2LTherefore:1 = A2121L32→ A121=r2L3(c) The expectation value of each of the three coordinates are:hxi =2L3LZ0x sin2πxLdx2LZ0sin2πyLdy2LZ0sin2πz2Ldz=2L3L2π2πZ0uxsin2(ux) dux· L · L =2Lπ2π24=L2hyi =2L3LZ0sin2πxLdx2LZ0y sin2πyLdy2LZ0sin2πz2Ldz=2L3L2·L2π22πZ0uysin2(uy) duy· L = Lhzi =2L3LZ0sin2πxLdx2LZ0sin2πyLdy2LZ0z sin2πz2Ldz=2L3·L2· L4L2π2πZ0uzsin2(uz) duz= LNotice that the average position is exactly at the middle (L2, L, L). This is true forparticle-in-box in any dimensionality, for all stationary