Lecture 1 – Fundamental Concepts in Quantum Physics



CHEM 3541 Physical ChemistryLecture NotesTextbooksAtkins’ Physical Chemistry, 7th ed., pp. 304-316Atkins’ Physical Chemistry, 10th ed., pp. 290, 292-305Table of Contents TOC \o "1-3" \h \z \u Lecture 1 – Fundamental Concepts in Quantum Physics PAGEREF _Toc24043308 \h 11.Schr?dinger Equation PAGEREF _Toc24043309 \h 12.Hermitian operator PAGEREF _Toc24043310 \h 33.Expectation value PAGEREF _Toc24043311 \h 54.Heisenberg’s uncertainty principle PAGEREF _Toc24043312 \h 5Lecture 2 – Translational Motion PAGEREF _Toc24043313 \h 81.One-dimensional particle-in-a-box model PAGEREF _Toc24043314 \h 82.Two-dimensional model PAGEREF _Toc24043315 \h 103.Three-dimensional model PAGEREF _Toc24043316 \h 114.Tunnelling PAGEREF _Toc24043317 \h 125.Particle in a finite square-well potential PAGEREF _Toc24043318 \h 13Lecture 3 – Vibrational Motion PAGEREF _Toc24043319 \h 151.One-dimensional harmonic oscillator PAGEREF _Toc24043320 \h 152.The vibration of a diatomic molecule PAGEREF _Toc24043321 \h 17Lecture 4 – Rotational Motion PAGEREF _Toc24043322 \h 191.Two-dimensional rotational motion PAGEREF _Toc24043323 \h patibility theorem PAGEREF _Toc24043324 \h 203.Three-dimensional rotational motion PAGEREF _Toc24043325 \h 20Lecture 5 – Hydrogen Atom PAGEREF _Toc24043326 \h 231.Separation of CM motion and relative motion PAGEREF _Toc24043327 \h 232.Separation of radial motion and rotational motion PAGEREF _Toc24043328 \h 24Lecture 1 – Fundamental Concepts in Quantum PhysicsSchr?dinger EquationTime-independent S.E.HΨ=EΨH: Hamiltonian operatorΨ: Wavefunction, Ψ?r1,r2,…,rnH=K+V: Kinetic energy operator and potential energy operator. In many situations, V operator is simply a function.Time-dependent S.E.If potential V and thus wavefunction depend on time explicitly, one need to solve time-dependent S.E.i??Ψr,t?t=HtΨr,tOne-dimensional system with one particleH=-?22m?d2dx2+Vxh is Planck constant, whereas ?=h2π is called reduced Planck constant. Vx operator is simply a function. The S.E. reads-?22m?d2dx2+VxΨx=EΨ(x)Wavefunction contains all the information of this system. The simplest one is probability density Ψx2, which tells that the probability to find the particle between x and x+dx is Ψx2dx.Since total probability to find the particle between -∞ and +∞ should be 1, we require the wavefunction to be normalized:-∞+∞Ψx2dx=1For an unnormalized wavefunction, it can be normalized asΨx=Ψ'x-∞+∞Ψ'x2dxExample 1: constant potentialFor Vx=const., the S.E. isd2Ψdx2=-2m?2E-VΨIf E>V, the general solution is Ψx=Aeikx+Be-ikx, where A,B∈C. The wave vector is defined as k=2mE-V/?.Kinetic energy K=E-V=?2k22m.Momentum p=?k=hλ. λ is de Broglie wavelength. To see the meaning of λ, let Ψ=eikx+e-ikx=2coskx, then the period of Ψ is 2πk=hp=λ.Example 2: normalization of wavefunctionQ. Given a wave function ψr∝e-r/a0 where a0 is a given constant, try to normalize it. Noted that r in the exponent is radial distance but not position vector.A. e-r/a02dr=4π0∞r2e-2r/a0dr=4πIk, where k=2/a0.Ik=0∞r2e-krdr=d2dk20∞e-krdr=d2dk21k=2k3Thus ψr=πa03-1/2e-r/a0.Three-dimensional system with one particleCartesian coordinatesWavefunction is Ψx,y,z. The probability to find the particle in x,x+dx∩y,y+dy∩z,z+dz is Ψx,y,z2dxdydz.H=-?22m?2+V(x,y,z), where ?2=?2?x2+?2?y2+?2?z2 is Laplacian operator. S.E. is -?22m?2+Vx,y,zΨx,y,z=EΨx,y,z. Wavefunction and energy are unknown variables.2362954203005Relationship to Cartesian coordinatesx&=rsinθcos?y&=rsinθsin?z&=rcosθr: radial distanceθ: polar angle?: azimuth angle00Relationship to Cartesian coordinatesx&=rsinθcos?y&=rsinθsin?z&=rcosθr: radial distanceθ: polar angle?: azimuth angleSpherical coordinates Wavefunction is Ψr,θ,?. The probability to find the particle in r,r+dr∩θ,θ+dθ∩?,?+d? is Ψr,θ,?2r2sinθdrdθd?.Laplacian operator is more complex in spherical coordinates?2=?2?r2+2r??r+Λ2r2where Λ2=?2?θ2+cotθ??θ+csc2θ?2??2.S.E. is -?22m?2+Vr,θ,?Ψr,θ,?=EΨr,θ,?Eigenequation, eigenfunction and eigenvalueThe solution of S.E. HΨ=EΨ is a set of eigenenergy/value Ei and eigenfunction Ψi, i=1,2,…,∞.For any operator Ω, if there exist some values satisfy Ωψ=Ωψ, ψ is then called eigen function of operator Ω.For example, since Ωψ=-αψ, ψ=e-αx is an eigenfunction of operator Ω=ddx with corresponding eigen value -α. But ψ=e-αx2 is not an eigenfunction of Ω.For one-dimensional system with one particle, we can choose two wavefunctions with different eigenvalues E1 and E2,Ψ1=eik1x, k1=2mE1-V?Ψ2=eik2x, k2=2mE2-V?Generally, Ψ'=AΨ1+BΨ2,?A,B≠0 is NOT a solution of S.E. HΨ=EΨ. Only when E1=E2=E, HΨ'=AE1Ψ1+BE2Ψ2=EAΨ1+BΨ2=EΨ'.Hermitian operatorFor any two functions, if the following equation holds, then Ω is called Hermitian operator.dτψj*Ωψi=dτψi*Ωψj*For example, ddx is not a Hermitian operator since-∞+∞dxψj*dψidx&=ψj*ψi-∞+∞--∞+∞dxψidψj*dx&=--∞+∞dxψi*dψjdx*&≠-∞+∞dxψi*dψjdx*But 1iddx is a Hermitian operator-∞+∞dxψj*1idψidx&=-1i-∞+∞dxψi*dψjdx*&=-∞+∞dxψi*1idψjdx*Eigenvalues of Hermitian operatorTheorem: Any eigenvalue of Hermitian operator is a real number.Proof:Given a Hermitian operator Ω with eigenvalue ωi and normalized eigenfunction ψi, i.e., Ωψi=ωiψi. Applying the definition of Hermitian operator, we haveωi&=dτψi*Ωψi&=dτψi*Ωψi*&=ωi*Thus ωi must be real.Q.E.D.All operators corresponding to physical observables (properties that can be measured) are Hermitian operators.Eigenfunctions of Hermitian operatorTheorem: Eigenfunctions of Hermitian operator with different eigenvalues are orthogonal.Proof:Using the definition of Hermitian operator, we havedτψi*Ωψj&=dτψj*Ωψi*&=ωi?τψj*ψi*&=ωidτψjψi*Also,dτψi*Ωψj=ωj?τψi*ψjSubtracting the first equation from the second one gives0=ωj-ωi?τψi*ψjFor ωj≠ωi, we have?τψi*ψj=0Q.E.D.Example1: one-dimensional momentum operatorFor particle in constant potential, take the wavefunction as ψk=eikx. Define the momentum operator aspx=?iddxwhich has been proved to be Hermitian. Apply it to above wavefunction,pxψk=?iddxψk=?kψkBy definition, p=?k, thus pxψk=pψk, and px is indeed the momentum operator.Example2: three-dimensional momentum operatorMomentum operator p=?i?=?iex??x+ey??y+ez??z is Hermitian since all components of it are Hermitian.Expectation valueFor a Hamiltonian with normalized wavefunctions Hψi=Eiψi, i=1,2,…, we construct a wavefunction ψ=c1ψ1+c2ψ2 where c12+c22=1. The expectation value, or the average value of energy is thenE&=dτψ*Hψ&=dτc1*ψ1*+c2*ψ2*c1E1ψ1+c2E2ψ2&=c12E1dτψ1*ψ1+c1*c2E2dτψ1*ψ2+&?????c2*c1E1dτψ2*ψ1+c22E2dτψ2*ψ2&=c12E1+c22E2Generally, for any operator Ω in a quantum state ψ, its expectation value isΩ=dτψ*ΩψHeisenberg’s uncertainty principleSuppose we have measured an observable ω N times (N?1), then the uncertainty of ω is defined as?ω=i=1Nωi-ω2NFor position and momentum, their uncertainties satisfy the following Heisenberg’s uncertainty principle?x??px≥?2Apply pxx to an arbitrary wavefunction ?x givespxx?=?iddxx?=?i?+x?id?dx=?i?+xpx?Thus, xpx-pxx?=i??. Since ?x is arbitrary, we havexpx-pxx=i?Define commutator of two operators A and B as A,B=AB-BA, the above equation can be rewritten asx,px=i?If ?→0, quantum mechanics will degenerate to classical mechanics.Heisenberg’s uncertainty principle: If the commutator of two operators A and B is A,B=iC, then their uncertainties satisfy the following inequality?A??B?≥12CProof:Given an arbitrary wave function ?, constructing a non-negative integral with real variable ξIξ=dτξA?+iB?2Noted that for a complex number ?, c2=c*c. Expanding above integral, we haveIξ&=ξ2dτA?2+dτB?2+iξdτA?*B?-iξdτB?*A?&=ξ2dτ?*A2?+dτ?*B2?+iξdτ?*A,B?&=ξ2dτ?*A2?-ξdτ?*C?+dτ?*B2?&≥0To ensure above quadratic function of ξ is non-negative, there must be4A2B2-C2≥0i.e.A2B2≥12CDefine ?A=A-A, ?B=B-B, apparently,?A,?B&=A-AB-A-AB-BA-A+BA-A&=AB-AB-BA+BA&=AB-BA&=A,BThus?A,?B=iCSubstitute A, B with ?A and ?B respectively in A2B2≥12C,?A2?B2≥12CNoted that ?A=?A2 and ?B=?B2, finally we have?A??B?≥12CQ.E.D.ExamplesA=x, B=px. x,px=i?, C=?. ?x??px≥?/2.ψk=eikx has definite momentum px=?k, so that ?px=0. ?x must be +∞, which means the particle is diffused in the whole x space.If ?x=0, we have ?px→+∞, thus H=px22m+V=px22m+V→+∞. This means we need infinite energy to constrain one quantal particle to a certain position.If two operators commute, i.e. AB=BA, C=0, we have ?A??B≥0. So A and B can be measured precisely at the same time.x,py=0A particle with mass m moves along x direction subjected to a potential Vx. H=-?22md2dx2+Vx, px=?iddx. H,pxψx=i?ψxdVxdx, i.e. H,px=i?dVxdxIf and only if Vx is equal to some constant will H and px commute, thus energy and x momentum can be measured precisely at the same time.From time-dependent S.E. i???tψ=Hψ, we define H=i???t. Since H,t?=Ht?-tH?=i??, ?E??t≥?/2.If a quantum state has definite energy, i.e. ?E=0, then the lifetime of this state will be ?t→∞.In reality, energy level is broadened, and ?t~?/?E is regarded as lifetime of the energy level.Lecture 2 – Translational MotionOne-dimensional particle-in-a-box modelSuppose we have one particle with mass m confined in a box 0,L, then its Hamiltonian isH=-?22md2dx2+Vxwhere the potential isVx=0,&?0<x<L+∞,&?x≤0?or?x≥LWavefunctions and energy levelsWithin 0,L, this S.E. has solutionψx=Aeikx+Be-ikx,?k=2mE/?Outside 0,Lψ=0Since wavefunction should be continuous, we impose following boundary conditionsψ0=ψL=0Plug x=0 and x=L into ψx=Aeikx+Be-ikx, we haveψ0=A+B=0?A=-BψL=-BeikL+Be-ikL=-2iBsinkL=0?kL=nπ,?n=1,2,…So that after normalization, within 0,L,ψnx=2LsinnπLx,?n=1,2,…Energy levels corresponding to each ψnx isEn=?2k22m=n2π2?22mL2, n=1,2,…Figure 1 First five normalized wavefunctionsOrthogonality of wavefunctionsFor n≠m,0Ldxψn*ψm&=2L0LdxsinnπxLsinmπxL&=-1L0Ldxcosn+mπxL-cosn-mπxL&=-1LLn+mπ?sinn+mπxL0L-Ln-mπ?sinn-mπxL0L&=0Uncertainty principle for ground stateNoted that ?A=?A2 and ?A=A-A, we can rewrite uncertainty as?A=A2-2AA+A2=A2-2AA+A2=A2-A2A2 and A are needed to calculate ?A.x&=2L0Lxsin2πxLdx=1L0Lx1-cos2πxLdx=L2-1L0Lxcos2πxLdx&=L2p=2?iL?πL0LsinπxLcosπxLdx=?πiL20Lsin2πxLdx=L2x2&=2L0Lx2sin2πxLdx=1L0Lx21-cos2πxLdx=L23-1L0Lx2cos2πxLdx&=L23-L22π2p2=2?2L?π2L20Lsin2πxLdx=π2?2L30L1-cos2πxLdx=π2?2L2?x=L23-L22π2-L22=L2ππ23-2?p=π?LThus?x??p=?2?π23-2≈1.136??2>?2Calculation detailsLet αk=0Lsinkxdx=1-coskLkdαkdk=0Lxcoskxdx=LsinkLk+coskL-1k2So0Lxcos2πxLdx=dαkdkk=2π/L=0Let βk=0Lcoskxdx=sinkLkd2βkdk2=-0Lx2coskxdx=-L2sinkLk-2LcoskLk2+2sinkLk3So0Lx2cos2πxLdx=-d2βkdk2k=2π/L=L32π2Example 8A.2 (pp. 321)Problem: β-Carotene is a linear polyene in which 10 single and 11 double bonds alternate along a chain of 22 carbon atoms. If we take each C-C bond length to be about 140 pm, then the length L of the molecular box in β-carotene is L = 2.94 nm. Estimate the wavelength of the light absorbed by this molecule from its ground state to the next higher excited state.Answer:?E=E12-E11=1.60?×?10-19?Jλ=h?E=1.24?μmTwo-dimensional model36766559632my=L2y=0x=0x=L100my=L2y=0x=0x=L1Hamiltonian operatorH=-?22m(?2?x2+?2?y2)+V(x,y)Vx,y=0,&?0<x<L1?and?0<y<L2+∞,&?otherwiseS.E. is Hψx,y=Eψx,y.Within 0<x<L1 and 0<y<L2, S.E. is-?22m?2ψx,y?x2+?2ψx,y?y2=Eψx,yBoundary conditions areψ0,y=ψL1,y=ψx,0=ψx,L2=0To solve this multivariable equation, we perform separation of variables. Let ψx,y=XxYy and plug this equation into S.E., we get-?22md2Xdx2Y+Xd2Ydy2=EXYDivide both sides by XY-?22m1Xd2Xdx2+1Yd2Ydy2=ETo ensure -?22m1Xd2Xdx2+1Yd2Ydy2=E, each term in the LHS should be some constant, viz.-?22m1Xd2Xdx2=E1, -?22m1Yd2Ydy2=E2with E1+E2=E.Now the 2D S.E. of ψx,y has been decomposed into two 1D S.E., thus its solution is just the product of two separated equations,ψn1,n2=2L1L2sinn1πxL1sinn2πyL2&,?within?2D?box0&,?outside?boxIts energy levels areEn1,n2=n12π2?22mL12+n22π2?22mL22where n1,?n2=1,2,3,….Three-dimensional modelHamiltonian:H=-?22m(?2?x2+?2?y2+?2?z2)+V(x,y,z)Vx,y,z=0,&?0<x<L1?and?0<y<L2?and?0<z<L3+∞,&?otherwiseS.E.: Hψx,y,z=Eψx,y,z, within 0<x<L1, 0<y<L2, and 0<z<L3-?22m?2ψ?x2+?2ψ?y2+?2ψ?z2=EψBoundary conditionsψ0,y,z=ψL1,y,z=0ψx,0,z=ψx,L2,z=0ψx,y,0=ψx,y,L3=0Now we perform similar procedures as 2D model. Let ψx,y,z=XxYyZ(z) and plug this equation into S.E., we get-?22md2Xdx2YZ+d2Ydy2XZ+d2ZdZ2XY=EXYZDivide above equation by XYZ-?22m1Xd2Xdx2+1Yd2Ydy2+1Zd2Zdz2=EEach term in the LHS should be some constant, viz.-?22m1Xd2Xdx2=E1, -?22m1Yd2Ydy2=E2, -?22m1Zd2Zdz2=E3where E1+E2+E3=E. So, the solution isψn1,n2,n3=8L1L2L3sinn1πxL1sinn2πyL2sinn3πzL3&,?within?3D?box0&,?outside?box?with energy levelsEn1,n2,n3=n12π2?22mL12+n22π2?22mL22+n32π2?22mL32where n1,?n2, n3=1,2,3,….Tunnelling33250922514VAA’AA’VAA’AA’In this quantum tunnelling model, potential is set to be zero in x<0 or x>L and be constant V in 0≤x≤L. The energy of incident wavefunction is E and E<V. Denote the amplitude of incident and exit wave functions as A and A' respectively and define transmission coefficient as T=A'A.In zero-potential region, S.E. and wavefunction are-?22m?d2ψdx2=Eψ,?ψ=e±ikx,?k=2mE/?For incident region x<0, we choose ψ1x=Aeikx+Be-ikx, where two terms stand for incident and reflection wavefunctions respectively. For x>L, ψ3x=A'eikx stands for tunnelling wavefunction.In potential barrier region, -?22m?d2ψdx2+Vψ=Eψ. The general solution isψ2x=Ceκx+De-κxwhere κ=2mV-E/?.At the two interfaces, wavefunction shall be smooth, i.e.ψ10=ψ20,?ψ2L=ψ3L, ψ1'0=ψ2'0,?ψ2'L=ψ3'Lthus we can obtain four equationsA+B=C+D#1ikA-ikB=κC-κD#2CeκL+De-κL=A'eikL#3κCeκL-κDe-κL=ikA'eikL#4C, D and A can all be expressed in A' asκ3+4: C=κ+ik2κeik-κLA', κ3-4: D=κ-ik2κeik+κLA'ik1+2: A=ik+κC+ik-κD2ik=A'4iκkκ+ik2eik-κL-κ-ik2eik+κLThen we try to solve transmission coefficient.AA'eikL&=14iκkκ+ik2e-κL-κ-ik2eκL&=14iκkκ2-k2e-κL-eκL+2iκke-κL+eκL&=12e-κL+eκL-iV-2E4EV-E?e-κL-eκLdenote ?=EV,AA'2&=14e-κL+eκL2+1161-4?1-??1-?e-κL-eκL2&=14e-κL+eκL2-14e-κL-eκL2+e-κL-eκL216?1-?&=1+e-κL-eκL216?1-?Finally,T=A'A=1+e-κL-eκL216?1-?-1/2If ??1, i.e. E?V: T≈4?1-?eκL-e-κL≈0.If κL?1, i.e. high, wide barrier: T≈4?1-?e-κL.The heavier the mass, the smaller the T.22860022675300Particle in a finite square-well potentialPotential is constant V in x<0 or x>L and zero in 0≤x≤L. The energy of particle is E and E<V. Denote k=2mE/? and κ=2mV-E/?.Similar to above tunnelling model, we use following wavefunctions x<0: ψ1x=Ceκx+C'e-κx0<x<L: ψ2x=Aeikx+Be-ikx x>L: ψ3x=De-κx-L+D'eκx-LAt infinity, wavefunction should vanish, thus C'=D'=0. At x=0, applying boundary conditions ψ10=ψ20 and ψ1'0=ψ2'0, we get C=A+B and κC=ik(A-B), i.e.A=ik+κ2ikC, B=ik-κ2ikCSimilarly, at x=L we have AeikL+Be-ikL=D and ikAeikL-Be-ikL=-κD, i.e.A=ik-κ2ike-ikLD, B=ik+κ2ikeikLDThen C can be expressed in terms of D as C=ik-κik+κe-ikLD or C=ik+κik-κeikLD. Use the first expression, we haveψx=D?&ik-κik+κe-ikLeκx&,x≤0&ik-κ2ike-ikLeikx+ik+κ2ikeikLe-ikx& , 0<x<L&e-κx-L&,?x≥LAt x=0, ψx should be continuousik-κik+κe-ikL=ik-κ2ike-ikL+ik+κ2ikeikLi.e.κ2-k2-2iκkcoskL-isinkL=?(κ2-k2+2iκk)coskL+isinkLReal parts of LHS and RHS are identical, and the imaginary parts should be equal-2κkcoskL+κ2-k2sinkL=2κkcoskL+κ2-k2sinkLi.e.4κkcoskL=2k2-κ2sinkLWhen coskL≠0 , we have tankL=2κkk2-κ2, i.e. tan2mEL?=2EV-E2E-V.When coskL=0 and k2=κ2, we have E=V2 and E=n+122π2?22mL2 where n=0,1,2,…. For a given V, if there is no such E satisfy these two equations, then this state is quantum forbidden.Lecture 3 – Vibrational Motion20568224447500One-dimensional harmonic oscillatorThe force of spring is F=-kfx, thus its potential can be calculated as Vx=-0xFdx'=12kfx. The Hamiltonian is then H=-?22md2dx2+12kfx2, and its corresponding S.E. is-?22md2ψxdx2+12kfx2ψx=EψxWe first change variable from x to y=x/α where α=?2mkf14. Under this operation, ψ(x) changes to ?y. After some algebra, we haved2?ydy2+λ-y2?y=0where λ=2E?ω and ω=kf/m.Now we take a look at asymptotic behaviour of above equation. When y→±∞,d2?dy2-y2?=0thus ?→e-y22. Rewrite ?y=N?Hye-y22, we have following Hermite equationd2Hdy2-2ydHdy+λ-1H=0Expand Hy as Hy=n=0∞cnyn and plug this expansion into Hermite equation,n=0∞yncn+2n+2n+1-2n-λ+1cn=0cn+2=2n+1-λn+2n+1cnBut when y→±∞, n=0∞cnyn→∞ as fast as ey2, causing ?~e-y22n=0∞cnyn→∞ as fast as ey22. To ensure ?±∞=0 the expansion of Hy must be truncated, i.e.cν≠0,?cν+2=0Thus we get quantized energy as2ν+1=λ=2E?ωEν=?ων+12,?ν=0,1,2,…It is worthwhile noting that the ground state energy, also called zero-point energy, E0=12?ω is non-zero.WavefunctionSolutions of Hermite equation are called Hermite polynomials. They satisfy following recursive and orthonormal relationsHν+1-2yHν+2νHν-1=0-∞+∞dyHν'Hνe-y2=0???????????&?,if?ν'≠νπ2νν!&?,?if?ν'=νFirst three Hermite polynomials are listed below.ν012Hν12y4y2-2Now we need to normalize wavefunction ?y=N?Hye-y22.-∞+∞dxψν*xψνx=Nν2α-∞+∞dyHν2e-y2=Nν2απ2νν!Thus Nν=απ2νν!-12 and the normalized result isψνx=απ2νν!-12Hνxαe-x22α2Uncertainty of positionx&=-∞+∞dxψν*xxψνx&=Nν2α2-∞+∞dyHνyyHνye-y2&=Nν2α2-∞+∞dyHνHν+1+2νHν-12e-y2&=0x2&=-∞+∞dxψν*xx2ψνx&=Nν2α3-∞+∞dyyHνyyHνye-y2&=Nν2α3-∞+∞dyHν+1+2νHν-122e-y2&=14Nν2α3-∞+∞dyHν+12+4νHν+1Hν-1+4ν2Hν-1e-y2&=α2π2ν+2ν!π2ν+1?ν+1!+νπ2ν+1?ν!&=α2ν+12Thus ?x=αν+12.Potential energyV=12kfx2=12kfα2ν+12=12?ων+12=12EνTunnelling – classically forbidden regionClassically forbidden region is where Vx>Eν. Quantal oscillator can reach classically forbidden region with some tunnelling probability.For ground state, ψ0=N0e-x22α2, E0=12?ω. Denote xL and xR as the negative and positive solutions of Vx=E0 respectively. The tunnelling probability is thenPx<xL+Px>xR=2xR+∞dxψ02x≈15.7%The vibration of a diatomic molecule26289016510m1kfx1x2m2m1kfx1x2m2Consider a diatomic molecule moving along x axis. The coordinates of two atoms with mass m1 and m2 are x1 and x2 respectively. The force constant is kf.H=-?22m1?2?x12-?22m2?2?x22+12kfx1-x22Classically, the kinetic energy of this system isK=12m1x12+12m2x22where xi=dxi/dt. Define the coordinate of centre-of-mass asxc=m1x1+m2x2m1+m2and the distance between two atoms asx=x1-x2It is easy to find the inverse transformations asx1=xc+m2m1+m2xx2=xc-m1m1+m2xThus the kinetic energy can be expressed in x,xc asK=12m1xc+m2m1+m2x2+12m2xc-m1m1+m2x2=12Mxc2+12μx2Here M=m1+m2 is the total mass and μ=m1m2/M is the reduced mass.Define the momentum of centre-of-mass and vibration asPc=Mxc, p=μxThe kinetic energy is thenK=Pc22M+p22μIts corresponding quantum Hamiltonian isH=-?22M?2?xc2-?22μ?2?x2+12kfx2The Schr?dinger equation Hψx1,x2=Eψx1,x2 is now separable. Assuming thatψx1,x2=Φxcφxwe have-?22Md2Φxcdxc2?φx+-?22μd2φxdx2+12kfx2φxΦxc=EΦxcφxDivide both sides by Φxcφx,-?22M1Φxcd2Φxcdxc2+-?22μ1φxd2φxdx2+12kfx2=EThus-?22Md2Φxcdxc2=E1Φxc-?22μd2dx2+12kfx2φx=E2φxE1+E2=EThe solution isΦxc=Aeikxc+Be-ikxc, k=2mE1?φx=απ2νν!-12Hνxαe-x22α2, α=?2μkf14E2=?ων+12, ω=kfμ,?ν=0,1,2,…Appendix: Direct coordinate transformations of HamiltonianIt’s straightforward to find that??x1=??x+m1M??xc, ??x2=-??x+m2M??xcThus?2?x12=?2?x2+2m1M?2?x?xc+m12M2?2?xc2?2?x22=?2?x2-2m2M?2?x?xc+m22M2?2?xc2Finally,H&=-?22m1?2?x12-?22m2?2?x22+12kfx1-x22&=-?221M?2?xc2+1m1+1m2?2?x2+12kfx2&=-?22M?2?xc2-?22μ?2?x2+12kfx2Lecture 4 – Rotational MotionTwo-dimensional rotational motion35329167547rm?00rm?A particle of mass m moves in a ring of radius r in the xy-plane with zero potential. We use cylindrical coordinates for convenience. Laplace operator has following form?2=1r??rr??r+1r2?2??2+?2?z2Since r,?z are fixed, it can be simplified to ?2=1r2d2d?2.S.E. is -?22mr2d2d?2ψ=Eψ. Denote I=mr2 as moment of inertia, we have-?22Id2d?2ψ?=Eψ?The solution is ψ?=Ne±i?? where ?=2IE?2. We then apply cyclic boundary condition ψ0=ψ2π, i.e. N=Ne±i2π?. e±i2π?=1 leads to ?=0,1,2,…, thusψ?=Neim?, m=0,±1,±2,?…After normalization, we have N=12π andψm?=12πeim?, m=0,±1,±2,?…EnergyHψm=?22Im2ψm, Em=m2?22I, m=0,±1,±2,?…Ground state energy is zero.First excited states are degenerated. E±1=?22I with degeneracy 2.Linear momentump=?i?=?iddr?=?irdd?pψm=m?rψm, pm=m?rThus ψm are also eigenfunctions of p.Angular momentumClassically, l=r×p. In quantum mechanics, for a particle in a circlel=rp=?idd?lψm=m?ψm, lm=m?Thus ψm are also eigenfunctions of l.These co-eigenfunction phenomenon are described by compatibility theorem next patibility theoremTheorem: Giving two Hermitian operators A and B, if A and B are commuting, viz A,B=0, we can conclude that A and B have a common eigen basis, i.e. we can find a set of ψi satisfying Aψi=aiψi and Bψi=biψiExamples: H,p=H,l=l,p=0, so ψm=12πeim? are their common eigenfunctions.Three-dimensional rotational motionA particle of mass m moves on the surface of a sphere of radius r with zero potential. Now we use spherical coordinates where?2=1r2??rr2??r+Λ2r2Λ2=1sin2θ?2??2+1sinθ??θsinθ??θSince r is fixed, Laplace operator is simplified to ?2=Λ2r2. The S.E. isHYθ,?=EYθ,?i.e. Λ2Y=-?Y with the same definition for ? as in 2D motion.Let Yθ,?=ΘθΦ?, we haveΘsin2θd2Φd?2+ΦsinθddθsinθdΘdθ=-?ΘΦDividing two sides by ΘΦ and rearranging the equation,1Φd2Φd?2+sinθΘddθsinθdΘdθ+?sin2θ=0Thus,1Φd2Φd?2=-βandsinθΘddθsinθdΘdθ+?sin2θ=βshould hold, where β is a constant.For 1Φd2Φd?2=-β, the solution isΦml=12πeiml?, ml=0,±1,±2,…For sinθΘddθsinθdΘdθ+?sin2θ=ml2, let u=cosθ, this equation can be rewritten as associated Legendre equation1-u2d2Θdu2-2udΘdu+?-ml21-u2Θ=0Its solutions Θlmlθ are called associated Legendre functions where l=0,1,2,… and ml=0,±1,±2,…,±l.The overall solutions Ylmlθ,?=ΘlmlθΦml? are called spherical harmonics which satisfy following equationΛ2Ylm=-ll+1YlmHereafter we will drop out subscript l from ml for simplicity. First few of them are listed below.l=0m=0: 1/4πl=1m=0: 3/4πcosθm=±1: ?3/8πsinθe±i?l=2m=0: 5/16π(3cos2θ-1)m=±1: ?15/8πcosθsinθe±i?m=±2: 15/32πsin2θe±2i?Energy and square of angular momentumThe Hamiltonian H=-?22m?Λ2r2,HYlm=?22I?ll+1Ylmgives out energy levelsEl=ll+1?22IFor square of angular momentum, L2, classically we have E=L2/2I and quantum mechanically L2=-?2Λ2.L2Ylm=ll+1?2YlmthusL2l=ll+1?2Angular momentumClassically, angular momentum is defined as L=r×p. In quantum mechanics, we change it into L=r×p, where r=xex+yey+zez, p=pxex+pyey+pzez and pμ=?i??μ for μ=x,y,z. Here ‘×’ means cross product. Some basic properties of cross product are shown belowa×kb=ka×b, a×a=0, a×b=-b×a, a×b+c=a×b+a×cAnd for bases, their cross products areex×ey=ez, ey×ez=ex, ez×ex=eyUsing these relations, we haveL&=xex+yey+zez×?i??xex+??yey+??zez&=?ix??y-y??xez+y??z-z??yex+z??x-x??zeyThe components of L are thenLx=?iy??z-z??y, Ly=?iz??x-x??z, Lz=?ix??y-y??xrespectively.We then calculate commutators between these operators since the commutation relation is a key feature of angular momentum. Take Lx and Ly as example,Lx,Ly&=-??2y??z-z??yz??x-x??z-z??x-x??zy??z-z??y&=-?2y??x+yz?2?z?x-yx?2?z2-z2?2?y?x+zx?2?y?z&?????+?2zy?2?x?z-z2?2?x?y-xy?2?z2+x??y+xz?2?z?y&=?2x??y-y??x&=i?LzSimilarly, we have Ly,Lz=i?Lx, Lz,Lx=i?Ly. For L2,Lz, we first decompose it into Lx2,Lz+Ly2,Lz+Lz2,Lz. Then it is straightforward to writeLz2,Lz=Lz3-Lz3=0Lx2,Lz&=Lx2Lz-LxLzLx+LxLzLx-LzLx2&=LxLx,Lz+Lx,LzLx&=-i?LxLy+LyLxLy2,Lz=i?LyLx+LxLyThus L2,Lz=0, and H,Lz=L22I,Lz=0. H, L2 and Lz are mutual commuting, which confirms that Ylm are the common eigenfunctions for them, i.e.HYlm=ll+1?22IYlmL2Ylm=ll+1?2YlmLzYlm=m?YlmLecture 5 – Hydrogen Atom-16510131533electronOpositronrerNr=re-rNmemNelectronOpositronrerNr=re-rNmemNIn this last lecture, we will try to solve a real system – hydrogen atom. The system is composed of two particles – one electron and one positron, and thus its total degree of freedom (DoF) is 6. Three of DoFs belong to translational motion, two of them rotational motion, and the last one is the relative radial motion.To begin with, we define following notations.Mass: nucleus mN, electron me, total mCM=mN+me, reduced μ=memNme+mNPosition vector: nucleus rN, electron re, centre of mass (CM) R=mere+mNrNme+mN, electron relative to nucleus r=re-rNR=Rxex+Ryey+Rzez, ?R2=?2?Rx2+?2?Ry2+?2?Rz2 (similar for ?r2)Classical momentum: nucleus pN=mNrN, electron pe=mere, CM pCM=mCMR, electron relative to nucleus pμ=μrVector without arrow means modulus r=r, etcThen we can separate out CM motion and relative motion in classical energy expression asE=pe22me+pN22mN-e2r=pCM22mCM+pμ22μ-e2rThus quantum Hamiltonian can also be written as sum of two partsH=-?22mCM?R2-?22μ?r2-e2rS.E.-?22mCM?R2-?22μ?r2-e2rΦr,R=EΦr,RSeparation of CM motion and relative motionLet Φr,R=χRψr,-?22mCM?R2χψ-?22μχ?r2ψ-e2rχψ=EχψDivide both sides by χψ:-?22mCM?R2χχ-?22μ?r2ψψ-e2r=EThus-?22mCM?R2χχ=ECM and -?22μ?r2ψψ-e2r=Eewhere ECM+Ee=E.Since mN?me, mCM≈me. Roughly, χR and ψr are nuclear and electron wavefunctions respectively.CM motion χR-?22mCM?R2χ=ECMχThis is just a free particle moving in 3D space. Its solution is plane waveχR=AeikCM?RIts wave vector kCM has modulus 2mCMECM? and is parallel to υp.Relative motion ψ-?22μ?r2ψ-e2rψ=Eeψ?r2 can be expressed in spherical coordinate system located at nucleus?r2=?2?r2+2r??r+Λ2r2From now on, we will omit the subscript of Ee for simplicity.Separation of radial motion and rotational motionLet ψr=RrYθ,?,-?22μr2d2Rdr2+2rdRdrY-e2rRY-Er2RY-?22μRΛ2Y=0Divide both sides by RY,-?2r22μRd2Rdr2+2rdRdr-e2r-Er2-?22μYΛ2Y=0Thus we have-?22μYΛ2Y=A-?2r22μRd2Rdr2+2rdRdr-e2r-Er2=-ARotational motion Yθ,?Rearrange -?22μYΛ2Y=A as Λ2Y=-2μA?2Y. The solution is apparently spherical harmonic functions, Y=Ylmθ,? with eigenvalues -2μA?2=-ll+1. Thus,A=ll+1?22μRadial motion Rr-?2r22μRd2Rdr2+2rdRdr-e2r-Er2=-ll+1?22μi.e.-?22μd2dr2+2rddr+ll+1?22μr2-e2rR=ERRemember L2=ll+1?2, here ll+1?2/2μr2 can be regarded as effective potential due to angular momentum.Solution for above equation isRnlr=NnlρlLn-l-12l+1ρe-ρ/2ρ=2r/na, a=?2/μe2 ≈0.529??.a0=?2/mee2, called Bohr radius, is the unit length in atomic unit.Lba: associated Laguerre polynomial.Normalization factor Nnl=2na3n-l-1!2nn+l!1/2.n=1,2,3,…; l=0,1,2,…,n-1.Energy En=-1n2e22a (or in SI -1n2e28πε0a).Overall, electronic wavefunction isψnlmr=RnlrYlmθ,?It depends on three quantum numbers,Principal quantum number: n=1,2,3,….Azimuthal quantum number: l=0,1,2,…,n-1.Magnetic quantum number: m=0,±1,±2,…,±l.But its energy levels En=-1n2e22a only depend on principal quantum number n. For ground state, n=1,l=0,m=0, it is non-degenerate. For first excited state it is 4-fold degenerated.n=2,l&=0,?m=0l&=1,m&=0m&=±1First few electronic wavefunctions are listed belowψ1s&=ψ100=1πa3e-raψ2s&=ψ200=18πa31-r2ae-r2aψ2pz&=ψ210=142πa5re-r2acosθψ2px&=ψ211+ψ21-12=142πa5re-r2asinθcos?ψ2py&=ψ211-ψ21-1i2=142πa5re-r2asinθsin? ................
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