Quantum Mechanics

Quantum MechanicsLuis A. AnchordoquiDepartment of Physics and AstronomyLehman College, City University of New YorkLesson IXApril 9, 2019L. A. Anchordoqui (CUNY)Modern Physics4-9-20191 / 54

Table of Contents1Particle in a central potentialGeneralities of angular momentum operatorSchrödinger in 3DInternal states of the hydrogen atomL. A. Anchordoqui (CUNY)Modern Physics4-9-20192 / 54

Particle in a central potentialGeneralities of angular momentum operatorOperators can be often constructedtaking corresponding dynamical variable of classical mechanicsexpressed in terms of coordinates and momentax x̂replacingp p̂Apply this prescription to angular momentumIn classical mechanics one defines angular momentum by L r pWe get angular momentum operator by replacing:vector r vector operator r̂ ( x̂, ŷ, ẑ)momentum vector p momentum vector operator p̂ i} ( x , y , z ) i / iL. A. Anchordoqui (CUNY)Modern Physics4-9-20193 / 54

Particle in a central potentialGeneralities of angular momentum operatorComplete fundamental commutation relationsof coordinate and momentum operators are:[ x̂, p̂ x ] [ŷ, p̂y ] [ẑ, p̂z ] i}and[ x̂, p̂y ] [ x̂, p̂z ] · · · [ẑ, p̂y ] 0It will be convenient to use following notationx̂1 x̂, x̂2 ŷ, x̂3 ẑ and p̂1 p̂ x , p̂2 p̂y , p̂3 p̂zSummary of fundamental commutation relations[ x̂i , p̂ j ] i}δijKronecker symbol:δij L. A. Anchordoqui (CUNY) 1 if i j0 if i 6 jModern Physics4-9-20194 / 54

Particle in a central potentialGeneralities of angular momentum operatorCommutation relations for components of angular momentum operatorConvenient to get at first commutation relations with x̂i and p̂iUsing fundamental commutation relationsL̂ x ŷ p̂z ẑ p̂y [ x̂, L̂ x ] 0L̂y ẑ p̂ x x̂ p̂z [ x̂, L̂y ] [ x̂, ẑ p̂ x ] [ x̂, x̂ p̂z ] i}ẑsimilarly [ x̂, L̂z ] i}ŷWe can ummarize the nine commutation relations[ x̂i , L̂ j ] i}eijk x̂kand summation over the repeated index k is impliedLevi-Civita tensor 1 if (ijk) (1, 2, 3) or (2, 3, 1) or(3, 1, 2) 1 if (ijk) (1, 3, 2) or (3, 2, 1) or (2, 1, 3)eijk 0 if i j or i k or j kL. A. Anchordoqui (CUNY)Modern Physics4-9-20195 / 54

Particle in a central potentialGeneralities of angular momentum operatorSimilarly we can show[ p̂i , L̂ j ] i}eijk p̂kNow it is straightforward to deduce:[ L̂i , L̂ j ] i}eijk L̂kImportant conclusion from this result:components of L̂ have no common eigenfunctionsMust show that angular momentum operators are hermitianThis is of course plausible (reasonable) since we know thatangular momentum is dynamical variable in classical mechanicsProof is left as exerciseL. A. Anchordoqui (CUNY)Modern Physics4-9-20196 / 54

Particle in a central potentialGeneralities of angular momentum operatorConstruct operator that commutes with all components of L̂L̂2 L̂2x L̂2y L̂2zIt follows that [ L̂ x , L̂2 ] [ L̂ x , L̂2x L̂2y L̂2z ] [ L̂ x , L̂2y ] [ L̂ x , L̂2z ]There is simple technique to evaluate commutator like [ L̂ x , L̂2y ]write down explicitly known commutator[ L̂ x , L̂y ] L̂ x L̂y L̂y L̂ x i} L̂zmultiply on left by L̂yL̂y L̂ x L̂y L̂2y L̂ x i} L̂y L̂zmultiply on right by L̂yL̂ x L̂2y L̂y L̂ x L̂y i} L̂z L̂yAdd these commutation relations to getL̂ x L̂2y L̂2y L̂ x i}( L̂y L̂z L̂z L̂y )Similarly L̂ x L̂2z L̂2z L̂ x i}( L̂y L̂z L̂z L̂y )All in all [ L̂ x , L̂2 ] 0 and likewise [ L̂y , L̂2 ] [ L̂z , L̂2 ] 0L. A. Anchordoqui (CUNY)Modern Physics4-9-20197 / 54

Particle in a central potentialGeneralities of angular momentum operatorSummary of angular momentum operator ˆ Lˆ rˆ pˆ i} rˆ L̂ xL̂yL̂z(1)in cartesian coordinates ŷ p̂z p̂y ẑ i} y z z y ẑ p̂ x p̂z x̂ i} z x x z x̂ p̂y p̂ x ŷ i} x y y x(2)commutation relations[ L̂i , L̂ j ] i} ε ijk L̂kand[ L̂2 , L̂ x ] [ L̂2 , L̂y ] [ L̂2 , L̂z ] 0(3)L̂2 L̂2x L̂2y L̂2zL. A. Anchordoqui (CUNY)Modern Physics4-9-20198 / 54

Particle in a central potentialSchrödinger in 3DPrescription to obtain 3D Schrödinger equation for free particle:substitute into classical energy momentum relationE p 22m(4)differential operatorsE i} t and p i} (5)resulting operator equation }2 2 ψ i} ψ2m t(6)acts on complex wave function ψ( x, t)Interpret ρ ψ 2 as probability density ψ 2 d3 x gives probability of finding particle in volume element d3 xL. A. Anchordoqui (CUNY)Modern Physics4-9-20199 / 54

Particle in a central potentialSchrödinger in 3DContinuity equationWe are often concerned with moving particlese.g. collision of particlesMust calculate density flux of particle beam From conservation of probabilityrate of decrease of number of particles in a given volumeis equal to total flux of particles out of that volume tZVρ dV ZS · n̂ dS ZV · dV (7)(last equality is Gauss’ theorem)Probability and flux densities are related by continuity equation ρ · 0 tL. A. Anchordoqui (CUNY)Modern Physics(8)4-9-201910 / 54

Particle in a central potentialSchrödinger in 3DFluxTo determine flux. . .First form ρ/ t by substracting wave equation multiplied by iψ from the complex conjugate equation multiplied by iψ} ρ ( ψ 2 ψ ψ 2 ψ ) 0 t2m(9)Comparing this with continuity equation probability flux density i} (ψ ψ ψ ψ )2m(10)Example free particle of energy E and momentum pψ Nei p· x iEt(11)has ρ N 2 and N 2 p/mL. A. Anchordoqui (CUNY)Modern Physics4-9-201911 / 54

Particle in a central potentialSchrödinger in 3DTime-independent Schrödinger equation for central potentialPotential depends only on distance from originV ( r ) V ( r ) V (r )(12)hamiltonian is spherically symmetricInstead of using cartesian coordinates x { x, y, z}use spherical coordinates x {r, ϑ, ϕ} defined by p 222 x r sin ϑ cos ϕ r x y pz y r sin ϑ sin ϕ ϑ arctan z/ x2 y2 z r cos ϑϕ arctan(y/x )(13)Express the Laplacian 2 in spherical coordinates 21 1 1 2 2r2 2sin ϑ 2 2(14)r r rr sin ϑ ϑ θr sin ϑ ϕ2L. A. Anchordoqui (CUNY)Modern Physics4-9-201912 / 54

Particle in a central potentialSchrödinger in 3DTo look for solutions.Use separation of variable methods ψ(r, ϑ, ϕ) R(r )Y (ϑ, ϕ) }2 Y d 2 Y R YR2 dR r sinϑ V (r ) RY ERY2m r2 drdr ϑr2 sin ϑ ϑr2 sin2 ϑ ϕ2Divide by RY/r2 and rearrange terms }2 1 ddR}21 Y1 2 Y r2 r 2 (V E ) sin ϑ 2m R drdr2mY sin ϑ ϑ ϑsin2 ϑ ϕ2} l ( l 1)Each side must be independently equal to a constant κ 2m2Obtain two equations1 dR dr r2 dRdr 2mr2(V E ) l ( l 1)}2 (15)1 2 Y l ( l 1 )Ysin2 ϑ ϕ2What is the meaning of operator in angular equation?1 sin ϑ ϑL. A. Anchordoqui (CUNY) Ysin ϑ ϑModern Physics(16)4-9-201913 / 54

Particle in a central potentialSchrödinger in 3DChoose polar axis along cartesian z directionAfter some tedious calculation angular momentum components L̂ x i} sin ϕ cot ϕ cos ϕ θ ϕ cot ϑ sin ϕL̂y i} cos ϕ ϑ ϕ L̂z i}(17) ϕForm of L̂2 should be familiar 1 1 2L̂2 }2sin ϑ sin ϑ ϑ ϑsin2 ϑ ϕ2(18)Eigenvalue equations for L̂2 and L̂z operators:L̂2 Y (ϑ, ϕ) }2 l (l 1)Y (ϑ, ϕ)L. A. Anchordoqui (CUNY)andModern PhysicsL̂z Y (ϑ, ϕ) }mY (ϑ, ϕ)4-9-201914 / 54

Particle in a central potentialSchrödinger in 3DWe can always know:length of angular momentum plus one of its componentsE.g. choosing the z-component3 2 - -2 -3 presentation of the angular momentum, with fixed Lz and L2 , but complete uncerL. A. Anchordoqui (CUNY)Modern Physics4-9-201915 / 54

Particle in a central potentialSchrödinger in 3DSolution of angular equation1 sin ϑ ϑ Y m (ϑ, ϕ)sin ϑ l ϑ 1 2 Ylm (ϑ, ϕ) l (l 1)Ylm (ϑ, ϕ) ϕ2sin2 ϑUse separation of variables Y (ϑ, ϕ) Θ(ϑ )Φ( ϕ)By multiplying both sides of the equation by sin2 ϑ/Y (ϑ, ϕ) 1ddΘ1 d2 Φ(19)sin ϑsin ϑ l (l 1) sin2 ϑ Θ(ϑ )dϑdϑΦ( ϕ) dϕ22 equations in different variables introduce constant m2 :d2 Φ m2 Φ ( ϕ )dϕ2 dΘdsin ϑsin ϑ [m2 l (l 1) sin2 ϑ]Θ(ϑ)dϑdϑL. A. Anchordoqui (CUNY)Modern Physics4-9-2019(20)(21)16 / 54

Particle in a central potentialSchrödinger in 3DSolution of angular equationFirst equation is easily solved to give Φ( ϕ) eimϕImposing periodicity Φ( ϕ 2π ) Φ( ϕ) m 0, 1, 2, · · ·Solutions to the second equation Θ(ϑ ) APlm (cos ϑ )Plm associated Legendre polynomialsNormalized angular eigenfunctionss(2l 1) (l m)! mP (cos ϑ )eimϕYlm (ϑ, ϕ) 4π (l m)! l(22)Spherical harmonics are orthogonal:Z π Z 2π0L. A. Anchordoqui (CUNY)00Ylm (ϑ, ϕ) Ylm0 sin ϑdϑdϕ δll 0 δmm0 ,Modern Physics(23)4-9-201917 / 54

1385Cuando ℓ 0, tenemos m 0. Ahora PY 1#φ)y la constantenormalización# de0,0(θ, ( 1 9 cos2 θ) sen3 θe3 i φParticlecentralSchrarmóödinger in 3D32potentialuaciónrepresentamoslas superficiesr5,3in a#Yφ)π# para estosl,m (θ,es c0,0 1/ 4π . Por tanto,TABLE I: Associated Legendre polynomials.éricosCuya representación es@m012Siendo las correspondientes gráficasl @1Y0,0 (θ, φ) .4π0P00 1 DiferencialesEcuacionesII011P1 cos #2P20 es(3 cos2 #La gráfica correspondiente3P30 (5 cos31)/21463 cos #)/2#P313P1 sin # Ecuaciones Diferenciales IIP21 3 cos # sin #P22 3 sin2 #Métodosy desarrolloautofunciones 3(5 cos2de#separación1)/2 desinvariables#P32 15encos# sin2 #We can then write the eigenvalue equations for these two operators: andl0 (#, ')L̂2 Y (#, ') }2 l(ll 1)Y[Capítulo3 sin3P33 15#ℓ 1 1L̂z Y (#, ') }mY (#, ')and(412)3.4.2.usedResoluciónla theyecuaciónwhere we alreadythe fact dethatshareradialcommon eigenfunctions. Then, by its very nature we can label theseeigenfunctions byl and m, i.e. Yl,m (#,según'). k sea nulo o no.Distinguimos dos casos§3.4]La ecuación de Helmholtz en coordenadas esféricas 122Y1 such(#, ')that l 0, 1, 2, · · · and0 that radialSi k 0ShowlaEXERCISE 10.142 z are integersY0ecuación(#, ') the allowed values for!0l and ! m!!!Y1, 1 (θ, φ)!Y1!the')Y(#,φ)!1,0 (θ,mz l, · · · , l 1, l. [Hint: This result can2 beinferredfromcommutationrelationship.]′′′r R 2rℓR 2 ℓ(ℓ 1)R 0Si ℓ 1 podemos tener tres casos: m 1, 0, 1. Debemos evaluar las funcionesWe Pnowgo back to the Schrödinger equation in spherical coordinatesde Legendre1,0 y P1,1 . Acudiendo a la fórmula (3.17) obtenemosl 2 Diferenciales IIEcuaciones147and we consider the angular and radiald(ξ 2 1) 2ξ,dξ!2dP1,1 (ξ) 1 ξ 2(ξ 2 1) 2 1 ξ 2Para. ℓ 2 es fácil obtenerd ξ2!2!!Y2,2 (θ, φ)!Y2 2(#, ')!!!!2!!!Y2,1 (θ,!2Y(θ,φ)2,0Y2 1φ)(#, ')Y20 (#, ')P1,0 (ξ) !Por ello,Y1,0 (θ, φ) Y1,1 (θ, φ) "3sen θ ei φ ,8π"Y1, 1 (θ, φ) Y2,0 (θ, φ) ""38πPor último, como ejercicio dejamos el cálculo deL. A. Anchordoqui (CUNY)"5( 1 3 cos2 θ),16π"1515iφY2,1 (θ, φ) sen θ cos θ e , Y2, 1 (θ, φ) sen θ cos θ e i φ ,8π8π"" iφ1515sen Yθ2,2e(θ,.22iφφ) sen θ e, Y2, 2 (θ, φ) sen2 θ e 2 i φ .32π32π3cos θ,4πModern Physics"4-9-201918 / 54

Particle in a central potentialSchrödinger in 3DSolution of radial equation d2mr22 dR (r )r 2 (V E ) l ( l 1) R (r )drdr}(24)to simplify solution u(r ) rR(r ) }2 d2 u} l ( l 1) V u(r ) Eu(r )2m dr22m r2(25)define an effective potentialV 0 (r ) V (r ) }2 l ( l 1 )2m r2(26)(25) is very similar to the one-dimensional Schrödinger equationWave function need 3 quantum numbers (n, l, m)ψn,l,m (r, ϑ, ϕ) Rn,l (r )Ylm (ϑ, ϕ)L. A. Anchordoqui (CUNY)Modern Physics(27)4-9-201919 / 54

Particle in a central potentialInternal states of the hydrogen atomInternal states of the hydrogen atomWe start with the equation forthe relative motion of electron and proton222rV r U rEH U rWe use the spherical symmetry of thisequationand change to spherical polar coordinatesFrom now on, we drop the subscript r in theoperator 2L. A. Anchordoqui (CUNY)Modern Physics4-9-201920 / 54