4_Schrodinger

4_Schrodinger - IV. Solving the Time Independent...

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Unformatted text preview: IV. Solving the Time Independent Schrödinger Equation Griffiths Chapters 2 and 4 1 A. The Operator Approach to the Harmonic Oscillator Griffiths 2.3 • The harmonic oscillator is represented by the force equation F = -kx. • This gives a potential energy of V(x) = -kx2/2. • Many potentials will reduce to this form for small values of x. • Therefore, as a first approximation, many physical phenomena can be understood using the harmonic potential. 2 A. The Operator Approach to the Harmonic Oscillator • Taylor expand V(x) about the point x0. x0 i s a minimum so the first derivative at x0 i s zero 1 ￿￿ V (x) = V (x0 ) + V (x0 )(x − x0 ) + V (x0 )(x − x0 )2 + · · · 2 ￿ drop constant term... doesn’t change the force k = V’’(x0) as long as (x-x0) is small, higher o rder terms will be negligible • For our quantum situation, we want to solve the Schrödinger ￿ k equation for the potential ω= m 1 V (X ) = mω 2 X 2 2 3 A. The Operator Approach to the Harmonic Oscillator 1. Wave Function and Energies • The time-independent Schrödinger equation will be ￿2 d2 ψ (x) 1 − + mω 2 X 2 ψ (x) = E ψ (x) 2m dx2 2 • This has an infinite number of discrete solutions. The ground state (lowest energy) of this quantum oscillator will be ψ0 (x) = ￿ mω ￿ 1 / 4 π￿ ω − m￿ x2 2 e ground state wave function • 1 E0 = ￿ω 2 ground state energy The higher energy solutions are ψn (x) = An (a+ ) ψ0 (x) n En = ￿ n+ 1 2 ￿ ￿ω 4 A. The Operator Approach to the Harmonic Oscillator • The higher energy solutions are ψn (x) = An (a+ ) ψ0 (x) n En = ￿ 1 n+ 2 ￿ ￿ω normalization c onstant • So if we know what the raising operator a+ is, then we can find all of the higher energy wave functions and their corresponding energies. • a+ and the related lowering operator a- are called ladder operators. See board IV.A.1 5 A. The Operator Approach to the Harmonic Oscillator 2. The Number Operator • The number operator is defined as N = a+a-. N ψn (x) = a+ a− ψn (x) = nψn (x) See board IV.A.2 3. Normalization • The ladder operators can be used to normalize the wave functions of the harmonic oscillator. 1 ψn (x) = √ (a+ )n ψ0 (x) n! and En = ￿ 1 n+ 2 ￿ ￿ω normalized See board IV.A.3 6 B. The Hydrogen Atom Griffiths 4.1 and 4.2 • The hydrogen atom is one of the few, realistic, systems that can be solved analytically. • The prescription is to write the time-independent Schrödinger equation in spherical coordinates and solve it using a Coulomb potential. • The math goes on a long time but involves mostly just separation of variables. • We use special functions such as Legendre polynomials, spherical harmonics and Laguerre polynomials to get our answer. • The principal, azimuthal and magnetic quantum numbers arise naturally in solving the differential equations. 7 B. The Hydrogen Atom 1. The Time Independent Schrödinger Equation in Spherical Coordinates • The time independent Schrödinger equation is H ψ (r) = E ψ (r) • For a potential V that doesn’t depend on time and is a central potential (depends only on radial distance r) we get ￿2 2 − ∇ ψ (r) + V (r)ψ (r) = E ψ (r) 2m • This spherically symmetric potential is most conveniently solved in spherical coordinates. The Laplacian ∇2 in spherical coordinates is ￿ ￿ ￿ ￿ ￿ 2￿ 1∂ 1 ∂ ∂ 1 ∂ 2 2∂ ∇=2 r +2 sin θ +2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 8 B. The Hydrogen Atom ￿2 2 − ∇ ψ (r) + V (r)ψ (r) = E ψ (r) 2m ￿ ￿ ￿ ￿ ￿ 2￿ 1∂ 1 ∂ ∂ 1 ∂ 2 2∂ ∇=2 r +2 sin θ +2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 • This partial differential equation can be split into three ordinary differential equations: one for r, one for θ and one for ϕ. • This is done using separation of variables where ψ(r,θ,ϕ) = R(r)Θ(θ)Φ(ϕ) Angular parts all together, Y(θ, ) = Θ(θ)Φ( ), are called spherical harmonics 9 B. The Hydrogen Atom Angular parts all together, Y(θ, ) = Θ(θ)Φ( ), are called spherical harmonics • The combined angular equation is ￿ ￿ ￿ ￿ 2 1 1∂ ∂ Y (θ, φ) 1 ∂ Y (θ, φ) sin θ + = −l(l + 1) 2 2 Y (θ, φ) sin θ ∂θ ∂θ ∂φ sin θ • The radial equation is ￿ ￿ 1d dR(r) 2mr2 r2 − [V (r) − E ] = l(l + 1) 2 R(r) dr dr ￿ These are equal to the same separation constant, which we choose to write in this strange way for later c onvenience. See board IV.B.1.a 10 B. The Hydrogen Atom • The combined angular equation is ￿ ￿ ￿ ￿ 2 1 1∂ ∂ Y (θ, φ) 1 ∂ Y (θ, φ) sin θ + = −l(l + 1) 2 2 Y (θ, φ) sin θ ∂θ ∂θ ∂φ sin θ • The angular equation can be further reduced using separation of variables again with Y(θ,ϕ) = Θ(θ)Φ(ϕ). We get two ordinary differential equations. ￿ ￿ ￿￿ 1 d dΘ(θ) sin θ sin θ + l(l + 1) sin2 θ = m2 Θ(θ) dθ dθ 1 d2 Φ(φ) = −m2 Φ(φ) dφ2 11 B. The Hydrogen Atom ￿ ￿ ￿￿ 1 d dΘ(θ) sin θ sin θ + l(l + 1) sin2 θ = m2 Θ(θ) dθ dθ 1 d2 Φ(φ) = −m2 Φ(φ) dφ2 • These have unnormalized solutions Φ(φ) = eimφ and Associated Legendre function Θ(θ) = APlm (cos θ) Normalization for both s olutions lumped together in A. 12 ASIDE: Plm (x) = (1 − x ) 1 Pl (x) = l 2 l! 2 |m|/2 ￿ d dx ￿l ￿ d dx ￿|m| Pl (x) Legendre polynomial (x2 − 1)l See Griffths 4.1.2 for more L egendre p roperties and discussion. 13 B. The Hydrogen Atom ￿ ￿ ￿￿ 1 d dΘ(θ) sin θ sin θ + l(l + 1) sin2 θ = m2 Θ(θ) dθ dθ 1 d2 Φ(φ) = −m2 Φ(φ) dφ2 • These have unnormalized solutions Φ(φ) = eimφ and Associated Legendre function Θ(θ) = APlm (cos θ) Normalization for both s olutions lumped together in A. See board IV.B.1.b 14 B. The Hydrogen Atom • These solutions combined and normalized give the spherical harmonics. ￿ (2l + 1) (l − |m|)! imφ m m Yl (θ, φ) = ε e Pl (cos θ) 4π (l + |m|)! where ε = (-1)m for m ≥ 0 and ε = 1 for m ≤ 0. • These have unnormalized solutions See G riffths 4.1.2 for more Spherical H armonic p roperties a nd discussion. 15 B. The Hydrogen Atom • • • The radial equation was ￿ ￿ 1d 2mr2 2 dR(r ) r − [V (r) − E ] = l(l + 1) 2 R(r) dr dr ￿ We can simplify this by changing variables to u(r) = rR(r). ￿ ￿ See board 22 2 ￿ d u(r) ￿ l(l + 1) − + V (r) + u(r) = Eu(r) IV.B.1.c 2 2 2m dr 2m r This is the same as the one dimensional Schrödinger equation but with an effective potential Veff(r) given by ￿2 l(l + 1) Veff = V (r) + 2m r2 and a normalization condition ￿∞ |u(r)|2 dr = 1 Centrifugal term 0 • We can’t go further without specifying the potential V(r). 16 B. The Hydrogen Atom 2. The Hydrogen Atom • The proton of the hydrogen atom is essentially motionless. We’ll put it at the origin of the spherical coordinate system. • The electron will feel a Coulomb attraction to the proton given by e2 1 V (r) = − 4π￿0 r • The radial equation for the electron will therefore be ￿ ￿ 2 2 ￿ d u(r) e1 ￿ l(l + 1) − +− + u(r) = Eu(r) 2 2 2m dr 4π￿0 r 2m r 2 • 2 We want to solve this equation for u(r) and E. 17 B. The Hydrogen Atom ￿ ￿ 2 2 ￿ d u(r) e1 ￿ l(l + 1) − +− + u(r) = Eu(r) 2 2 2m dr 4π￿0 r 2m r 2 2 • We want to solve this equation for u(r) and E. • We can simplify the equation by making the substitutions √ me2 −2mE ρ0 ≡ κ≡ ρ ≡ κr 2π￿0 ￿2 κ ￿ We are interested in bound electron s tates, where E<0, so this argument o f the square will be positive • This gives ￿ ￿ d u(ρ) ρ0 l(l + 1) = 1− + u(ρ) 2 2 dρ ρ ρ 2 See board IV.B.2.a 18 B. The Hydrogen Atom ￿ ￿ d u(ρ) ρ0 l(l + 1) = 1− + u(ρ) 2 2 dρ ρ ρ 2 • Next we change the dependent variable too. u(ρ) = ρl+1 e−ρ v (ρ) • This gives d2 v (ρ) dv (ρ) ρ + 2(l + 1 − ρ) + [ρ0 − 2(l + 1)]v (ρ) = 0 2 dρ dρ See board IV.B.2.b • We try a power series solution for this form of the radial equation. ∞ ￿ v (ρ) = cj ρj j =0 19 B. The Hydrogen Atom v (ρ) = • • • ∞ ￿ cj ρj j =0 This is a solution provided the coefficients obey the recursion relation ￿ ￿ 2(j + l + 1) − ρ0 See board cj +1 = cj IV.B.2.c (j + 1)(j + 2l + 2) In order for this solution to be finite, the series must terminate at some maximum value of j, call it jmax. jmax must obey the relationship See board IV.B.2.d 2(jmax + l + 1) − ρ0 = 0 • Define the principal quantum number n = jmax + l + 1. The allowed energies are therefore ￿ ￿ 2 ￿2 ￿ Bohr E1 m e 1 En = 2 = − , n = 1, 2, 3, · · · 2 2 Formula n 2￿ 4π￿0 n 20 B. The Hydrogen Atom ￿ E1 m En = 2 = − n 2￿ 2 ￿ e2 4π￿0 ￿2 ￿ 1 n2 , 4π￿0 ￿2 Bohr −10 a≡ = 0.529 × 10 m 2 Radius me Bohr n = 1, 2, 3, · · · Formula See board IV.B.2.e • We have the allowed energy levels for the hydrogen atom. Now we need the corresponding wave functions. • The radial part of the wave function is 1 l+1 −ρ Rnl (r) = ρ e v (ρ) r where v(ρ) is a polynomial of degree jmax = n - l - 1with coefficients given by the previous recursion formula. ￿ ￿ cj +1 = 2(j + l + 1) − ρ0 (j + 1)(j + 2l + 2) cj 21 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre polynomials. +1 v (ρ) = L2l−l−1 (2ρ) n 22 ASIDE: Associated Laguerre p olynomial Lp−p (x) q Lq (x) ≡ e x ￿ d dx ≡ (−1) ￿q p ￿ d dx ￿p Lq (x) Laguerre polynomial (e−x xq ) See Griffths pages 152 and 153 for more L aguerre properties a nd discussion. 23 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre polynomials. +1 v (ρ) = L2l−l−1 (2ρ) n • The hydrogen wave functions are the combination of the radial and angular solutions. ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ) • Substituting in all our results we get the wave functions ψnlm (r, θ, φ) = ￿￿ 2 na ￿3 (n − l − 1)! −r/na e 2n[(n + l)!]3 ￿ 2r na ￿l ￿ 2l+1 ￿m Ln−l−1 (2r/na) Yl (θ, φ) Hydrogen atom wave functions See board IV.B.2.f 24 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre polynomials. +1 v (ρ) = L2l−l−1 (2ρ) n • The hydrogen wave functions are the combination of the radial and angular solutions. ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ) • Substituting in all our results we get the wave functions ψnlm (r, θ, φ) = ￿￿ 2 na ￿3 (n − l − 1)! −r/na e 2n[(n + l)!]3 ￿ 2r na ￿l ￿ 2l+1 ￿m Ln−l−1 (2r/na) Yl (θ, φ) Hydrogen atom wave functions See board IV.B.2.f 25 B. The Hydrogen Atom 26 B. The Hydrogen Atom 3. The Spectrum of Hydrogen • The hydrogen wave functions we have found are stationary states. In principle, the hydrogen electron, if left alone, will never change states. • In reality, there are always collisions between atoms or with photons (light). These interactions are beyond our present time-independent discussion. • The result of such interactions however, is that the hydrogen electron may undergo a transistion, that is, change states. It will absorb or emit energy in the process. • The energy the hydrogen atom releases (or absorbs) is equal to the energy difference between the bound states. 27 B. The Hydrogen Atom ground state e nergy ￿ Eγ = Ei − Ef = −13.6eV energy of e mitted photon energy of i nitial state energy of f inal state f re • 1 1 2 − n2 ni f Recall that Eν = hν and λ = c/ν. Planck Formula q ue ￿ principal q uantum n umbers n cy wave length 1 =R λ t an con st ￿ 1 1 2 − n2 nf i ￿ g e2 be r R = m d Ry 4π c￿3 4π￿0 ￿2 ￿ = 1.097 × 107 m−1 28 B. The Hydrogen Atom ground state e nergy ￿ Eγ = Ei − Ef = −13.6eV energy of e mitted photon energy of i nitial state energy of f inal state f re • Recall that Eν = hν and λ = c/ν. Planck Formula q ue principal q uantum n umbers n cy wave length 1 =R λ t an con st ￿ 1 1 2 − n2 nf i ￿ g e2 be r R = m d Ry 4π c￿3 4π￿0 • 1 1 2 − n2 ni f ￿ ￿2 ￿ = 1.097 × 107 m−1 Bohr’s theory of the hydrogen atom was able to explain these experimental results. 29 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.

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