Lecture27.pptx

# Lecture27.pptx - 3/23/11 The Hydrogen Atom: Insert...

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Unformatted text preview: 3/23/11 The Hydrogen Atom: Insert the Coulomb PotenPal PHYS 360 Quantum Mechanics Wed Mar 23, 2011 Lecture 27: Back to operator algebra and commutators: What are the allowed angular momentum values? V (r ) = − e2 1 4πε 0 r The goal: solve for the eigenfuncPons and eigenvalues, i.e. the wave funcPons and the energies. The radial equaPon: d ⎛ 2 dR ⎞ 2 mr 2 ⎣ ⎦ ⎜r ⎟ − 2 ⎡V ( r ) − E ⎤ R = l ( l + 1) R dr ⎝ dr ⎠ Midterm #2 on Mon Mar 28 The exact soluPon: ⎛ 2 ⎞ ( n − l − 1) ! − r / na ⎛ 2r ⎞ 2 l +1 m ψ nlm = ⎜ ⎟ e ⎜ ⎟ ⎡ L ( 2r / na ) ⎤ Yl (θ , φ ) ⎦ ⎝ na ⎠ 2 n[( n + l ) !]3 ⎝ na ⎠ ⎣ n−l −1 1 2 3 l Note: n=1 requires that l=m=0. The ground state wave funcPon: hUp://www.falstad.com/qmatom/ hUp://homepages.ius.edu/kforinas/physlets/quantum/hydrogen.html hUp://www.phy.davidson.edu/StuHome/cabell_f/Radial.html ψ 100 ( r ,θ , φ ) = R10 ( r )Y00 (θ , φ ) R10 ( r ) = c0 a e −r/a ψ 100 ( r ,θ , φ ) = 1 π a3 e− r / a These are web sites that graphically display the wave funcPons. 3 Recap – the harmonic oscillator: Write the Hamiltonian in terms of operators: 3 1 ˆ ˆ ˆ H= [ p 2 + ( mω x )2 ]. 2m a+ ≡ 1 2mω 1 2mω ˆ ˆ ( − ip + mω x ) 1 V ( x ) = mω 2 x 2 2 2 1 Try factoring the Hamiltonian in terms of two new operators: 0 1 2 raising: lowering: a− ≡ ˆ ˆ ( + ip + mω x ) -2 -1 − 2 2 d ψ n 1 + mω 2 x 2ψ n = Enψ n 2 m dx 2 2 UPlize the commutator of the posiPon & momentum operators: ˆˆ [ x , p] = i 1 3/23/11 The “factorized” Schrödinger EquaPon: ⎛ 1⎞ ω ⎜ a± a ± ⎟ ψ n = Enψ n 2⎠ ⎝ This all leads to the eigenvalues and eigenfuncPons of the Schrödinger EquaPon: Claim: Assume we have a soluPon ψn with energy En. Consider the new funcPon you get by operaPng on ψn with the raising operator: a+ψn . This new funcPon will also be a soluPon of the Schrödinger EquaPon, with energy En+ħω. That is, it raised the energy by one unit (ħω). We will use this property to learn a lot about this system…. 7 ψ n ( x ) = An ( a+ ) n ψ 0 ( x ), ⎛ 1⎞ with En = ⎜ n + ⎟ ω 2⎠ ⎝ Note: 1.  The harmonic oscillator has equally ­spaced energy levels. 2.  The lowest energy is NOT zero. 3.  We don’t have the excited state wavefuncPons explicitly, but we do have the recipe. New material…. We now consider the angular momentum associated with the states of the hydrogen atom: L=r×p Recall that for the successful Bohr model of the hydrogen atoms, it was the angular momentum of the orbitals that was quanPzed, so we know this plays a special role in hydrogen. 9 10 We already know the eigenfuncPons and eigenvalues of the Hamiltonian operator for hydrogen, from the Schrödinger equaPon. What eigenfuncPons and eigenvalues are there for angular momentum? Answer: it turns out that everything we need is in the spherical harmonics already part of the hydrogenic wave funcPons. L=r×p The components of the vector cross product in Cartesian coordinates are: Lx = ypz − zp y , Ly = zpx − xpz , Lz = xp y − ypx We are now going to look at commutators involving these components. Why? Remember this: ⎛1 ˆ ˆ ⎞ 2 σ 2σ B ≥ ⎜ [ A, B] ⎟ A ⎝ 2i ⎠ 2 ⎛ 2 ⎞ ( n − l − 1) ! − r / na ⎛ 2r ⎞ 2 l +1 m ψ nlm = ⎜ ⎟ e ⎜ na ⎟ ⎡ Ln−l −1 ( 2r / na ) ⎤ Yl (θ , φ ) ⎦ ⎝ na ⎠ 2 n[( n + l ) !]3 ⎝ ⎠⎣ 11 3 l This is the generalized uncertainty principle. The commutator tells us if two quanPPes can be exactly determined at the same Pme. 12 2 3/23/11 Expand: ⎡ Lx , Ly ⎤ = ⎡ ypz − zp y , zpx − xpz ⎤ ⎣ ⎦⎣ ⎦ = ⎡ ypz , zpx ⎤ − ⎡ ypz , xpz ⎤ − ⎡ zp y , zpx ⎤ + ⎡ zp y , xpz ⎤ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦ Plug into uncertainty principle: ⎛1 ⎞ ⎛ ⎞ σ2 σ2 ≥ ⎜ iLz ⎟ ≥ ⎜ ⎟ Lx Ly ⎝ 2i ⎠ ⎝ 2⎠ 2 2 Lz 2 σL σL ≥ x y L 2z Collect terms: ⎡ Lx , Ly ⎤ = ypx [ pz , z ] + xp y [ z , pz ] = i( xp y − ypx ) = iLz ⎣ ⎦ Any two components fail to commute! ⎡ Lx , Ly ⎤ = iLz ; [ Ly , Lz ] = iLx ; [Lz , Lx ] = iLy ⎣ ⎦ No two components can have precisely determined values at the same Pme. Not just that you can’t measure this; they do not exist. (IncompaPble observables….) 13 14 What is wrong with this picture? But, consider the commutators of L2: L2 ≡ L2 + L2y + L2 x z ⎡ L2 , Lx ⎤ = 0, ⎡ L2 , Ly ⎤ = 0 , ⎡ L2 , Lz ⎤ = 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ L2 , L ⎤ = 0 ⎣ ⎦ σL σL ≥ x y L 2z So, L2 is compaPble with any component of L. 15 16 Using “algebraic” arguments based on raising and lowering operators (similar to harmonic oscillator), the eigenvalues of L2 and Lz are determined: L2 f l m = 2 l l + 1 f l m ; Lz f l m ; = mf l m ( ) l = 0, 1 / 2, 1, 3 / 2, ...; m = − l , − l + 1, ... , l − 1, l (Note: half ­integer values of l are allowed.) L2 f l m = 2 l l + 1 f l m ; Lz f l m ; = mf l m l = 0, 1 / 2, 1, 3 / 2, ...; m = − l , − l + 1, ... , l − 1, l 17 ( ) Note: We sPll don’t know what the eigenfuncPons are. 18 3 ...
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## This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue.

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