Lecture26.pptx

Lecture26.pptx - 3/21/11 PHYS 360 Quantum Mechanics...

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Unformatted text preview: 3/21/11 PHYS 360 Quantum Mechanics Mon Mar 21, 2011 Lecture 26: How much did you forget about the hydrogen atom over Spring Break? Note: 1.  HW 8 is posted: Ch 4, #24, 26, 26, & 28 2.  Late HW policy: If you send me email before class on Monday, you can turn it in without penalty on Wednesday. 3.  Lecture quesTon policy: There is no excused grade for occasional absences. If there is a reported illness, or if you noTfy me in advance of absences due to interviews etc, the absence will be excused. 4.  Last year’s Midterm 2 has been posted. Midterm #2 one week from today – Mon Mar 28 1 2 Midterm #2 one week from today. Topics to be studied: Sec 2.3 Harmonic Oscillator Sec 2.4 Free ParTcle Sec 2.5,2.6: 1 ­Dim. barriers, tunneling, sca]ering states Chapter 3 Formalism: Hilbert space, Hermite operators, “Generalized StaTsTcal InterpretaTon of QM,” Uncertainty Principle. Chapter 4: Sec 4.1 QM in Spherical coordinates Sec 4.2 Hydrogen – angular coordinates, radial equaTon, energies, wave funcTons… (not angular momentum). 3 Chapter 4: QM in Three Dimensions Some review…. 4 Use the explicit form of the momentum operators: px → ∂ ∂ ∂ , py → , pz → i ∂x i ∂y i ∂z As before, if the potenTal does not depend on Tme we can employ the separaTon of variables trick, and let: Ψn r ,t () − iE t / = ψ n ( r )e n Now use the notaTon of vector calculus: p→ ∇ i This leads to the three ­dimensional, Tme ­independent Schrödinger equaTon: Then… i ∂Ψ ( r , t ) 2 2 =− ∇ Ψ ( r , t ) + V ( r , t )Ψ ( r , t ) ∂t 2m ∇2 ≡ ∂2 ∂2 ∂2 + + ∂x 2 ∂y 2 ∂ 2 Z − 2 2 ∇ ψ + V ψ = Eψ 2m …and the general soluTon to the Tme ­dependent equaTon is: …where − iE t / Ψ ( r , t ) = ∑ cnψ n ( r )e n 5 6 1 3/21/11 Consider Problem 4.2: Use separaTon of variables in Cartesian coordinates to solve the infinite cubical well. ⎧ 0 if 0 ≤ x, y, z ≤ a ⎪ V ( x, y, z ) = ⎨ otherwise ⎪∞ ⎩ − 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞ + + 2 ⎟ = Eψ ⎜ 2 m ⎝ ∂x 2 ∂y 2 ∂z ⎠ SeparaTon of variables… − 2 ⎛ d 2 X d 2Y d2Z ⎞ YZ + X 2 Z + XY 2 ⎟ = EXYZ 2 m ⎜ dx 2 dy dz ⎠ ⎝ Ψn r ,t () − iE t / = ψ n ( r )e n ⎛ 1 d 2 X 1 d 2Y 1 d 2 Z ⎞ 2m ⎜ X dx 2 + Y dy 2 + Z dz 2 ⎟ = − 2 E ⎝ ⎠ d2X 2 = − kx X , dx 2 d 2Y 2 = − kx Y dy 2 3/ 2 d2Z 2 = − kx Z dz 2 E= 2 2 2 kx + ky + kz2 2m ( ) 2 2 − ∇ ψ n n n + V ψ n n n = En n n ψ n n n 123 123 123 123 2m ⎛ 2⎞ ψ ( x, y, z ) = ⎜ ⎟ ⎝ a⎠ E= sin( n yπ x n πx nxπ x ) sin( ) sin( z ) a a a nx , ny , nz = 1, 2, 3.... Note: The number of “quantum numbers” depends on the potenTal. 7 π 22 2 2 (nx + ny + nz2 ), 2 ma 2 8 Now, assume we deal with a central potenTal and use spherical coordinates: SeparaTon of variables: If the potenTal is only a funcTon of distance from the origin, i.e. V (r ) = V ( r ) = V (r ), …then it is best to use spherical coordinates: ⎡1 ∂⎛ ∂⎞ ⎛ ∂2 ⎞ ⎤ 1 ∂⎛ ∂⎞ 1 ∇2 = ⎢ 2 ⎜ r 2 ⎟ + 2 ⎜ sin θ ∂θ ⎟ + r 2 sin 2 θ ⎜ ∂φ 2 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎢ ⎥ ⎣ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝ Hence: ⎛ ∂ 2ψ ⎞ ⎤ 2 ⎡ 1 ⎛ 2 ∂ψ ⎞ 1 ∂⎛ ∂ψ ⎞ 1 r + sin θ + ⎢ ⎥ + V ( r )ψ = Eψ 2 m ⎢ r 2 ∂r ⎜ ∂r ⎟ r 2 sin θ ∂θ ⎜ ∂θ ⎟ r 2 sin 2 θ ⎜ ∂φ 2 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎣ − 9 10 i ∂Ψ ( r , t ) 2 2 =− ∇ Ψ ( r , t ) + V ( r , t )Ψ ( r , t ) ∂t 2m Yl m θ , φ = ε () ( 2l + 1)( l − m ) ! imφ m e Pl cos θ 4π ( l + m ) ! ( ) Ψn r ,t () − iE t / = ψ n ( r )e n − 2 2 ∇ ψ + V ψ = Eψ 2m ψ ( r ,θ , φ ) = R( r )Y (θ , φ ) = R( r )Θ(θ )Φ(φ ) The Radial EquaTon: Theta equaTon: Phi equaTon: 1 d ⎛ 2 dR ⎞ 2 mr 2 r ⎣ ⎦ ⎟ − 2 ⎡V ( r ) − E ⎤ = l l + 1 R dr ⎜ dr ⎠ ⎝ ( ) 1⎡ d⎛ dΘ ⎞ ⎤ 2 2 ⎢sin θ ⎜ sin θ ⎥ + l l + 1 sin θ = m Φ⎣ θ⎝ dθ ⎟ ⎦ ⎠ ( ) 1 d 2Φ = − m2 Φ dφ 2 Φ φ = eimφ , () m = 0, ± 1, ± 2, .... 11 Orthogonality: ∫∫ 0 2π π 0 [Yl m (θ , φ )]* [Yl m ' θ , φ ] sin θ dθ dφ = δ ll 'δ mm ' ' 12 () 2 3/21/11 The Radial EquaTon: d ⎛ 2 dR ⎞ 2 mr 2 ⎣ ⎦ ⎜r ⎟ − 2 ⎡V ( r ) − E ⎤ R = l ( l + 1) R dr ⎝ dr ⎠ The Hydrogen Atom: Insert the Coulomb PotenTal V (r ) = − e2 1 4πε 0 r SubsTtute: u( r ) ≡ rR( r ) 2 d 2u ⎡ 2 l l + 1 − + ⎢V + 2 m dr 2 ⎢ 2m r 2 ⎣ 2 d 2ψ New radial equaTon: ( ) ⎤ u = Eu ⎥ ⎥ ⎦ The goal: solve for the eigenfuncTons and eigenvalues, i.e. the wave funcTons and the energies. The radial equaTon: The exact soluTon: d ⎛ 2 dR ⎞ 2 mr 2 ⎣ ⎦ ⎜r ⎟ − 2 ⎡V ( r ) − E ⎤ R = l ( l + 1) R dr ⎝ dr ⎠ Looks like 1 ­D Schrödinger equaTon, − 2 m 2 + Veff ψ = Eψ dx 2 l ( l + 1) 2m r 2 …with: Veff = V + (centrifugal term) 13 ⎛ 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 14 3 l We have three quantum numbers for the wave funcTon: ψ nlm ( r ,θ , φ ) = Rnl ( r )Yl m (θ , φ ) Energy, however, only depends on n. Set n=1 for the lowest, ground state energy: ⎡ m 2 ⎛ e2 ⎞ 2 ⎤ ⎥ −13.6 eV E1 = − ⎢ ⎢ 2 ⎜ 4πε 0 ⎟ ⎥ ⎝ ⎠ ⎣ ⎦ Note: n=1 requires that l=m=0. The ground state wave funcTon: 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 Also, determined Bohr radius: a≡ 4πε 0 2 me2 = 0.529 x 10−10 m 15 16 h]p://www.falstad.com/qmatom/ h]p://homepages.ius.edu/kforinas/physlets/quantum/hydrogen.html h]p://www.phy.davidson.edu/StuHome/cabell_f/Radial.html These are web sites that graphically display the wave funcTons. 3 ...
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