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Unformatted text preview: 3/30/11 PHYS 360 Quantum Mechanics Wed Mar 30, 2011 Lecture 29: Spin: How do we do this without knowing the wave funcFons? ⎛ 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 3 l These are the eigenfuncFons of the Hamiltonian for the hydrogen atom. By axiom, they form a complete set, which means they are the basis onto which any other funcFon can be expanded: ∞ − iE t / ∞ Ψ ( r , t ) = ∑ cnψ n ( r )e n =∑ cn Ψ n ( r , t ). n=1 n−1 HW #8 due Monday: Ch 4, #24, 26, 27, 28 However, we need beTer notaFon: Dirac notaFon. 1 2 ⎛ 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 3 l Also, Once the basis funcFons for a parFcular potenFal are known, you can skip all the details and just write: ψ nlm (r,θ , φ ) → nlm . H = ∫ ψ * m (r,θ , φ ) Hψ nlm (r,θ , φ )r 2 sin θ dr dθ dφ nl = nlm H nlm
Orthonormality: For example, ∫ψ nlm (r,θ , φ ) r 2 sin θ dr dθ dφ = ∫ ψ * m (r,θ , φ )ψ nlm (r,θ , φ )r 2 sin θ drdθ dφ nl
2 n′l ′m′ nlm = δ n , n ′δ l , l ′δ m , m ′ = nlm nlm 3 4 4.4 SPIN Classically, ﬁnd both orbital angular momentum as well as spin angular momentum. Fundamentally they are the same thing. Quantum mechanically, orbital angular momentum arose naturally in solving the hydrogen atom. Spin, however, was ﬁrst discovered and then put into QM by hand. We call this intrinsic angular momentum (S), as opposed to extrinsic or orbital angular momentum (L). Why is it called “intrinsic?” Because the magnitude of an electron’s spin can never change (only its direcFon). Also, the magnitude is the same for every electron in the Universe, and for every proton, and for every muon, every photon, etc…. 5 6 1 3/30/11 We assume that spin angular momentum follows all of the rules that we found for the algebraic (operator) theory of angular momentum: ⎡ Lx , Ly ⎤ = iLz ; [ Ly , Lz ] = iLx ; [Lz , Lx ] = iLy ⎣ ⎦
⎡ L , Lx ⎤ = 0, ⎡ L , Ly ⎤ = 0 , ⎡ L , Lz ⎤ = 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2 2 2 ⎡ S x , S y ⎤ = iS z , ⎡ S y , S z ⎤ = iS x , ⎡ S z , S x ⎤ = iS y ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ S 2 s m = s s + 1 2 s m ; S z s m = m s m ( ) 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 ( ) S ± sm = s s + 1 − m m ± 1 s m ± 1
13 s = 0, ,1, , ...; m = − s, − s + 1, ..., s − 1, s 22 ( ) ( )( ) In fact, we make this the deﬁni3on of angular momentum in quantum mechanics: any operator that saFsﬁes these commutaFon relaFons is an angular momentum operator. 7 8 4.4.1 Spin ½ The electron has spin one
half, and is the most important example. There are only two eigenstates: sm = sm =
11 22 = ↑↑ 1 2 (− 1 ) = ↑↓ 2 These are the only two basis funcFons in the enFre Hilbert space for spin ½. 9 2 ...
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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 UniversityWest Lafayette.
 Spring '11
 DURBIN,STEPHEN
 mechanics

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