ECE 162A: Lecture 15 - Spin in Quantum Mechanics, Study notes of Materials science

Download ECE 162A: Lecture 15 - Spin in Quantum Mechanics and more Study notes Materials science in PDF only on Docsity! ECE 162A Mat 162A Lecture #15:Spin E/R Chapter 8: John Bowers Bowers@ece.ucsb.edu ECE/Mat 162A Accomplished • Chapter 1: Thermal radiation, Planck’s postulate • Chapter 2: Light: Wavelike and particlelike (photon) • Chapter 3: Matter: Wavelike and particlelike • Chapter 4: Thompson, Rutherford, Bohr Model of Atom • Chapter 5: Schroedinger Theory: Time dependent, Time independent Schroedinger Equations • Chapter 6: Solutions of Time Independent SE: Free particle, Step t ti l B i t ti l I fi it ll fi it llpo en a , arr er po en a , n n e square we , n e square we , harmonic oscillator. • 3D Solutions. Ch t 7 H d t Q t b d d• ap er : y rogen a om. uan um num ers an egeneracy. Angular momentum. Commutator. Simultaneous eigenvalues. 2D harmonic oscillator. ECE/Mat 162A • If two operators commute, then the eigenvalues associated with those operators are simultaneous eigenvalues • If two operators do not commute, then the eigenvalues associated with those two operators typically exhibit an uncertainty relation. • In general, for every system one may identify at l l f ieast one comp ete set o commut ng observables. ECE/Mat 162A Specific Case: 2D Harmonic Oscillator MCV 22222 )(1)(1)( EyxM M yxyxyx =++ ∂ ∂ + ∂ ∂− +≡+= 222 2 2 2 22 )( 2 1)( 2 22 , ψψωψψ ω h gf ygxfyx yx ∂∂− = 222 222 1 )()(),(ψ h EyMfxMf EfgfgyxM y f x g M = ∂− ++ ∂− =++ ∂ + ∂ 22 22 22 22 22 )1()1( )( 2 )( 2 ωω ω hh EtConstCons yMfxMf =+ − ∂∂ 22 22 tantan 2222 g fEfxM x f M x ∂− =+ ∂ ∂− 22 22 22 2 1 2 1 2 ω h h ECE/Mat 162A EEE gEgyM yM yx y =+ =+ ∂ 2 22 ω 2D Harmonic Oscillator Solutions 222xx )()( 2/)(= +− ayxnnnn ea H a H xxyx ψ 210 )1( = ++= yx n nnE ωh ,...2,1,0 ,...,, =y x n ECE/Mat 162A N=1 Solutions 22 / 10 2011 arexnnn −==== ψ 22 /2101 ar yx y a 01yx ea nnn −==== ψ These are not solutions that satisfy: ψψ ∂ = L LL zz hˆ ˆ ECE/Mat 162A φ∂ = iz N=1 Solutions 22 / 10 2 2011 aryx y e a xnnn −==== ψ 22 / 01 : 101 aryx Note e a nnn −==== ψ sincos sincos i i iyxirrre iyxirrre − −=−= +=+= φ φ φφ φφ 2222 // 0101 2)(2 ariar eereiyxi So −− = + =+= φψψψ 2222 // 0101 2)(2 ariar ee a re a iyxi aa −−− = − =−= φψψψ ECE/Mat 162A These are both solutions with Lz= +1 and -1 respectively. Dirac Notation yxnn ψ Is represented by the Dirac ket vector yx nn >,| This notation is a useful shorthand: imn >+>>=== 1,0|0,1|1,1| The projection of onto all possible positions is yxnnyx nnyx ψ>=< ,|, the wave function ECE/Mat 162A Bohr Magneton • Classically, mrvL = m eLevrμ == 22 • (Bohr Magneton) h Ae If 22310927 L m m b b μμ μ ×=≡ −. 2 h = ECE/Mat 162A Bohr Magneton • Classically, mrvL = m eLevrμ == 22 • (Bohr Magneton) Am e If b hμ ×=≡ − 22310927. L m b h μμ = 2 • The correct quantum mechanical result is • (gl is the orbital g factor and gl=1) ECE/Mat 162A Lg bl r h r μμ −= Dipole in a Magnetic Field • The effect of a magnetic field on a magnetic dipole is to exert a torque rr • The potential energy is lowest when the dipole BT r×= μ is aligned with the magnetic moment BE rr •−=Δ μ ECE/Mat 162A Stern Gerlach Experiment • Stern Gerlach Exp: Pass a beam of silver atoms through a nonuniform magnetic field and record deflections • Classical prediction: a range of deflections corresponding to μz ranging from + μ to - μ. • Quantum mechanical prediction: discrete deflections corresponding to ml=–l,…0…l ECE/Mat 162A • Result: 2 discrete components: one positive, one negative. Phipps/Taylor (1927) Experiment • Repeat Stern Gerlach Exp with hydrogen atoms in the ground state l=0 m =0 . . l . • Quantum mechanical prediction: No deflection corresponding to ml=0 ECE/Mat 162A Phipps/Taylor (1927) Experiment • Repeat Stern Gerlach Exp with hydrogen atoms in the ground state l=0 m =0 . . l . • Quantum mechanical prediction: No deflection corresponding to ml=0 • Result: 2 beams, one deflected positive, one negative. • Something is missing in the theory. • Size of deflection: 2000x bigger than Bohr magneton for a proton. • The atom is not responsible for the deflection. The electron is! ECE/Mat 162A Total Angular Momentum SLJ rrr += •Due to the spin orbit interaction, L and S are not independent. The spin orbit interaction causes a coupling between L and S and a i b t thprecess on a ou e axes. •The total angular momentum is fixed and quantized. •The z component of total angular momentum is quantized. ECE/Mat 162A Total Angular Momentum SLJ += rrr jjJ )1( += h The total angular momentum in terms of quantum number j mJ = h The z component of angular momentum is jz Where the quantum number mj is jjm j ...0,...−= ECE/Mat 162A How does j relate to l and s? )1()1()1()1()1( +−+≥+≥+++ −≥≥+ sslljjssll SLJSL rrrrr 2/12/1+ llj The result of this inequality is that j can have two values: 2/1 , −= j When l=0, then there is only one value of j: = S J S ECE/Mat 162A L L J