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Transcript
PH 401 Dr. Cecilia Vogel Review Commutators and Uncertainty Angular Momentum Radial momentum Outline Spherically Symmetric Hamiltonian H-atom for example Eigenstates of H, Lz, L2 Degeneracy Spherically Symmetric Problem Suppose the potential energy depends only on r, not q or f. such as for hydrogen atom ke2 V (r ) r then the Hamiltonian looks like. Hˆ = 2 2m 2 V (r ) Spherically Symmetric Problem The Hamiltonian in spherical coordinates, if V is symmetric Hˆ = 2 2m 2 V (r ) 2 1 1 2 ˆ H = r sin q 2 V (r ) 2 2 2mr r r sin q q q sin q f Written in terms of pr, L2 and Lz: 2 2 2 p L Hˆ = r V (r ) 2 2m 2mr Commutation Components of angular momentum commute with r, pr, and L2. Therefore Lz and L2 commute with H We can (and will) find set of simultaneous eigenstates of L2, Lz, and H Let Y(r,q,f) = R(r)f(q)g(f) Let quatum numbers be n, ℓ, mℓ Separation of variables In the TISE, Hy=Ey, where 2 1 1 2 ˆ H = r sin q 2 V (r ) 2 2 2mr r r sin q q q sin q f 2 there is only one term with f or f derivatives So this part separates 2 y constant *y 2 f Eigenstate of Lz In fact ˆ Lz = i f So eigenvalue eqn for Lz is Lz|nlm> = m|nlm> means 2y 2 m y 2 f Eigenstate of L2 Also, angular derivatives only show up in L2 term, which also separates: 2 1 1 ˆ2y = 2 L sin q 2 2 q sin q f sin q q y FYI solution is Spherical Harmonics Note that L2=Lx2+Ly2+Lz2 n m | Lˆ2 | n m n m | Lˆ2x Lˆ2y Lˆ2z | n m so ( 1) 2 2 2 2 ˆ ˆ Lx Ly m 2 ( 1) m therefore | m | 2 Eigenstate of H Radial part of eqn ˆpr 2 ( 1) 2m R 2mr 2 2 R V ( r ) R En R As r goes to infinity, V(r) goes to zero For hydrogen atom So solution is (polynomial*e-r/na.) lowest order in polynomial is , highest order in polynomial is n-1, so is a non-negative integer, n is a positive integer, & n Degeneracy of Eigenstates Consider n=5 4th excited state of H-atom What are possible values of ? For each , what are possible values of m? for each n & , how many different states are there? “subshell” for each n, how many different states are there? “shell” what is the degeneracy of 4th excited state? Spin Quantum Number Actually there turns out to be twice as many H-atom states as we just described. Introduce another quantum number that can have two values spin can be up or down (+½ or -½) It is called spin, but experimentally the matter of the electron is not spinning. It is like a spinning charged object, though, in the sense that it acts like a magnet, affected by B-fields it contributes to the angular momentum, when determining conservation thereof. Quantized Lz g (f ) i m g (f ) f Solution g (f ) eim f Impose boundary condition g(2p)=g(0) requires that ml=integer Lz is quantized or Lx or Ly, but not all Eigenstate of Lx, Ly, and Lz Recall [Ly,Lz]=iLx is not zero so there’s not complete set of simultaneous eigenstates of Ly and Lz What if Lx=0? OK, but then also simultaneous eigenstate of Lx, and [Lx,Lz]=-iLy is not zero unless Ly=0 Eigenstate of Lx, Ly, and Lz Similarly [Lx,Ly]=iLz is not zero unless Lz=0 So we can have a simultaneous eigenstate of Lx, Ly, Lz, and L2 if the eigenvalues are all zero |n 0 0> is an eigenstate of all components of L Component and Magnitude The fact that | m | is a consequence of the fact that a component of a vector ( m ) can’t be bigger than magnitude of the vector ( ( 1) )