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PH 401. Dr. Cecilia Vogel. Review. Commutators and Uncertainty Angular Momentum Radial momentum. Spherically Symmetric Hamiltonian H-atom for example Eigenstates of H, L z , L 2 Degeneracy. Outline. Spherically Symmetric Problem. Suppose the potential energy depends only on r,

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ph 401
PH 401

Dr. Cecilia Vogel

review
Review
  • Commutators and Uncertainty
  • Angular Momentum
  • Radial momentum
  • Spherically Symmetric Hamiltonian
    • H-atom for example
    • Eigenstates of H, Lz, L2
    • Degeneracy

Outline

spherically symmetric problem
Spherically Symmetric Problem
  • Suppose the potential energy
    • depends only on r,
    • not q or f.
    • such as for hydrogen atom
  • then the Hamiltonian looks like.

=

spherically symmetric problem1
Spherically Symmetric Problem
  • The Hamiltonian in spherical coordinates, if V is symmetric
  • Written in terms of pr, L2 and Lz:

=

=

=

commutation
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
Separation of variables
  • In the TISE, Hy=Ey, where
    • there is only one term with f or f derivatives
    • So this part separates

=

eigenstate of lz
Eigenstate of Lz
  • In fact
    • So eigenvalue eqn for Lz is
    • Lz|nlm> = m|nlm>
    • means

=

eigenstate of l 2
Eigenstate of L2
  • Also, angular derivatives only show up in L2 term, which also separates:
    • FYI solution is Spherical Harmonics
  • Note that L2=Lx2+Ly2+Lz2
    • so

y

y =

eigenstate of h
Eigenstate of H
  • Radial part of eqn
    • 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, &
degeneracy of eigenstates
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
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
Quantized Lz
  • Solution
  • 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
Eigenstate of Lx, Ly, and Lz
  • Recall
    • [Ly,Lz]=iLx 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]=-iLy is not zero
    • unless Ly=0
eigenstate of lx ly and lz1
Eigenstate of Lx, Ly, and Lz
  • Similarly
    • [Lx,Ly]=iLz 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
Component and Magnitude
  • The fact that
    • is a consequence of the fact that
    • a component of a vector ( )
    • can’t be bigger than
    • magnitude of the vector ( )