# PH 401 Dr. Cecilia Vogel. Review Outline  Spherically Symmetric Hamiltonian  H-atom for example  Eigenstates of H, L z, L 2  Degeneracy  Commutators.

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PH 401 Dr. Cecilia Vogel

Review Outline  Spherically Symmetric Hamiltonian  H-atom for example  Eigenstates of H, L z, L 2  Degeneracy  Commutators and Uncertainty  Angular Momentum  Radial momentum

Spherically Symmetric Problem  Suppose the potential energy  depends only on r,  not  or .  such as for hydrogen atom  then the Hamiltonian looks like.  =

Spherically Symmetric Problem  The Hamiltonian in spherical coordinates, if V is symmetric    Written in terms of p r, L 2 and L z :  = = =

Commutation  Components of angular momentum commute with r, p r, and L 2.  Therefore L z and L 2 commute with H  We can (and will) find set of simultaneous eigenstates of  L 2, L z, and H  Let  (r,  ) = R(r)f(  )g(  )  Let quatum numbers be n, ℓ, m ℓ

Separation of variables  In the TISE, H  =E  where  there is only one term with  or  derivatives  So this part separates  =

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

Eigenstate of L 2  Also, angular derivatives only show up in L 2 term, which also separates:   FYI solution is Spherical Harmonics  Note that L 2 =Lx 2 +Ly 2 +Lz 2  so  = 

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  Consider n=5  4 th 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 4 th 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  Solution   Impose boundary condition  g(2  )=g(0)  requires that m l =integer  Lz is quantized  or Lx or Ly, but not all

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 Lz  Similarly  [Lx,Ly]=i  Lz is not zero  unless Lz=0  So we can have a simultaneous eigenstate of Lx, Ly, Lz, and L 2  if the eigenvalues are all zero  |n 0 0> is an eigenstate of all components of L

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 ( )

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