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PH 401 Dr. Cecilia Vogel

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

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

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

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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

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

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

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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 =

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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, &

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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?

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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.

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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

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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

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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

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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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