PS 6 Sol

# PS 6 Sol - CHEM 120A Problem Set 6 Solutions David Homan...

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CHEM 120A Problem Set 6 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 3/17/2010 1. The solutions can be found in the McQuarrie and Simon Solutions manual. Note that the units used (e.g. for the gyromagnetic ratio) are different from the convention adopted in class. 2. We want to find the energy expectation values for the (2 , 1 , 1), (2 , 1 , 0) and (2 , 1 , - 1) states of the hydrogen atom when exposed to a magnetic field in the z -direction. Our perturbation is thus: V = E mag = - μ z B = - e 2 m e c ˆ L z B . (1a) And the total Hamiltonian for the system is: H = H (0) + V (1b) Lets stop and think about the effect of the pertubation on the system. Our 0 th order states (the hydrogen atom wavefunctions of form R ( r ) Y m l ( θ, φ )) just happen to be eigenstates of the ˆ L z operator. All off-diagonal terms of the Hamiltonian matrix are zero, h l 0 , m 0 , n 0 |H| l, m, n i = E (0) n h l 0 , n 0 , m 0 | l, m, n i + - e 2 m e c B h l 0 , m 0 , n 0 | ˆ L z | l, m, n i = 0 for n 0 , l 0 , m 0 6 = n, l, m (1c) and the 1 st order energies correspond directly to the diagonal matrix elements. For a (2 , 1 , m ) state, the energy is: E = H (0) + V (1d) = h 2 , 1 , m |H (0) | 2 , 1 , m i + h 2 , 1 , m | V | 2 , 1 , m i (1e) = E (0) 2 + h 2 , 1 , m | - e 2 m e c ˆ L z B | 2 , 1 , m i (1f) Where E (0) 2 is the unperturbed energy of a state with n = 2. The ˆ L z operator acts on a general ( n, l, m ) wave function as follows: ˆ L z ψ nlm ( r, θ, φ ) = m ~ ψ nlm ( r, θ, φ ) . (1g) 1

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E | ~ B | E 2 , 1 , 0 E 2 , 1 , - 1 E 2 , 1 , 1 E 2 , 1 , 0 μ B B E 2 , 1 , 1 μ B B E 2 , 1 , - 1 Figure 1: Depiction of splitting due to an applied magnetic field, Zeeman split- ting . Allowing us to write the perturbed energy as: hHi = E (0) 2 + e B 2 m e c m ~ h 2 , 1 , m | 2 , 1 , m i = E (0) 2 + μ B B m. (1h) Subbing in the values of m = - 1 , 0 , 1 we see that the (2 , 1 , 0) state is unperturbed by the magnetic field while the (2 , 1 , 1) state is raised in energy and the (2 , 1 , - 1) state is lowered in energy. See the diagram in Figure 1 . 3. (a) In the absence of a perturbation field, the n = 2 states of the hydrogen atom are all degenerate, with energy E (0) 2 . As we saw in lecture, this means that the total Hamiltonian matrix, expressed in the basis of the unperturbed ψ (0) states, will have non-zero off-diagonal elements. Our job is to figure out these elements,
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