- #1
Marco99
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- Homework Statement
- Hello everyone,
So I have a QM assignment in which I have a s=1/2 particle with an unperturbed hamiltonian H diagonal in the |l,l_{z}>|s, s_{z}> base of the form H = A*L^2 + B*S_{z} (with 0 < B << A), and a perturbation term H' of the form H'=const * < L | S >, which can be shown to be diagonal in the |l, s; j, j_{z}> base. I am asked to calculate the energy corrections for the three lowest energy levels in the presence of the perturbation, and I applied time independent perturbation theory.
- Relevant Equations
- None
The lowest two energy level corrections (l=0, s_{z}=-1/2 and l=0, s_{z}=1/2) are easy to work out since the eigenvalues are not degenerate and the unperturbed energy levels also happen to be eigenstates of H'.
However I have three degenerate energy levels for the third eigenvalue of the form |l=1, l_{z}=0, +1, -1>|s=1/2, s_{z}=-1/2>. The l_{z}=-1 case is trivial, since it corresponds to the |l=1, s=1/2; j=3/2, j_{z}=-3/2> state, which is also an eigenstate of H'.
The l_{z}=0 and l_{z}=1 case is less trivial, so I used time independent perturbation theory in the degenerate case to diagonalise the H' matrix in the unperturbed state base |l=1, l_{z}=0, +1>|s=1/2, s_{z}=-1/2>.
The thing is in this base H' is already represented by a diagonal matrix since, calling |n_0}> and |n_1}> the two unperturbed states with l_{z}=0 and l_{z}=1 and expressing these in the |j, j_{z}> base, it happens that the off-diagonal terms <n_0 | H' | n_1> = <n_1 | H' |n_0> = 0, because |j, j_{z}> states with different j_{z} values are orthonormal.
Which means that the unperturbed states |n_0> and |n_1> already correspond to the perturbed eigenstates, which makes no sense since these two are not H' eigenstates at all. What am I missing?
However I have three degenerate energy levels for the third eigenvalue of the form |l=1, l_{z}=0, +1, -1>|s=1/2, s_{z}=-1/2>. The l_{z}=-1 case is trivial, since it corresponds to the |l=1, s=1/2; j=3/2, j_{z}=-3/2> state, which is also an eigenstate of H'.
The l_{z}=0 and l_{z}=1 case is less trivial, so I used time independent perturbation theory in the degenerate case to diagonalise the H' matrix in the unperturbed state base |l=1, l_{z}=0, +1>|s=1/2, s_{z}=-1/2>.
The thing is in this base H' is already represented by a diagonal matrix since, calling |n_0}> and |n_1}> the two unperturbed states with l_{z}=0 and l_{z}=1 and expressing these in the |j, j_{z}> base, it happens that the off-diagonal terms <n_0 | H' | n_1> = <n_1 | H' |n_0> = 0, because |j, j_{z}> states with different j_{z} values are orthonormal.
Which means that the unperturbed states |n_0> and |n_1> already correspond to the perturbed eigenstates, which makes no sense since these two are not H' eigenstates at all. What am I missing?
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