How to select the good basis for the special Hamiltonian?

In summary, the conversation discusses how to select the most appropriate operators to work with for the Hamiltonian, which includes spin-orbit coupling. The set of operators J^2, J_z, L^2, and S^2 are found to be the most suitable, as they commute with the first two terms of the Hamiltonian and are conserved quantities in the presence of spin-orbit coupling. This is determined through analyzing the Hamiltonian in the absence and presence of spin, and finding the appropriate quantum numbers to describe the eigenstates.
  • #1
hokhani
504
8
How to select the good basis for the special Hamiltonian??

For the Hamiltonian [itex] H=\frac{P^2}{2\mu} -\frac{Ze^2}{r}+ \frac{\alpha}{r^3} L.S[/itex] (which we can use [itex] L.S=\frac{1}{2} (J^2-L^2-S^2)[/itex]in the third term) how to realize that the third term,[itex]\frac{\alpha}{r^3} L.S[/itex], commutes with sum of the first two terms,[itex] \frac{P^2}{2\mu} -\frac{Ze^2}{r}[/itex], and then conclude that the set of the operators [itex] J^2, J_z, L^2, S^2[/itex] are the best to work with? ([itex] \alpha, Z, e, \mu[/itex] are constants)
 
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  • #2
Usually, you arrive at it by steps. You first have the Hamiltonian in the absence of spin, from which you find that you have quantum numbers ##n##, ##l##, and ##m_l##, the latter usually chosen to be the projection along z. When you add spin, you find that the Hamiltonian is independent of spin, so you describe the eigenstates with
$$
\left| n, l, m_l, s, m_s \right\rangle
$$

When you add spin-orbit coupling, you realize that ##m_l## and ##m_s## are no longer good quantum numbers (conserved quantities), and you try to figure out what is still conserved. the total angular momentum ##j## and its projection ##m_j## are, so you switch to
$$
\left| n, l, s, j, m_j \right\rangle
$$

Note that ##j## and ##m_j## were already good quantum numbers, but since they are redundant with ##l##, ##m_l##, ##s##, and ##m_s## in the absence of spin-orbit coupling, they are not explicitely mentionned.
 
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FAQ: How to select the good basis for the special Hamiltonian?

What is the importance of selecting a good basis for the special Hamiltonian?

The basis for the special Hamiltonian is important because it affects the accuracy and efficiency of the calculations. A good basis can lead to more reliable results and a faster computation time.

How do I determine the best basis for a special Hamiltonian?

The best basis for a special Hamiltonian can be determined by considering the symmetry and structure of the system. The basis should also be chosen to minimize the number of basis functions needed to accurately describe the system.

Can I use a pre-existing basis for a special Hamiltonian?

Yes, pre-existing bases such as plane waves or localized orbitals can be used for a special Hamiltonian. However, they may not be the most efficient or accurate choice for every system.

How do I know if my chosen basis is sufficient for the special Hamiltonian?

A common method for determining the sufficiency of a basis is to increase the size of the basis and check if the results significantly improve. Additionally, comparing the results with experimental data can also indicate if the basis is sufficient.

What are the trade-offs between using a small or large basis for a special Hamiltonian?

A small basis may reduce the computation time, but it may also sacrifice accuracy. On the other hand, a large basis may provide more accurate results, but it may also increase the computation time. The choice ultimately depends on the specific system and desired level of accuracy.

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