I am reading a text about the splitting of the energy levels in crystals caused by the spin orbit interaction. In particular, the argument is treated from the point of view of the group theory.(adsbygoogle = window.adsbygoogle || []).push({});

The text starts saying that a representation (TxD) for the double group can be obtained from the direct product of an irreducible representation T of the single group and an irreducible representation of the full rotational group D. Then, it concludes that:

_If the spin orbit term is neglected in the Hamiltonian, then (TxD) is irreducible, and every eigenvalue that was p-degenerate without considering the spin becomes 2p-degenerate.

_If the spin orbit term is included in the Hamiltonian, then (TxD) can be in general reducible, and the number and dimensions of the irreducible representations contained in this reducible representation determines the splitting of the energy levels due to the spin-orbit interaction.

Now the question is: given that the double group of the Hamiltonian operator and its representation (TxD) are the same with and without considering the spin orbit term, how can it be that (TxD) is irreducible neglecting the spin orbit term, and can become reducible by considering it?

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# Spin orbit and double group representations

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