iOpenShell » Electronic Structure and Quantum Chemistry » Symmetry in exited states

Symmetry in exited states

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Member
Registered: Nov 2012
Posts: 3
Hi,

Since I guess my last question was written in a too confused way, here's the question that's bothering me boiled down.

Why is symmetry important when dealing with open shell systems?

All I find is how it can be treated, but I don't get why the (spacial and spin) symmetry has to be included into calculations. I always thought using symmetries just reduces calculation time by eliminating the corresponding parts, but it seems here it has an actual influence on the result.
How do I know an excited state isn't just completely asymmetrical?

Please advice, this is driving me crazy :'(

Greetings,
Sera
Member
Registered: Nov 2012
Posts: 3
Ah, finally found it. Degeneracies. Koch-Holthausen chapter 5.3.6.
Member
Registered: Aug 2015
Posts: 1
Glad to hear from you. :)

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