What are the criteria that one has to consider while chosing the number of roots in a CASSCF calculation?

  • $\begingroup$ It all depends on your desired precision and the computational budget you have available. Can you provide more on the problem that you want to study? $\endgroup$ Jan 20 at 15:47
  • $\begingroup$ @AndreaPellegrini Sorry I didn't see your comment when I wrote my answer! The answer was posted 4 minutes after your comment, so I must have been working on it a couple dozen minutes earlier. I figured that the user simply needed to know that "number of roots" means "number of electronic states" meaning that you're correct about the computational budget being the main consideration to take into account. If I had infinite resources, I'd always include a huge amount of roots! $\endgroup$ Jan 21 at 4:05
  • $\begingroup$ Sure @NikeDattani. The only thing, I think that if you want to simulate an UV-Vis spectra it requires a certain number of "roots" to be taken into consideration, if you know that you to try find the conic intersaction of a [2π,2π] cycloaddiction you can "simply" consider just 2 "roots" $\endgroup$ Jan 21 at 9:08
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    $\begingroup$ @AndreaPellegrini Although it is supposed to be a genreal question, I am focused in the problems with open shell transition metal complexes in different spin state and their relative energies. With the increase in the nuber of the unoccupied active orbitals, the CI-expansions become practically unachiveable even with a decent amount of computational resources. I was looking for any kind of formalism that can guide through while chosing affordable number of electronic states for a CAS space. $\endgroup$ Jan 24 at 5:45

1 Answer 1


The term "roots" refers to the number of "electronic states". If you are only interested in the ground state, then you are seeking 1 root. If you want the ground state and the first 4 excited states, then your number of roots is 5. Basically the only factor that I consider when choosing the number of roots, is the computational cost (the calculation will take longer and need more computational resources if you want a lot of roots).

When I searched "roots casscf" on Google, the first 4 results from software package documentation were from (in order of appearance):


"Additional roots have to be included for excited state calculations. Then the number of roots (which corresponds to the number of states) is assigned by the nroots keyword (see State-averaging section below)."


"Calculations on excited states of molecular systems may be requested using the NRoot option. Note that a value of 1 specifies the ground state, not the first excited state (in contrast to usage with the CIS or TD keywords). State-averaged CASSCF calculations may be performed using the StateAverage and NRoot options to specify the states to be used."

Molcas 5 (this page, hosted by the lead author of Psi4, contains a lot useful examples and detailed explanations about choosing the number of roots!):

"we expect to compute six A$_1$ states and therefore we include six roots in the CASSCF state-average input. "

Molcas 8 (official current documentation, this page also contains examples and some explanations about how to work with roots)

"Keyword [CiRoot= 5 5 1] informs the program that we want to compute a total of five states, the ground state and the lowest four excited states at the CASSCF level and that all of them should have the same weight in the average procedure."

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    $\begingroup$ One caveat is that, if one of the degenerate components of a degenerate (or approximately degenerate) state is calculated, the other degenerate components of the same state must also be calculated. For example, to calculate the ground state of [Ti(H2O)6]3+ (3d1) one should include at least 3 roots, so as to fully capture the triply degenerate ground state without introducing any artificial symmetry breaking, even though the particular geometry may not be exactly symmetrical (therefore giving only an approximate degeneracy). Similar for excited states. $\endgroup$
    – wzkchem5
    Jan 21 at 9:51
  • $\begingroup$ Thank you Nike Dattani and @wzkchem5 for the clarification. $\endgroup$ Jan 24 at 5:49

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