== Input == {{{ $COMPASS Title CH2O+ Molecule test run Basis ano-rcc-vtzp Geometry I 0.0 0.0 0.0 Br 0.0 0.0 2.4 END geometry group c(1) $END $XUANYUAN scalar heff 3 soint hsoc 2 $END $SCF RHF charge 0 spin 1 $END $TRAINT orbi hforb $END $CCSDSO MEMMEGA 40000 itriple 0 ifdebug 0 maxTcyc 100 threshT 10 $END $EOMIPSO MEMMEGA 40000 nroots 8 0 0 0 0 0 0 0 ifdebug 0 $END }}} == Output == grep 'EXCITED STATE' in the output, the following results will be found {{{ EXCITED STATE 1 ENERGY: 0.353732546294 -0.000000000000 9.625556430678eV EXCITED STATE 2 ENERGY: 0.372675469829 -0.000000000000 10.141019826274eV EXCITED STATE 3 ENERGY: 0.445807213072 -0.000000000000 12.131036659141eV EXCITED STATE 4 ENERGY: 0.460122479736 0.000000000000 12.520575050608eV EXCITED STATE 5 ENERGY: 0.504585100956 -0.000000000000 13.730465048273eV EXCITED STATE 6 ENERGY: 0.567408973367 -0.000000000000 15.439990324973eV EXCITED STATE 7 ENERGY: 0.572574989992 0.000000000000 15.580564849613eV EXCITED STATE 8 ENERGY: 0.600837510259 -0.000000000000 16.349627483387eV }}} which gives relative energies (eV) for states of IBr+ as: {{{ 0.0 0.515463396 2.505480228 2.89501862 4.104908618 5.814433894 5.955008419 6.724071053 }}} This could be compared with TD-DFT results for the first few states, which share the same configurations: {{{ 1 -4.1081 eV 49.9% Spin: |So,1> 1-th B2 -3.8266 -0.2815 0.0000 0.00 2 -4.1081 eV 49.9% Spin: |So,2> 1-th B2 -3.8266 -0.2815 0.0000 0.00 3 -3.5878 eV 49.5% Spin: |So,1> 1-th B1 -3.8266 0.2388 0.5203 4196.81 4 -3.5878 eV 49.5% Spin: |So,2> 1-th B1 -3.8266 0.2388 0.5203 4196.81 5 -1.6230 eV 49.9% Spin: |So,1> 2-th B2 -1.4243 -0.1987 2.4851 20043.67 6 -1.6230 eV 49.9% Spin: |So,2> 2-th B2 -1.4243 -0.1987 2.4851 20043.67 7 -1.2286 eV 49.9% Spin: |So,1> 2-th B1 -1.4243 0.1957 2.8795 23224.68 8 -1.2286 eV 49.9% Spin: |So,2> 2-th B1 -1.4243 0.1957 2.8795 23224.68 9 -0.0915 eV 94.6% Spin: |Gs,1> 0-th A1 0.0000 -0.0915 4.0166 32395.92 10 -0.0915 eV 94.6% Spin: |Gs,2> 0-th A1 0.0000 -0.0915 4.0166 32395.92 11 1.0921 eV 50.0% Spin: |S+,4> 1-th B2 1.3992 -0.3071 5.2002 41942.64 12 1.0921 eV 50.0% Spin: |S+,1> 1-th B2 1.3992 -0.3071 5.2002 41942.64 13 1.2737 eV 49.5% Spin: |S+,3> 1-th B2 1.3992 -0.1255 5.3818 43406.97 14 1.2737 eV 49.5% Spin: |S+,2> 1-th B2 1.3992 -0.1255 5.3818 43406.97 15 1.4780 eV 49.0% Spin: |S+,2> 1-th B1 1.3992 0.0788 5.5862 45055.34 16 1.4780 eV 49.0% Spin: |S+,3> 1-th B1 1.3992 0.0788 5.5862 45055.34 }}} The results show that for the first few states, both methods give very similar energies.