[QE-users] Why do I get multiple imaginary frequencies using nat-todo in QE 6.4
Yang-Caiyu
yangcaiyu at buaa.edu.cn
Fri Nov 18 10:08:34 CET 2022
从 Windows 版邮件发送
Dear all,
I have been running the frequency calculation of transition state of NH3* → NH2* + H* (* means the substances adsorbed on the Pt(111)). In the calculation process, I fixed Pt atoms by using nat-todo except the NH3 molecule and 2 Pt atoms bonded with it. But I get the results obtained 18 real frequencies and 18 imaginary frequencies. The results are as follows:
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freq ( 1 - 1) = -494.9 [cm-1] --> A I+R
freq ( 2 - 2) = -224.9 [cm-1] --> A I+R
freq ( 3 - 3) = -204.8 [cm-1] --> A I+R
freq ( 4 - 4) = -127.3 [cm-1] --> A I+R
freq ( 5 - 5) = -102.5 [cm-1] --> A I+R
freq ( 6 - 6) = -77.1 [cm-1] --> A I+R
freq ( 7 - 7) = -66.7 [cm-1] --> A I+R
freq ( 8 - 8) = -48.3 [cm-1] --> A I+R
freq ( 9 - 9) = -38.3 [cm-1] --> A I+R
freq ( 10 - 10) = -12.9 [cm-1] --> A I+R
freq ( 11 - 11) = -6.2 [cm-1] --> A I+R
freq ( 12 - 12) = -2.8 [cm-1] --> A I+R
freq ( 73 - 73) = 86.1 [cm-1] --> A I+R
freq ( 74 - 74) = 221.7 [cm-1] --> A I+R
freq ( 75 - 75) = 313.7 [cm-1] --> A I+R
freq ( 76 - 76) = 439.3 [cm-1] --> A I+R
freq ( 77 - 77) = 504.5 [cm-1] --> A I+R
freq ( 78 - 78) = 692.8 [cm-1] --> A I+R
freq ( 79 - 79) = 745.6 [cm-1] --> A I+R
freq ( 80 - 80) = 833.3 [cm-1] --> A I+R
freq ( 81 - 81) = 1123.9 [cm-1] --> A I+R
freq ( 82 - 82) = 1434.7 [cm-1] --> A I+R
freq ( 83 - 83) = 3250.8 [cm-1] --> A I+R
freq ( 84 - 84) = 3377.1 [cm-1] --> A I+R
I want to know what’s wrong in my calculation process? And my input files are available by the following link.
Thanks in advance
Caiyu Yang
Beihang University
https://www.aliyundrive.com/s/MeXbhArWjLL
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