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gongchangjie

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[交流] 【讨论】讨论一下,用ms计算系统能量问题 已有4人参与

用ms计算系统能量问题,计算得出的能量值有三个,
Final energy E=....
Final free energy E-TS=......
NB est.0K energy(E-0.5TS)=......
,这三项各代表什么意思?
如果选择系统最小能量,应该是那一项?谢谢,欢迎讨论,拍砖

[ Last edited by gongchangjie on 2010-6-2 at 10:08 ]
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夕阳西下

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ym23(金币+1):欢迎常来交流 2010-09-21 23:27:58
系统最小能量应该是Final free energy E-TS
4楼2010-09-21 22:28:12
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gongchangjie

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555,怎么没有高手回应我一下啊
物理改变世界!世界因物理而精彩!
2楼2010-06-03 09:17:52
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guolianshun

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好像听人家说要看最后一个前边的运算过程中的变化值吧,我也想学习,高手呢?
3楼2010-09-21 15:59:25
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wanglianli136

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aylayl08(金币+1):谢谢信息 2010-09-22 09:12:37
zxzj05(金币+1):谢谢回帖交流 2010-09-22 10:51:03
help里有
The CASTEP output is as follows. The line

Final energy, E             =  -861.8315373731     eV

gives the Kohn-Sham energy, EKS, which ignores the entropy. However, it is the total Mermin free energy that is minimized by CASTEP and reported in the SCF cycles in the line

Final free energy (E-TS)    =  -861.8315374521     eV

If you were actually interested in the electronic states at some reasonable nonzero temperature, you would smear the Fermi level with a thermal (i.e., Fermi-Dirac) smearing method and the free energy reported by CASTEP would be the value you require.

However, it is often the T = 0 K value that is required and the smearing is necessary to improve the convergence of the numerical algorithm. As the smearing ('temperature') is increased, the TS term increases and the free energy, EKS - TS, goes down. At the same time, as the entropy contributes more and more to the free energy, the electronic minimization pays less attention to the EKS term, so the final EKS increases. Because the dependence of the free energy and the Kohn-Sham energy on the smearing width is the same (to first order), averaging the two values gives a much better approximation to the T = 0 K result:

NB est. 0K energy (E-0.5TS) =  -861.8315374126     eV
下面两个能量应该是因为smearing的原因
5楼2010-09-22 09:09:17
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