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xiuhui木蟲 (正式寫手)
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[求助]
算頻率出錯(cuò)。
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最近,我用G03計(jì)算一系列有機(jī)分子體系,當(dāng)烷基鏈增大時(shí),如C21,計(jì)算頻率就會(huì)出現(xiàn) CPHF failed to converge in LinEq2.錯(cuò)誤。不知道該如何解決,請(qǐng)高手指教。 計(jì)算方法如下。 ****************************************** %Nproc=4 Will use up to 4 processors via shared memory. %nproc=4 Will use up to 4 processors via shared memory. %chk=C11-xianan-C11.chk %mem=10GB ----------------------------------------------------- # b3lyp/6-311+g(d,p) opt(tight,nolinear) freq=noraman 錯(cuò)誤信息: 1 vectors were produced by pass991. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass992. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass993. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass994. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass995. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass996. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass997. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass998. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 13. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass999. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. PrismC: NFx= 2048 NFxT= 14 NFxU= 14. 1 vectors were produced by pass***. CPHF failed to converge in LinEq2. |
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地溝油冶煉專家
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參見第九項(xiàng) http://people.smu.edu/wzou/qcsc/faq/gaussian.html |

木蟲 (著名寫手)
歷史控+原著控

木蟲 (正式寫手)
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第九項(xiàng)是這樣的 9. 問(wèn): HF/DFT解析頻率計(jì)算不收斂,該如何解決? 答: 在原子數(shù)很少的情況下(通常5個(gè)原子以下),可以考慮改用數(shù)值頻率(Freq=Numerical或Freq=DoubleNumer);蛘咝薷目刂艭PHF/CPKS方程收斂的IOP選項(xiàng),有以下幾個(gè)。在單點(diǎn)能輸出之后,找到類似于下面的輸出(如果計(jì)算加上了#P開關(guān),位于"(Enter .../g09/l1002.exe)"的下一行): Minotr: Closed shell wavefunction. IDoAtm=111 Direct CPHF calculation. Differentiating once with respect to electric field. with respect to dipole field. Differentiating once with respect to nuclear coordinates. Using symmetry in CPHF. Requested convergence is 1.0D-10 RMS, and 1.0D-09 maximum. Secondary convergence is 1.0D-12 RMS, and 1.0D-12 maximum. 最后兩行定義了解CPHF/CPKS方程的四個(gè)收斂標(biāo)準(zhǔn)。 第一個(gè)標(biāo)準(zhǔn)由IOp(10/7=N)控制:10^(-N),默認(rèn)N=8 第二個(gè)標(biāo)準(zhǔn)由系統(tǒng)設(shè)置,等于第一個(gè)標(biāo)準(zhǔn)乘10 第三、四個(gè)標(biāo)準(zhǔn)通常相等,由IOp(10/16=N)控制:10^(-N),默認(rèn)N=12。用于溶劑計(jì)算。 如果是氣態(tài)分子計(jì)算,把IOp(10/7)改小一點(diǎn)就行了。其它可能用到的還有: IOp(10/73):可以增大CPHF方程的迭代次數(shù),默認(rèn)1000 IOp(10/18=3):CPHF方程僅迭代一次,不檢查收斂?梢杂脕(lái)檢查定性結(jié)果,通常會(huì)帶來(lái)幾個(gè)到幾十個(gè)波數(shù)的振動(dòng)頻率誤差,不能用于正式計(jì)算。 另外還需要注意,IOP不能在多步任務(wù)中傳遞,因此如果使用上面的IOP選項(xiàng),結(jié)構(gòu)優(yōu)化和頻率計(jì)算要分開算。 我算的大分子體系,針對(duì)大分子有什么解決方案嗎? |
木蟲 (正式寫手)
專家顧問(wèn) (職業(yè)作家)
地溝油冶煉專家
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專家經(jīng)驗(yàn): +458 |
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頻率計(jì)算加上開關(guān)iop(10/33=1),可以判斷屬于以下情況中的哪一種。 如果CPHF有收斂趨勢(shì),可以降低收斂閾值IOp(10/7),或者增加迭代次數(shù)IOp(10/73),也可以同時(shí)設(shè)置這兩個(gè)參數(shù)。 如果CPHF沒(méi)有收斂趨勢(shì)(通常迭代是發(fā)散的)。少數(shù)情況下是數(shù)值誤差造成的,用同樣的結(jié)構(gòu)重新做頻率計(jì)算,運(yùn)氣好的話也許能避免這個(gè)問(wèn)題。一般的解決辦法是換理論方法(泛函和基組)重新優(yōu)化結(jié)構(gòu)和計(jì)算頻率。如果原子數(shù)在20以下,也可以算半數(shù)值頻率(Freq=Numerical),但這通常比解析頻率耗時(shí)。 |

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