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北京石油化工学院2026年研究生招生接收调剂公告
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[求助] 已知动力学矩阵求声子振动频率

QE中有一个例子计算Si的声子谱,下面是si.dyn_G文件,请问如何手工实现动力学矩阵对角化得到声子振动频率呢?(声子频率在下方)

Dynamical matrix file
                                                                           
  1    2  2 10.2000000  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000
           1  'Si  '    25598.3672624059     
    1    1      0.0000000000      0.0000000000      0.0000000000
    2    1      0.2500000000      0.2500000000      0.2500000000

     Dynamical  Matrix in cartesian axes

     q = (    0.000000000   0.000000000   0.000000000 )

    1    1
  0.27605479  0.00000000    0.00000000  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.27605479  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.00000000  0.00000000    0.27605479  0.00000000
    1    2
-0.27549986  0.00000000    0.00000000  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000   -0.27549986  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.00000000  0.00000000   -0.27549986  0.00000000
    2    1
-0.27549986  0.00000000    0.00000000  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000   -0.27549986  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.00000000  0.00000000   -0.27549986  0.00000000
    2    2
  0.27605479  0.00000000    0.00000000  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.27605479  0.00000000    0.00000000  0.00000000
  0.00000000  0.00000000    0.00000000  0.00000000    0.27605479  0.00000000

     Diagonalizing the dynamical matrix

     q = (    0.000000000   0.000000000   0.000000000 )

**************************************************************************
     omega( 1) =       0.484378 [THz] =      16.157120 [cm-1]
(  0.254905  0.000000  0.128027  0.000000 -0.647018  0.000000 )
(  0.254905  0.000000  0.128027  0.000000 -0.647018  0.000000 )
     omega( 2) =       0.484378 [THz] =      16.157120 [cm-1]
(  0.659031  0.000000 -0.021585  0.000000  0.255367  0.000000 )
(  0.659031  0.000000 -0.021585  0.000000  0.255367  0.000000 )
     omega( 3) =       0.484378 [THz] =      16.157120 [cm-1]
(  0.026485  0.000000 -0.695085  0.000000 -0.127104  0.000000 )
(  0.026485  0.000000 -0.695085  0.000000 -0.127104  0.000000 )
     omega( 4) =      15.270849 [THz] =     509.380687 [cm-1]
( -0.489933  0.000000 -0.509869  0.000000  0.000000  0.000000 )
(  0.489933  0.000000  0.509869  0.000000  0.000000  0.000000 )
     omega( 5) =      15.270849 [THz] =     509.380687 [cm-1]
(  0.214690  0.000000 -0.206295  0.000000  0.641366  0.000000 )
( -0.214690  0.000000  0.206295  0.000000 -0.641366  0.000000 )
     omega( 6) =      15.270849 [THz] =     509.380687 [cm-1]
( -0.462465  0.000000  0.444383  0.000000  0.297740  0.000000 )
(  0.462465  0.000000 -0.444383  0.000000 -0.297740  0.000000 )
**************************************************************************
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