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木虫 (正式写手)

应助: 10 (幼儿园)
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注册: 2008-11-21
专业: 金属材料的凝固与结晶学

[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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