one question about the result of self-consistent calculation

[Tom]

hello everyone
  I calculate the spin system --A  two-probe system of the graphene nanoribbon .In the out file list some datas,for example

 0  C     q =  2.02095 [ s:  0.491, py:  0.474, pz:  0.507, px:  0.514, d(-2):  0.003, d(-1):  0.006, d( 0):  0.009, d( 1):  0.011, d( 2):  0.005 ]

 I want to know  the meaning about s,kx,ky,kz , d(-2)......
 If s,kx,ky,kz represent the spin of the orbit,what's the meaning of d() ?

[zh]

Please see here for more details of atomic orbital:
http://en.wikipedia.org/wiki/Atomic_orbital
s orbital: l=0, m=0;
p orbitals: l=1, m =0 for p_z; l=1, m= -1 for p_x; l=1, m=+1 for p_z;
d orbitals: l=2, m = 0, +-1; +-2;  i.e, m =0 for d_z²; m = -1 for d_xz; m = +1 for d_yz; m = -2 for d_xy; m= +2 for d_x²-y².

0  C     q =  2.02095 [ s:  0.491, py:  0.474, pz:  0.507, px:  0.514, d(-2):  0.003, d(-1):  0.006, d( 0):  0.009, d( 1):  0.011, d( 2):  0.005 ]

These are the Milliken population charge for each orbital of C atom. "s" means the s orbital, "py" for the ^p_y orbital; and so on. In particular, the number 'x" in "d(x)"  stands for the magnetic quantum number of d orbital. As explained above, "d(-2)" stands for the d_xy orbital.

[Tom]

Thank you for your answer.But C atom doesn't have d orbital.How to explain this phenomena?

[zh]

Do you observe that in the calculated results the components for the d orbitals are quite small? They can be ignored. The 3d orbital of carbon atom are unbound states.