Multiwfn forum

ridstk

Member

Registered: 2022-03-03

Posts: 3

spin density

Hi,
I am interested to find out the spin density on each fragment of the molecule. Is this possible to calculate the value of spin density from each fragment?

sobereva

Tian Lu (Multiwfn developer)

From: Beijing

Registered: 2017-09-11

Posts: 1,468

Website

Re: spin density

Please distinguish between spin density and spin population, they are very different. What you want to obtain is spin population of fragments.

There are many different ways to compute spin population. As an example, we calculate spin population of OH fragment of examples\ethanol_triplet.fch, boot up Multiwfn and input

examples\ethanol_triplet.fch
15  //Fuzzy atomic space analysis
-5  //Define the atoms to be considered in options 1, 2, 13
8,9  //The atoms in the fragment
1  //Perform integration in fuzzy atomic spaces for a real space function
5  //Spin density

Then spin density is integrated within spaces of atoms 8 and 9, you can see

  Atomic space        Value                % of sum            % of sum abs
    1(C )            0.00000000             0.000000             0.000000
    2(H )            0.00000000             0.000000             0.000000
    3(H )            0.00000000             0.000000             0.000000
    4(H )            0.00000000             0.000000             0.000000
    5(C )            0.00000000             0.000000             0.000000
    6(H )            0.00000000             0.000000             0.000000
    7(H )            0.00000000             0.000000             0.000000
    8(O )            1.06453241            69.848730            69.848730
    9(H )            0.45952165            30.151270            30.151270
Summing up above values:          1.52405406
Summing up absolute value of above values:          1.52405406

The spin population of the fragment, namely the sum of spin population of O8 and H9, is the 1.524 shown above.

ridstk

Member

Registered: 2022-03-03

Posts: 3

Re: spin density

I got it. Thank you for your reply.