Dear
Ankur, I
found a mistake/typo in my previous post to you, and I hope I haven’t
confused you with this. What I meant is that in the case of
the openshell diradical, the pure S = 0 state (not the the brokensymmetry
determinant) is
a mixture of 50% a(1)b(2)> and 50% b(1)a(2)> (a and b symbolising alpha
and betaspin for the first and second particle in different
spatial orbitals, respectively). The openshell diradical is therefore a linear
combination
of
two determinants. The brokensymmetry determinant, which only uses say
thea(1)b(2)> part, is likewise
a linear combination of the singlet and triplet states with equal weights.
Hence a(1)b(2)> = SQRT(1/2)0,0> +
SQRT(1/2)1,0>. The
coefficients SQRT(1/2) in this case (and therefore weights) can be obtained from
tables (Wigner coefficients,
ClebschGordon coefficients). Similar arguments then apply to spin states other
than the one discussed above, ie. the
Ms
= ½ (not Ms = 3/2 as I wrote) brokensymmetry doublet. This is what I meant
to say, hope
this makes more sense now. Best, Tobias Dr.
Tobias Krämer Lecturer
in Inorganic Chemistry Department
of Chemistry Maynooth
University, Maynooth, Co. Kildare, Ireland. E:
tobias.kraemer
#at# mu.ie
T:
+353 (0)1 474 7517 From:
ownerchemistry+tobias.kraemer==mu.ie #at# ccl.net
<ownerchemistry+tobias.kraemer==mu.ie #at# ccl.net>
On Behalf Of Tobias Kraemer Tobias.Kraemer],[mu.ie Dear
Ankur, I
guess the formulation in the book is a bit misleading, and I would rather call
this the weighted average. You can derive the eigenvalue
of S^2 from the individual contributions of the uncoupled electronic states to
the brokensymmetry wavefunction. The
contribution of each of these states is given by the Wigner coefficients (or
ClebschGordon coefficients), which tell you how these
spin vectors are coupled together. In the case of the openshell diradical, the
Ms = ½ state (the brokensymmetry determinant) is
a mixture of 50% a(1)b(2)> and %50 b(1)a(2)> (a and b symbolising alpha
and betaspin for the first and second particle, respectively).
This stems from the fact that the alphabeta combination is: a(1)b(2)> =
SQRT(1/2)0,0> + SQRT(1/2)1,0> (see Tables
for vector coupling) This
indicates that the BS solution is a 50:50 mixture of the Ms=0 components of the
singlet and triplet states. So the expectation
value of <S^2> could be obtained from 0.5x0.0 + 0.5x2.0 =
0.5(0.0+2.0) = 1.0, which is the average of both. When more than 2 spin
centres
are
involved, things become somewhat more complex, but follow the same procedure.
For the Ms=3/2 case (the brokensymmetry doublet)
the expansion (using the vector coupling tables) involves 66% 1/2,1/2>
(for which S^2 of the pure doublet state is 0.75), mixed with 33%
3/2,1/2> (with
3.75 for the ‘pure’ quartet state). Hence, this is not simply a
50% mix any longer, but you should be able to work it out from
here. At least I think this is the way the author in the book chapter you have
cited looks at this. Hope
this clarifies it. Tobias Dr.
Tobias Krämer Lecturer
in Inorganic Chemistry Department
of Chemistry Maynooth
University, Maynooth, Co. Kildare, Ireland. E:
tobias.kraemer #at# mu.ie
T: +353 (0)1 474 7517 From:
ownerchemistry+tobias.kraemer==mu.ie #at#
ccl.net <ownerchemistry+tobias.kraemer==mu.ie #at# ccl.net>
On Behalf Of Ankur Gupta ankkgupt,,iu.edu Dear Dr. Krämer Thank you for your reply. I was aware of the equation for
S^2 (UHF) in Szabo. However, I was a bit confused by the language used in a few
books and papers that I came across. Specifically, the paragraph in the
following book (Page 238 of
Molecular Water Oxidation Catalysis: A Key Topic for New Sustainable Energy
Conversion Schemes, Editor Antoni Llobet), explicitly states the following, "For a
brokensymmetry singlet contaminated by a triplet state, we would expect
HS<
S2
>
to be 2 (the
correct eigenvalue for a triplet state) and
BS<
S2
>
to be
about 1 (the average of a singlet and a triplet). In the same way, for a
brokensymmetry doublet contaminated by a quartet state,
we would expect HS<
S2
>
to be 3.75
(the correct eigenvalue for a quartet) and
BS<
S2
>
to be 1.75
(the
average of a doublet and a quartet)." Best, Ankur On Tue, Jun 18, 2019 at 3:26 PM Tobias Kraemer
Tobias.Kraemer]![mu.ie
<ownerchemistry..ccl.net> wrote:
