C.W. Ufford and G.H. Shortley in Physical Review 1932, Vol. 42, pg. 167, separate the two $^2$D's of $d^3$. On page 173 (pg 7 of the PDF) of this article they determine the matrix of the electrostatic energy in eqn. 3.3 as: $$\begin{array}{ccc} - &a & b \\ a & 3F_0+7F_2+63F_4 & 3(21)^{1/2}(F_2-5F_4) \\ b & 3(21)^{1/2}(F_2-5F_4) & 3F_0+3F_2-57F_4 \\ \end{array}$$

I'm trying to figure out how they actually determine this matrix. How do they do it?

Here is what I understand:

This matrix comes from a description of the two terms, $$^2D^a_{2,1/2} = 1/2 [-A+B-C+E]$$ $$^2D^b_{2,1/2} = (84)^{-1/2}[-5A-3B-C+4D-3E-2(6)^{1/2}F]$$

where $A,B,C,D,E,F$ are zero-order states of $d^3$ which are $$\begin{array}{cccc} A &\mathcal{A}u_1(2^+)u_2(2^-)u_3(-2^+) & B &\mathcal{A}u_1(2^+)u_2(1^+)u_3(-1^-) \\ C &\mathcal{A}u_1(2^+)u_2(1^-)u_3(-1^+) & D &\mathcal{A}u_1(2^-)u_2(1^+)u_3(-1^+) \\ E &\mathcal{A}u_1(2^+)u_2(0^+)u_3(0^-) & F &\mathcal{A}u_1(1^+)u_2(1^-)u_3(0^+) \\ \end{array}$$

In this notation $u_i(a^j)$ indicates the one-electron central field eigenfunction for electron $i$ with the $j^{th}$ set of numbers $n^il^im_l^im_s^i$. $\mathcal{A}$ represents the antisymmetrizing operator $(N!)^{-1/2}\sum_P (-1)^p P$.

We can say that $(ab|1/r_{12}|cd)$ depends, in part, on $\sum_{k=0}^\infty c^k (l^a m_l^a, l^c m_l^c) c^k(l^dm_l^d,l^bm_l^b)$

There are tables of these values already determined, notably as $a^k$ and $b^k$, which can be used to evaluate the Coulomb and exchange integrals for $k=0,2,4$.

E.U. Condon then extended Slater's method to determine the non-diagonal matrix elements in a paper (Physical Review, 1930, Vol. 36, pg. 1121) for separating multiplets of the same kind, as we have in our case.

In case you are confused by the notation, $J(a,b)=\sum a^k$ and $K(a,b)=\sum b^k$ and $a^k(l^am_l^a,l^bm_l^b)=c^k(l^am_l^a,l^am_l^a)c^k(l^bm_l^b,l^bm_l^b)$ and $b^k(l^am_l^a,l^b m_l^b) = [c^k(l^am_l^a,l^bm_l^b)]^2$.

When we get to dealing with the doublet D's, we see that they depend on the following wavefunctions: $(2^+,2^-,-2^+), (2^-,1^+,-1^+), (2^+,0^+,0^-), (2^+,1^+,-1^-), (1^+,1^-,0^+), (2^+,1^-,-1^+) $ -- or $A,B,C,D,E,F$.

Using Slater's method we find that $E(^2H)+ E(^2G)+E(^4F)+ E(^2F)+E(^2D)+E(^2D)=18F_0-13F_2-152F_4$ and,

$E(^2H)+ E(^2G)+E(^4F)+ E(^2F)=12F_0 -23F_2 - 158F_4$

Therefore, $E(^2D)+E(^2D)= 18F_0-13F_2-152F_4 - (12F_0 -23F_2 - 158F_4) = 6F_0+10F_2+6F_4$. Thus the mean energy is $3F_0 + 5F_2 + 3F_4$.

But I don't understand how to get to the non-diagonal elements yet...

  • $\begingroup$ what are D and d? we don't have the article, so if you provided some more context that might be useful $\endgroup$ – Timtam Aug 10 '11 at 21:59
  • 1
    $\begingroup$ "$^2$D" is a Russell-Saunders term symbol, a common description of the angular momentum quantum numbers in spectroscopy. $d^3$ as in for an $d$ orbital. $\endgroup$ – Chris Aug 11 '11 at 20:21

We'll look at just the $aa$ case.

$ ^2D^a$ by $^2D^a$ $= 1/4[(A|F|A)+(B|F|B)+(C|F|C)+(E|F|E) -2(A|G|B) + 2(A|G|C)$ $ - 2(A|G|E) - 2(B|G|C) + 2(B|G|E) - 2(C|G|E)]$

We'll convert to two-electron operators ( for example, $((2^+,2^-,-2^+)|G|(2^+,1^+,-1^-))$ becomes only $((2^-,-2^+)|G|(1^+,-1^-)$ ). Recall that for $(ab|F|cd) = (ab|F|cd) - (ab|F|dc)$. Mind you, which integrals to actually evaluate depends on the spins, so ++ = ++, -- = --, +-=+- and -+=-+ are non-zero whereas the rest are 0 for each of these two electron-operators. This gives us (not showing diagonal element calculation since you seem to have that under control):

$=1/4[3F_0+12F_2-67F_4+3F_0-9F_2+3F_4+3F_0-3F_2-27F_4+3F_0-8F_2+33F_4$ $+$ non-diagonal elements

$= 1/4[12F_0-8F_2-58F_4$

$ -2((2^-,-2^+)|G|(-1^-,1^+)) $






Now we evaluate each integral,



$((2^-,-2^+)|G|(0^-,0^+)) =c(2,0)c(0,-2)=-(4F_2+15F_4)$




Substituting in these values gives us a grand total of $3F_0+7F_2+63F_4$


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