2
$\begingroup$

While studying notes from a couple years ago, I encountered a diagram that I don't have further information on but would like to understand.

While I believe I managed to find sources to guide me through the details of the calculation, there are a few conceptual questions I had, which are summarized at the bottom.

The diagram is for a $d^1$ octahedral complex under a magnetic field:

enter image description here

a) ligand field splitting

b) Spin-Orbit coupling

c) First-order Zeeman splitting

d) second-order Zeeman splitting

a)

The ligand-field splitting is straightforward to me.

b)

The spin-orbit coupling I believe I understand.

The $E_g$ level does not have an electron, and therefore does not undergo any spin-orbit coupling.

For the $t_{2g}$ energy level splitting, I followed the procedure covered on pages 9 and 10 here (the formula is also given on the wiki). The formula for spin-orbit coupling is:

$$\Delta E=\frac{B}{2}[j(j+1)-l(l+1)-s(s+1)]$$

For a $^2D$ term, $l=2$ and $s=\frac{1}{2}$. Then $\Delta E$ would be:

$$j=\frac{5}{2}\therefore \Delta E=\frac{3}{2}B$$.

$$j=\frac{3}{2}\therefore \Delta E=-B$$

I assume in the diagram, $\zeta=B$.

  • Why does the diagram obtain different coefficients ($-\frac{1}{2}\zeta$ and $\zeta$)? Did I do this wrong?

c)

I think I might have understood this one, using the following link. I'll be referencing equations from this link for these two sections. I'm substituting the Bohr Magneton, $\mu _B$ with simply $\mu$, as it renders better this way, and replacing $B_{ext}$ with $B$ for the same reason.

In deriving equation 8.15 from the link, I am confused what is meant by the line

let the hermitian, unperturbed hamiltonian $H_0^\dagger =H_0$ "act to the left" on $\langle n^0|$

I have never used an operator to "act to the left." I've always applied it toward the right.

Assuming it somehow works out to $E_n^0$, I understand how theorem 8.1 is obtained:

$$E_n^1=\langle n^0|H_I|n^0\rangle$$

They apply the above theorem in equation 8.25 with $H_I=H_z$, where

$$H_z=\frac{\mu}{\hbar}(L_z+2S_z)B$$

to obtain:

$$\Delta E =\mu (m_l+2m_s)B$$

In my diagram above, $m_l=2$ and $m_s=\frac{1}{2}$, so

$$\Delta E = \mu (2\pm 1)B=3\mu B-\mu B$$

$$= 2\mu B$$

That seems to match the diagram (where $B=H$).

  • Why does the $E_g$ level split into two double degenerate levels, and the ground state $\Gamma _8$ (4) not split at all?

d)

For second-order splitting, the linked PDF provided this formula (theorem 8.3):

$$E_n^{(2)}=\sum_{m\ne n}\frac{\left| \langle m^0|H_I|n^0\rangle \right| ^2}{E_m^0-E_n^0}$$

I was unsure how this relation was true:

$$(\langle n^0|H_I|m^0\rangle )(\langle m^0|H_I|n^0\rangle )=\left| \langle m^0|H_I|n^0\rangle \right| ^2 $$

but otherwise followed the derivation. Accordingly, the numerator indicates the 2nd order zeeman correction is simply the square of the first order correction; however, the denominator I did not follow, namely:

$$E_m^0-E_n^0=3\zeta$$

  • Where does this $3\zeta$ come from?
  • Similar to the first-order splitting, why do the energy levels experience different types of splitting?

summary of questions

  1. In section b), why are my splitting coefficients different from the diagram?

  2. In section c) of the diagram, why is the highest-lying energy level in the diagram split into two double degenerate levels, the middle-lying energy level split into two singly degenerate levels, and the lowest-lying energy level not split at all? I assume the answer to this question will also explain why the second-order splittings in section d) are also inconsistent with each other.

  3. In section d) of the diagram, how does the $3\zeta$ term in the denominator for the energy splitting come about?

  4. What is responsible for certain energy levels being doubly degenerate after following splitting from the Zeeman Effect?

  5. Where does the term symbol splitting according to the projection of total angular momentum along the axis of the magnetic field, ie. $j_z=j,j-1,...-j$ fit in here? Aren't these also separated in a magnetic field?

$\endgroup$

closed as too broad by Avnish Kabaj, A.K., Todd Minehardt, Waylander, LDC3 Aug 26 '18 at 23:11

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ this might be a bit too broad due to the number of questions. I think they would be better split up and linked together. $\endgroup$ – Tyberius Aug 25 '18 at 17:42
1
$\begingroup$

To answer part b it is necessary to consider the symmetry properties in an octahedral field. The spin orbit Hamiltonian is $H_i=\zeta\sum_i\bar l_i\cdot \bar s_i$ where $\zeta$ is a constant characteristic of the radial part of the d orbitals. The orbital angular momentum, which may be written in the form (for the $x$ component) as $\displaystyle l_x =-i\hbar \left(y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y}\right)$ (and similarly for $l_y$ and $l_x$ by rotating subscripts) transforms as does the $T_{1g}$ irreducible representation in an $O_h$ field.

If a term has an orbital symmetry $X$ it will only be susceptible to spin orbit coupling if the symmetry product $X\cdot T_{1g}\cdot X$ contains the totally symmetric representation $A_{1g}$. In other notation $\langle i|H |j\rangle \ne 0$ or $\int \psi_i H \psi_f d\tau \ne 0$). Thus the $E_g$ levels are unaffected in point group $O_h$ because $E_g \cdot T_{1g}\cdot E_g$ does not contain $A_{1g}$ but the $T_{2g}$ interact ($T_{2g} \cdot T_{1g}\cdot T_{2g}$ does contain $A_{1g}$).

Notes: It is assumed that the potential energy effect of the ligands is only small compared to that of the central field of the ion, thus the spin-orbit coupling is a small perturbation on the energies.

The full derivation is complex and involves using double symmetry groups.

The free ions have SO couplings $^2\mathrm D_{3/2} =-3\zeta_d/2$ and $^2\mathrm D_{5/2} =\zeta_d$. (The formula you use if for the free ion). In the octahedral field the calculation is rather complicated but the result is that $T_{2g}$ levels are split by $3\zeta/2$ in the ratio you show.

In the remaining parts of your question similar symmetry arguments should indicate what levels are split, but only a detailed calculation will produce the energies that they are split by.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.