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I am taking an introductory course in symmetry and bonding and have so far learnt about SALCs and various point groups etc... one of my problems asks me to show a $\mathrm{^{2}T_{2g}\rightarrow\ ^{2}E_g}$ is allowed in a $\mathrm d^1$ metal. I am not sure how my knowledge of symmetry can help me answer this, all my searching flags up difficult quantum mechanics/group theory I don't understand! I think a relevant example I could use is $\ce{[Ti(H2O)6]^{3+}}$ as this is an octahedral, $\mathrm d^1$ system.

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For a state-to-state transition coupled by an operator to be allowed, the direct product of the irreducible representations (irreps) of all three components must contain the completely symmetric irrep of the point group you're working in. That is, we are determining whether the following integral is forced to be zero by symmetry:

$$ \int \Psi_{i}(\tau) \times \hat{\mathbf{A}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \overset{?}{=} 0 $$

where $\Psi_{i}$ and $\Psi_{f}$ are the initial and final states, $\hat{A}$ is the operator (electric dipole, magnetic dipole, one-electron spin-orbit, there are many more), and $\tau$ is the integration coordinate, usually over electron position. Why bother with this if the integral can be evaluated exactly?

  1. It's no more complicated than this pencil-and-paper calculation to gain qualitative understanding about whether or not a system with some symmetry will have a certain kind of spectroscopic response.
  2. Why bother calculating the integral if it's going to be zero anyway? It's just extra work, though it usually isn't very expensive.

For $O_{\mathrm{h}}$, the completely symmetric irrep is $\mathrm{A_{1g}}$. It's always listed as the first irrep in a character table.

$$ \small\begin{array}{c|cccccccccc|cc}\hline O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline \mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2+z^2 \\ \mathrm{A_{2g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & -1 & & \\ \mathrm{E_g} & 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\ \mathrm{T_{1g}} & 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & -1 & (R_x,R_y,R_z) & \\ \mathrm{T_{2g}} & 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 & & (xy,xz,yz) \\ \mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\ \mathrm{A_{2u}} & 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 & & \\ \mathrm{E_u} & 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 & & \\ \mathrm{T_{1u}} & 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 & (x,y,z) & \\ \mathrm{T_{2u}} & 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 & & \\ \hline \end{array} $$

The magnetic dipole operator is the angular momentum operator with some constants in front:

$$ \hat{\mathbf{m}} = \gamma_{\mathrm{e}} \hat{\mathbf{l}} = -\frac{e}{2m_{\mathrm{e}}} \hat{\mathbf{l}} \approx \left(\hat{l}_{x}, \hat{l}_{y}, \hat{l}_{z}\right) $$

The angular momentum operator components have rotational symmetry, so $\hat{l}_{y}$ would use the irrep corresponding to $R_{y}$. Constant prefactors don't affect the symmetry of operators, so all three components of the magnetic dipole operator have $\mathrm{T_{1g}}$ symmetry.

Now we must show that

$$ \left< \Psi_{i} | \hat{\mathbf{m}} | \Psi_{f} \right> \equiv \int \Psi_{i}(\tau) \times \hat{\mathbf{m}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \rightarrow \mathrm{T_{2g}} \times \mathrm{T_{1g}} \times \mathrm{E_{g}} \overset{!}{\ni} \mathrm{A_{1g}}, $$

and to do this we need a product table.

$$\begin{array}{c|ccccc} \hline & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline \mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\ \mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\ \mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\ \mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline \end{array}$$

Product tables are symmetric, so only the lower or upper triangle needs to be shown, and inversion symmetry is left out here as long as you consider the rule

$$ \begin{align} \mathrm{g} \times \mathrm{g} &= \mathrm{g} \\ \mathrm{g} \times \mathrm{u} &= \mathrm{u} \\ \mathrm{u} \times \mathrm{u} &= \mathrm{g}, \end{align} $$

just like multiplying even and odd integers. Some product tables show all of the redundant information.

Working from left to right, find the product of the symmetry of the initial state with the symmetry of the operator:

$$ \mathrm{T_{2g}} \times \mathrm{T_{1g}} = \mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}. $$

The product of degenerate irreps with each other leads to multiple irreps, hence the sum. You might think that the next step is

$$ \left(\mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}\right) \times \mathrm{E_g} = \dots, $$

however we take advantage of two nice rules:

  1. For the transition to be allowed, the final product only needs to contain the completely symmetric irrep; irreps other than $\mathrm{A_{1g}}$ are allowed to be present as long as $\mathrm{A_{1g}}$ is there.
  2. The product of any irrep with itself always gives the completely symmetric irrep; see the diagonal elements in any product table.

From the last equation, we see that $\mathrm{E_g}$ is in the sum, so the other terms can safely be ignored because $\mathrm{E_g} \times \mathrm{E_g} = \mathrm{A_{1g}}$. Additionally, because all three components of the magnetic dipole operator transform as the same irrep, this electronic transition will be allowed if you apply a magnetic field along any of the three Cartesian axes in the molecular frame.

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  • $\begingroup$ Thank you for your detailed explanation...what does the line "⟨Ψi|m̂ |Ψf⟩→A1g∈!T2g×T1g×Eg" mean? Does it interpret as a magnetic dipole transition from Ψi to Ψf in A1g symmetry? What about the last part? $\endgroup$ – gamma1 Sep 20 '17 at 15:32
  • $\begingroup$ It's shorthand for how that integral leads to this product of three irreps, and $\mathrm{A_{1g}}$ must be contained in that product. If there's a better way of representing that, please tell me. $\endgroup$ – pentavalentcarbon Sep 20 '17 at 17:07
  • $\begingroup$ and why are we taking a triple product? are these the "3 components" mentioned earlier and match the integrand? $\endgroup$ – gamma1 Sep 20 '17 at 18:24
  • $\begingroup$ Ah, are you familiar with Dirac braket notation? I'll expand on in the question if not. It's a simplified way of writing integrals. But yes, it corresponds to the three things you multiply together before integrating (the two wavefunctions and the operator). $\endgroup$ – pentavalentcarbon Sep 20 '17 at 18:30
  • $\begingroup$ i am not but i will research this as it is definitely worth it-seeing it a lot! $\endgroup$ – gamma1 Sep 20 '17 at 19:03

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