# How do I show that a transition is electric dipole allowed with group theory/symmetry?

This is actually a follow up of this question

The follow-up is not because of the electric instead of magnetic dipole (this is trivial). It is because I'm interested in extra info.

Suppose I have a system with $$O_\mathrm{h}$$ symmetry. The electric dipole allowed transitions from the ground state belong to the $$\mathrm{T_{1u}}$$ irrep. Following the argument in the answer to the previous question, the ground state (or initial state) is $$\mathrm{A_{1g}}$$, the electric dipole $$\mathrm{T_{1u}}$$ and states which belong to $$\mathrm{T_{1u}}$$ satisfy the requirement that $$\mathrm{A_{1g}} \in (\mathrm{A_{1g}} \otimes \mathrm{T_{1u}} \otimes \mathrm{T_{1u}})$$. Nice.

Now suppose that I create an excited state with an electric field. Then I will end up in a state which belong to the $$\mathrm{T_{1u}}$$ irrep. To know which transitions are electric dipole allowed from such states (i.e. which can be probed via an electric field), I need to find the irreps $$Y$$ for which $$\mathrm{A_{1g}} \in (\mathrm{T_{1u}} \otimes \mathrm{T_{1u}} \otimes Y)$$. Since $$(\mathrm{T_{1u}}\otimes \mathrm{T_{1u}}) = (\mathrm{A_{1g}} \oplus \mathrm{E_g} \oplus \mathrm{T_{1g}} \oplus \mathrm{T_{2g}})$$ and $$A_{1g} \in (Y \otimes Y)$$ is always verified, any state belonging to any of $$\mathrm{A_{1g}}$$, $$\mathrm{E_g}$$, $$\mathrm{T_{1g}}$$, $$\mathrm{T_{2g}}$$ can be bright.

Here the new part. In a real experiment, which transitions are possible also depends on the relative polarization of the pump and the probe electric field. Say that the pump selects the $$x$$-polarization (let me call it $$\mathrm{T}^x_\mathrm{1u}$$ state) and the probe can be polarized along $$x$$ or along $$y$$. Can I say something about $$(\mathrm{T}^x_\mathrm{1u}\otimes \mathrm{T}^x_\mathrm{1u}\otimes Y)$$ vs $$(\mathrm{T}^x_\mathrm{1u}\otimes \mathrm{T}^y_\mathrm{1u}\otimes Y)$$?

• The xyz in the point group also describes the photon's polarisation as it were, the probability of absorbing a photon goes as polarisation direction of photon as a dot product with the dipole direction in the molecule $q_{photon} \cdot q_{dipole}$ where $q$ is any of $xyz$ in same frame of reference Jun 29, 2022 at 7:26
• @porphyrin. Sure, what you writeis completly general. However it does not reply to the question above, i.e. how this can be translated in terms of the irreps. Jun 30, 2022 at 13:36
• The T$_{1u}$ symmetry species is the same in x, y or z, in Oh point group so there is no difference, i.e. any polarisation of your radiation will be absorbed depending in what Y is, e.g. ground state A$_{1g}$. You will need to look up the direct product of T$_{1u}$ (or work it out from irreps) and multiply each of products by your choice of Y. So if you want to probe from the state T$_{1u}$ you have produced it will depend on what the final state symmetry is and on dipole direction you choose. Jun 30, 2022 at 14:28
• Yeah. This is the point. My question can be maybe changed like that. I know that $T_{1u}\times T_{1u} = (A_{1g}+E_g+T_{1g}+T_{2g})$ where $T_{1u}$, as you say, does not distinguish between x, y ,a z. Now, suppose I define a way to distinguish the direction, notation $T_{1u}^i$ with $i=x,y,z$ (not sure if this is possible). Can I say for example something like $T^x_{1u} \times T^x_{1u} = A_{1g}+E_g$ while $T^x_{1u} \times T^y_{1u} = T_{1g}+T_{2g}$ ? Jun 30, 2022 at 14:40
• No: I don't see how this would work. The irreps are the irreps, thats the point about group theory there are only so many ways of arranging things. Jun 30, 2022 at 14:47

$$T_{1u}$$ transforms like $$(x,y,z)$$ I assume that selecting the pump polarization along $$x$$, I select the state which transforms like $$x$$.
Sending a probe polarized along $$x$$ I obtain something that transforms like $$x^2$$, i.e. which belong to $$A_{1g}$$ ($$A_{1g}$$ transforms like $$x^2+y^2+z^2$$), or $$E_g$$ ($$E_g$$ transforms like combinations of $$x^2$$, $$y^2$$, and $$z^2$$).
As a result, for pump and probe with same polarization, these are the allowed transitions: $$T_{1u}\rightarrow A_{1g},E_g$$.
For orthogonal pump and prove I obtain something which belongs to $$T_{2g}$$ ($$T_{2g}$$ transforms like $$xy$$, $$xz$$, $$yz$$) or $$T_{1g}$$ ($$T_{1g}$$ transforms like $$R_x$$, $$R_y$$, $$R_z$$, since i.e. like $$x p_y - y p_x$$).
As a result, for pump and probe with orthogonal polarization, these are the allowed transitions: $$T_{1u}\rightarrow T_{1g}, T_{2g}$$.
• If you pump with any polarisation from the ground state you will get $T_{1u}$ and so probing with any polarisation will measure ground state bleaching, if in region of gs absorption, or gain if in emission region, or absorption from excited state if elsewhere and the final state must be $A_{1g}$ in each case as molecule has such high symmetry, i.e. only $A_{1g}\mu_{x,y,z}T_{1u}$ contains $A_{1g}$. Operators $x^2,xy\cdots$ are for Raman transitions not dipole as Raman measured polarisability which is a measure of volume and so cross section is a squared coordinate term. Nov 15, 2022 at 8:39
• Here I'm discussing absorption from excited state. As explained in the answer (And I've also verified numerically), the final state is either $A_{1g}$ or $E_{g}$ for probe polarization parallel to the pump polarization, or $T_{1g}$ or $T_{2g}$ for probe polarization perpendicular to the pump polarization. I'm going to upload soon an arxiv with the numerical results on LiF and hBN. I'll add the link to the reference here. Nov 16, 2022 at 9:18