# Relationship between Eigenvectors of Hamiltonian vs function of the Representations of the Group

I am trying to understand the relationship between the eigenvectors obtained from a diagonalizing a Hamiltonian and the basis functions of the Representations of the Group, $$G$$, used to build the basis for a Hamiltonian.

primary Ref Material:

Yu and Cardona: Fundamentals of Semiconductors
Heine: Group Theory in Quantum Mechanics


Some Context:

Say you are building a basis for a Hamiltonian in some representation for a specified Group. For example, you know your Hamiltonian has $$O_h%$$ (Diamond point Group, space group 227) symmetry and you want to use the basis functions corresponding to atomic orbitals (s,p,d,...).

Using the functions listed on http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=904&option=4 , we see that the functions associated with the s,p and d orbitals are:

$$A_{1g}: x^2+y^2+z^2 \;(l=0,s=0;s) \\ T_{1u}: \{x,y,z\}\;\;\;\;\;\;\;(l=1,s=\pm1,0;p)\\ T_{2g}: \{xz,yz,zx\}\;\;\; (l=2,s=\pm1,0;d)\\ E_g\,:\{3z^2-r^2,x^2-y^2\}\;(l=2,s=\pm2;d)$$

Here is maybe where I lose traction.

If we want to build our Hamiltonian out of these functions for a two atom unit system, then the Group of the Hamiltonian is:

$$H=(A_{1g}\bigoplus T_{1u}\bigoplus T_{2g}\bigoplus E_{g})_{atom1}\bigoplus(A_{1g}\bigoplus T_{1u}\bigoplus T_{2g}\bigoplus E_{g})_{atom2}$$

An eigenvector in this basis would look like (with numverical coeff. instead of orbital identifiers obv.):$$(s_{1},x_{1},y_{1},z_{1},xy_{1},yz_{1},zx_{1},3z^2-r^2_{1},x^2-y^2_{1},s_{2},x_{2},y_{2},z_{2},xy_{2},yz_{2},zx_{2},3z^2-r^2_{2},x^2-y^2_{2})$$

My understanding is that since I am using an irreducible representation, they are orthogonal. Thus there should be no off block-diagonal elements in my Hamiltonian (No mixing of different representations).

Example: $$T_{1u} \bigotimes H \bigotimes T_{2g}=0$$

However, this I am not getting zero (The answer does not contain $$A_{1g}$$ which is the scalar representation, so I guess numerically its zero???).

Since:

$$H \bigotimes T_{2g}=T_{2g}\bigotimes T_{2g}= A_{1g}\bigoplus E_g \bigoplus T_{1g}\bigoplus T_{2g}$$ (via the Great Orthoganality Theorem)

so the element $$T_{1u}\bigotimes H \bigotimes T_{2g}=T_{1u}\bigotimes(A_{1g}\bigoplus E_g \bigoplus T_{1g}\bigoplus T_{2g})$$

$$T1u\bigotimes T2g=A2u\bigoplus Eu\bigoplus T1\bigoplus T2u\\ T1u\bigotimes T1g=A1u \bigoplus Eu\bigoplus T1u\bigoplus T2u\\ T1u\bigotimes Eg=T1u\bigoplus T2u\\ T1u\bigotimes A1g=T1u$$

This is where I definitely need help.

So to summarize thus far, I expect blocks from different irreducible representations to of no coupling/off diagonal matrix elements in my Hamiltonian. As a result of having no coupling, I should expect that the eigenvectors should contain character of only one Representation.

ie, There should be no eigenvectors that mix say p and d which belong to $$T_{1u}$$ and $$T_{2g}$$, respectively.

When I work through my actual Hamiltonian, I get states that commute with symmetry operations, however they are mixing Representations.

As to have a specific question to answer...

Can the Eigenvectors of the Hamiltonian mix irreducible representations of the Group of the Hamiltonian?

I would assume that if they do not mix then my rep. is somehow reducible, but confirmation of that would also be appreciated.

I have been learning Group Theory along side my research as necessary, so I am sure I have some misunderstanding of the concepts presented.

Thanks!

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