2
$\begingroup$

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!

New contributor
user3282375 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

Your Answer

user3282375 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.