Projection operators in a three-state quantum system

[el_hijoeputa]

This might sound silly... I was on a travel, and arrived late at my class of Quantum Mechanics 2. The proffesor was in the middle of a disscusion, and I'm trying to figure what was it. I asked one of my classmates and said he was talking about something with Projection operators. He was discussing a problem of a system with two states, and then started to discuss a system of three states, two states with strong interaction, and another one of less interaction (?).

-------------- (P)



--------------
-------------- (Q)

He then defined:
$$P = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$

$$Q = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$$

All seems to get down to claculate P G(z) P

Anyway, I won't see the professor until next thursday. If someone can figure out the topic so I can start studying to get up to date this weekend, I will appreciate it.

 

[marlon]

according to me, your professor was talking about irreducible representations of symmetry-groups and how you can construct operators that will extract the parts of some physical system (ie the operators on the wavefunction) that correspond to each irreducible representation. In order to do so, the socalled partnerfunctions can be used. These functions can also generate a representation for certain symmetry-groups. These are all applications of group theory in QM.

could it be something like this ?

marlon

 

[marlon]

Each interaction can be described using such symmetry-groups and therefore it is always valid to know what parts of the wavefunction correspond to one specific interaction. you have parts coming from L-S-coupling , parts coming from the Zeemann-effect and so on...until quarks and their colours

marlon