Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: lostinspace on 13/05/2013 11:51:39

I'm struggling with some aspects of group theory required for quantum mechanics  in particular when operators are bracketed then (AB)C= A(BC) is required for associativity. This is explained as the end result must be the same for both sides , but the operators are done from right to left so for both sides of the equation the order will be C then B then A. This obviously gives the same result for both sides! but I thought the brackets imply these operators are carried out first  if you ignore them and carry out the operators right to left it is bound to be equal. If the operators are for example rotations and reflections, then how can you group them with brackets then ignore the brackets when you perform the operations?

You have it all back to front. It is because of the associative law that we can afford to do this.
In reality we should do the operation inside the parentheses first, then the other one (in the correct order), before applying the operation to the object, but it does not really matter.

I do not have full command of bbl codes, so I will use s instead of sigma
I am taking an example from C_{3v}[/i] symmetry group.
C_{3}^{1} * s(1) = s(2); s(2)*s(2) = E
s(1)*s(2) = C_{3}^{2}; C_{3}^{1} * C_{3}^{2} = E
Thus we have {C_{3}^{1} * s(1)} * s(2) = C_{3}^{1} * {s(1) * s(2)}