Linear operators and a change of basis

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
wakko101
Messages
61
Reaction score
0
So...I've got an operator.

Omega = (i*h-bar)/sqrt(2)[ |2><1| + |3><2| - |1><2| - |2><3| ]

Part a asks if this is Hermitian, and my answer, unless I'm missing something, is no. Because the second part in square brackets is |1><2| + |2><3| - |2><1| - |2><3| which is not the same as Omega.

The second part asks to construct the matrix Omega(ij) = <i|Omega|j> that represents this operator in the basis {|1>, |2>, |3>}. I figured it out to be
0 1 0
1 0 1 all multiplied by that funky complex factor at the beginning.
0 1 0

Or so I think.

The third part is where it gets tricky. It gives a new basis {|1'>, |2'>, |3'>} that's somewhat complicated looking and written in terms of the original basis vectors. It asks to write out the operator

U = (from i=1 to 3) SUM(|i'><i|)

in terms of the kets {|1>, |2>, |3>} and bras {<1|, <2|, <3|}. And it asks if the operator is unitary. My problem is this...how do I go about writing out this operator?
 
Physics news on Phys.org
I don't think your reason for it being non-Hermitian makes sense. The part in brackets can't be the same as Omega because there's a factor out the front, could you clarify what you meant? I would say that Omega is not because <2|O(1)> does not equal <O(2)|1>.