# If K=|a><b| where |a> and |b>

1. Oct 1, 2005

### cire

if K=|a><b| where |a> and |b> are two vectors of the state space.
I'm trying to show that K can always be written in the form K=gPQ where g is a constant and P and Q are projectors.
this is what I get:
K=|a><b|
<a|b>K=<a|b>|a><b| multiplying by the number <a|b>
<a|b>K=|a><a|b>|<b| inserting that number in between
P=|a><a|
Q=|b><b|
g=<a|b>^-1
so I proved if the vectors are not orthogonal, how to prove it when they are????

2. Oct 1, 2005

### vanesch

Staff Emeritus
Intuitively I'd have a hard time imagining how you should be able to do this!
After all, what the original operator does, is to take a random vector, find its b component, and then reproduce an a component with the same length.
What you did in the non-orthogonal case was in fact to use Thales' theorem: you found the b component, which has (because of the non-orthogonality) itself of course still a small a component ; so you re-projected on a, and adjusted for the loss (the angle between a and b).

With 3 operators, I can see a solution: you project first on b, then on something like (a+b) and then finally on b, adjusting for the losses. But I have a hard time imagining, with 2 projectors, how to do so. I'm not saying it is impossible, just that I don't intuitively see how to do so.
You could also simply try to write down an equation, having each of the projectors to have as general a form as possible, like
P = ( u|a) + v|b) ) (u*(a| + v*(b| ) etc... and see if you can find a condition that works out.

cheers,
Patrick.

3. Oct 2, 2005

### CarlB

Dear cire,

Your work is correct, and your comment that there is a problem when the vectors is orthogonal is also correct. If <a| and <b| are orthogonal, there is no possible way to solve the problem as stated.

Carl