# Is the unitary operator unique?

1. Sep 14, 2011

### Demon117

So there is a theorem at the beginning of section 1.5 in Sakurai that states the following:

Given two sets of base kets, both satisfying orthonormality and completeness. there exists a unitary operator $U$ such that

$|b^{(1)}> = U|a^{(1)}>,|b^{(2)}> = U|a^{(2)}>,...,|b^{(n)}> = U|a^{(n)}>$

By a unitary operator we mean an operator fulfilling the conditions

$U^{t}U=1$

as well as

$UU^{t}=1$

So this is not difficult to prove. But my real question is can we prove that $U$ is unique or is that just not the case and why?

2. Sep 14, 2011

### Physics Monkey

The elements of U are $U_{nm} = \langle a_n | U | a_m \rangle = \langle a_n | b_m \rangle,$ so if you fix the basis sets then you fix U.

3. Sep 14, 2011

### Demon117

That is what I thought, but I wanted a second opinion. Thank you.