I'm working my way through Shankar's Quantum Mechanics (7th printing, and I'm doing it alone, so I apologize if I have core concepts completely wrong).
He has a section on Active and Passive Transformations (which seems to be slightly different from what I know as Active & Passive Transformations, but that would be a different question 🙂 ), and it reads like this:
Suppose we subject all the vectors $|V\rangle$ in a space to a unitary transformation
$$ |V\rangle \rightarrow U|V\rangle $$
Under this transformation, the matrix elements of any operator $\Omega$ are modified as follows:
$$ \langle V'|\Omega|V\rangle \rightarrow \langle UV'|\Omega|UV\rangle = \langle V'|U^\dagger\Omega U|V\rangle $$
It is clear that the same change would be effected if we left the vectors alone and subjected all operators to the change
$$ \Omega\rightarrow U^\dagger\Omega U$$
I was trying to wrap my head around this, so I was working through some examples I made up and realized that the formula $\Omega\rightarrow U^\dagger\Omega U$ only works when $U$ is unitary. In my own examples I worked it out for an arbitrary (viz. non-unitary) change of basis T to be $$\Omega\rightarrow T^{-1}\Omega T$$ Like I said, I worked this out myself, so it may be wrong, but it checks out for the unitary case because when $U$ is unitary, $U^{-1} = U^\dagger$.
My question is: what part of Shankar's logic above makes use of the fact that U is unitary? It seems like his logic holds fine even without that constraint, but then it gives us a formula that isn't true for non-unitary U.
Best Answer
If we rewrite the bra-ket expression in terms of matrix multiplication you can say
$$\langle V | \Omega | W \rangle \equiv V^{\dagger} \cdot \Omega \cdot W $$
where $\cdot$ means matrix multiplcation. Now suppose we transform $V$ and $W$ by a transformation $T$ (avoiding $U$ because I'm not saying anything about unitary-ness). Then we find
\begin{eqnarray} \langle TV | \Omega | TW \rangle & \equiv & (T \cdot V)^{\dagger} \cdot \Omega \cdot (T \cdot W) \\ &=& V^{\dagger} \cdot T^{\dagger}\cdot \Omega \cdot T \cdot W \\ &=& V^{\dagger} \cdot (T^{\dagger}\cdot\Omega\cdot T )\cdot W \\ &\equiv& \langle V | T^{\dagger}\Omega T | W \rangle. \end{eqnarray}
So actually, the daggers are the general case and it's only in the case where $T$ is unitary that you can write the transformation of $\Omega$ as $\Omega \rightarrow T^{-1}\Omega T$.