Show that $(d^{1/2}d^{1/2}V^{-1}\vec{x},V^{-1}\vec{y})=(d^{1/2}V^{-1}\vec{x},d^{1/2}V^{-1}\vec{y})$ for self adjoint matrix $d^{1/2}$ and unitary $V$

Question

I am reading "Applications of group theory in quantum mechanics" by M.I. Petrashen & E.D. Trifonov. In chapter 3, section 3.4 - I'm struggling to understand part of the proof of Existence of an equivalent unitary representation, which starts on page 28.

This is a 'follow-up' question from my previous question on the same topic (but a different part of the proof, hence the new question).

On page 30, Petrashen writes

It is clear that $d^{1/2}d^{1/2}=d$, and if we use the self-adjoint property of the matrix $d^{1/2}$, we have $$\sum_{i=1}^m \left({\bf{x}}^{i},{\bf{y}}^{i}\right)=\left(VdV^{-1}{\bf{x}},{\bf{y}}\right)$$$$=\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)=\left(d^{1/2}V^{-1}{\bf{x}},d^{1/2}V^{-1}{\bf{y}}\right)\tag{3.37}$$

On page 29, Patrashen states that $V$ is a unitary transformation (I'm sorry, I searched hard for an online pdf version of this book for which I could link the relevant pages, but could not find one).

My first problem is that I don't understand the logic behind this part of expression $(3.37)$:

$$\left(VdV^{-1}{\bf{x}},{\bf{y}}\right)=\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)\tag{1}$$

Petrashen denotes the scalar product as $\bbox[5px,border:2px solid blue]{({\bf{x}},{\bf{y}})=x_1\overline{y_1}+ x_2\overline{y_2}+\cdots + x_n\overline{y_n}}\,$, where $\overline{x_i}={x_i}^\ast$.

I find it easier to understand the scalar product (LHS of $(1)$) by writing out a few of the terms explicitly, $$(VdV^{-1}{\bf{x}},{\bf{y}})=VdV^{-1}x_1y_1^\ast+VdV^{-1}x_2y_2^\ast+\cdots+VdV^{-1}x_my_m^\ast\tag{2}$$ Where I have also made use of the result in the blue box. Considering one of these terms, say, $VdV^{-1}x_iy_i^\ast$. I now multiply both sides of this term by $V^{-1}$, then, I find that $dV^{-1}x_iy_i^\ast V^{-1}=d^{1/2}d^{1/2}V^{-1}x_i\color{red}{y_i^\ast V^{-1}}$, which is close but not the same as the required term, $d^{1/2}d^{1/2}V^{-1}x_i V^{-1}y_i^\ast$.

So my question is why is the author commuting the $y_i^\ast$ past the $V^{-1}$ (the part I marked in red)? Or, put in another way, what operation was carried out in $(1)$ to go from $\left(VdV^{-1}{\bf{x}},{\bf{y}}\right)\to\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)$?

---

My second problem is that (for very similar reasons) I don't understand why $$\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)=\left(d^{1/2}V^{-1}{\bf{x}},d^{1/2}V^{-1}{\bf{y}}\right)\tag{3}$$

Following the same reasoning as before and writing out the terms of the scalar product on the LHS of $(3)$, $$\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)=d^{1/2}d^{1/2}V^{-1}x_1\Big(V^{-1}\Big)^{\ast}{y_1}^{\ast}+d^{1/2}d^{1/2}V^{-1}x_2\Big(V^{-1}\Big)^{\ast}{y_2}^{\ast}+\cdots +d^{1/2}d^{1/2}V^{-1}x_m\Big(V^{-1}\Big)^{\ast}{y_m}^{\ast}$$

Now considering one of these terms, say, $d^{1/2}d^{1/2}\color{green}{V^{-1}x_i}\Big(V^{-1}\Big)^{\ast}{y_i}^{\ast}$, since when can I possibly commute $d^{1/2}$ past $V^{-1}x_i$ marked in green? As this is what (I thought) was necessary to get the required term, $d^{1/2}V^{-1}x_i d^{1/2}\Big(V^{-1}\Big)^\ast{y_i}^\ast$ which sum to give

$$\begin{align}\left(d^{1/2}V^{-1}{\bf{x}},d^{1/2}V^{-1}{\bf{y}}\right) &&\text{RHS of Eqn. (3.37)}\\ \end{align}$$

So my second question is why is the author (Petrashen) commuting the $d^{1/2}$ past $V^{-1}x_i$. Or, put in another way, what operation was carried out to show that $$\left(d^{1/2}d^{1/2}V^{-1}{\bf{x}},V^{-1}{\bf{y}}\right)=\left(d^{1/2}V^{-1}{\bf{x}},d^{1/2}V^{-1}{\bf{y}}\right)?$$

---

A few remarks:

In the answer given to my previous question it is written that

Presumably the Hermitian form is defined so that $⟨u,Mv⟩=⟨M^\dagger u,v⟩$ for any matrix $M$. In that case you just have $$\sum_{i=1}^m \langle D(g_i) \mathbf{x}, D(g_i) \mathbf{y}\rangle = \sum_{i=1}^m \langle D(g_i)^\dagger D(g_i) \mathbf{x}, \mathbf{y}\rangle = \left\langle \sum_{i=1}^m D(g_i)^\dagger D(g_i) \mathbf{x}, \mathbf{y}\right\rangle$$

After reading this answer I looked at page 126, of the book by Michael Tinkham called "Group Theory and Quantum Mechanics" which I found a pdf for it here (which is on page 136 if viewed online) and Tinkham denotes the Hermitian scalar product as $$\sum_{\mu}{V_\mu}^{\ast}W_{\mu}$$ but even with this knowledge does not help me understand the 2 questions I have above. I did some digging around, but could not find anything on self-adjoint properties that could help me understand the two questions I have above.

Answer 1

1. Your unnumbered expanded form of $(VdV^{-1}\mathbf{x},\mathbf{y})$ continues to have a key error. You write $VdV^{-1}x_1 y_1^*$ when you mean $(VdV^{-1}x)_1 y_1^*$. This error bites you a moment later when you say you multiply the expression by $V^{-1}$ and cancel the initial $V$. This simply does not make sense when the proper grouping symbols are used.

2. As for (1), since $V$ is unitary, by definition $V^\dagger = V^{-1}$. We have $(Mu, v) = (u, M^\dagger v)$ [and $(M^\dagger u, v) = (u, Mv)$] for any $M$ as I noted in my earlier answer. Hence \begin{align*} (VdV^{-1}\mathbf{x}, \mathbf{y}) = (dV^{-1}\mathbf{x}, V^\dagger\mathbf{y}) = (dV^{-1}\mathbf{x}, V^{-1}\mathbf{y}) = (d^{1/2}d^{1/2}V^{-1}\mathbf{x}, V^{-1}\mathbf{y}). \end{align*}

3. As for (2), this is the exact same computation as in my previous line except using the fact that $d^{1/2}$ is self-adjoint, so $(d^{1/2})^\dagger = d^{1/2}$.

4. You seem to be confusing yourself quite a lot by mixing many different sources with different conventions and perspectives. I've seen students do this many times, especially as the internet has made information easily accessible. It can easily lead to the "illusion of fluency". The fix is to find a unified approach with an actual expert. Best of luck.