From my lecture notes, I have that
Where the missing equations are $$\hat P {\bf{x}}=-{\bf{x}}\tag{6.23}$$
and $${\hat P}^2 {\bf{x}}=\hat P\left(\hat P {\bf{x}}\right)=\hat P\left(-{\bf{x}}\right)={\bf{x}}\tag{6.24}$$
From my notes at the top, I would like to know why
$$\langle {\bf{x}^{\prime}} |\hat P^{\dagger}\hat P|{\bf{x}}\rangle=\langle-{\bf{x}^{\prime}}|-{\bf{x}}\rangle\tag{?}$$
I asked my professor this question and this is the email response I received:
Beginning with an operator $\hat P$ acting on a ket $|x\rangle$, $\hat P|x\rangle$, we can take the Hermitian conjugate of both to obtain $\langle x|\hat P^\dagger$. $\hat P^\dagger$ now acts to the left on the bra $\langle x|$. This can be written as $\langle x|\hat P^\dagger$, or as $\langle \hat P x|$. Try this with matrices to convince yourself.
So trying to convince myself with matrices, I let $\hat P=\begin{pmatrix}8 & 6 \\ 1 & 3\end{pmatrix}$ and column vector (ket), $|x\rangle=\begin{pmatrix}3 \\ 2\end{pmatrix}$.
$$\langle x|\hat P^\dagger=\begin{pmatrix}3 & 2\end{pmatrix}{\begin{pmatrix}8 & 6 \\ 1 & 3\end{pmatrix}}^\dagger=\begin{pmatrix}3 & 2\end{pmatrix}{\begin{pmatrix}8 & 1 \\ 6 & 3\end{pmatrix}}\\=\begin{pmatrix}36 & 9\end{pmatrix}$$
But the problem is that
$$\langle \hat P x\lvert=\begin{pmatrix}8 & 6 \\ 1 & 3\end{pmatrix}\begin{pmatrix}3 & 2\end{pmatrix}=???$$
I have just shown that I am unable to convince myself that $\langle x|\hat P^\dagger=\langle \hat P x|$ using simple matrices. It's the words of the last sentence (I put them in bold) of the quote that is confusing me here, what does "This" pertain to?
Also in the first sentence of that quote, what does "both" pertain to? Does it mean $\left(\hat P|x\rangle\right)^\dagger$ or does it mean $\left(|x\rangle\hat P|x\rangle\right)^\dagger$?
Basically, I want to know why $$\langle {\bf{x}^{\prime}} |\hat P^{\dagger}\hat P|{\bf{x}}\rangle=\langle\hat P{\bf{x}^{\prime}}|\hat P{\bf{x}}\rangle\tag{!}$$ which must be the case, as it is the only way to ensure $(\mathrm{?})$ is true.
Just a small note – I have already checked this website thoroughly for answers, and have seen many similar questions, but I am not asking for algebraic proof of $(\mathrm{!})$, I just want to see a simple matrix example that shows $\langle x|\hat P^\dagger=\langle \hat P x|$.
Best Answer
Your interpretation of $\langle Px |$ is incorrect. For a ket $|x \rangle$ corresponding to the vector $\mathbf x \in \Bbb C^n$, the bra $\langle x|$ corresponds to the Hermitian conjugate $\mathbf x^\dagger$, which is a row vector.
In this case, $| Px \rangle$ corresponds to the vector $$ \mathbf y = \pmatrix{8&6\\1&3} \pmatrix{3\\2} = \pmatrix{36\\9}. $$ It follows that $\langle Px |$ corresponds to the row-vector $$ \pmatrix{36\\9}^\dagger = \pmatrix{36 & 9}, $$ which matches your result for $\langle x | P^\dagger$.