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The condition for normalization for a ket vector is $$\langle A \mid A\rangle = 1.$$ However, to test if ket $\mid A \rangle$ is normalized, should I form the inner product with its complex conjugate $\langle A^* \mid$? How do I obtain the $\langle A \mid$ in the equation?

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Here $A$ is just a quantum number (or a set of quantum numbers) labeling a vector, so it need not be conjugated. Products such as $\langle A^*|A\rangle$ do happen, e.g., when dealing with coherent states, but in this case $|A\rangle$ and $|A^*\rangle$ are different states.

The exact way of evaluating the product $\langle A|A\rangle$ depends on the representation that you work with. E.g., if these are two states in the coordinate representation, $\varphi_n(x), \varphi_m(x)$, we have $$\langle n|m\rangle = \int dx \varphi_n^*(x)\varphi_m(x).$$ If these are column vectors, then $$ |\uparrow\rangle =\begin{bmatrix}\alpha\\\beta\end{bmatrix}, \langle\uparrow|=\begin{bmatrix}\alpha^*&\beta^*\end{bmatrix},$$ and $$ \langle\uparrow|\uparrow\rangle = \begin{bmatrix}\alpha^*&\beta^*\end{bmatrix}\cdot \begin{bmatrix}\alpha\\ \beta\end{bmatrix}.$$

Finally, as it may have been clear by now from the examples, the normalization factor can be included as a simple numerical factor. I.e., if $|A\rangle$ is an unnormalized state, then its normalized version is $|B\rangle=c|A\rangle$, and this factor is indeed conjugated: $$\langle B|B\rangle = |c|^2\langle A|A\rangle=1 \Rightarrow |c|^2=\frac{1}{\langle A|A\rangle}.$$

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The easy way to manipulate kets is to use the resolution of unity as much as possible $$\int d^3x |x\rangle \langle x| = 1 $$ Similarly in momentum space $$\int d^3p |p\rangle \langle p| = 1 $$ Then you have $$ \langle A |A \rangle = \int d^3x \langle A|x\rangle \langle x|A \rangle $$

and since $ \langle A|x\rangle $ is complex conjugate of $\langle x|A \rangle$ it is clear that complex conjugation applies to the wave functions, not to the quantum numbers

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