The wave function $\Psi_2(1,2)$ given by Equation 9.39 is not normalized as it stands. Determine the normalization constant of $\Psi_2(1,2)$ given that the "1s" parts are normalized.
We want to find the constant $c$ such that
$$
I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=1
$$
First notice that $\Psi_2(1,2)$ can be factored into the product of a spatial part and a spin part:
$$
\begin{aligned}
\Psi_2(1,2) & =1 s(1) 1 s(2)[\alpha(1) \beta(2)-\alpha(2) \beta(1)] \\
& =1 s\left(\mathbf{r}_1\right) 1 s\left(\mathbf{r}_2\right)\left[\alpha\left(\sigma_1\right) \beta\left(\sigma_2\right)-\alpha\left(\sigma_2\right) \beta\left(\sigma_1\right)\right]
\end{aligned}
$$
The normalization integral becomes the product of three integrals:
$$
I=c^2\langle 1 s(1) \mid 1 s(1)\rangle\langle 1 s(2) \mid 1 s(2)\rangle\langle\alpha(1) \beta(1)-\alpha(2) \beta(1) \mid \alpha(1) \beta(2)-\alpha(2) \beta(1)\rangle
$$
The spatial integrals are equal to 1 because we have taken the $1 s$ orbitals to be normalized. Now let's look at the spin integrals. When the two terms in the integrand of the spin integral are multiplied, we get four integrals. One of them is
$$
\begin{aligned}
\iint \alpha^*\left(\sigma_1\right) \beta^*\left(\sigma_2\right) \alpha\left(\sigma_1\right) \beta\left(\sigma_2\right) d \sigma_1 d \sigma_2 & =\langle\alpha(1) \beta(2) \mid \alpha(1) \beta(2)\rangle \\
& =\langle\alpha(1) \mid \alpha(1)\rangle\langle\beta(2) \mid \beta(2)\rangle=1
\end{aligned}
$$
where once again we point out that integrating over $\sigma_1$ and $\sigma_2$ is purely symbolic; $\sigma_1$ and $\sigma_2$ are discrete variables. Another is
$$
\langle\alpha(1) \beta(2) \mid \alpha(2) \beta(1)\rangle=\langle\alpha(1) \mid \beta(1)\rangle\langle\beta(2) \mid \alpha(2)\rangle=0
$$
The other two are equal to 1 and 0 , and so
$$
I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=2 c^2=1
$$
or $c=1 / \sqrt{2}$.