Demonstrating orthonormality for wave functions for ground state and excited state of He The electron configuration of the ground state $\ce{He}$ atom is $\ce{1s^2}$, and the excited state configuration under study is $\ce{1s^2 2s^1}$. Show that these states are orthonormal, i.e.,
  
  $$\langle\Psi_0 \vert\Psi_0\rangle = 1$$
  
  $$\langle\Psi_1 \vert\Psi_1\rangle = 1$$
  
  $$\langle\Psi_0 \vert\Psi_1\rangle = 0$$


I start off by defining the wave functions as Slater determinants, 

$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) &amp; \phi_1(1)\beta(1) \\
\phi_1(2)\alpha(2) &amp; \phi_1(2)\beta(2)
\end{vmatrix}
$$

$$
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) &amp; \phi_2(1)\beta(1) \\
\phi_1(2)\alpha(2) &amp; \phi_2(2)\beta(2)
\end{vmatrix}
$$

I am able to show that the wave functions are normalized, but I have trouble showing that they are also orthogonal. I get that $\langle\Psi_0 \vert\Psi_1\rangle = 1$. I set up the integral over all coordinates, carry out the multiplications, and add together the terms that are equivalent due to indistinguishable electrons (identical when all coordinates are interchanged). I end up with the expression

$$
\int d\tau \left[  \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(1)\alpha(1)\phi_2(2)\beta(2) - \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(2)\alpha(2)\phi_2(1)\beta(1)   \right]
$$

where $d\tau$ represents all coordinates. Separating the spin coordinates from the spatial coordinates, I get that

$$
\langle\Psi_0 \vert\Psi_1\rangle = \\
\sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(1)\beta(2)\beta(2) \cdot \langle\phi_1(1) \vert\phi_1(1)\rangle \cdot \langle\phi_2(2) \vert\phi_2(2)\rangle \\
- \sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(2)\beta(2)\beta(1) \cdot \langle\phi_1(1) \vert\phi_2(1)\rangle \cdot \langle\phi_2(2) \vert\phi_1(2)\rangle \\
= 1 - 0 = 1
$$

The first term reduces to $1$, I find, while the second term reduces to $0$. To get $\langle\Psi_0 \vert\Psi_1\rangle = 0$, either both terms need to be $1$, or both terms need to be $0$. Something must be incorrect. Is there some obvious flaw in my derivations?
First prove a well-known fact that any two spin orbitals $\psi_i$ and $\psi_j$ are orthogonal if the corresponding spatial orbitals from which they are constructed are themselves orthogonal. 


For two spin orbitals constructed from two different spatial orbitals orthogonality follows from orthogonality of spatial orbitals regardless of spin components.
For two spin orbitals constructed from the same spatial orbital orthogonality follows from orthogonality of spin functions.


Then you can work with spin orbitals to (re)establish a well-known result that for two Slater determinants differ in just one spin orbital, the overlap integral $\langle \Psi_0 \vert \Psi_1 \rangle$ is zero. 
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) &amp; \psi_2(1) \\
\psi_1(2) &amp; \psi_2(2)
\end{vmatrix} \, ,
\quad
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) &amp; \psi_3(1) \\
\psi_1(2) &amp; \psi_3(2)
\end{vmatrix} \, ,
$$
where
$$
\psi_1(1) = \phi_1(1) \alpha(1) \, ,
\quad
\psi_2(1) = \phi_1(1) \beta(1) \, ,
\quad
\psi_3(1) = \phi_2(1) \beta(1) \, .
$$
Algebraically, it is very easy to show that,
$$
  \langle \Psi_0 \vert \Psi_1 \rangle =
  \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \langle \psi_1(1) \psi_2(2) - \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) - \psi_3(1) \psi_1(2) \rangle \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \psi_2(2) \vert \psi_1(1) \psi_3(2) \rangle
  - \langle \psi_1(1) \psi_2(2) \vert \psi_3(1) \psi_1(2) \rangle \\
  - \langle \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) \rangle
  + \langle \psi_2(1) \psi_1(2) \vert \psi_3(1) \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \vert \psi_1(1) \rangle
    \langle \psi_2(2) \vert \psi_3(2) \rangle 
  - \langle \psi_1(1) \vert \psi_3(1) \rangle
    \langle \psi_2(2) \vert \psi_1(2) \rangle \\
  - \langle \psi_2(1) \vert \psi_1(1) \rangle 
    \langle \psi_1(2) \vert \psi_3(2) \rangle
  + \langle \psi_2(1) \vert \psi_3(1) \rangle
    \langle \psi_1(2) \vert \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(c_1 \cdot 0 - 0 \cdot 0 - 0 \cdot 0 + 0 \cdot c_1 \Big) = 0 \, .
$$
In the case when spin orbitals are normalized (to one) $c_1 = 1$, but it is not required here: the overlap integral will vanish for spin orbitals that are just orthogonal and not necessarily normalized (to one) as well.