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,

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

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

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?

  • $\begingroup$ page 8 of the pdf at this link discusses this problem, but it has been so long since I did this that I don't really remember how to do the math. chem.tamu.edu/rgroup/hughbanks/courses/673/handouts/… Obviously the gist is that $1s^12s^1$ to $1s^2$ is a forbidden transition. $\endgroup$
    – MaxW
    Commented Oct 31, 2015 at 19:39

1 Answer 1


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. $$ \Psi_0 = \frac{1}{\sqrt{2}} \begin{vmatrix} \psi_1(1) & \psi_2(1) \\ \psi_1(2) & \psi_2(2) \end{vmatrix} \, , \quad \Psi_1 = \frac{1}{\sqrt{2}} \begin{vmatrix} \psi_1(1) & \psi_3(1) \\ \psi_1(2) & \psi_3(2) \end{vmatrix} \, , $$ 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.


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