# Modern Quantum Chemistry

Chapter02 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41

Exercise. This is the solution to exercise 2.1 in the book.

Solution. Consider the following five cases.

Case 1. Both $i$ and $j$ are even and $i\ne j$, $i=2k$, and $j=2m$. Then

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx=\int dr{\psi }_{k}^{\beta }\left(r\right){\psi }_{m}^{\beta }\left(r\right)d\omega {\beta }^{2}\left(\omega \right)=\int dr{\psi }_{k}^{\beta }\left(r\right){\psi }_{m}^{\beta }\left(r\right)\cdot 1=0$

because ${\psi }_{k}^{\beta }\left(r\right)$ and ${\psi }_{m}^{\beta }\left(r\right)$ are orthonormal.

Case 2. Both $i$ and $j$ are even and $i=j$, $i=2k$, and $j=2k$. Then

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx=\int dr{\psi }_{k}^{\beta }\left(r\right){\psi }_{k}^{\beta }\left(r\right)d\omega {\beta }^{2}\left(\omega \right)=\int dr{\psi }_{k}^{\beta }\left(r\right){\psi }_{k}^{\beta }\left(r\right)\cdot 1=1$

because ${\psi }_{k}^{\beta }\left(r\right)$ are normalized.

Case 3. Both $i$ and $j$ are odd and $i\ne j$, $i=2k-1$, and $j=2m-1$. Then

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{m}^{\alpha }\left(r\right)d\omega {\alpha }^{2}\left(\omega \right)=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{m}^{\alpha }\left(r\right)\cdot 1=0$

because ${\psi }_{k}^{\alpha }\left(r\right)$ and ${\psi }_{m}^{\alpha }\left(r\right)$ are orthonormal.

Case 4. Both $i$ and $j$ are odd and $i=j$, $i=2k$, and $j=2k$. Then

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{k}^{\alpha }\left(r\right)d\omega {\alpha }^{2}\left(\omega \right)=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{k}^{\alpha }\left(r\right)\cdot 1=1$

because ${\psi }_{k}^{\alpha }\left(r\right)$ are normalized.

Case 5. If $i$ is odd and $j$ is even, for example $i=2k-1$ and $j=2m$, then

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{m}^{\beta }\left(r\right)d\omega \alpha \left(\omega \right)\beta \left(\omega \right)=\int dr{\psi }_{k}^{\alpha }\left(r\right){\psi }_{m}^{\beta }\left(r\right)\cdot \int d\omega \alpha \left(\omega \right)\beta \left(\omega \right)={S}_{km}\cdot 0=0$

because $\alpha \left(\omega \right)$ and $\beta \left(\omega \right)$ are orthonormal.

Therefore

$\int {\chi }_{i}\left(x\right){\chi }_{j}\left(x\right)dx={\delta }_{ij}$

so the set $\left\{{\chi }_{i}\right\}$ of $2K$ spin orbitals is orthonormal.