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The value turns out to be zero . This is understandable since they are completely independent of each other and also perpendicular in real space. But also turns out to be zero . How do I justify the orthogonality of the s and the p levels?

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The mathematical answer to this specific question lies in the fact that s and p orbitals are simultaneous, non-degenerate eigenfunctions of the quantum mechanical $\hat{L}^2$ operator, which is hermitian:

$$\begin{align} \hat{L}^2\left|\mathrm{s}\right> = [l(l+1)\hbar]^2 \left|\mathrm{s}\right> &= 0 \\ \hat{L}^2\left|\mathrm{p}\right> = [l(l+1)\hbar]^2 \left|\mathrm{p}\right> &= 2\hbar^2 \left|\mathrm{p}\right> \end{align}$$

since $l = 0$ for s orbitals and $l = 1$ for p orbitals.

It is a simple proof that the eigenstates of hermitian operators that correspond to different eigenvalues are necessarily orthogonal. Suppose, for a hermitian operator $A$, we have $A\left|m\right> = a_m\left|m\right>$ and $A\left|n\right> = a_n\left|n\right>$ with $a_m \neq a_n$. Then:

$$\begin{align} \left\langle m \middle| A \middle| n\right\rangle &= \left\langle n \middle| A \middle| m\right\rangle^* & &\text{(definition of hermiticity)} \\ a_n\left\langle m \middle| n\right\rangle &= (a_m\left\langle n \middle| m\right\rangle)^* & &\text{(extracting the eigenvalues)} \\ &= a_m\left\langle n \middle| m\right\rangle^* & &\text{(eigenvalues of hermitian operator are real)} \\ &= a_m\left\langle m \middle| n\right\rangle & &\text{(definition of inner product)} \\ (a_n - a_m)\left\langle m \middle| n\right\rangle &= 0 \\ \left\langle m \middle| n\right\rangle &= 0 & &(a_m \neq a_n) \end{align}$$

You can likewise show that any pair of atomic orbitals in the hydrogen atom are orthogonal, using whichever of the three operators $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ that you need. All atomic orbitals are simultaneous eigenfunctions of these three operators, and have different eigenvalues which correspond to the quantum numbers $n$, $l$, and $m_l$:

$$\begin{align} \hat{H}\left|n,l,m_l\right> &= E_n\left|n,l,m_l\right> = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 n^2}\left|n,l,m_l\right> \\ \hat{L}^2\left|n,l,m_l\right> &= L^2\left|n,l,m_l\right> = [l(l+1)\hbar]^2\left|n,l,m_l\right> \\ \hat{L}_z\left|n,l,m_l\right> &= L_z\left|n,l,m_l\right> = m_l\hbar\left|n,l,m_l\right> \end{align}$$

Since no two orbitals have the same three quantum numbers, no two orbitals can have the same eigenvalues for all three operators. Consequently, there must be at least one operator which can be used to show that those two orbitals are orthogonal.

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  • $\begingroup$ Thanxx for the explanation! Another thing I would like to ask is that since px and py are perpendicular in real space...can this be used to say that their wave functions will be orthogonal? $\endgroup$ – CodeMaxx Nov 13 '15 at 20:18
  • $\begingroup$ Yes, you can. Although do note that the $\mathrm{p_x}$ and $\mathrm{p_y}$ orbitals are no longer eigenfunctions of the $\hat{L}_z$ operator (they are linear combinations of the eigenfunctions). To prove it mathematically, you would need the explicit form of the wavefunctions. $\endgroup$ – orthocresol Nov 13 '15 at 20:23
  • $\begingroup$ Thanxx @orthocresol $\endgroup$ – CodeMaxx Nov 17 '15 at 6:07

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