# Quantum numbers and radial probability of the electrons

In this book it has been written:

The $ns$, $(n − 1)d$, and $(n − 2)f$ orbitals are so close to one another in energy, and interpenetrate one another so extensively.

And in the wikipedia article of Pauli exclusion principle it has been written:

The Pauli exclusion principle is the quantum mechanical principle that states that two identical fermions (particles with half-integer spin) cannot occupy the same quantum state simultaneously. In the case of electrons in atom, it can be stated as follows: it is impossible for two electrons of a poly-electron atom to have the same values of the four quantum numbers.

Does this mean that two electrons of an atom can have significant radial probability at the same location even if they are defined by different set of quantam numbers?

• Yes of course! Pauli's exclusion principle is only about the 'quantum numbers', more correctly, it states that 'A system containing several electrons must be described by an antisymmetric total eigenfunction', which is the stronger statement. The weaker statement is that no two electrons can have two identical set of quantum numbers. It doesnt say anything about the probability or the energy or any other obesrvable. Any property deduced satisfying the condition thereof is purely a mathematical result and entirely in accordance with the principle Oct 17, 2016 at 7:14
• @PrasadMani,Thank you very much.Please write this as an answer so that you may get deserved reputation.
– user132865
Oct 17, 2016 at 7:34

If we consider space variables of two electrons (identical particles) to have almost the same values, then their wavefunctions are 'almost' identical if they are in the same quantum state, ie, $\psi_{a}(1)~ \simeq~\psi_{a}(2)$ and $\psi_{b}(1)~\simeq~\psi_{b}(2)$ [the label 1 and 2 denote the spatial co-ordinates of the electron '1' and '2' i.e. ($x_1,y_1,z_1$) and ($x_2,y_2,z_2$), and the labels a and b for the wavefunction denote the three quantum numbers $n,l,m$ of two different quantum states].
$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) - \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) - \psi_{b}(1) \psi_{a}(2)]\simeq 0$$