Are the inner circular rings in higher 's' sub shells part of inner 's' sub shells, as the diagrams suggests?

The diagram of, for example, 2s and 3s, is the innermost circle showing electron probability has to be 1s subshell. How could it be the part of 2s or 3s, since it is a complete circle around nucleus like 1s sub shell, and they obviously cannot overlap


[![Diagram of 1s,2s,3s subshells](https://i.stack.imgur.com/koe3w.jpg)](https://i.stack.imgur.com/koe3w.jpg)


Although I do not know the Schrödinger equation, I saw the graph of variation of wave function with distance, and graph of each higher 's' sub shell seems to add on to the previous one. What I mean is that the 2s looks like the continuation of 1s graph and 3 the continuation of 2s, and so on.
[![Graphs](https://i.stack.imgur.com/iImL3.png)](https://i.stack.imgur.com/iImL3.png)


In short, my question is that are the inner rings of electron probabilities in for example 3s sub shell that of 1s and 2s sub shells?




Spherical nodes of higher s orbitals and related inner spheres are not lower s orbitals nor their parts. They just partially occupy the common place.


Different orbitals mean different states, not necessarily places. There is generally large overlap of orbitals. Lower orbitals occupy, in sense of significant probability regions, a subset of the space of upper orbitals.


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The angular part of the $s$ orbital is always spherical, these pictures are misleading because they include radial nodes, only the radial functions are different. A wave function has no physical meaning, the overlap in this case is different from the overlap between the plotted functions. The plot of $\psi^2$ has no physical meaning a well, the radial probability density is $r^2\psi^\*\psi$. The overlap gives the probability of two electrons occupying $1s$ and $2s$ states by $<1s|2s> = \int\_{-\infty}^{0} \psi\_{1s}(r) \psi\_{2s}(r) r^2 sin(\theta)drd\theta d\phi =0$ (in spherical coordinates) since this probability is zero, $1s$ and $2s$ are orthogonal even though the probability density may overlap.


In the case of a simple molecular orbitals like $H\_2$, the overlap in the plot is proportional to the overlap integral. There is another case of fully overlapping probability density without overlap : the case of two plane waves $\psi\_1 = \frac{1}{V} e^{ikr}$, $\psi\_2 = \frac{1}{V} e^{ik'r}$ confined in a volume $V$ (to be normalizable), the square modulus gives 1 for both inside the volume, so there is a complete overlap, but $\langle\psi\_1|\psi\_2\rangle=\delta\_{k,k'}=0$ because $k\neq k'$. To summarize, there is $3s$ wave function overlapping with $1s$ and $2s$ in the plot, although an electron cannot have $1s$ and $2s$ states at the same time.


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