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5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...

1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?

2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged

4

The graph shows the probability of finding the electron between the distances $r$ and $r + dr$. This probability is given by: $$P = \psi^* \psi dV$$ where $dV$ is the volume element: $$dV = 4\pi r^2 dr$$ So we get the probability: $$P(r,r+dr) = \psi^* \psi 4\pi r^2 dr$$ and therefore when $r = 0$ the probability $P = 0$. It isn't that the ...

0

Yes, it is the same as for every solution to the Schroedinger equation. The trick is that the more fundamental one is the time-dependent Schroedinger equation. However, since often (or perhaps always in introductory quantum mechanics) we do not care about a global phase, we throw it away. That is the essence, the enabling step, to derive the time-independent ...

2

I pulled most of this from Wikipedia here. A stationary state is called ''stationary'' because the system remains in the same state as time elapses, in every observable way. For a single-particle Hamiltonian, this means that the particle has a constant probability distribution for its position, its velocity, its spin, etc. A stationary state is not ...

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