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If I am given the initial stationary spatial wavefunction of a hydrogen atom, how does it change through time? I'm wondering if it is the same as the time evolution of any old stationary state, that you tack on the time dependence $e^{iE_nt/\hbar}$

Thank you!

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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 mathematically constant: $$|\Psi(t)\rangle = e^{-iE_{\Psi}t/\hbar}|\Psi(0)\rangle$$ However, all observable properties of the state are in fact constant. For example, if $|\Psi(t)\rangle$ represents a simple one-dimensional single-particle wavefunction $\Psi(x,t)$, the probability that the particle is at location $x$ is:

$$|\Psi(x,t)|^2 = \left| e^{-iE_{\Psi}t/\hbar}\Psi(x,0)\right|^2 = \left| e^{-iE_{\Psi}t/\hbar}\right|^2 \left| \Psi(x,0)\right|^2 = \left|\Psi(x,0)\right|^2$$

which is independent of the time $t$. Note, we are able to write

$$ \left| e^{-iE_{\Psi}t/\hbar}\right|^2 \left| \Psi(x,0)\right|^2 = \left|\Psi(x,0)\right|^2$$ because $e^{-iE_{\Psi}t/\hbar}$ is a unitary operator. The constancy of the probability distribution follows from that fact by simple rearrangement.

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    $\begingroup$ I think a proper answer to this question would address the reason why the exponential factor arises, which will immediately make it clear why and when one can or cannot simply "tack on an exponential". $\endgroup$ – Danu Apr 5 '15 at 8:55
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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 from the time-dependent Schroedinger equation.

If you understand what exactly you are doing, you can obviously undo this simplification. If you do not understand it, it will look like "tacking on" an exponential $\exp(\pm i \Delta E t / \hbar)$. The details, like the sign, will depend on the sign for your energy (difference) $\Delta E$. Frankly, if any of this raises any question or doubt, there is really only one authoritative answer: Rederive your result without premature simplifications (like applying the time-independent Schroedinger equation when you actually are interested in the dependency on time). However, you will find the obvious, that e.g. the energy $\Delta E$ arbitrarily depends on how you define it. The physically relevancy of it is simply how it compares to other energies in your system. That is another way of saying it only matters in a larger system where your hydrogen atom may compare to or interact with other atoms.

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