**Based on comments, let's clarify things first.** The H-atom (or any other quantum system) is not in an "eigenvalue" of a measurable and observable quantity (call these observables from now on). Observables can be position, momentum, energy, angular momentum, etc. A quantum system may be in the *eigenstates* of these observables, to which certain eigenvalues correspond.

If you measure an observable on the system, it will "jump" into one of the observable's eigenstates, and the result of the measurement will be the eigenvalue corresponding to this state. A system may be in the superposition of an observable's eigenstates, and the coefficient of an eigenstate in the superposition determines the probability of the system jumping into that eigenstate upon measurement (and thus the probability of the measurement apparatus showing the eigenvalue corresponding the that state). Actually most often the state of the system is always in the superpositon of some observable's eigenstates. E.g. the hydrogen may be in an energy eigenstate, but it is in the superposition of the eigenstates of the position operator (that is, it has a certain energy, but an uncertain position).

When you solve the Schrödinger equation for the hydrogen, you will receive the energy eigenstates and the corresponding energy eigenvalues. The Scrödinger equation for the hydrogen is a second order partial differential (and non-linear) equation, for the solving of which you need initial conditions, that is how the hydrogen looked like in $t=0$. Whether the hydrogen was in a superposition of energy eigenstates, depends on the initial conditions, which is determined by you, the solver of the equation. E.g. you may say, that initially, its position is completely determined, thus it is in a certain eigenstate of the position operator correspondng to eigenvalue $\vec{r}_0$. Thus if you measure the position of you system in $t=0$, it will certainly be $\vec{r}_0$. You may also say that you want it initially be in an energy eigenstate, corresponding to energy eigenvalue $E_i$. This way, at $t=0$, you will certainly measure its energy to be $E_i$.

**Now comes the answer to your question.** Suppose you solve the equation for the hydrogen in a way that it may interact with its environment (actually, this isn't the Schrödinger equation, but e.g. the Lindblad equation. which is able to treat open systems like this, that interact with their environment-the Lindblad gives the Schrödinger equation as a special case, when the interaction between system and environment is zero). Suppose also that this environment is a thermal bath at $T=0$, and you have specified an intial condition that the H-atom be in an energy eigenstate $E_i>E_0$ where $E_0$ is the ground state energy. Then you will receive the result that the H-atom will quickly be in the superposition of energy eigenstates, including in the superposition the ground state, and all the coefficients in the superposition decay exponentially (*asymptotically* to zero), except the coefficient of the ground state (it rises *asymptotically* to one). The probability of measuring a certain energy eigenvalue being proportionate to the coefficient in the superposition gives you that in just a few milliseconds it will be in the ground state with probability *almost* one, but not exactly one (as I have said, it may happen with neglectable probability, that it is in an excited state, even after a million years).

Now suppose that you solve the Lindblad equation for the hydrogen, again with a thermal environment, where $T=0$. But let the initial condition be that the hydrogen is in the superposition of energy eigenstates. If the ground state is not present in the initial superposition (it's coefficient is zero), it will quickly be, and its coefficient will start to rise to one, and the other superposition coefficients will quickly become zero (again, just *asymptotically*). So you get the same result.

**As a summary**, it doesn't matter, if the hydrogen started out in a superposition of energy eigenstates (like it usually starts out in the superpostion of position eigenstates) or it started out in a certain energy eigenstate, it will exponentially decay into the ground state.

**As a sidenote**, if the intial condition you specify were that the hydrogen should start out in the ground state, it will remain there forever.

**Also as a sidenote**, a real thermal environment is not $T=0$, but e.g. $T=2000~K$ (or even higher, I don't know at what temperature hydrogen molecules disintegrate into atoms), and if you take lots of hydrogen atoms (forming a gas), there will be all kinds of atoms in it. I.e. some in the ground state, some in excited states, some in superpositions, etc., partly because the environment with high temperature not only dissipates but invests energy too.

**Third sidenote** It emerged as a question why the position doesn't tend to reach some eigenstate, while the energy does. First of all I emphasize again that the energy state *will not be determined* if one leaves the system to evolve, it will just go to the ground state with *high probaility*. But it won't be surely the ground state, only after measurement, or if the system started out in the ground state. Secondly, a quantum system has this behaviour if and only if it is joined with a thermal bath havin a temperatur around $0~K$. Otherwise, an esmeble with many subsystems in it joined with a thermal bath of high temperature will have all kind of subsytems, some in the superposition of energy eigenstates (thus having undetermined energy), some being in some excited state, and only a few being in the ground state. Having said that, let's take a single quantum system joined with a thermal bath at $T=0~K$. It started out in some excited state, or superposition of energy eigenstates, and it tends to reach ground state. The position operator's eigenstates however doesn't show this kind of tendency. This is because a fundamental law of nature is that enthropy of a system should be maximized, and this is reached by getting as close to thermodynamic equilibrium with the environment (in this case the thermal bath) as possible. This makes energy go to minimum, which is the ground state energy. However energy and position are non-commuting observables. \begin{equation}\hat{H}=\hat{K}(\vec{r})+\hat{V}(\vec{r})=\frac{\hat{p}^2}{2m}+V(\vec{r})=\frac{-\hbar^2}{2m}\bigtriangleup+V(\vec{r})\end{equation}
\begin{equation}\hat{\vec{r}}=\vec{r}\end{equation}
\begin{equation}\left[\hat{H},\hat{\vec{r}}\right]\psi(\vec{r})=\left(\frac{-\hbar^2}{2m}\bigtriangleup+V(\vec{r})\right)(r\psi)-r\left(\frac{-\hbar^2}{2m}\bigtriangleup+V(\vec{r})\right)(\psi)=(Vr+V)\psi\end{equation}
This is true for all $\psi$, thus
\begin{equation}\left[\hat{H},\hat{\vec{r}}\right]=Vr+V\end{equation}
This means that due to the Heinserberg uncertainity a quantum system cannot be in the eigenstate of the position operator and in the eigenstate of the energy operator at the same time. That is why if the energy tends to be very ceratin, the position tends to be uncertain. Note however that this is only true for quantum systems in a potential, if $V=0$ the two observables (that is energy and position) commute, and a simultaneous eigenstates of them exist, and system is allowed to be in one of them.

## Best Answer

This is just an additional remark – the other answers are fine.

However, you can get all states from the variational method, just not with the trial function method (unless your trial function space contains the exact excited states).

While the excited states are usually not extrema of the variation functional, they are stationary points! That is, by varying the functional you get the excited states as solutions, by requiring the variation vanish. Consider the variation functional with a Lagrange multiplier to enforce the norm of the states: $$E\big[\left|\psi\right> , \left<\psi\right|\big] = \left< \psi \middle| H \middle| \psi \right> - \lambda \big( \left< \psi \middle| \psi \right> -1 \big).$$ Computing the variation one gets: $$\delta E = \left< \delta \psi \middle| H \middle| \psi \right> + \left< \psi \middle| H \middle| \delta \psi \right> - \lambda \left< \delta \psi \middle| \psi \right> - \lambda \left< \psi \middle| \delta \psi \right> $$ (Note, that we consider $\left| \psi \right>$ and $\left< \psi \right|$ to be independently varied.)

From this we can see, that the functional is stationary for: $$H \left| \psi \right> = \lambda \left| \psi \right>.$$ Which is the time independent Schrödinger equation, which gives all the stationary states. The Langrange multiplier turns out to be the eigenenergy $E$ of the state.

In other words, when viewed from this angle, the variational principle is equivalent to the time independent Schrödinger equation.