# The state of a hydrogen electon ?

1. May 31, 2014

### siddharth5129

The eigenstates of a hydrogen atom are stationary states with definite values of energy. Now, as I understand it, the quantum mechanical state of the electron in the hydrogen atom is really a linear superposition of all these energy eigenstates. So this should mean that there is a finite probability of getting a higher energy value ( higher than the ground state energy value ) for any hydrogen atom, because the only thing that defines an hydrogen atom is the proton's coulomb field and the fact that it is a single electron system. So does that mean that there is no such thing as 'ground state' hydrogen ? At least not until after you've made an energy measurement on the system.
Or does the lowest possible potential energy requirement force the electron to stay in the ground state ? How is this 'forced to stay in the ground state' condition realized through quantum theory? I remember reading somewhere that electrons in atoms always occupy stationary states. Is this true ? I was always under the impression that the general solution to the schrodinger equation is a linear superposition of stationary states.

2. May 31, 2014

### hilbert2

The state of the hydrogen atom can be any superposition of the energy eigenstates, including the superposition in which the coefficients of all eigenstates except the lowest energy one are zero (that is what we call the ground state). If an H atom is put in the ground state, it will stay there forever unless somehow perturbed from the outside (it's a stationary state and cannot emit photons).

If we consider a system of interacting particles, such as an electron and a proton, we should not be talking about the state of the electron or the state of the proton, but about the state of the system (the hydrogen atom) as a whole. The full wavefunction of the hydrogen atom is a function of six variables, the x,y and z coordinates of both the electron and the proton, but it can be reduced to three coordinates by eliminating center-of-mass motion.

3. May 31, 2014

### Simon Bridge

In addition to, and without detracting at all from, above:
Provided the coefficient of the state is non-zero, this is true.

The ground state exists whether or not an H atom is in that state.

An atom prepared in a state will stay in that state until some interaction forces another one. An example would be an interaction which measures the energy of the system ... so that a state which is a superposition of energy eigenstates becomes one with a definite eigenvalue. Subsequent measuremenets of energy will produce the same eigenvalue.

... in real life, atoms are not isolated. They are constantly interacting with some sort of environment. Lots of different kinds of interactions. The result being that the available energy is distributed among lots of atoms. The net effect shows up as heat.

It is the background interactions that force atoms to seek the lowest supportable energy state.
At the level of an individual atom, we would model this by saying that the system is in contact with a heat bath (or something like that).

No. It is just very common. There are lots of processes that effectively measure energy, thus energy eigenstates are likely. The description you are likely to have seen, where electrons occupy "shells" in order etc, is an approximation that works very well. In complicated systems like atoms, we use a lot of approximations. Wait till you see solid state.

This is correct. But the particular solution belonging to the state of a particular atom need not be.

You seem to be conflating ideas that belong to different situations.

These individual H atom equations and models you've been learning about basically assume that the Universe has only one H atom in it and nothing else.

Once you've got those ideas, then you can consider what happens if there are other H atoms in the Universe. You get a theory of the H2 molecule with only one electron. Once you are used to that you can add another electron... and so on, building up until you can cope with large numbers of interacting particles and the kinds of mental gymnastics you have to do to cope with that.

The trouble is that we usually have to teach you the approximations for the later stages at the same time. For now - keep them in separate boxes. The connections between them will come.

4. May 31, 2014

### siddharth5129

Wow. Thanks a ton. That cleared up the whole mess of ideas in my head for me.

5. May 31, 2014

### dextercioby

For stationary states only, the general solution of the SE for a wave function is the product between a time-dependent complex exponential times a time-independent wave function which is solution of the spectral equation of the Hamiltonian in a position or momentum representation.

But the 'general solution to the Schroedinger equation' for an arbitrary time-depending Hamiltonian is actually approximately described through a Dyson series (integral).