Normalizing Wave-functions

I am trying to understand particle in 1-D box problem.Schrodinger's equation is a Linear differential equation. If ψ1, ψ2..are solutions, general solution is ψ=c1*ψ1+c2*ψ2+c3*ψ3+… So my question is, why is the normalization condition applied to individual ψn? and not to ψ?Can a particle in box have multiple modes at the same time? Like a guitar string?
- Alok (age 26)
Kingston, RI

Actually, the normalization condition is applied to that whole state:

|c₁|²+ |c₂​|²+...|c_n​|²=1.

That makes the spatial integral of the absolute square of ψ come out 1, because the different  ψ_j are orthogonal to each other and thus their products drop out of that integral. Here I'm assuming that your ψ_j's are meant to be the energy eigenstates.

Mike W.

posted without vetting until Lee gets back.

(published on 01/19/2015)

You explained earlier that the normalization condition for wave functions is applied to whole state.|c1|2+ |c2​|2+...|cn​|2=1.So, if we say that a hydrogen atom is in the ground state, does that mean it is most likely in the ground state, but has some small chance to be in higher states? But then it means the atom does not have a definite energy. It has many energies at the same time. How is that possible?
- Alok (age 26)
Kingston, RI, USA

If the atom is in the ground state, it's just plain in the ground state and not a superposition of states with different energies. More general states do indeed have amplitudes for states with different energies. Despite how quantum mechanics is often taught, there's nothing more or less mysterious about having states with a range of different energies than there is about having states with a range of different positions and a range of different momenta. Those spreads are of course basic properties of all quantum states. States with a single fixed energy (energy eigenstates) play a special role because their physical properties don't change in time. So you know right away that such states must be atypical, because things happen.

Mike W. 

posted without vetting until Lee returns

(published on 01/20/2015)

OK.So suppose we have a hydrogen atom having a spread of energy over different levels at a given temperature.Its said that it has to emit or absorb light only at specific frequencies- those corresponding to the differences in the energy levels. Why is this true? If an atom can have a spread over the energy levels at the same time, why can it not emit/absorb an arbitrary amount of energy and redistribute the energy among different states?
- Alok (age 26)

The Schroedinger equation is completely linear. That means that the output state after some time has passed is exactly the sum of the output states of the various components of the initial state. So all we have to think about is what becomes of each of those various energy eigenstates separately, since we can always add up things like that to get what becomes of the initial state that is a superposition of them. 

Each so-called energy eigenstate (other than the ground state) is actually not quite an eigenstate of the full Hamiltonian, which includes the quantized electromagnetic field.  Nonetheless, the electromagnetic field will not end up in a lower energy state than it started in, so conservation of energy tells us that the atom will not end up in a higher energy state than it started in. If the atom changes, it can only change to some superposition of the discrete batch of lower-energy quasi-eigenstates. For each of those components energy conservation requires that the EM field carry off the particular energy lost by the atom going from the initial to the final state. So that leaves a discrete set of energies for the field. Generally, the fastest processes involve emitting a single photon, so those discrete energy differences give the set of single-photon energies. For rarer multi-photon processes only the sum of the photon energies is set by that constraint. So those processes can give a smooth spectrum, rather than sharp emission lines.

A quick search turns up an accessible article on this effect from quantum wells (), rather than atoms, but the basic ideas should be the same. The two-photon emission spectrum is very broad.  Note that even the single-photon spectrum already has a bit of width. That's a reminder that the electron eigenstates aren't quite true eigenstates once the quantized field is included.

Mike W.

posted without vetting until Lee returns

(published on 01/20/2015)