Time evolution of hydrogen wave function that is not an energy eigenfunction

[bier0134]

I recently had a probelm in QM to find the time evolution of a hydrogen prepared in a state with a wave function that is not an energy eigenfunction: specifically, psi = Y21*R2p where Y is then the D spherical harmonic. Of course, n=2 hydrogen doesn't have d oribtals.

So the problem is I can't simply apply the time evolution operator e^(-iHt/hbar) to the wave function and replace it with e^(-iEt/hbar), where E is the energy eigenvalue for the particular energy eigenfunction the hydrogen atom is prepared in, because it isn't in an energy eigen state.

So how do I time-develop [A> to [A(t)>?

 

[nnnm4]

Edit: that actually can't be done. I'm at a loss as well.

 

[phyzguy]

Your initial state can be written as a superposition of energy eigenfunctions:
[tex]|A_i> = \sum<u_n|A_i>|u_n>[/tex]
where Ai is the initial wave function and un are the energy eigenfunctions. Then with time, each energy eigenfunction will evolve with time as [tex]e^{-i E_n/\hbar}[/tex], so:
[tex]|A_i(t)> = \sum<u_n|A_i(0)>|u_n> e^{-i E_n/\hbar}[/tex]

 

[nnnm4]

Phyzguy, that's what I initially said but it may be more subtle. The thing is it depends on your hamiltonian. If it's just a coulomb hamiltonian and we're neglecting fine-structure, etc, then you can just apply the time evolution operator to the above state, the energy eigenvalue only depends on the radial part of the wave function. So your time-evolution operator will have the n=2 energy.

 

[phyzguy]

But it is not in an energy eigenstate, as you said. So it doesn't have a definite value of energy. The Coulomb interaction does not depend on the angular part, but the [tex]\frac{-\hbar^2}{2 m}\nabla^2[/tex] part of the Hamiltonian certainly does depend on the angular part. To find the time evolution, you need to do what I said. Each component of the wave function evolves according to its energy.

 

[kuruman]

But it is not in an energy eigenstate, as you said. So it doesn't have a definite value of energy. The Coulomb interaction does not depend on the angular part, but the [tex]\frac{-\hbar^2}{2 m}\nabla^2[/tex] part of the Hamiltonian certainly does depend on the angular part. To find the time evolution, you need to do what I said. Each component of the wave function evolves according to its energy.

That may be so, but the state "psi = Y21*R2p", assuming that it follows the standard notation,

ψ_nlm = R_nlY_lm

implies that n =2 and l = 2 which can't be because l_max = n - 1.

However, I am not sure what "p" in R_2p means. What does it mean bier0134?

 

[phyzguy]

That may be so, but the state "psi = Y21*R2p", assuming that it follows the standard notation,
ψ_nlm = R_nlY_lm
implies that n =2 and l = 2 which can't be because l_max = n - 1.

When you say "it can't be", you are referring to rules which apply only to the energy eigenstates. This state is not an energy eigenstate. The initial wave function can be anything - any combination of states that I want, as long as it is normalized. The question is to determine how the initial state changes with time.

 

[kuruman]

When you say "it can't be", you are referring to rules which apply only to the energy eigenstates.

Yes, I am.

This state is not an energy eigenstate. The initial wave function can be anything - any combination of states that I want, as long as it is normalized.

Correct, but it has to be a linear combination of legitimate eigenstates. Any hydrogenic wavefunction that has n = l is illegitimate.

The question is to determine how the initial state changes with time.

Yes indeed, but what do you do if the initial state does not make sense to begin with?

 

[phyzguy]

Yes, I am.

Correct, but it has to be a linear combination of legitimate eigenstates. Any hydrogenic wavefunction that has n = l is illegitimate.

Yes indeed, but what do you do if the initial state does not make sense to begin with?

This initial state does "make sense" The only requirement is that it be normalized to 1, which this state satisfies. It is a linear combination of eigenstates. But the initial state does not have a definite value of either n or l. I suggest you go back and review the basics of QM.

 

[kuruman]

This initial state does "make sense" The only requirement is that it be normalized to 1, which this state satisfies. It is a linear combination of eigenstates. But the initial state does not have a definite value of either n or l. I suggest you go back and review the basics of QM.

Please help me here. Specifically, what is the normalized linear combination that characterizes

psi = Y21*R2p

In other words, what are the coefficients and eigenstates that make up the linear combination? What do you think subscript "p" (and all the other subscripts) stand for?

 

[phyzguy]

Please help me here. Specifically, what is the normalized linear combination that characterizes

psi = Y21*R2p

In other words, what are the coefficients and eigenstates that make up the linear combination? What do you think subscript "p" (and all the other subscripts) stand for?

As you said in post 6, the hydrogen eigenfunctions are written:
[tex] \psi_{nlm} = R_{nl} Y_{lm}[/tex]
I assume (bier0134 will have to confirm) that R2p means n=2, l=1, using the spdf notation. So the initial wave function is given by:
[tex] \psi(t=0) = R_{21} Y_{21}[/tex]
As has been said, this is not an energy eigenfunction. To expand this in energy eigenfunction, we use the normal process of expanding a wave function in energy eigenstates, remembering the the Ylm spherical harmonics are all orthogonal, giving:
[tex] \psi(t=0)= \sum_{nlm}<R_{nl}Y_{lm}|R_{21}Y_{21}> |R_{nl}Y_{lm}> = \sum_{n}<R_{n2}|R_{21}> |R_{n2}>[/tex]
So you need to calculate the radial integrals of each Rn2 with the initial radial wave function R21. This will give the component of the initial wave function contained in each Rnd eigenstate. Do you see?

 

[kuruman]

I see.