• A
Malamala
Hello! I have a 2 level system given by:

\begin{pmatrix}
0 & A \\
A & Bcos(\omega t)
\end{pmatrix}

If I populate only one of the states initially, at ##t=0## the 2 states are B apart, while after half a period they are -B apart. Basically the system went from one side of the avoided crossing to the other. If I am in the regime where ##A<<B<\omega##, based on some numerical calculations I made, the population transfer between the states is well described by ##sin^2(At)##. I thought that if you move very fast through the avoided crossing, you get full population transfer with a period of ##~1/\omega##. Why does it take so much longer ##~1/A## to get full population transfer. Also why is the population transfer even happening? Given that I go back and forth around the crossing I would have expected to populate and depopulate the states with a rate of ##~1/\omega##. Can someone help me understand this behaviour? Thank you!

• BillKet

Hyperfine
Your readers would benefit from you telling them explicitly what representation you are using and explicitly defining the terms in the matrix.

Malamala
Your readers would benefit from you telling them explicitly what representation you are using and explicitly defining the terms in the matrix.
What do you mean by representation? That is the Hamiltonian of the system. You can use any representation you want to solve the Schrodinger equation. Also why would the meaning of the terms matter in solving the equation?

• BillKet
Hyperfine
What do you mean by representation?
Is the representation, the Hamiltonian, adiabatic or diabatic?
You can use any representation you want to solve the Schrodinger equation.
You can use any complete set of basis functions.
Also why would the meaning of the terms matter in solving the equation?
You posed a question. Part of posing any good question is to provide an explicit definition of the terms within an equation that is part of that question. It would require nothing more than two lines of type and it would provide your readers with the basic knowledge they need to assist you.

• BillKet and Malamala
BillKet
Hello! I have a 2 level system given by:

\begin{pmatrix}
0 & A \\
A & Bcos(\omega t)
\end{pmatrix}

If I populate only one of the states initially, at ##t=0## the 2 states are B apart, while after half a period they are -B apart. Basically the system went from one side of the avoided crossing to the other. If I am in the regime where ##A<<B<\omega##, based on some numerical calculations I made, the population transfer between the states is well described by ##sin^2(At)##. I thought that if you move very fast through the avoided crossing, you get full population transfer with a period of ##~1/\omega##. Why does it take so much longer ##~1/A## to get full population transfer. Also why is the population transfer even happening? Given that I go back and forth around the crossing I would have expected to populate and depopulate the states with a rate of ##~1/\omega##. Can someone help me understand this behaviour? Thank you!
That is an interesting question. Most probably @Twigg might have some nice insight into it.

@Hyperfine I think your questions are confusing the OP, more than helping. The Hamiltonian of the problem is shown explicitly. That IS the representation the OP is using (a 2x2 matrix representation of the Hamiltonian operator). And in terms of the parameters it's quite easy to guess: ##B\cos(\omega t)## is an oscillatory signal with amplitude ##B## and frequency ##\omega##, while ##A## is a term connecting the 2 levels. Knowing the particularities of the problem (i.e. whether it is a 2 level atom, molecule, solid state system, SHO) doesn't seem relevant to me given the OP specific question (as I said, it might even add extra confusion).

• Malamala
Malamala
Is the representation, the Hamiltonian, adiabatic or diabatic?

You can use any complete set of basis functions.

You posed a question. Part of posing any good question is to provide an explicit definition of the terms within an equation that is part of that question. It would require nothing more than two lines of type and it would provide your readers with the basic knowledge they need to assist you.
I am not sure what you mean by adiabatic or diabatic. You take the hamiltonian, plug it in the Schrodinger equation, and get the bahavior of the system. How would this be different in a diabatic vs adiabatic case?

Hyperfine
I am not sure what you mean by adiabatic or diabatic. You take the hamiltonian, plug it in the Schrodinger equation, and get the bahavior of the system. How would this be different in a diabatic vs adiabatic case?
I assumed, perhaps incorrectly, that the system under consideration was molecular. In that case, in the region of an avoided crossing one usually casts the problem in terms of a diabatic representation in which the nuclear kinetic energy operator is diagonal. That serves to simplify the treatment.

Malamala
I assumed, perhaps incorrectly, that the system under consideration was molecular. In that case, in the region of an avoided crossing one usually casts the problem in terms of a diabatic representation in which the nuclear kinetic energy operator is diagonal. That serves to simplify the treatment.
I am really confused. The Schrodinger equation is:

$$i\frac{\partial\psi}{\partial t} = H\psi$$
I gave H explicitly above. One would solve this differential equation and get the time evolution. For example ##\psi## can be a column vector with 2 elements. What do you mean by kinetic energy? There is no kinetic energy, just that Hamiltonian I showed. I mean if you want to change the H by applying a unitary transformation you totally can, that won't change the evolution of the system. But again, I am not sure how is that relevant?

• BillKet
Hyperfine
It is directly relevant to one who is attempting to help you understand the basics rather than to solve one immediate problem.

Malamala
It is directly relevant to one who is attempting to help you understand the basics rather than to solve one immediate problem.
What do you mean by kinetic energy in this setup? Where would that come into play?

BillKet
I assumed, perhaps incorrectly, that the system under consideration was molecular. In that case, in the region of an avoided crossing one usually casts the problem in terms of a diabatic representation in which the nuclear kinetic energy operator is diagonal. That serves to simplify the treatment.
As I said, your questions are quite confusing. Adiabatic and diabatic come into play (in molecular case), when comparing the relative effects of potential and kinetic energy vibrational terms in a molecule and thus deciding how to diagonalize the hamiltonian. This is waaay more complex that needed for the OP question. If he doesn't know much about molecules that would just confuse him. Also is seems from the original question that there is no kinetic term in the original Hamiltonian, no?

• Malamala
Hyperfine
What do you mean by kinetic energy in this setup? Where would that come into play?
It is question of how the nuclear kinetic energies are treated: how you construct the molecular wave function and how you separate the molecualr geometry variables and the electronic degrees of freedom.

The nuclear kinetic energies cannot be neglected in the region of an avoided level crossing. Recasting the Hamiltonian into a diabatic representation makes the nuclear kinetic energy operator diagonal, thus simplifying the analysis.

Malamala
It is question of how the nuclear kinetic energies are treated: how you construct the molecular wave function and how you separate the molecualr geometry variables and the electronic degrees of freedom.

The nuclear kinetic energies cannot be neglected in the region of an avoided level crossing. Recasting the Hamiltonian into a diabatic representation makes the nuclear kinetic energy operator diagonal, thus simplifying the analysis.
What nuclear kinetic energies? There is no nucleus involved in my question. Why would you assume there is a nuclear kinetic energy involved.

Hyperfine
What nuclear kinetic energies? There is no nucleus involved in my question. Why would you assume there is a nuclear kinetic energy involved.
In your original post, you explicitly mentioned an avoided crossing. In such a region the Born-Oppenheimer approximation does not apply and as I stated explicitly above, the kinetic energies of nuclear motion within the molecule cannot be neglected.

I am sorry if my comments have lead you astray, but you did opt to label this thread as a category A.

• Malamala
Malamala
In your original post, you explicitly mentioned an avoided crossing. In such a region the Born-Oppenheimer approximation does not apply and as I stated explicitly above, the kinetic energies of nuclear motion within the molecule cannot be neglected.

I am sorry if my comments have lead you astray, but you did opt to label this thread as a category A.
But my system is not a molecule. I never said, at any point, it is a molecule. There are so many systems in which you can have avoided crossings (and my question should be applicable to all of them?). I don't understand why you are going into so many details about molecular physics, without that being mentioned at any point (my setup is actually a 2 level atom in the presence of an electric field).

Hyperfine
Again, you demand that your reader guess. Better, more complete questions lead to better, less confusing replies.

Malamala
Again, you demand that your reader guess. Better, more complete questions lead to better, less confusing replies.
Ok... could you explain to me (and I really want to understand it, this is not sarcastic), how would you answer the above question assuming it was a molecule and how would you do, assuming it is an atom? I really don't see how the Schrodinger equation solution, given an explicit Hamiltonian, changes, in the 2 cases. I don't mean to necessarily give me an explanation to the original question, but starting from the same initial conditions, and evolving the 2x2 Hamiltonian in time, how would an atom and a molecule give you different results?

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