This is the Jaynes-Cummings Hamiltonian in the interaction picture: Hni=ℏg√n+1(0exp(−iδt)exp(iδt)0)
I want to transform it into another basis, so that it looks like this:
Hni2=ℏ(−δ/2g√n+1g√n+1δ/2)
I present the solution that I found with help from DanielSank and march below.
Answer
Following DanielSank's suggestion, I tried writing both Hamiltonians in terms of Pauli operators: Hni=ℏg√n+112[(σx+iσy)e−iδt+(σx−iσy)eiδt],
So I'm basically looking for a transformation that converts the content of the brackets into σx and also makes this other term appear.
If I want to get rid of the time dependence, I can transform everything back into the Schroedinger picture by using U=(exp(iδt/2)00exp(−iδt/2)).
This gives: UHniU+=ℏ√n+1σx=H∗
I try to apply this U to the Schrodinger equation from the left side:
UHni|Ψ⟩=iℏU∂t|Ψ⟩
Then I insert U−1U before the state |Ψ⟩ on both sides:
UHniU−1U|Ψ⟩=iℏU∂tU−1U|Ψ⟩
Then I rename U|Ψ⟩≡|Ψ′⟩
Now product rule, as march suggested:
UHU−1|Ψ′⟩=iℏUU−1∂t|Ψ′⟩+iℏU∂t(U−1)|Ψ′⟩
The part of the left side is now exactly what I was looking for!
Update
A follow-up question generalizes this special case.
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