Physics Asked on July 12, 2021
I want to apply a transformation to the rotating frame of a two level system such that a state in the transformed frame is $ |hat{phi} rangle = U |phi rangle$, where U is the generator of rotations $ U = e^{iomega J_{z}t}$ with its Hermitian conjugate $U^dagger $.
Given the Hamiltonian in the stationary frame $H = omega_0 J_z + epsilon (J_{+}e^{-iomega t} +J_{-} e^{iomega t})$, I want to derive the expression for the Hamiltonian in the rotating frame, which should be $ hat{H} = (omega_0 -omega)J_z + epsilon(J_{+} +J_{-})$.
So far, I proceeded as follows:
Demanding that both$|phirangle$ and $|hat{phi} rangle$ satisfy the time-dependent Schroedinger equation, we may write (let $hbar =1$) :
$i frac{d}{dt}|hat{phi} rangle = hat{H}|hat{phi} rangle$,
where the LHS evaluates to $i frac{d}{dt}|hat{phi} rangle = i frac{dU}{dt}|phirangle + U i frac{d}{dt}|phirangle = – omega J_{z} e^{iomega J_z t}|phirangle + UH|phirangle$.
Since U is unitary, we can also write $|phirangle = U^dagger|hat{phi}rangle $ and substitute this into the expression to above to find
$i frac{d}{dt}|hat{phi} rangle = (- omega J_{z} + UHU^{dagger})|hat{phi} rangle $.
This would imply that $hat{H} = (- omega J_{z} + UHU^{dagger}) $
I would highly appreciate any help how to get to the desired result.
You got $$ ifrac{d}{dt}|hat{phi}rangle = -omega P_z U |phirangle + UH|phirangle$$ but you want the equation for $|hat{phi}rangle$. That's no problem - you just plug in $U^{dagger}U=1$ to get $$ ifrac{d}{dt}|hat{phi}rangle = -omega P_z U |phirangle + UHU^{dagger}U|phirangle = left[UHU^{dagger}-omega P_zright]|hat{phi}rangle$$ and you can identify $$ hat{H} = UHU^{dagger}-omega P_z$$ as $UP_{pm}U^{dagger} = e^{pm iomega t}P_{pm} $ (you can get that from Baker-Hausdorff or just by checking for the two possible values of $P_z$ by hand) you get the desired Hamiltonian.
(p.s. here I assumed that $P_z$ has eigenvalues $pm 1/2$, if that's not the case then you have to adjust the transformation accordingly)
Answered by user245141 on July 12, 2021
I suppose this is not a homework problem I'm denying you anymore. There is no CBH rearrangement involved.
You wish to evaluate $$hat{H} = - omega J_{z} + e^{iomega t J_{z}} He^{-iomega t J_{z}} , $$ given $$ [J_z,J_{pm}]= pm J_{pm}, $$ whence you can easily prove $$ J_{pm}~ f(J_z) = f(J_zmp 1) ~J_{pm}, $$ for any function f.
This expression, then readily reduces to $$ H = omega_0 J_z - omega J_z + epsilon e^{iomega t J_{z}} (J_{+}e^{-iomega t} +J_{-} e^{iomega t})e^{-iomega t J_{z}} = (omega_0 - omega ) J_z + epsilon (J_{+} +J_{-} ) . $$
Answered by Cosmas Zachos on July 12, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP