Physics Asked on December 13, 2020
The model considered
Consider an atom modeled by a two level system of energy $hbar omega$. We assume this atom is interacting with an electric field through electric-dipole interaction.
The full Hamiltonian is thus this one ($|erangle$ is the excited state and $|grangle$ the ground one):
$$ H = hbar omega_0 |erangle langle e | -mathbf{d} cdot mathbf{E_0} cos(omega_L t) , .$$
We can rewrite it as
$$ H = frac{hbar omega}{2} mathbb{1} + hbar omega_0 S_z -2 hbar Omega_1 cos(omega_L t) S_x $$
where $S_i = sigma_i/2$ ($sigma_i$ is the i$^text{th}$ Pauli matrix), $Omega_1 = langle e|mathbf{d}|grangle cdot mathbf{E_0}=langle g|mathbf{d}|erangle . mathbf{E_0}$ (I assumed for simplicity that the two dipole elements are positive numbers).
The analogy with a spin-magnetic field interaction
This Hamiltonian, if we put away the energy constant $frac{hbar omega}{2}$ can be interpreted as a spin interacting with the magnetic field via $H=-mathbf{B} cdot mathbf{S}$ where $[1]$
begin{align}
mathbf{B} &= mathbf{B_0} + mathbf{B_1}
mathbf{B_0} &= hbar omega_0 mathbf{U_z}
mathbf{B_1} &= 2 hbar Omega_1 cos(omega_L t) mathbf{U_x}=hbar Omega_1 mathbf{U_+}+hbar Omega_1 mathbf{U_-}
end{align}
where $mathbf{U_+}$ is a unitary vector rotating around the z-axis at frequency $omega_L$ and $mathbf{U_-}$ rotated in the opposite direction at the same frequency.
What we usually do in quantum mechanics is to go in the interacting picture, which mean to go in the rotating frame at frequency $omega_L$ around z-axis.
In this frame, the magnetic field $hbar Omega_1 mathbf{U_+}$ is now constant, $hbar Omega_1 mathbf{U_-}$ turns at frequency $2 omega_L$ I will neglect the latter corresponding to the rotating wave approximation.
Result of the model from a purely quantum mechanical treatment
The “real” quantum mechanical treatment would tell me that the rotating frame Hamiltonian is now
$$H=hbar (omega-omega_L) S_z – hbar Omega_1 S_x , .$$
Result of the model with the spin-magnetic field analogy
However, using this magnetic field vision, I would end up with
$$H=hbar omega S_z – hbar Omega_1 S_x , .$$
Indeed, the field $mathbf{B_0}$ doesn’t change with this rotation because I do a rotation around its axis. And as I only kept the part of $mathbf{B_1}$ that rotates in the counter-clockwise direction, $mathbf{B_1}$ is now constant in this frame $[2]$.
My question
Where is the problem in the “magnetic field” vision of it? Why doesn’t the analogy seem to work here?
$[1]$: My magnetic field are not in SI dimensions here, you can add an appropriate dimensional constant if you wish.
$[2]$: Actually it wouldn’t surprise me if $B_0$ would change in the rotating frame, but here http://puhep1.princeton.edu/~mcdonald/examples/rotatingEM.pdf it seems that the magnetic field should stay the same.
The rotating coordinate system method is equally applicable to classical and quantum-mechanical systems.
In the classical formulation the equation of motion of the system in a stationary coordinate system is
$d vec J / dt = vec mu times vec B = gamma vec J times vec B$
where:
$vec mu = gamma vec J$ magnetic moment
$vec J$ angular momentum
$gamma$ gyromagnetic ratio
$vec B$ magnetic field
In a rotating coordinate system with angular velocity $vec omega$, we have
$d vec J / dt = partial vec J / partial t + vec omega times vec J$
where $vec J$ on both sides is the angular momentum measured in the stationary frame while $partial vec J / partial t$ measures the change of $vec J$ in the rotating frame.
Rearranging
$partial vec J / partial t = gamma vec J times vec B_{rotating}$
where $vec B_{rotating} = vec B + vec omega / gamma$
In most of the cases $vec B$ is given by a constant field $vec B_0$ plus a (usually much weaker) $vec B_1$ field perpendicular to $vec B_0$ and rotating with angular velocity $- omega$. In a coordinate system rotating with $vec B_1$ both the fields are constant, so the axes of the rotating system can be chosen that
$vec B_0 = B_0 vec k$
$vec B_1 = B_1 vec i$
$vec omega = - omega vec k$
Then , in the rotating coordinate system
$vec B_{rotating} = (B_0 - omega / gamma) vec k + B_1 vec i$
The Hamiltonian in the rotating coordinate system is
$H_{rotating} = -vec mu cdot vec B_{rotating} = -gamma vec J cdot ((B_0 - omega / gamma) vec k + B_1 vec i)$
If $vec J = J_x vec i + J_y vec j + J_z vec k$, we have
$H_{rotating} = -(gamma B_0 - omega) J_z - gamma B_1 J_x$
As you can read the classical treatment shows formally identical to the quantum mechanical treatment.
Note: In my demonstration $omega$ corresponds to $omega_L$ in your post.
Answered by Michele Grosso on December 13, 2020
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