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 ($|e\rangle$ is the excited state and $|g\rangle$ the ground one):

$$ H = \hbar \omega_0 |e\rangle \langle e | -\mathbf{d} \cdot \mathbf{E_0} \cos(\omega_L t) \, .$$

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), $\Omega_1 = \langle e|\mathbf{d}|g\rangle \cdot \mathbf{E_0}=\langle g|\mathbf{d}|e\rangle . \mathbf{E_0}$ (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.

  • 1
    $\begingroup$ what is the unitary transformation to go to the rotating frame in each version? $\endgroup$
    – wcc
    Apr 14 '19 at 14:03
  • $\begingroup$ @IamAStudent In the Q.M treatment we swich to interacting picture through $U=e^{+\frac{i}{ \hbar} \hbar \omega_L |e\rangle \langle e|}$. Indeed we have rewritten the Hamiltonian as : $H=\hbar \omega_L |e\rangle \langle e| + \hbar (\omega_0 - \omega_L) |e\rangle \langle e| -d.E_0 cos(\omega t)$. Thus we have $|\psi\rangle \rightarrow |\psi_I\rangle = U |\psi\rangle$. In the analogy we don't really apply any unitary transformation. We just rewrite the "analogy" magnetic field in the rotating frame at frequency $\omega_L$ and it gives us our new Hamiltonian. $\endgroup$
    – StarBucK
    Apr 14 '19 at 16:22
  • 1
    $\begingroup$ well, I don't think that is true. Your operator $S$ in the Zeeman Hamiltonian $-S \cdot B$ will be affected by the same kind of unitary transformation. Pauli matrices span two-level system Hamiltonians, so there is no reason the magnetic dipole and B-field analogy should behave somewhat different from the actual two-level sytem. $\endgroup$
    – wcc
    Apr 14 '19 at 17:41
  • $\begingroup$ @IamAStudent I agree and this is my question : why do I end up with two different results ? Where is my mistake in the analogy ? $\endgroup$
    – StarBucK
    Apr 14 '19 at 17:51
  • $\begingroup$ I believe the mistake is that your claim that the magnetic field analogy doesn't involve unitary transformation. $\endgroup$
    – wcc
    Apr 14 '19 at 19:47

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
$$d \vec J / dt = \vec \mu \times \vec B = \gamma \vec J \times \vec B$$
$\vec \mu = \gamma \vec J$ 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.
$$\partial \vec J / \partial t = \gamma \vec J \times \vec B_{rotating}$$
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 such 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.