If we consider a two level spin system with frequency $\omega_0$ in the presence of an oscillatory force with frequency $\omega$ chosen off-resonance. The force is designed so that a state $| \downarrow \rangle$ feels an equal but opposite force to a particle in state $| \uparrow \rangle$. How would it follow that after a period of $t = \frac{2 \pi}{\omega_0 - \omega}$, that the drive force has completely dephased and rephased with the oscillating two level system, thus accelerating and decelerating it to its initial motional state?

Thanks for any assistance.

  • $\begingroup$ Could you expand your question a bit and provide a model Hamiltonian? Do you mean something like $H_0=\omega_0\sigma_z$ and some external $H_1=\omega \sigma_y$? $\endgroup$ Apr 18, 2017 at 16:58
  • $\begingroup$ @ZeroTheHero Yes definitely an $H_0$ as you stated and some external oscillating electric field I would assume (even though the paper states in simply as an oscillating force). Does this idea make any sense to you how? $\endgroup$
    – user100411
    Apr 18, 2017 at 18:33
  • $\begingroup$ Come to think of it it's probably the reverse: $H_0=\omega_0\sigma_y$ and $H_1=\omega \sigma_z$. This way the contribution of $H_1$ is such that the contribution of $\sigma_z$ to $\vert \uparrow\rangle$ and $\vert\downarrow\rangle$ is equal but opposite. I'm not sure I understand the use of "dephased" and "rephased" though. $\endgroup$ Apr 18, 2017 at 20:18
  • $\begingroup$ @ZeroTheHero What about $H_0 = \omega_0 \sigma_{z}$ and $H_{1} = \omega \sigma_{x}$? $\endgroup$
    – user100411
    Apr 18, 2017 at 20:37
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    $\begingroup$ @JohnDoe Sorry couldn't get back sooner. JDR's answer is on the right track, but if you'd like a more intuitive picture you should look into using the Heisenberg (or interaction) picture and the eom-s for the averages of spin components. You get a linear system with time-dependent coeffs that can be solved fairly easily and you can map the solution onto the Bloch sphere. I can try to write something later, but may take a while. I see JDR refers to a paper, but I don't see a link. Would help if you could provide one. $\endgroup$
    – udrv
    Apr 19, 2017 at 16:32

1 Answer 1


I think there might still some potential ambiguities in the original question, but I'm going to give it a small go and let others chime in with extra contributions. I'll assume you're referring to a Hamiltonian like $H=\omega_0I_z+\omega_1 \cos(\omega t)I_y$ (where $I_\phi=\frac{1}{2}\sigma_\phi$ in natural units), since you imply a transverse field oscillating in magnitude.

This might be best treated by going to the "rotating frame" (we're going to rotate around the z axis at the same frequency as the $I_y$ field to make it look constant), in which case you have a transformed Hamiltonian $\tilde{H}=(\omega_0-\omega)I_z+\omega_1I_y$.

Aside: Rotating Frame

You can think of the rotating frame by just spinning your eigenvectors with a rotation operator, $\tilde{\left|\psi\right\rangle}=R(-\omega t)\left|\psi\right\rangle$, then asking how the Schrodinger equation must behave. By applying the chain rule and Schrodinger equation to $\frac{d}{dt}(R(-\omega t)\left|\uparrow\right\rangle)$, we end up with a "rotating frame Schrodinger equation" $i\frac{d}{dt}\tilde{\left|\psi\right\rangle}=\tilde{H}\tilde{\left|\psi\right\rangle}$, where $\tilde{H}=R(-\omega t)HR(\omega t)-\omega I_z$. It's the chain rule that gives us the "$-\omega I_z$" term, which is important here.

See e.g. Levitt, Spin Dynamics, p.241

So the time evolution operator for this time-independent Hamiltonian is just $U(t)=\exp (-i\tilde{H} t)$, and for any wavefunction $\left|\psi(t)\right\rangle=U(t)\left|\psi(0)\right\rangle$. Plugging in the rotating frame Hamiltonian and the time you gave (call it $T$), we've got

$U(T)=\exp [-i((\omega_0-\omega)I_z+\omega_1I_y)(\frac{2\pi}{\omega_0-\omega})]=\exp [-2\pi i(I_z+\frac{\omega_1}{\omega_0-\omega}I_y)]$

And this could be a mere net $2 \pi$ rotation, so any "dephasing" is "rephased" by the time you've hit period T.

However, this is also where I get confused by your question. I can see a possibility that this time evolution operator will end up just being a net $2 \pi$ rotation around some axis, but perhaps the magnitude $\omega_1$ needs to be specially defined? Is this what you mean by "the force is designed so that a state ... feels an equal but opposite force..."?

I'm not sure where else to go from here, perhaps you could provide a link to the paper you're referring to, or someone could pick up my thread where I'm leaving off and take the credit for a satisfactory answer, if this one didn't get to what you were wondering :)

edit: for clarity and just so many mistakes.

  • $\begingroup$ The only thing that I'm able to add from reading the paper is to look at pg. 12, and note that a detail I excluded above is that the $\cos(\omega t)$ term is actually considered as two rotating fields in opposite directions $\frac{1}{2}(e^{i\omega t} + e^{-i \omega t})$, like their expression written in terms of raising and lowering operators. Sorry I can't help more than that $\endgroup$
    – JDR
    Apr 18, 2017 at 21:00
  • $\begingroup$ Looks like that part is also along $z$. $\endgroup$ Apr 18, 2017 at 21:19
  • $\begingroup$ @JDR Thanks for your response. Just a few questions: 1. Could we have considered instead $\hat{H} = \omega_0I_z + \omega_1 \cos(\omega t)I_x$, would that be fine as well? 2. Are you chossing $R(t) = e^{\frac{i H t}{\hbar}}$ as your rotation operator? 3. Would we require that $U(T) = e^{-2 \pi \sigma_{z}}$ to get my quoted result, thanks. $\endgroup$
    – user100411
    Apr 19, 2017 at 18:29
  • $\begingroup$ 1. Certainly using $I_x$ would work fine as well (or even any combination of $I_x$ and $I_y$, the point is having a field in the transverse plane). 2. A rotation using the spin operators will always look like $\exp [i I_{\phi} \theta]$ for a rotation of angle $\theta$ around axis $\phi =x,y,z$. So if the $H$ you've written were $\hbar \omega I_y$ for example, you can see you would have a rotation by angle $\omega t$ around the y axis. $\endgroup$
    – JDR
    Apr 20, 2017 at 14:22

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