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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.

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.

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, causing 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

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.

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 is 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 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 :)

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, causing 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