In studying Mach-Zehnder and Ramsey interferometers, I came across the expression "$\pi/2$ pulse". What does it mean exactly? I am working with a Bloch vector representation $(u,v,w)$ of a 2 state system. We have a Rabi frequency $\Omega_0$ and a detuning parameter $\delta$ to the $|1\rangle\rightarrow|2\rangle$ transition frequency. In those conditions, I think the "$\pi/2$ pulse" is a rotation around the $(-\Omega_0,0,-\delta)$ axis for a duration $\tau=\frac{\pi/2}{\sqrt{\Omega_0^2+\delta^2}}$. Is that correct?
[Physics] Exact meaning of “pi/2 pulse”
bloch-spherequantum mechanicsquantum-opticsspectroscopy
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A magnetic dipole transition can be modelled as a time-dependent perturbation $V_{\text{md}}(t) = {e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}e^{-i \omega t}$. Fermi's Golden Rule tells us that the transition rate for $b-X,1$ is proportional to the matrix element of the perturbation between the initial and final states,
$$W \propto \langle \psi_b|{e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}|\psi_{X}\rangle,$$
where $|\psi_b\rangle$ is the excited state and $|\psi_{X}\rangle$ is the ground state (with three possible $M_S$ values.)
The effect of $\vec L$ and $\vec S$ will be to turn the final state into some combination of the triplet states, but it won't change $J$. Therefore we might expect the transition to be 'spin-forbidden':
$$W \propto \langle b^1\Sigma_g^+ |X^3\Sigma_{g,M_S=0,\pm1}^-\rangle = \langle J=0 | J=1 \rangle = 0.$$
This is where the spin-orbit coupling comes into play. Spin-orbit coupling is the reason why the singlet $(b)$ state has a higher energy than the triplet $(X)$ states. It is a perturbation of the form $V_\text{SO}={\mu\over\hbar}\vec{L}\cdot\vec{S}$, which can be rewritten as $V_\text{SO}={\mu \over 2\hbar}(J^2-L^2-S^2)$. In a spherically symmetric system like the helium atom, this perturbation commutes with the Hamiltonian, so all you get is a shift in the energy of the triplet (L=1) and singlet (L=0) states. However, in a linear molecule like $O_2$ you lose the spherical symmetry, so $[L^2,H]\neq0$ and in addition to an energy shift, you also get some mixing of the unperturbed eigenstates, so that the excited state is not exactly $|b^1\Sigma_g^+\rangle$, but rather $|\psi_b\rangle = c_1|b^1\Sigma_g^+\rangle + c_2|X^3\Sigma_{g,M_S=0}^-\rangle$. This mixing of J=0 and J=1 states is what allows $W$ to have a nonzero value. Since we can write $S_x = S_+ + S_-$, there will be a term in the transition rate like
$$W\propto c_2^*\langle X^3\Sigma_{g,M_S=0}^-|S_{\pm}|X^3\Sigma_{g,M_S=\mp1}^-\rangle+\cdots \neq 0.$$
Does this help? I know this is a bit hand-wavy so let me know if I can clarify anything.
(I hope I did this right...)
Start from your original $H$ and $\vert\psi(t)\rangle$ and make the change $\vert \phi(t)\rangle = U(t)\vert\psi(t)\rangle$ as you suggested. The Schrodinger equation then becomes \begin{align} i\hbar \frac{\partial }{\partial t}\vert \psi(t)\rangle &= H(t) \vert \psi(t)\rangle\\ i\hbar \frac{\partial }{\partial t} U^{\dagger}(t)\vert\phi(t)\rangle &= H(t) U^{\dagger}(t)\vert \phi(t)\rangle \end{align} which in turn yields $$ i\hbar U^{\dagger}(t) \frac{-i\omega}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right) \vert \phi(t)\rangle + i\hbar U^\dagger(t)\frac{\partial }{\partial t}\vert\phi(t) \rangle = H U^\dagger(t)\vert \phi(t)\rangle\, . $$ Multiplying the $U(t)$ from the left gives \begin{align} i\hbar\frac{\partial}{\partial t}\vert\phi(t)\rangle&= -\frac{\hbar\omega}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right)\vert\phi(t)\rangle + H'\vert\phi(t)\rangle\, , \\ &= \frac{\hbar\omega}{2}\left(\begin{array}{cc} -1 & 0 \\ 0 &1 \end{array}\right)\vert\phi(t)\rangle + \frac{\hbar}{2} \left(\begin{array}{cc}\omega_0 &\omega_r \\ \omega_r &-\omega_0 \end{array}\right)\vert\phi(t)\rangle\, ,\\ &=\frac{\hbar}{2} \left(\begin{array}{cc}\omega_0-\omega &\omega_r \\ \omega_r &-(\omega_0 -\omega) \end{array}\right)\vert\phi(t)\rangle \end{align} As you can see, there is an additional term, which I eventually moved to the right, that comes from the time derivative of $U(t)$.
Nota: this is similar to the Coriolis-like terms that occur when going to a rotating frame, the origins of which are also in the time derivative of the rotation matrix linking the space-fixed and the body-fixed frames.
Best Answer
$\frac{π}{2}$ pulse means that all the particles in the system have gone to the higher level. $π$ pulse excites all particles in the first half time and de-exites in the second, so all particles are in lower level.