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A spin-$1/2$ particle is subject to an external magnetic field $$\mathbf{B}\left(t\right)=B\left(\mathbf{i}\cos{\omega t}-\mathbf{j}\sin{\omega t}\right) + B_0\mathbf{k} \; \left(B,B_0\in\mathbb{R^+}\right).$$ The state ket obeys the equation $$i\hbar\frac{\rm{d}}{\rm{d}t}\left|\alpha,t\right\rangle=-\gamma\mathbf{S\cdot B}\left|\alpha,t\right\rangle, \; \left(\gamma\in\mathbb{R^+}\right)$$ with initial condition $$\left|\alpha,0\right\rangle=\left|\frac{1}{2}\frac{1}{2}\right\rangle\equiv\left|+\right\rangle$$ in the $\left\{\mathbf{S}^2,S_z\right\}$ basis. Calculate $\left|\alpha,t\right\rangle$.

Writing the Hamiltonian as $$H\left(t\right)=-\gamma\mathbf{S\cdot B}\equiv H_0 + V\left(t\right),$$ I tried to use interaction picture, $$\left|\alpha,t\right\rangle_I\equiv e^{i H_0 t/\hbar}\left|\alpha,t\right\rangle.$$Now the ket obeys the differential equation $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\left|\alpha,t\right\rangle_I=V_I\left(t\right)\left|\alpha,t\right\rangle_I,$$ with initial condition $$\left|\alpha,0\right\rangle_I =\left|\alpha,0\right\rangle=\left|+\right\rangle$$ and $$V_I\left(t\right)\equiv e^{iH_0 t/\hbar}V\left(t\right)e^{-iH_0 t/\hbar}.$$ So, since $$H_0\equiv -\gamma B_0 S_z=-\omega S_z, \;\; V\left(t\right)\equiv-\gamma B \left(S_x \cos{\omega t} - S_y \sin{\omega t}\right),$$ I got the coupled system of ODEs $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\left|\alpha,t\right\rangle_I=-\frac{\gamma B\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\left|\alpha,t\right\rangle_I$$ for some ket of the form $$\left|\alpha,t\right\rangle_I=\begin{pmatrix}a\left(t\right)\\b\left(t\right)\end{pmatrix}.$$ Differentiating again with respect to time, I got $$\frac{\mathrm{d}^2}{\mathrm{d}t^2}a\left(t\right)+\omega_1^2 a\left(t\right)=0,$$ whose solution, taking initial condition into account, is $$a\left(t\right)=a_0 \cos{\omega t}.$$ So, $$b\left(t\right)=\frac{2i\omega a_0}{\gamma B}\sin{\omega t.}$$ To deal with a normalized state ket, one must have $$1=\left|a\left(t\right)\right|^2 + \left|b\left(t\right)\right|^2=\left|a_0\right|^2\left(\cos^2{\omega t}+\left(\frac{2\omega}{\gamma B}\right)^2\sin^2{\omega t}\right),$$ which, because $\omega=\gamma B_0$, brings the conditions $$\left|a_0\right|^2=1, \;\; B=2 B_0.$$ But I find very strange the second condition. Is there any problem with my idea or with my calculations?

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  • $\begingroup$ This is a pretty standard problem. How does your solution compare with the usual textbook solutions? $\endgroup$
    – ZeroTheHero
    Feb 12, 2018 at 22:06
  • $\begingroup$ @ZeroTheHero, never read anything about this before. This is just what I've got and I don't know if there are usual textbook solutions. $\endgroup$
    – Vincenzo Ventriglia
    Feb 12, 2018 at 22:10
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    $\begingroup$ A couple of comments. From the computations that follow later I assume the $\omega$ in the original problem is the same as $\gamma B_0$, as you mention later in the question, so it would have been helpful to mention that from the beginning. In any case, be careful about the labeling because one can use several $\omega$'s here. Second, notice that in that second order equation for $a(t)$ you have quantities with different units... maybe that $\hbar$ should not be there? Third, the $\omega$ in the $\cos$ and $\sin$ after that is not the same $\omega=B_0 \gamma$; There you should have $B \gamma$. $\endgroup$
    – secavara
    Feb 12, 2018 at 23:44
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    $\begingroup$ For reference, I give here the solution for arbitrary $\omega$ not necessarily equal to $\gamma B_0$. The state in Schrodinger picture is $|\psi\rangle_{\mathrm{S}} = a \, |+\rangle+ b \, |-\rangle$ with $a=\exp\left(\frac{i t \omega}{2}\right) \left[\cos\left(\frac{t \, \omega_r}{2}\right) + i \frac{B_0 \gamma-\omega}{\omega_r} \sin\left(\frac{t \, \omega_r}{2}\right)\right]$, $b=i\frac{B \gamma}{\omega_r}\exp\left(-\frac{i t \omega}{2}\right)\sin\left(\frac{t \, \omega_r}{2}\right)$, and $\omega_r=\sqrt{B^2\gamma^2+\left(B_0 \gamma-\omega\right)^2} \,$. $\endgroup$
    – secavara
    Feb 12, 2018 at 23:51
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    $\begingroup$ I think you are doing fine. The conceptual part is all well. just check the computations step by step and be careful when distinguishing distinct $\omega$s. The only thing I did in addition was use Mathematica for the algebra. Similar problems are discussed in section 5.5 of Sakurai's book, Modern Quantum Mechanics. $\endgroup$
    – secavara
    Feb 13, 2018 at 11:09

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Checking calculations (thanks to @secavara) I've found that the correct equation is $$\ddot{a}\left(t\right) + \varpi^2 a\left(t\right)=0,\\ \varpi\equiv\frac{\omega_1}{\hbar}\equiv-\frac{\gamma B}{2},$$ whose solution, taking initial condition into account, is given by $$a\left(t\right)=c\cos{\varpi t}.$$ Differentiating with respect to time, immediately gives $$b\left(t\right)=-ic\sin{\varpi t}.$$ The constant $c$ can be choosen with normalization condition, i.e $$1=\left|a\left(t\right)\right|^2+\left|b\left(t\right)\right|^2=\left|c\right|^2.$$ Fixing the phase factor so that $c=1$, I finally got $$\left|\alpha,t\right\rangle_I=\begin{pmatrix} \cos{\varpi t}\\-i\sin{\varpi t}\end{pmatrix}.$$ Now it's easy to write the state back in the Schrödinger picture, because $$\left|\alpha,t\right\rangle=e^{i\omega S_z t/\hbar}\left|\alpha,t\right\rangle_I= \begin{pmatrix}\cos{\varpi t}\,e^{i\omega t/2}\\-i\sin{\varpi t}\, e^{-i\omega t/2}\end{pmatrix}.$$

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