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At $t=0$, a spin-$1/2$ particle has the wave function

$$\psi=\frac{R(r)}{\sqrt{3}}\big[\sqrt{2} Y_1^{-1}(\theta,\phi) + Y_1^0(\theta,\phi) \big] \chi_{1/2}^{1/2},$$ that, in Dirac notation and using Clebsch-Gordan coefficients, I rewrite in the $\left|J,M \right\rangle$ basis as $$\psi=\frac{R(r)}{\sqrt{3}}\Big[\sqrt{\frac{2}{3}}\left|\frac{3}{2},-\frac{1}{2}\right\rangle -\sqrt{\frac{4}{3}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle + \sqrt{\frac{2}{3}}\left|\frac{3}{2},\frac{1}{2}\right\rangle - \frac{1}{\sqrt{3}} \left|\frac{1}{2},\frac{1}{2}\right\rangle \Big].$$ Given the hamiltonian $$\mathcal{H}=gJ_z,$$ how do you write the $J_z$ and the time evolution of the state, namely $\psi(t)$?

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    $\begingroup$ Do you know of any equations that relate the time dependence of a wavefunction to the Hamiltonian operating on that wavefunction? Have you heard of Schrödinger, for example? $\endgroup$ – Mike Dec 17 '17 at 2:16
  • $\begingroup$ Sure. But how do you write $J_z$? $\endgroup$ – Vincenzo Ventriglia Dec 17 '17 at 10:33
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    $\begingroup$ In your Dirac basis, as applied to your state, $J_z$ is but diag(-1/2,1/2), since you never had to consider the ±3/2 eigenvalues. $\endgroup$ – Cosmas Zachos Dec 17 '17 at 14:49
  • $\begingroup$ For example, $J_z \lvert 3/2, 1/2 \rangle = \hbar/2\, \lvert 3/2, 1/2 \rangle$. Note that $J_z r = 0$. Now give each spin ket a time dependence like $e^{\pm i \omega t}$ (with the sign based on the sign of $M$) and solve for $\omega$. Note that it's just $\pm \omega$ because each of these kets is an energy eigenstate with the same magnitude (but opposite signs) of energy. $\endgroup$ – Mike Dec 17 '17 at 15:02
  • $\begingroup$ @CosmasZachos, i really thank you! I know realize it was a silly question. That's because I don't have enough familiarity with exercises of this kind. $\endgroup$ – Vincenzo Ventriglia Dec 17 '17 at 20:18
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In general, the full time dependent solutions are $$ \vert \psi_n(t)\rangle =e^{-itE_n/\hbar}\vert \psi_n\rangle $$ with $\vert\psi_n\rangle$ the $n$'th eigenket of the time-independent equation.

As the Schrodinger equation is linear, a sum of solutions is also a solution; thus the general form of a solution is $$ \vert\Psi(t)\rangle =\sum_n c_n e^{-itE_n/\hbar}\vert \psi\rangle\, . $$ Since the Schrodinger equation is first order in time, one can determine the coefficients $c_n$ from the initial state $\vert\Psi(0)\rangle$.

It is then a simple exercise for you to find the $c_n$'s given your initial state.

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  • $\begingroup$ Yes, sure. But how do you write $J_z$ to answer the question? I'm not sure, but it should be a diagonal $6\times 6$ matrix in this basis $\endgroup$ – Vincenzo Ventriglia Dec 17 '17 at 10:32
  • $\begingroup$ @VincenzoVentriglia your individual kets are eigenkets of $J_z$. $\endgroup$ – ZeroTheHero Dec 17 '17 at 15:28

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