What i'm struggling with is this question:

Consider a spin S in an external magnetic field along the quantization axis so that the $|S_{z}>$ states are eigenstates with equally spaced eigenvalues E = $-h$$S_{z}$. For this Hamiltonian describe the time evolution of a spin-coherent state and relate it to the classical angular momentum dynamics.

Attempt at a solution:

So i've previously derived that when the time evolution operator $U(t)$ acts on a coherent state i.e. $U(t)|\alpha(0)>$ = $e^{\frac{-iHt}{\hbar}}|\alpha(0)>$ this becomes $U(t)|\alpha(0)>$ = $e^{\frac{-i\omega t}{2}}|\alpha(0)e^{-i\omega t}>$ = $e^{\frac{-i\omega t}{2}}|\alpha(t)>$, which i'm not sure whether will help or not.

I'm guessing here that $\vec{B} = B\vec{z}$, and that possibly the hamiltonian is $H=− μ⃗ · B⃗ = −g q S⃗ · B⃗ = -gqS_{z}B_{z}$ but I don't know where to go from there.

Any help would be amazing, thank you!


1 Answer 1


I will refer to your previous question for some basics but will use a different notation. Let $z=\tan(\theta/2)e^{i\phi}$. The evolution of the spin coherent state \begin{align} \vert z\rangle&:=\frac{1}{(1+\vert z\vert^2)^J} \sum_{M=-J}^J z^{J-M}{{2J}\choose{J+M}}^{1/2}\vert JM\rangle\, , \end{align} should be something like \begin{align} e^{-i\alpha S_zt}\vert z\rangle&= \frac{1}{(1+\vert z\vert^2)^J} \sum_{M=-J}^J z^{J-M}{{2J}\choose{J+M}}^{1/2}e^{-i\alpha M t}\vert JM\rangle\, ,\\ &=e^{-i\alpha J t}\frac{1}{(1+\vert z\vert^2)^J} \sum_{M=-J}^J z^{J-M}{{2J}\choose{J+M}}^{1/2}e^{-i\alpha (M-J) t}\vert JM\rangle\, ,\\ &=e^{-i\alpha J t}\vert e^{i\alpha t}z\rangle\, . \end{align} Please check carefully and make adjustments to your notation but this should give you enough of a hint.

  • $\begingroup$ Thanks for this! Could you explain what alpha is in your equations? $\endgroup$ Apr 21, 2017 at 21:44
  • $\begingroup$ Your Hamiltonian is something like $H=-gqB_z S_z$ so I guess this makes my $\alpha$ your $-gqB_z$ $\endgroup$ Apr 21, 2017 at 21:50
  • $\begingroup$ Ah great I see. Is there also supposed to be a $1/\hbar$ in the leftmost exponential as well? $\endgroup$ Apr 21, 2017 at 21:56
  • $\begingroup$ Probably yes. You should double check but I think if you take the eigenvalues of $S_z$ to be $\hbar M$ then the $\hbar$s cancel. $\endgroup$ Apr 21, 2017 at 21:59

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