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The state ket of a spin-1 particle with orbital angular momentum is given by $$\left|\alpha,0\right\rangle=\frac{f\left(r\right)}{4\sqrt2}\left[\sqrt2\left|1-1\right\rangle_L \left|11\right\rangle_S + 2\left(\left|10\right\rangle_L\left|10\right\rangle_S +\sqrt6\left|00\right\rangle_L\left|10\right\rangle_S\right)- \sqrt2\left|11\right\rangle_L\left|1-1\right\rangle_S\right].$$ The Hamiltonian is $$\mathcal{H}=\frac{1}{\hbar}\left(\beta\mathbf{L}^2+2\gamma\mathbf{J\cdot S}\right).$$

Using CGCs, I wrote the state in the $\left|\mathbf{J}^2,J_z\right\rangle$ basis as $$\left|\alpha,0\right\rangle=\frac{f\left(r\right)}{2\sqrt2}\left[\sqrt{\frac{2}{3}}\left|20\right\rangle + \left(\sqrt6 -1\right)\left|10\right\rangle - \frac{1}{\sqrt3}\left|00\right\rangle\right].$$ Since $\mathcal{H}$ is diagonal, because $$\mathcal{H}=\frac{1}{\hbar}\left(\beta\mathbf{L}^2+\gamma\left(\mathbf{J}^2 -\mathbf{L}^2+\mathbf{S}^2 \right)\right)=\frac{1}{\hbar}\left(\gamma\mathbf{J}^2 +\left(\beta -\gamma\right)\mathbf{L}^2 +\gamma\mathbf{S}^2 \right),$$ I wrote the energies as $$E_{jls}=\hbar\left[\gamma j\left(j+1\right) + \left(\beta-\gamma\right)l\left(l+1\right) +\gamma s\left(s+1\right)\right].$$ Is it correct the following? $$\left\{\begin{matrix} \left|20\right\rangle&&E_{211}=\hbar\left(6\gamma+2\beta\right)\\ \left|10\right\rangle&&E_{111}=\hbar\left(2\gamma+2\beta\right),\;E_{101}=\hbar\left(4\gamma\right)\\ \left|00\right\rangle&&E_{011}=\hbar\left(2\beta\right) \end{matrix}\right.$$ So, the time-evoluted state must be $$\left|\alpha,t\right\rangle=\frac{f\left(r\right)}{2\sqrt2}\left[\sqrt{\frac{2}{3}}\left|20\right\rangle e^{-i\left(6\gamma+2\beta\right)t} + \sqrt{6}\left|10\right\rangle e^{-i4\gamma t} -\left|10\right\rangle e^{-i\left(2\gamma+2\beta\right)t} - \frac{1}{\sqrt3}\left|00\right\rangle e^{-i2\beta t}\right] \,?$$

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    $\begingroup$ Seems fine. I was concerned about the possibility that you had mixed the contributions from states with different $l$ but it turns out you had kept them separate, without mentioning it explicitly. $\endgroup$ – secavara Feb 14 '18 at 20:42
  • $\begingroup$ Yes, @secavara. Do you think I should write the two $\left|J,M\right\rangle=\left|10\right\rangle$ states (coming from different $l$) with different marks? I.e., are these states really the same or not? It's only due to $\mathcal{H}$ (so, it's eigenvalues) that we can distinguish these states. $\endgroup$ – Vincenzo Ventriglia Feb 15 '18 at 11:02
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    $\begingroup$ I think in this case it would be appropriate to label them differently because they do have different eigenvalues for $L^2$. I guess this is when the more convoluted notation $|J,M,L,S\rangle$ becomes relevant. $\endgroup$ – secavara Feb 15 '18 at 11:09

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