自旋1粒子的时间演化：利用CG系数求解相关问题

Let's work through this time evolution problem for a spin-1 particle with orbital angular momentum step by step. First, let's recap our given information clearly:

Given #

• Initial state vector:
  $$
  \left|\alpha,0\right\rangle=\frac{f\left(r\right)}{4\sqrt2}\left[\sqrt2\left|1-1\right\rangle_L \left|11\right\rangle_S + 2\left(\left|10\right\rangle_L\left|10\right\rangle_S +\sqrt6\left|00\right\rangle_L\left|10\right\rangle_S\right)- \sqrt2\left|11\right\rangle_L\left|1-1\right\rangle_S\right]
  $$
• Hamiltonian:
  $$
  \mathcal{H}=\frac{1}{\hbar}\left(\beta\mathbf{L}^2+2\gamma\mathbf{J\cdot S}\right)
  $$
  where $\mathbf{J} = \mathbf{L} + \mathbf{S}$ is the total angular momentum.

Step 1: Rewrite the Hamiltonian using conserved total angular momentum operators #

Since $\mathbf{J} = \mathbf{L} + \mathbf{S}$, we can expand the squared total angular momentum:
$$
\mathbf{J}^2 = \mathbf{L}^2 + \mathbf{S}^2 + 2\mathbf{L}\cdot\mathbf{S}
$$
Rearranging this gives us an expression for $\mathbf{J}\cdot\mathbf{S}$ (which equals $\mathbf{L}\cdot\mathbf{S}$):
$$
\mathbf{J}\cdot\mathbf{S} = \frac{1}{2}\left(\mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2\right)
$$

Substitute this into the Hamiltonian to rewrite it in terms of mutually commuting conserved operators:
$$
\begin{align*}
\mathcal{H} &= \frac{1}{\hbar}\left[\beta\mathbf{L}^2 + 2\gamma \cdot \frac{1}{2}\left(\mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2\right)\right] \
&= \frac{1}{\hbar}\left[ (\beta - \gamma)\mathbf{L}^2 + \gamma\mathbf{J}^2 - \gamma\mathbf{S}^2 \right]
\end{align*}
$$

Step 2: Calculate energy eigenvalues for joint eigenstates #

For a state $|j,l,s,m_j\rangle$ (our particle has $s=1$), the operator eigenvalues are:

• $\mathbf{L}^2 |j,l,s,m_j\rangle = l(l+1)\hbar^2 |j,l,s,m_j\rangle$
• $\mathbf{S}^2 |j,l,s,m_j\rangle = s(s+1)\hbar^2 = 2\hbar^2 |j,l,s,m_j\rangle$ (since $s=1$)
• $\mathbf{J}^2 |j,l,s,m_j\rangle = j(j+1)\hbar^2 |j,l,s,m_j\rangle$

Substitute these into the Hamiltonian to get the energy eigenvalue $E_{j,l}$:
$$
\begin{align*}
E_{j,l} &= \langle j,l,s=1,m_j | \mathcal{H} | j,l,s=1,m_j \rangle \
&= (\beta - \gamma)l(l+1)\hbar + \gamma j(j+1)\hbar - 2\gamma\hbar \
&= \hbar\left[ (\beta - \gamma)l(l+1) + \gamma(j(j+1)-2) \right]
\end{align*}
$$

Step 3: Decompose the initial state into joint eigenstates #

The initial state has $m_j = 0$ (every $|m_L\rangle_L|m_S\rangle_S$ term satisfies $m_L + m_S = 0$). We convert each product state to $|j,l,s=1,m_j=0\rangle$ using Clebsch-Gordan (CG) coefficients, then simplify the coefficients:

1. $|2,1,1,0\rangle$ (j=2, l=1): Coefficient simplifies to $\frac{f(r)\sqrt{3}}{6}$
2. $|1,1,1,0\rangle$ (j=1, l=1): Coefficient simplifies to $\frac{f(r)\sqrt{2}}{4}$
3. $|0,1,1,0\rangle$ (j=0, l=1): Coefficient simplifies to $-\frac{f(r)\sqrt{6}}{12}$
4. $|1,0,1,0\rangle$ (j=1, l=0): This term is already a joint eigenstate, with coefficient $\frac{f(r)\sqrt{3}}{2}$

The sum of squared coefficients confirms the state is normalized (assuming $f(r)$ is a normalized radial wave function).

Step 4: Time evolution of the state #

Time evolution of a stationary state $|j,l,s,m_j\rangle$ follows $e^{-iE_{j,l}t/\hbar}|j,l,s,m_j\rangle$. First compute each component's energy:

• $E_{2,1} = 2\hbar(\beta + \gamma)$
• $E_{1,1} = 2\hbar(\beta - \gamma)$
• $E_{0,1} = 2\hbar(\beta - 2\gamma)$
• $E_{1,0} = 0$ (the $\mathbf{L}^2$ term vanishes, and $\mathbf{J}^2$ and $\mathbf{S}^2$ contributions cancel)

Applying the time evolution operator to each component, the final state at time $t$ is:
$$
\begin{align*}
|\alpha,0;t\rangle &= \frac{f(r)\sqrt{3}}{6}e^{-i2(\beta+\gamma)t}|2,1,1,0\rangle + \frac{f(r)\sqrt{2}}{4}e^{-i2(\beta-\gamma)t}|1,1,1,0\rangle \
&\quad - \frac{f(r)\sqrt{6}}{12}e^{-i2(\beta-2\gamma)t}|0,1,1,0\rangle + \frac{f(r)\sqrt{3}}{2}|1,0,1,0\rangle
\end{align*}
$$

If needed, you can convert these back to $|m_L\rangle_L|m_S\rangle_S$ product states using inverse CG coefficients, but this form is optimal for analyzing time-dependent behavior.

内容的提问来源于stack exchange，提问作者Vincenzo Ventriglia