π键角动量：量子化学中角动量沿核间轴投影的技术问询

Alright, let's break this down clearly—this is a core concept in molecular quantum chemistry, especially for diatomic systems. First off, we anchor this to diatomic (or linear) molecules, where the internuclear bond axis is a natural symmetry axis; we typically set this as the z-axis in our coordinate system to make calculations easier.

1. 轨道角动量沿键轴的分量：$\Lambda$量子数 #

For the total electronic orbital angular momentum vector $\vec{L}$, its projection onto the bond axis ($L_z$) follows quantum quantization:
$$L_z = \Lambda \hbar$$
where $\hbar = h/(2\pi)$ (the reduced Planck constant). The rules for $\Lambda$ are:

• $\Lambda$ is an integer, taking values $0, \pm1, \pm2, ..., \pm L$ (here $L$ is the total electronic orbital angular momentum quantum number, just like in atomic systems where $L=0,1,2$ maps to S, P, D states).
• To simplify notation, we use non-negative $\Lambda$ values to label molecular electronic states: $\Lambda=0$ corresponds to $\Sigma$ states, $\Lambda=1$ to $\Pi$ states, $\Lambda=2$ to $\Delta$ states, and so on (mirroring how atomic states use S/P/D for $L=0/1/2$).

2. 自旋角动量沿键轴的分量：$\Sigma$量子数 #

Next, the projection of the total electronic spin angular momentum $\vec{S}$ onto the bond axis ($S_z$) is:
$$S_z = \Sigma \hbar$$
Note: This $\Sigma$ (uppercase sigma) is a quantum number, not the $\Sigma$ electronic state we mentioned earlier. Its rules:

• $\Sigma$ is either an integer or half-integer, depending on the total spin quantum number $S$. It takes values $S, S-1, ..., -S$.
• For example: if the molecule has a total spin $S=1/2$ (like a radical with one unpaired electron), $\Sigma$ can be $+1/2$ or $-1/2$. If $S=1$ (two unpaired electrons with parallel spins), $\Sigma$ takes $+1, 0, -1$.

3. 总电子角动量沿键轴的分量：$\Omega$量子数 #

The total electronic angular momentum is $\vec{J} = \vec{L} + \vec{S}$. Its projection onto the bond axis ($J_z$) is:
$$J_z = \Omega \hbar$$
where $\Omega$ is the sum of $\Lambda$ and $\Sigma$:
$$\Omega = \Lambda + \Sigma$$

• The possible values of $\Omega$ range from $\Lambda+S$ down to $|\Lambda-S|$, in integer steps.
• Since $\Lambda$ is non-negative, the absolute value of $\Omega$ tells us the magnitude of the total electronic angular momentum's projection along the bond axis.

Quick Comparison to Atomic Systems #

As you noted, this mirrors atomic quantum mechanics:

• In atoms, we take an arbitrary axis (usually an external magnetic field z-axis) and the projection of total angular momentum $\vec{J}$ is $m_J \hbar$, where $m_J$ ranges from $-J$ to $+J$.
• In molecules, the bond axis is a built-in symmetry axis, so we use $\Lambda$ (orbital projection), $\Sigma$ (spin projection), and $\Omega$ (total projection) instead of $m_L, m_S, m_J$. The core idea is identical: angular momentum projections onto a fixed axis are quantized in multiples of $\hbar$.

A quick note to avoid confusion: Sometimes people use $J$ for the total angular momentum of the entire molecule (electronic + rotational), but your question focuses on angular momentum associated with "rotation around the bond"—so we're sticking to electronic angular momentum projections here.

内容的提问来源于stack exchange，提问作者Sergio