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I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not konw why.


I currently have a preliminary understanding of this:
$/pi$ orbit
In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+\rangle-|-\rangle)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.
Is my understanding right?

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Please, do not substitute “>” for $\rangle$ — it makes typesetting errors. Also, one should not italicize spectroscopic symbols (s, p, d…) especially p that could be mistaken for momentum. –  Incnis Mrsi Aug 20 at 17:14
    
What kind of typesetting errors? I find > to be aesthetically nigh-unacceptable. –  Danu Aug 20 at 18:16
    
@Danu: invalid spacing. Examples like $|Φ><Ψ|x>=|T|x>$ produce a barely legible mess; compare to $|Φ\rangle\langle Ψ|x\rangle = |T|x\rangle$. –  Incnis Mrsi Aug 21 at 7:26
    
@IncnisMrsi Right, that too. It is ugly to the point of illegibility ;) –  Danu Aug 21 at 7:28

1 Answer 1

The famous “two-lobed” p eigenstate has the angular momentum projection to its symmetry axis equal to 0, and all other projections of said momentum are u n c e r t a i n . Let x denote original poster’s «z'». Although $L_x$ is uncertain, the two-lobed state is a linear combination of $L_x = 1$ and $L_x = -1$ eigenstates; it was mentioned by original poster. If we glue a molecular orbital from two such atomic orbitals (it is incorrect but, as a dirty approximation, one can try it), then its $L_x$ is uncertain too, but again, it will be a combination of $L_x = 1$ and $L_x = -1$ states. If we turn it by 90° around x, we can hope to obtain another combination of the same states, so we would find two quantum states that correspond to $L_x = 1$ and $L_x = -1$. That’s how the childish four-lobed picture above caused us to make extra steps towards understanding π bonds.

Generally, there are two “styles” for eigenfunctions pertaining to ℓ ≠ 0 orbitals (such as p): real-valued ones and complex-valued ones, where only complex-valued can serve $m_ℓ ≠ 0$ cases. I’d suggest to start from an $ℓ = 1,\ L_x = 1$ eigenstate (it is a complex linear combination of said z-elongated two-lobed state and a similar y-elongated one) and go directly to an $L_x = 1$ π orbital state. It will be already axially symmetric, up to complex argument.

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