Do All Electrons in a Molecular Orbital Share the Same Magnetic Quantum Number?

grunf
Messages
3
Reaction score
0
I have one question about wave functions and molecular orbitals (MO). In the computational quantum chemistry scientists often use Linear Combination Atomic Orbitals (LCAO) to construct MO.

\varphi_{i\lambda\alpha}(\vec{r})=<br /> \sum\limits_{p=1}^{N}\chi_{p\lambda\alpha}(\vec{r})<br /> C_{i\lambda p},
where
\chi_{p\lambda\alpha}(\vec{r})=2^{-1/2}<br /> \left[\chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{A})<br /> +\sigma_{\lambda}\chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{B})\right].

Here \chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{J}) (J=A,B) are the usual Slater-type functions (STF-s) centered on A and B, respectively. \lambda is the symmetry species (\textrm{for example } \sigma or \pi) and \alpha (for example g or u) is the subspecies of symmetry \lambda.

Note, that in the above sum's there is no dependence m_{\lambda\alpha}
from p!?

The question is: Does it means that in some MO (\sigma or \pi, for example) all electrons must have the same magnetic quantum number? Why is that?
Is there some symmetry, some rule, some law or some common property for all electrons in the same MO.

In some papers I have found that for LCAO molecule N_{2} all electrons have
m=0 while in the case of molecule O_{2} m=1. Why is that? Is it possible that some electrons have m=-1?

If somebody have any sugestions about this question, I will be very appreciate.

Regards
 
Last edited:
Physics news on Phys.org
I have absolutely no idea at all about computational chemistry or STFs I'm afraid... (which I think is what may have put a lot of physicists off trying to answer!)
What I can tell you is that magnetic quantum numbers can range over the integers (-L, L) where L is the orbital quantum number, so yes electrons can have a magnetic moment of minus one. See here for a little more info, or google 'magnetic quantum numbers'.
 
grunf said:
I have one question about wave functions and molecular orbitals (MO). In the computational quantum chemistry scientists often use Linear Combination Atomic Orbitals (LCAO) to construct MO.

\varphi_{i\lambda\alpha}(\vec{r})=<br /> \sum\limits_{p=1}^{N}\chi_{p\lambda\alpha}(\vec{r})<br /> C_{i\lambda p},
where
\chi_{p\lambda\alpha}(\vec{r})=2^{-1/2}<br /> \left[\chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{A})<br /> +\sigma_{\lambda}\chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{B})\right].

Here \chi_{n_{\lambda p} l_{\lambda p}m_{\lambda\alpha}}(\vec{r}_{J}) (J=A,B) are the usual Slater-type functions (STF-s) centered on A and B, respectively. \lambda is the symmetry species (\textrm{for example } \sigma or \pi) and \alpha (for example g or u) is the subspecies of symmetry \lambda.

Note, that in the above sum's there is no dependence m_{\lambda\alpha}
from p!?

The question is: Does it means that in some MO (\sigma or \pi, for example) all electrons must have the same magnetic quantum number? Why is that?
Is there some symmetry, some rule, some law or some common property for all electrons in the same MO.

In some papers I have found that for LCAO molecule N_{2} all electrons have
m=0

all? really? All fourteen electrons? :wink:

while in the case of molecule O_{2} m=1. Why is that? Is it possible that some electrons have m=-1?

If somebody have any sugestions about this question, I will be very appreciate.

Regards
 
olgranpappy said:
all? really? All fourteen electrons? :wink:

I was quite unaccurate. In this case I was thinking on the electron in the Highest Ocuppied Molecular Orbital (HOMO). My mistake :frown:
 
muppet said:
I have absolutely no idea at all about computational chemistry or STFs I'm afraid... (which I think is what may have put a lot of physicists off trying to answer!)
What I can tell you is that magnetic quantum numbers can range over the integers (-L, L) where L is the orbital quantum number, so yes electrons can have a magnetic moment of minus one. See here for a little more info, or google 'magnetic quantum numbers'.

Thanx for the tips. I already know that. I am sure that is the key in the some molecular symmetries, but I am not sure. I am looking for some rule, law... Thanx again
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top