LS coupling for identical electrons

mk_gm1
Messages
10
Reaction score
0
Let's say I'm considering the 3p^2 electrons. From the Pauli Exclusion Principle, we know that two electrons cannot have the same state, which in this case means ml and ms cannot both be the same for each electron.

What this means is that the following 6 terms must not be allowed:

m_{l1} \hspace{0.1 in} m_{l2} \hspace{0.1 in} m_{s1} \hspace{0.1 in} m_{s2}

-1 \hspace{0.1 in} -1 \hspace{0.1 in} \downarrow \hspace{0.3 in} \downarrow
-1 \hspace{0.1 in} -1 \hspace{0.1 in} \uparrow \hspace{0.3 in} \uparrow
0 \hspace{0.4 in} 0 \hspace{0.3 in} \downarrow \hspace{0.3 in} \downarrow
0 \hspace{0.4 in} 0 \hspace{0.3 in} \uparrow \hspace{0.3 in} \uparrow
+1 \hspace{0.1 in} +1 \hspace{0.1 in} \downarrow \hspace{0.3 in} \downarrow
+1 \hspace{0.1 in} +1 \hspace{0.1 in} \uparrow \hspace{0.3 in} \uparrow

These correspond to M_L=\sum m_{li}<br /> = -2, -2, 0, 0, 2, 2 and M_S = \sum m_{si} = -1, 1, -1, 1, -1, 1 respectively.

My question is this - how does this lead to the conclusion that the allowed terms are 1S, 1D and 3P ? For example, there's a ML = 0, MS= -1 term in both 3S and 3P - why do we disallow one and not the other?

Also, what leads us to disallow 1P (for which ML=-1, 0, 1 and MS=0)? Surely the only way to have MS = 0 is to have \downarrow_1 \hspace{0.1 in} \uparrow_2 or vice versa, and hence m_{s1} \neq m_{s2} and we have no violation of the Pauli Exclusion Principle?
 
Physics news on Phys.org
What do you mean by ''why do we disallow one and not the other?''
 
Ok let me try and clarify what I'm confused by. Have a look at the following table which I have taken from my notes:

UZYKl.png


In this table we are looking at part of the identical (np)2 configuration. There are 15 possible configurations, all of which I haven't posted.

Take a look at the 2nd row. I can see why this state corresponds to the 3P term. For it to be a P term, it must have L = 1 where L \geq |M_L|. It clearly has this. For it to be a 3P term, the multiplicity 2S+1=3 i.e. S=1 where S \geq | M_S |. This is obvious.

Now if you look at the 4th row, I am not sure why it also has to be a 3P state . I understand that it can be (i.e. that the 3P term includes this state), but this could equally well be a 1P state, no? This is because here S=0. Therefore why do we choose one and not the other?
 
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
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Back
Top