Understanding Coupled Spin Operators

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Discussion Overview

The discussion revolves around the mathematical treatment of coupled spin operators, particularly in the context of a hydrogen atom consisting of a proton and an electron. Participants explore the implications of non-commuting spin operators and the formulation of the total spin operator.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the application of non-commuting spin operators in deriving the total spin operator S^2, questioning the validity of the dot product approach.
  • Another participant seeks clarification on whether the non-commutativity applies to the spin components of the proton and electron.
  • A participant reiterates the concern about evaluating non-commuting components in the dot product, suggesting it seems illogical.
  • There is a proposal to express the total spin operator as S = Sz + Sy + Sx, leading to a discussion about the implications of treating these components as vectors.
  • A later reply emphasizes the distinction between the vector representation of spin and the spin magnitude, suggesting that the direction of the vector is not invariant under rotations.
  • Participants engage in a back-and-forth about the implications of these mathematical representations and their physical interpretations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the approach to the total spin operator or the implications of non-commutativity. The discussion remains unresolved with differing viewpoints on the mathematical treatment of spin operators.

Contextual Notes

There are limitations related to the assumptions made about the commutation relations of the spin operators and the implications of treating them as vectors. The discussion does not resolve these mathematical complexities.

cooev769
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Trying to get my head around this one. Given that you can have a proton and an electron in a hydrogen atom for example, and they can create a singlet or triplet configuration, with spin 1 and spin 0 respectively. The total spin operator can be derived as:

S^2 = (Se + Sp)^2 = Se^2 + Sp^2 + 2Se.Sp

Where the dot is the dot product, but then what you get is a bunch of Sx,Sz,Sy acting on the tensor products to provide eigenvalues and the eigenvectors out. The math is all well and good, but the problem I have is that the spin operators Sz,Sy and Sx don't commute, so how come you can just apply the operators to the respective spins to get out an answer, seems odd to me. Any help would be appreciated.

Cheers.
 
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What, are you saying that the components of the spin of the proton don't commute with those of the electron?
 
No. I'm saying that the component of the electron don't commute with each other yet they're evaluated in the dot production order to solve for the S^2 operator. But this seems daft to me given they do not commute.
 
cooev769 said:
No. I'm saying that the component of the electron don't commute with each other yet they're evaluated in the dot production order to solve for the S^2 operator. But this seems daft to me given they do not commute.
Write out the dot product. Do you get any terms that are affected by noncommutativity?
 
I understand what you're saying strange rep, but by that logic wouldn't that mean I can just write out the total spin operator as:

S = Sz + Sy + Sx

And apply this to any state to get the total spin. It's effectively the same thing.
 
cooev769 said:
[...] wouldn't that mean I can just write out the total spin operator as:

S = Sz + Sy + Sx
If each of your terms on the rhs are vectors, then S is also a vector. If that's what you meant, then let's write them properly, in bold, i.e.,

S = Sz + Sy + Sx.

Now draw a diagram with 3 orthogonal axes in 3D space and draw a vector from the origin to the point (1,1,1), i.e., x=1, y=1, z=1.

The vector S corresponds to spin-projection along that direction. This is quite different from the spin magnitude, i.e., the length of the vector.

The direction of a vector is not invariant under rotations, but the length is.

[...] And apply this to any state to get the total spin. It's effectively the same thing.
Hopefully you can now see that it's not the same thing?
 
Okay I'll keep contemplating your answer. Thanks.
 
Ah I get it now, thank you.
 

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