Why is the spin of this state equal to one?

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Discussion Overview

The discussion revolves around the spin states of a two-electron system, specifically addressing why a particular state is assigned a total spin quantum number of s=1 and a magnetic quantum number m_s=0. Participants explore the implications of quantum spin, the application of Hund's rules, and the calculations involved in determining total spin.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion regarding the assignment of s=1 to the state and questions the calculation of spin values.
  • Another participant corrects the misunderstanding, noting that the sum of spins should not be confused with the sum of the z-components of spins.
  • A different participant emphasizes that the calculated value is not the total spin but rather the projection along the z-axis.
  • Questions arise about the application of Hund's rules in determining total spin.
  • One participant provides a mathematical formulation for total spin, suggesting that the operator acting on the state yields a value consistent with s=1.
  • Another participant clarifies the distinction between total spin and individual particle spins, noting that while the z-component can sum to zero, it does not imply that other components are also zero.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of spin calculations and the application of Hund's rules. There is no consensus on the correct understanding of these concepts, and the discussion remains unresolved.

Contextual Notes

Participants highlight potential misunderstandings regarding the definitions of total spin and its components, as well as the implications of quantum mechanics on the calculations presented.

shedrick94
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I've just come across the spin states of a two electron system. There are 4 states possible and I am a little confused as to why the state below has values of s=1 m_s=0?

[1/√2]{α(1)β(2)+α(2)β(1)}

where α(i) and β(i) tell us if the particle has +ve or -ve z component of spin respectively.

I don't quite understand why s=1 as the sum of the spins for this would be (+1/2)(-1/2)+(+1/2)(-1/2)

http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec11_compressed.pdf a similar thing can be seen here on slide 10.
 
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shedrick94 said:
I don't quite understand why s=1 as the sum of the spins for this would be (+1/2)(-1/2)+(+1/2)(-1/2)
That's not correct. First, you are mixing up the total spin and the projection of the spin along z (the spin of an electron is always 1/2, never -1/2).

Second, what is the result of ##\hat{S} \alpha## or ##\hat{S} \beta##?
 
What you calculated, is not the sum of the spins but the sum of the z-component of the spins. The z-component can take on the values -1, 0, 1 for a particle with spin 1.
 
Do hunds rules not say that the S= sum of all m_s values?
 
Why is the spin of this state s=1 then? What am I missing?
 
If we let \vec{S} = \vec{S_1} + \vec{S_2}, where \vec{S} is the total spin, and \vec{S_1} and \vec{S_2} are the spins of the two electrons, then we have:

S^2 = (S_1)^2 + (S_2)^2 + 2 \vec{S_1} \cdot \vec{S_2}

If you let this operator act on the state, you will find that yields value 2 \hbar^2, which is consistent with an angular momentum of 1 \hbar. (Remember, the total angular momentum is quantized to have value S^2 = s(s+1) \hbar^2, so S^2 = 2 \hbar^2 \Rightarrow s = 1)
 
shedrick94 said:
Do hunds rules not say that the S= sum of all m_s values?

No, you have the total spin, \vec{S}, which has components S_x, S_y, S_z. You have the individual particle spins \vec{S_1} and \vec{S_2}, which have components (S_1)_x, (S_1)_y, (S_1)_z, (S_2)_x, (S_2)_y, (S_2)_z.

It is true that S_z = (S_1)_z + (S_2)_z = 0, but that doesn't imply that S_x and S_y are zero.
 

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