# Why is the spin of this state equal to one?

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• shedrick94
The sum of the m_s values is not the total spin, it is the projection of the total spin along a particular axis (in this case, the z-axis). In summary, the conversation discusses the spin states of a two electron system, where the possible states are determined by the values of s=1 and m_s=0. The individual particle spins and the total spin are discussed, with the total spin being quantized to a value of S^2 = s(s+1) \hbar^2, resulting in a spin of s=1 for this particular state.

#### shedrick94

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.

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.