# Total spin of triplet and singlet states

I'm a litte confused about spin triplet and singlet states. How do we know that for ↑↓+↓↑ the total spin S is 1, and for ↑↓-↓↑ the total spin S is 0?
Also, how is total ms computed for these two states? (I understand that they are both 0, but not sure where that comes from)

Thank you very much for the help!

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blue_leaf77
Homework Helper
How do we know that for ↑↓+↓↑ the total spin S is 1, and for ↑↓-↓↑ the total spin S is 0?
Apply the (squared) total spin operator ##\mathbf{S}^2 = (\mathbf{S_1}+\mathbf{S_2})^2## to those states.
Also, how is total ms computed for these two states
Apply the operator ##S_z = S_{1z} + S_{2z}## to those states. Alternatively, upon following the theorem of the addition of angular momenta, you will find that the z component of the resultant spin state is equal to the sum of the z components of the individual states appearing in the resultant state's representation in the individual spin state basis.

Apply the (squared) total spin operator ##\mathbf{S}^2 = (\mathbf{S_1}+\mathbf{S_2})^2## to those states.

Apply the operator ##S_z = S_{1z} + S_{2z}## to those states. Alternatively, upon following the theorem of the addition of angular momenta, you will find that the z component of the resultant spin state is equal to the sum of the z components of the individual states appearing in the resultant state's representation in the individual spin state basis.