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I am currently stuck on the following notation:

$\frac{1}{2}\otimes\frac{1}{2} = 0 \text{ (antisym) } \oplus 1 \text{ (sym) }$

No matter what I tried, I couldn't derive the identity. I am sure that it is trivial, but I can't figure how to treat the notation. It would be great, if somebody could write this in matrix notation or in bra/ket notation with explicit spins ($\uparrow\downarrow$).

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Do you know about the rules for addition of angular momenta in quantum mechanics? I assume that in your formula $n$ is a space of states for a spin $n$ particle, where $n=0,\frac{1}{2},1$. The equation says that the state space for two spin $\frac{1}{2}$ particles can be decomposed into a sum of eigenspaces for the total spin operator $S=S^{(1)}\otimes \mathbb{I}+\mathbb{I}\otimes S^{(2)}$. This can be done using Clebsch-Gordan coefficients. If you want more details let me know and I'll put them in an answer! –  Edward Hughes Jan 2 '13 at 13:56
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3 Answers

up vote 3 down vote accepted

The basis states on the left are given by $$|{\uparrow\uparrow}\rangle, |{\uparrow\downarrow}\rangle, |{\downarrow\uparrow}\rangle,\text{ and }|{\downarrow\downarrow}\rangle.$$

On the right, you are supposed to symmetrize these states with respect to exchanging the first and second spin (that is what sym and antisym stand for). There is only a single antisymmetric combination (try to see why there is only this one up to multiplication with a complex-constant) $$|S\rangle = 2^{-1/2}(|{\uparrow \downarrow}\rangle - | {\downarrow \uparrow }\rangle)$$ where $2^{-1/2}$ is for normalization purposes. Because this state is alone it is called singlet.

The orthogonal complement of the four states are three states (triplet) which are symmetric under exchange $$|T,1\rangle = |{\uparrow\uparrow }\rangle, |T,0\rangle = 2^{-1/2}(|{\uparrow \downarrow}\rangle +| {\downarrow \uparrow}\rangle), \text{ and } |T,1\rangle = |{\downarrow\downarrow}\rangle.$$

This is written out explicitly what the notation $$\frac{1}{2}\otimes\frac{1}{2} = 0 \text{ (antisym) } \oplus 1 \text{ (sym) }$$ means.

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Thanks very much for this answer. It almost satisfied me ;-). The only thing I am still wondering about is how you explicitly calculate the direct sum on the right side. I read about the direct sum in hilbert spaces on en.wikipedia.org/wiki/…, but was unfortunatelly unable to apply the given formulas to the states given in your answer. It would be great if you could add how to explicitly calculate the right side. Thanks a lot. –  ftiaronsem Jan 2 '13 at 14:58
    
A direct sum $H=V_1 \oplus V_2$ is nothing but a statement that the Hilberspace $H$ consists of $V_1$ and $V_2$, i.e., a basis in $H$ is the combined basis of $V_1$ and $V_2$. –  Fabian Jan 2 '13 at 15:01
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The notation on the left hand side refers to the set of tensor product of two spin-1/2 states (two components) and the right hand side refers to the direct sum of a spin-0 (single component) and spin-1 (three components). The antisym. and sym. refer to taking symmetric and antisymmetric combinations of spins.

Since I'm not sure immediately how to do bra/ket's in Mathjax I'll use matrix notation. Let $\psi_i$ and $\chi_i$, $i=1,2$ represent two two-component spinors. The states on the left hand side are $\psi_i \chi_j$. There are four of them corresponding to the combinations $(i,j)=(1,1),(1,2),(2,1),(2,2)$. These can be rewritten in terms of the linear combinations

$$ \psi_1 \chi_2 - \psi_2 \chi_1 $$

and

$$ \psi_1 \chi_1, \psi_1 \chi_2 + \psi_2 \chi_1, \psi_2 \chi_2 $$

where I've left out normalizing factors. These are the singlet and triplet on the right hand side, respectively. Notice that the singlet is antisymmetric and the triplet is symmetric under swapping of indices.

To prove that the triplet and singlet have the claimed values of the angular momentum operate on the states with the $\vec{J}^2 = (\vec{J}_1 + \vec{J}_2)^2$ operator, where $\vec{J}_1$ and $\vec{J}_2$ are the angular momentum operators for the $\psi$ and $\chi$ components respectively. The only nontrivial piece is the term involving $\vec{J}_1 \cdot \vec{J}_2$. You should find that it doesn't mix any of the combination states I've written.

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using group theory language $(\frac{1}{2},0)\bigotimes(\frac{1}{2},0)=(\frac{1}{2},\frac{1}{2};0,0)=(1,0)\bigoplus(0,0) $ For Su(2), (0,0) is an invariant, $\epsilon_{ij}$ and it is a antisymmetric tensor. (1,0) is a symmetric tensor. The number of tensor indices is $2s$ for $(s,0)\equiv(s)$.

references:

(1) Coleman's lecture: An introduction to unitary symmetry

(2) Georgi Lie algebras in particle physics (2nd edition) chapter 10

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