How Does Tensor Product Decomposition Work in SU(2) Representations?

[CAF123]

Homework Statement

Construct the decompositions $\mathbf 2 \otimes \mathbf 2 = \mathbf 3 \oplus \mathbf 1$, where $\mathbf N$ is the representation of su(2) with $\mathbf N$ states and thus spin j=1/2 (N-1).

Homework Equations

Substates within a state labelled by j can take on values -j to j in integer steps

3. The Attempt at a Solution
I think I get the idea but was hoping someone could just make sure I understand things correctly.
So we consider some states in the $\mathbf 2$ representation of SU(2), labelled as $|j_1, m_1 \rangle$ and take the tensor product of this with another state $|j_2, m_2 \rangle$. If N=2, then j=1/2. So states are |1/2, 1/2> and |1/2,-1/2>, So out of these two states can form four possible tensor products. Take for example, $$|1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle$$ Then by Clebsch Gordan, possible states are $|J.M\rangle$ where $|j_1 - j_2| < J < j_1 + j_2$ and $-J < M < J$? So the r.h.s is $|0,0\rangle + |1,0\rangle + |1,-1 \rangle + |1,1\rangle$ which is exactly those states in $\mathbf 3 \oplus \mathbf 1$?

I am just wondering how the |0,0> state is part of $\mathbf 3 \oplus \mathbf 1$ on the r.hs? $\mathbf 1$ contains |0,0> but $\mathbf 3$ is always of the form $|1,-1>, |1,0>$ or $|1,1>$.

Thanks!

 

[fzero]

You will need to use the angular momentum generators to solve this problem, specifically in the basis $J_z, J_\pm$. You will have to determine the eigenstates of the total angular momentum $J_i = J^{(1)}_i + J^{(2)}_i$ for the states of the form $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$. The properties under $J_\pm$ will tell you whether a state belongs to the singlet or triplet. For instance, you should find two eigenfunctions with $J_z|j,m\rangle =0$. One eigenstate will satisfy $J_\pm |j,m\rangle= 0$, but the other will give another eigenstate,

 

[CAF123]

Hi fzero,

You will need to use the angular momentum generators to solve this problem, specifically in the basis $J_z, J_\pm$. You will have to determine the eigenstates of the total angular momentum $J_i = J^{(1)}_i + J^{(2)}_i$ for the states of the form $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$. The properties under $J_\pm$ will tell you whether a state belongs to the singlet or triplet. For instance, you should find two eigenfunctions with $J_z|j,m\rangle =0$. One eigenstate will satisfy $J_\pm |j,m\rangle= 0$, but the other will give another eigenstate,

Ok, so $J_z |j,m\rangle = m |j, m \rangle$ and $J_{\pm} |j,m\rangle = \sqrt{(j \mp m)(j \pm m -1)} |j, m\pm 1 \rangle$. The two eigenfunctions with $J_z |j,m\rangle = 0$ are |0,0> and |1,0>. |0,0> satisfies $J_{\pm} |0,0> = 0$, while $J_{\pm} |1,0\rangle = \sqrt{2} |1, \pm 1 \rangle$. So acting with the $J_{\pm}$ operators on the |0,0> does not take you to another state in the representation labelled by j=0 (since there is no other state), so |0,0> is a singlet. Similarly, acting with the step operators on |1,0>, |1,-1> or |1,1> take us to another state within the representation labelled by j=1 with the property that $J_{-} |1,-1\rangle =0$ and $J_{+} |1,1\rangle = 0$.

So I conclude that $\mathbf 1 = \left\{|0,0\rangle \right\}$ and $\mathbf 3 = \left\{|1,-1 \rangle, |1,0 \rangle, |1,1\rangle \right\}$.

I am not seeing why this helps me to solve the decomposition problem $\mathbf 2 \otimes \mathbf 2 = \mathbf 3 \oplus \mathbf 1$?
Thanks!

 

[fzero]

Hi fzero,

Ok, so $J_z |j,m\rangle = m |j, m \rangle$ and $J_{\pm} |j,m\rangle = \sqrt{(j \mp m)(j \pm m -1)} |j, m\pm 1 \rangle$. The two eigenfunctions with $J_z |j,m\rangle = 0$ are |0,0> and |1,0>. |0,0> satisfies $J_{\pm} |0,0> = 0$, while $J_{\pm} |1,0\rangle = \sqrt{2} |1, \pm 1 \rangle$. So acting with the $J_{\pm}$ operators on the |0,0> does not take you to another state in the representation labelled by j=0 (since there is no other state), so |0,0> is a singlet. Similarly, acting with the step operators on |1,0>, |1,-1> or |1,1> take us to another state within the representation labelled by j=1 with the property that $J_{-} |1,-1\rangle =0$ and $J_{+} |1,1\rangle = 0$.

So I conclude that $\mathbf 1 = \left\{|0,0\rangle \right\}$ and $\mathbf 3 = \left\{|1,-1 \rangle, |1,0 \rangle, |1,1\rangle \right\}$.

I am not seeing why this helps me to solve the decomposition problem $\mathbf 2 \otimes \mathbf 2 = \mathbf 3 \oplus \mathbf 1$?
Thanks!

You want to find the linear combinations of the product states $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$ that are eigenstates of the total angular momentum. So you're going to solve
$$ | j,m\rangle = \sum_{j_1,m_1;j_2,m_2} C^{jm}_{j_1 m_1 j_2 m_2} |j_1,m_1\rangle\otimes|j_2,m_2\rangle,$$
where the coefficients are the Clebsch-Gordan coefficients. You already understand the algebra for the left-hand side of this, so now you want to work it out on the right-hand side.

 

[CAF123]

You want to find the linear combinations of the product states $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$ that are eigenstates of the total angular momentum. So you're going to solve
$$ | j,m\rangle = \sum_{j_1,m_1;j_2,m_2} C^{jm}_{j_1 m_1 j_2 m_2} |j_1,m_1\rangle\otimes|j_2,m_2\rangle,$$
where the coefficients are the Clebsch-Gordan coefficients. You already understand the algebra for the left-hand side of this, so now you want to work it out on the right-hand side.

I see, ok so I guess it would be best to work with the state $|1,1\rangle = C|1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle$ since then C=1 and subsequent states can be found by applying $J_-$ to this. But how does $J_{-} (|1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle)$ work? Do we apply the operator in turn to each of the states, like $J_{-} (A \otimes B) = (J_{-}A) \otimes B + A \otimes (J_{-} B)$?

Thanks!

 

[fzero]

I see, ok so I guess it would be best to work with the state $|1,1\rangle = C|1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle$ since then C=1 and subsequent states can be found by applying $J_-$ to this. But how does $J_{-} (|1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle)$ work? Do we apply the operator in turn to each of the states, like $J_{-} (A \otimes B) = (J_{-}A) \otimes B + A \otimes (J_{-} B)$?

Thanks!

Yes, and since the total angular momentum is $J_i=J^{(1)}_i+J^{(2)}_i$, we can write this as $J_{-} (A \otimes B) = (J_{-}^{(1)}A) \otimes B + A \otimes (J_{-}^{(2)} B)$.