SU(2) representation

This article is one of Lie Group & Representation contents.

The $SU(2)$ algebra is $$[J_i,J_k]=i\epsilon_{ikl} J_l$$

$J_3$ eigenstates

goal: reduce the Hilbert space to block diagonal form.

We can diagonalize only one element, choose $J_3$.

Then pick the states with the highest value of $J_3$, and call that value as $j$.

$$ J_3 \left| i, \alpha\right\rangle = j\left| j,\alpha \right\rangle$$ where $\alpha$ is another label.

Also we can choose the states so that $$\left\langle j,\alpha | j,\beta \right\rangle = \delta_{\alpha \beta}$$

Raising and Lowering Operators

Define raising and lowering operators, $$J^\pm = (J_1 \pm J_2) / \sqrt{2}$$ satisfying $$[J_3, J^\pm]=\pm J^\pm,\ [J^+, J^-]=J_3$$ so they raise and lower the value of $J_3$ on the states. If $$J_3 \left| m \right\rangle = m \left| m \right\rangle$$ then $$J_3 J^\pm \left| m \right\rangle = J^\pm J_3 \left| m \right\rangle \pm J^\pm \left| m \right\rangle = (m \pm 1) J^\pm \left| m \right\rangle$$

The key idea is that we can use the raising and lowering operators to construct the irreducible representations.

There is no state with $J_3 = j+1$ because we have assumed that $j$ is the highest value of $J_3$. Thus is must be that $$J^+ \left| j, \alpha \right\rangle = 0\ \forall \alpha$$ because any non-zero states would have $J_3 = j+1$. THe states obtained by acting with the lowering operator have $J_3=j-1$, so it makes sense to define $$J^- \left| j,\alpha \right\rangle \equiv N_j (\alpha) \left| j-1, \alpha \right\rangle$$ where $N_j(\alpha)$ is a normalization factor. But we easily see that states with different $\alpha$ are orthogonal, because $$N_j(\beta)^*N_j(\alpha)\left\langle j-1, \beta | j-1, \alpha \right\rangle = \left\langle j, \beta \right| J^+ J^- \left| j,\alpha \right\rangle = \left\langle j, \beta \right| [J^+ J^-] \left| j, \alpha \right\rangle = \left\langle j, \beta \right| J_3 \left| j,\alpha \right\rangle = j\left\langle j, \beta | j, \alpha \right\rangle = j \delta_{\alpha \beta}$$

Thus we can choos the states $\left| j-1, \alpha \right\rangle$ to be the orthonormal by choosing $$N_j(\alpha) = \sqrt{j} \equiv N_j$$

Then we have $$J^+ \left| j-1, \alpha \right\rangle = \frac{1}{N_j} J^+ J^- \left| j,\alpha \right\rangle = \frac{1}{N_j} [J^+, J^-] \left| j, \alpha \right\rangle = \frac{j}{N_j} \left| j, \alpha \right\rangle = N_j \left| j, \alpha \right\rangle$$

Now there are orthonormal states $\left| j-2, \alpha \right\rangle$ satisfying $$J^- \left| j-1, \alpha \right\rangle = N_{j-1} \left| j-2, \alpha \right\rangle \\ J^- \left| j-2, \alpha \right\rangle = N_{j-1} \left| j-1, \alpha \right\rangle$$

Continuing the process, we find a whole tower of orthonormal states, $\left| j - k, \alpha \right\rangle$ satisfying $$ J^- \left| j-k, \alpha \right\rangle = N_{j-K} \left| j-k-1, \alpha \right\rangle \\ J^+ \left| j-k-1, \alpha \right\rangle = N_{j-k} \left| j-k, \alpha \right\rangle$$

The $N$s can be chosen to be real, and because of the algebra, they satisfy $$N_{j-k}^2 = \left\langle j-k,\alpha \right| J^+ J^- \left| j - k,\alpha \right\rangle = \left\langle j-k, \alpha \right| [J^+, J^-] \left| j-k, \alpha \right\rangle + \left\langle j - k, \alpha \right| J^-J^+ \left| j-k, \alpha \right\rangle = N_{j-k+1}^2 + j-k$$

This is a recursion relation for the $N$s which is easy to solve by starting with $N_j$: $$\begin{align*} N_j^2 & &=j \\ N_{j-1}^2 &- N_j^2 &= j-1 \\  \vdots & & \vdots \\ N_{j-k}^2 &- N_{j-k+1}^2 &= j-k \\ \hline  N_{j-k}^2 & = (k+1)j & -  k(k+1)/2 \\ & = \frac{1}{2}(k+1)(2j-k) & \end{align*}$$ or setting $k=j-m$ $$ N_m = \frac{1}{\sqrt{2}} \sqrt{(j+m)(j-m+1)}$$

Because the representation is finite dimensional, there must be some maximum number of lowering operators, $l$, that we can apply to $\left| j,\alpha \right\rangle$. We must eventually come to some $m=j-l$ such that applying any more lowering operators gives $0$. Then $l$ is a non-negative integer specifying the number of times we can lower the states with highest $J_3$. Another lowering operator annihilates the state $$J^- \left| j-l, \alpha \right\rangle = 0$$ But then the norm of $J^- \left| j-l, \alpha \right\rangle$ must vanish, which means that $$N_{j-l} = \frac{1}{\sqrt{2}} \sqrt{(2j-l)(l+1)} = 0$$ the factor $l+1$ cannot vanish, thus we must have $$l=2j$$

Finally, we don't need $\alpha$, because generator don't touch $\alpha$. $\alpha$ become label of each irreducible representation.

The Standard Notation

Standard notation of irreducible representations label highest $J_3$ value in the representation and the $J_3$ value: $$\left| j,m \right\rangle$$ and the matrix elements of the generators are determined by the matrix elements of $J_3$ and the raising and lowering operators, $J^\pm$: $$\left\langle j,m' \right| J_3 \left| j,m \right\rangle = m\delta_{m'm}\\ \left\langle j,m' \right\rangle J^+ \left| j,m \right\rangle = \sqrt{(j+m+1)(j-m)/2} \delta_{m',m+1}\\ \left\langle j,m' \right| J^- \left| j,m \right\rangle = \sqrt{(j+m)(j-m+1)/2} \delta_{m',m-1}$$

These matrix elements define the spin $j$ represenatation of the $SU(2)$ algebra: $$[J^j_a]_{kl} = \left\langle j,j+1-k \right| J_a \left| j,j+1-l\right\rangle$$

If we use different convention with $m$ and $m'$, $$[J_a^j]_{m'm} = \left\langle j,m' \right| J_a \left| j,m\right\rangle$$ where $m$ and $m'$ run from $j$ to $-j$ in steps of $-1$. 

For example, for $1/2$, this gives the spin 1/2 representation $$J_1^{1/2} = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} = \frac{1}{2} \sigma_1 \\ J_2^{1/2} = \frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix} = \frac{1}{2}\sigma_2 \\ J_3^{1/2} = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \frac{1}{2} \sigma_3$$ where the $\sigma$s are the Pauli matrices.

$$\sigma_1 = \begin{pmatrix}0&1\\1&0\end{pmatrix}, \sigma_2 = \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ satrisfying $$ \sigma_a \sigma_b = \delta_{ab} + i\epsilon_{abc} \sigma_c$$

The spin $1/2$ representation is the defining(or fundamental) represeantation of $SU(2)$

Then group element can represented by $$e^{i\vec{\alpha} \cdot \vec{\sigma}/2}$$ which are the most general $2\times 2$ unitary matrices with determinant 1.

For another example, spin 1 representation looks like $$J_1^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&1\\1&0&1\\0&1&0\end{pmatrix},\ J_2^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\  J_3^1=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$.

This is adjoint representation with $f_{abc} = \epsilon_{abc}$.

In general, the $J_3$ values are called weights, and this process is highest weight construction.

1. Diagonlaize $J_3$.

2. Find the states with the highest $J_3$ value $j$.

3. For each such state, explicitly construct the states of the irreducible spin $j$ representation by applying the lowering operator to the states with highest $J_3$.

4. Now set aside the subspace spanned by these representations, which is now in canonical form, and concentrate on the subspace orthogonal to it.

5. THake these remaining states, go to step 2 and start again with the states with next highest $J_3$ value.

The end result will be the construction of a basis for the Hilbert space of the form $$\left| j,m,\alpha \right\rangle $$ where $m$ and $j$ refer to the $J_3$ value and the representation as usual and $\alpha$ refers to all the other observables that can be diagonalized to characterize the state. These satify $$\left\langle j', m', \alpha' | j,m,\alpha \right\rangle = \delta_{m'm} \delta_{j'j} \delta_{\alpha' \alpha}$$

They are also required by Schur's lemma, becauses the matrix elements satisfy $$\left\langle j',m',\alpha' \right| J_a \left| j,m,\alpha \right\rangle = [J_a^{j'}]_{m'm''} \left\langle j',m'',\alpha' | j,m,\alpha\right\rangle = \left\langle j',m',\alpha' | j,m'',\alpha\right\rangle [J_a^j]_{m''m}$$ because we can insert a complete set of intermediate states on either side of $J_a$. Thus $\left\langle j',m',\alpha'|j,m,\alpha\right\rangle$ commutes with all the elements of an irreducible representation, and is either $0$ if $j\ne j'$ or proportional to the identity, $\delta_{m'm}$ if $j=j'$.

Tensor Products

This is just addition of angular momentum.

Product of two different kinds of kets is $$\left| i, x \right\rangle \equiv \left| i \right\rangle \left| x \right\rangle$$ where the first states, $\left| i \right\rangle$ transforms under representation $D_1$ of the group and the second, $\left| x \right\rangle$, under $D_2$. Then the product, called the tensor product, transforms as follows: $$D(g)\left| i,x\right\rangle = \left| j,y \right\rangle [D_{1\otimes 2} (g)]_{iyix}=\left| j \right\rangle \left| y \right\rangle [D_1(g)]_{ji}[D_2(g)]_{yx}=(\left| j \right\rangle [D_1(g)]_{ji}) (\left| y \right\rangle [D_2(g)]_{yx})$$

Let's look at this near the identity, for infinitesimal $\alpha_a$, $$(1+i\alpha_a J_a)\left| i,x\right\rangle = \left| j,y\right\rangle \left\langle j,y \right| (1+i\alpha_a J_a)\left| i,x \right\rangle = \left| j,y \right\rangle ( \delta_{ji} \delta_{yx} + i\alpha_a [J_a^{1\otimes 2} (g)]_{iyix})=\left| j,y \right\rangle (\delta_{ji} + i\alpha_a [J_a^1]+{ji})(\delta_{yx} + i\alpha_a [J_a^2]_{yx})$$ Thus identifying first powers of $\alpha_a$: $$[J_a^{1\otimes 2}(g)]_{jyix} = [J^1_a]_{ji}\delta_{yx} + \delta_{ji}[J_a^2]_{yx}$$

We can understand $J_a$ operator operate each space, $$J_a(\left| j \right\rangle \left| x \right\rangle) = (J_a \left| j \right\rangle ) \left| x \right\rangle + \left| j \right\rangle (J_a \left| x \right\rangle)$$

$J_3$ value add

The $J_3$ values of tensor product states are just the sums of the $J_3$ values of the factors: $$J_3(\left| j_1, m_1\right\rangle \left| j_2, m_2 \right\rangle ) = (m_1+m_2)(\left| j_1, m_1\right\rangle \left| j_2, m_2\right\rangle)$$ 

For example, the tensor product of a spin 1/2 and spin 1 represenation.

There is a unique highest weight state, $$\left| 3/2, 3/2 \right\rangle = \left| 1/2, 1/2\right\rangle \left| 1,1\right\rangle$$

We can now construct the rest of the spin 3/2 states by applying lowering operators to both sides. For example, $$J^-\left| 3/2, 3/2 \right\rangle = J^- (\left| 1/2, 1/2\right\rangle \left| 1,1\right\rangle )=\sqrt{\frac{3}{2}}\left| 3/2,1/2\right\rangle = \sqrt{\frac{1}{2}} \left| 1/2,-1/2\right\rangle \left| 1,1\right\rangle + \left| 1/2,1/2\right\rangle \left| 1,0\right\rangle$$ or $$\left| 3/2,1/2 \right\rangle = \sqrt{\frac{1}{3}} \left| 1/2,-1/2\right\rangle \left| 1,1 \right\rangle + \sqrt{\frac{2}{3}} \left\| 1/2, 1/2\right\rangle \left| 1,0\right\rangle$$

Continuing the process gives $$\left| 3/2, -1/2 \right\rangle = \sqrt{\frac{2}{3}}\left| 1/2, -1/2\right\rangle \left| 1,0\right\rangle + \sqrt{\frac{1}{3}}\left| 1/2,1/2\right\rangle \left| 1,-1\right\rangle \\ \left| 3/2,-3/2\right\rangle = \left| 1/2, -1/2\right\rangle \left| 1,-1\right\rangle$$

Then the remaining states are orthogonal to these $$ \sqrt{\frac{2}{3}}\left| 1/2, -1/2\right\rangle \left| 1,1\right\rangle - \sqrt{\frac{1}{3}}\left| 1/2,1/2\right\rangle \left| 1,0\right\rangle \\ \sqrt{\frac{1}{3}}\left| 1/2, -1/2\right\rangle \left| 1,0\right\rangle - \sqrt{\frac{2}{3}}\left| 1/2,1/2\right\rangle \left| 1,-1\right\rangle$$

applying the highest weight scheme to this reduces space gives $$\left| 1/2, 1/2 \right\rangle = \sqrt{\frac{2}{3}}\left| 1/2, -1/2\right\rangle \left| 1,1\right\rangle - \sqrt{\frac{1}{3}}\left| 1/2,1/2\right\rangle \left| 1,0\right\rangle \\ \left| 1/2, -1/2 \right\rangle = \sqrt{\frac{1}{3}}\left| 1/2, -1/2\right\rangle \left| 1,0\right\rangle - \sqrt{\frac{2}{3}}\left| 1/2,1/2\right\rangle \left| 1,-1\right\rangle$$

Reference

Howard Georgi - Lie Algebras in Particle Physics