Aug 7, 2013

Many-Body Method for Angular Momentum Addition

Let's start from a simple problem: the total spin of two spin-1/2 particles.

The two particles have total spin $j_1$ and $j_2$, as well as spin projection $m_1$ and $m_2$. Similarly, the composite system will also have a total spin $j$ and a spin projection $m$. The total spin is the sum of the two $\mathbf{J} = \mathbf{J}_1 + \mathbf{J}_2$, according to angular momentum conservation. The (scalar) total spin $J = \sqrt{\mathbf{J}^2}$. In some interactions, such as the $\mathbf{L}\cdot\mathbf{S}$ coupling, total $J$ is conserved, instead of $L$ or $S$. It can be seen from \[
\mathbf{L}\cdot\mathbf{S} = \frac{1}{2}\left[ \mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2 \right] \\
 = \frac{1}{2}\left[ j(j+1) - l(l+1) - s(s+1) \right]
 \] A notable example of the $\mathbf{L}\cdot\mathbf{S}$ coupling is the spin-orbit coupling in the hydrogen-like atom caused by the relativistic correction: \[
V(r) = \frac{Ze^2}{8\pi\varepsilon_0 m^2 r^3}\mathbf{L}\cdot\mathbf{S}.
\] However, a product state like $\left.|1/2,+1/2\right>\left.|1/2,-1/2\right>$ (often also denoted as $\left.|\uparrow\downarrow\right>$ for simplicity) is usually not a mutual eigenstate of the total spin operator $J$ and spin projection operator $J_z$ due to the non-commuting nature of the angular momentum operator. According to quantum mechanics, it is in fact the superposition of several eigenstates with various total spin $j$ ( and in principle spin projection $m$ ). The exact expression relating the product states and the eigenstates of the total spin operator is given by Clebsch-Gordan (CG) decomposition:\[ \left.|j_1,m_1\right>\left.|j_2,m_2\right> = \sum_{j,m} \left( j_1,m_1,j_2,m_2|j,m\right) \; \left.|j,m\right> \] or \[
\left.|j,m\right> = \sum_{j_1,m_1,j_2,m_2} \left(j,m|j_1,m_2,j_2,m_2\right) \;  \left.|j_1,m_1\right>\left.|j_2,m_2\right>
In the case of the spin-1/2 particles, there are four eigenstates categorized into a spin singlet: \[
\left.|0,0\right> = \frac{1}{\sqrt{2}}\left( \left.| \uparrow\downarrow\right> - \left.|\downarrow\uparrow\right>  \right)
\] and three spin triplets: \[
\left.|1,+1\right>      = \left.| \uparrow\uparrow\right>    \\
\left.|1,0\right> = \frac{1}{\sqrt{2}}\left( \left.| \uparrow\downarrow\right> + \left.|\downarrow\uparrow\right>  \right)    \\
\left.|1,-1\right> = \left.| \downarrow\downarrow  \right>
CG coefficients give the analytic solutions to the eigenvalue problem of the two-body total angular momentum operator $J$. In a more general many-body system, it may be more convenient to solve the eigenvalue problem directly (including using numeric methods).

The total spin squared operator
\mathbf{J}^2 = \left( \sum_a \mathbf{J}_a \right)^2
= \sum_i \mathbf{J}_a^2 + 2 \sum_{a<b} \mathbf{J}_a\cdot \mathbf{J}_b.
\] where $a, b$ enumerate the constituents. Noting that $2\mathbf{J}_a \cdot \mathbf{J}_b =2 J_a^x J_b^x + 2 J_a^y J_b^y +2 J_a^z J_b^z = J_a^+ J_b^- + J_a^- J_b^+ + 2 J_a^z J_b^z$, where $J^\pm = J_x \pm i J_y$ is the raising/lowering operator. Then the total $J$ operator becomes \[

\mathbf{J}^2 = \sum_a \mathbf{J}_a^2 + 2 \sum_{a<b} J^z_a \cdot J_b^z + J_a^+ J_b^- + J_a^- J_b^+.

\] The raising/lowering operator acts on spin states as \[
J^\pm \left.| j, m\right> = \sqrt{j(j+1) - m(m \pm 1)}\left.|j, m\pm 1\right>.
\] Suppose $\alpha=(\{j_1,m_1\}, \{j_2,m_2\}, \cdots  )$ and $\beta=(\{j'_1,m'_1\}, \{j'_2,m'_2\}, \cdots)$ are many-body basis states. Then the matrix element is \[
\left< \beta \left| \mathbf{J^2}\right| \alpha \right> = \delta_{\alpha, \beta} \sum_a  j_a (j_a+1) + \delta_{\alpha, \beta} 2  \sum_{a<b} m_a m_b  \\ + \sum_{a<b} \delta_{\hat{\alpha}_{ab}, \hat{\beta}_{ab}} \delta_{j_a,j'_a} \delta_{j_b,j'_b} \delta_{m_a, m'_a+1} \delta_{m_b, m'_b-1}\sqrt{\left[j_a(j_a+1)-m_a(m_a+1) \right] \left[j_b(j_b+1) - m_b(m_b-1) \right]}
 \] where $\hat{\alpha}_{a,b}$ is the list of quantum numbers except the $a$-th and $b$-th.

We have constructed the matrix elements of the total spin squared operator for the many-body basis. The total $j$ states $\left.|j, m\right>$ are merely the eigenstates of the total spin squared operator. The CG coefficients are an analytical solution of the eigenvalue problem. In general, we obtain the total spin $j$ along with the coefficients by diagonalize the total spin operator. With the many-body matrix in hand, the diagonalization can be done numerically.

This method is particularly suitable for many-body system (particle number larger than 2). It is also superior in system with identical particles. In that case, the basis space shrinks to symmetric/anti-symmetric basis.

The many-body method can also be generalized to $SU(N_c)$ problem, in which the corresponding quantity is often termed as "color". The total color squared operator is the Casimir operator \[
\lambda^2 = \sum_c^{N_c^2 - 1} \sum_i \lambda^c_i \sum_j \lambda^c_j = \sum_i \sum_c^{N_c^2-1} \lambda_i^c \lambda_i^c + 2 \sum_{i<j} \sum_c^{N^2_c-1} \lambda^c_i \lambda_j^c
where $\lambda^c, c=1,2,\cdots N_c^2-1$ is the generalized Gell-Mann matrix. The matrix elements can be constructed.

It should be mentioned that another systematic method is the Young tableau.

It is also not a surprise that we are looking for irreducible representations from fundamental and/or induced representations.


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