LS-couplings

LS-couplings#

The spin_conservation() rule is one of the more complicated checks in the conservation_rules module. It provides an implementation of \(LS\)-couplings, which is a procedure to determine which values for total angular momentum \(L\) and coupled spin \(S\) are allowed in an interaction node. In this notebook, we illustrate this procedure with the following decay chain as an example:

\[ J/\psi \to \Sigma^+ \bar\Sigma(1670)^-, \quad \bar\Sigma(1670)^- \rightarrow \bar p K^0. \]

In this decay chain, there are two decay nodes that we investigate separately. In addition, both decays are mediated interactions by the strong force, which means there is also parity conservation.

In the following derivations, the Particle.spin and Particle.parity values are of importance:

\[\displaystyle J/\psi(1S)=1^-\qquad \Sigma^{+}=\tfrac{1}2^+\qquad \overline{\Sigma}(1670)^{-}=\tfrac{3}2^+\qquad \overline{p}=\tfrac{1}2^-\qquad K^{0}=0^-\]

Procedure#

Imagine we have a two-body decay of \(p_0\rightarrow p_1p_2\). We denote the Spin.magnitude of each particle \(p_i\) as \(s_i\) and their parity as \(\eta_i\). The values for \(L\) and \(S\) can now be determined as follows:

Determine all values for \(S\) that satisfy \(\left| s_1-s_2 \right| \le S \le s_1+s_2\). The difference between each value for \(S\) has to integer, so \(S = \left| s_1-s_2 \right|, \left| s_1-s_2 \right|+1, \dots, s_1+s_2\).
Determine all values for \(L\) that satisfy \(\left| L-S \right| \le s_0 \le L+S\), with \(L\) being a non-negative integer.
If there is parity conservation, \(L\) has to satisfy an additional constraint: \(\eta_0 = \eta_1\cdot\eta_2\cdot(-1)^L\).

\(J/\psi \to \Sigma^+\bar\Sigma(1670)^-\)#

The spin and parity of each particle in the first transition can be summarized as \(1^-\to\frac{1}{2}^+\frac{3}{2}^+\). Following step 1 in the procedure, we get:

\[\begin{split} \begin{aligned} \left|s_{\Sigma^+} - s_{\bar\Sigma(1670)^-}\right| & \le S & \le s_{\Sigma^+} + s_{\bar\Sigma(1670)^-} \\ \left|\tfrac{1}{2}-\tfrac{3}{2}\right| & \le S & \le \tfrac{1}{2} + \tfrac{3}{2} \\ 1 & \le S & \le 2 \end{aligned} \end{split}\]

\[ \Rightarrow S=1 \quad \text{or} \quad S=2 \]

Next, we determine the allowed total angular momentum values \(L\) with step 2:

\[\begin{split} \begin{aligned} |L-S| & \le s_{J/\psi} & \le L+S \\ |L-S| & \le 1 & \le L+S \end{aligned} \end{split}\]

\[\begin{split} \Rightarrow \begin{cases} L=0,1,2 & \text{if} & S=1\\ L=1,2,3 & \text{if} & S=2. \end{cases} \end{split}\]

So in total, we have 6 \(LS\)-combinations:

\[ (L,S) = (0,1), (1,1), (2,1), (1,2), (2,2), (3,2). \]

This decay however goes via the strong force. This means that parity has to be conserved and we have to follow step 3:

\[\begin{split} \begin{aligned} \eta_{J/\psi} & = & \eta_{\Sigma^+} \cdot \eta_{\bar\Sigma(1670)^-} \cdot(-1)^L \\ (-1) & = & (+1)\cdot(+1)\cdot(-1)^L \\ (-1) & = & (-1)^{L}. \end{aligned} \end{split}\]

From this, we can easily see that only odd \(L\) values are possible, which leaves us with 3 \(LS\)-combinations:

\[ (L,S) = (1,1), (1,2), (3,2). \]

\(\bar \Sigma(1670)^-\to \bar pK^0\)#

The second part of the decay chain can be expressed as \(\frac{3}{2}^+ \to \frac{1}{2}^- 0^-\). Following step 1, we see:

\[\begin{split} \begin{aligned} |s_{\bar p} - s_{K^0}| & \le S & \le s_{\bar p} + s_{K^0} \\ \left|\tfrac{1}{2}-0 \right| & \le S & \le \tfrac{1}{2} + 0 \end{aligned} \end{split}\]

\[ \Rightarrow S = \tfrac{1}{2}. \]

This time, only one coupled spin value is allowed. That allows for the following values of \(L\):

\[\begin{split} \begin{aligned} |L-S| & \le s_0 & \le L+S \\ \left|L-\tfrac{1}{2}\right| & \le \tfrac{3}{2} & \le L+\tfrac{1}{2}. \end{aligned} \end{split}\]

\[ \Rightarrow L = 1,2 \]

By now, only two \(LS\)-combinations are possible:

\[ (L,S)=\left(1,\tfrac{1}{2}\right), \left(2,\tfrac{1}{2}\right). \]

This again is a strong interaction, which means we have to check for parity conservation.

\[\begin{split} \begin{aligned} \eta_{\bar \Sigma(1670)^-} & = & \eta_{\bar p}\cdot\eta_{K^0}\cdot(-1)^L\\ (+1) & = & (-1)\cdot(-1)\cdot(-1)^L\\ (+1) & = & (-1)^L. \end{aligned} \end{split}\]

Again, it is clear that only even \(L\)’s are allowed. This means that only one \(LS\)-combination is possible:

\[ (L,S)=\left(2,\tfrac{1}{2}\right) \]

Check with QRules#

Finally, let’s use generate_transitions() to check whether the allowed \(LS\)-couplings are found by qrules as well. Note that we have to increase the maximum angular momentum to find the \((L,S)=(3,2)\) combination as well.

        flowchart LR
    T0_0["$$0: K^{0}$$"]
    T0_1["$$1: \Sigma^{+}$$"]
    T0_2["$$2: \overline{p}$$"]
    T0_A["$$J/\psi(1S)$$"]
    T0_N0(("$$\begin{gathered} L = 3 \\\ S = 2 \\\ P = \text{+}1 \end{gathered}$$"))
    T0_N1(("$$\begin{gathered} L = 2 \\\ S = \frac{1}{2} \\\ P = \text{-}1 \end{gathered}$$"))
    T0_3("$$\overline{\Sigma}(1670)^{-}$$")
    T0_A --- T0_N0
    T0_N0 --- T0_3
    T0_3 --- T0_N1
    T0_N0 --- T0_1
    T0_N1 --- T0_0
    T0_N1 --- T0_2
    T1_0["$$0: K^{0}$$"]
    T1_1["$$1: \Sigma^{+}$$"]
    T1_2["$$2: \overline{p}$$"]
    T1_A["$$J/\psi(1S)$$"]
    T1_N0(("$$\begin{gathered} L = 1 \\\ S = 2 \\\ P = \text{+}1 \end{gathered}$$"))
    T1_N1(("$$\begin{gathered} L = 2 \\\ S = \frac{1}{2} \\\ P = \text{-}1 \end{gathered}$$"))
    T1_3("$$\overline{\Sigma}(1670)^{-}$$")
    T1_A --- T1_N0
    T1_N0 --- T1_3
    T1_3 --- T1_N1
    T1_N0 --- T1_1
    T1_N1 --- T1_0
    T1_N1 --- T1_2
    T2_0["$$0: K^{0}$$"]
    T2_1["$$1: \Sigma^{+}$$"]
    T2_2["$$2: \overline{p}$$"]
    T2_A["$$J/\psi(1S)$$"]
    T2_N0(("$$\begin{gathered} L = 1 \\\ S = 1 \\\ P = \text{+}1 \end{gathered}$$"))
    T2_N1(("$$\begin{gathered} L = 2 \\\ S = \frac{1}{2} \\\ P = \text{-}1 \end{gathered}$$"))
    T2_3("$$\overline{\Sigma}(1670)^{-}$$")
    T2_A --- T2_N0
    T2_N0 --- T2_3
    T2_3 --- T2_N1
    T2_N0 --- T2_1
    T2_N1 --- T2_0
    T2_N1 --- T2_2