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:

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from IPython.display import Math

import qrules

PDG = qrules.load_pdg()
particle_names = [
    "J/psi(1S)",
    "Sigma+",
    "Sigma(1670)~-",
    "p~",
    "K0",
]
latex_expressions = []
for name in particle_names:
    particle = PDG[name]
    parity = "+" if particle.parity > 0 else "-"
    if particle.spin.is_integer():
        spin = int(particle.spin)
    else:
        nominator = int(particle.spin * 2)
        spin = Rf"\tfrac{{{nominator}}}{2}"
    latex_expressions.append(f"{particle.latex}={spin}^{parity}")
Math(R"\qquad ".join(latex_expressions))

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

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:

1. 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 \(S = \left| s_1-s_2 \right|, \left| s_1-s_2 \right|+1, \dots, s_1+s_2\).

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.

3. 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:

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

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

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

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

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

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:

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

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:

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

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

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

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

\[ \Rightarrow L = 1,2 \]

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.

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

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.

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import logging

import graphviz

LOGGER = logging.getLogger()
LOGGER.setLevel(logging.ERROR)

reaction = qrules.generate_transitions(
    initial_state="J/psi(1S)",
    final_state=["K0", "Sigma+", "p~"],
    allowed_intermediate_particles=["Sigma(1670)"],
    allowed_interaction_types="strong",
    max_angular_momentum=3,
)
dot = qrules.io.asdot(reaction, render_node=True, strip_spin=True)
graphviz.Source(dot)