A Problem evaluating an anticommutator in supersymmetric quantum mechanics

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The discussion revolves around verifying equation 5.31 from a paper on N = 4 gauge quantum mechanics derived from N = 1 gauge theory. The author is struggling to confirm the anticommutator relation involving the supercharges Q and Q-bar, specifically the terms related to the Pauli matrices and gauge fields. Despite extensive efforts, the author suspects a conceptual misunderstanding in their calculations, particularly in manipulating the Pauli matrices and their commutation relations. They seek assistance in resolving discrepancies in the derived terms, especially concerning the correct formulation of the relation involving the Pauli matrices. The conversation highlights the complexity of supersymmetric quantum mechanics and the challenges in reproducing established results.
Gleeson
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I am trying to reproduce the results of a certain paper here. In particular, I'm trying to verify their eqn 5.31.

The setup is N = 4 gauge quantum mechanics, obtained by the dimensional reduction of N = 1 gauge theory in 4 dimensions. ##\sigma^i## denotes the ith pauli matrix. ##\lambda_{A \alpha}## is a two component complex fermion (or rather its ##\alpha##th component). ##A## labels the generators of the gauge group.

\begin{align*}
H &= \frac{1}{2}\pi^m_A \pi^m_A + \frac{1}{4} g^2 (f_{ABC}\phi^m_B \phi^n_C)^2 + igf_{ABC}\bar{\lambda}_A \sigma^m\phi^m_B \lambda_C \\
Q_{\alpha} &= (\sigma^m \lambda_A)_{\alpha}(\pi^m_A - iW^m_A)\\
\bar{Q}_{\beta} &= (\bar{\lambda}_B\sigma^n)_{\beta}(\pi^n_B + iW^n_B)\\
W&= \frac{1}{6}g f_{ABC} \epsilon_{mnp}\phi^m_A \phi^n_B \phi^p_C \\
W^m_A &= \frac{\partial W}{\partial \phi^m_A} \\
[\phi^m_A, \pi^n_B] &= i \delta_{AB}\delta^{mn} \\
\{\lambda_{A \alpha}, \bar{\lambda}_{B \beta} \} &= \delta_{AB} \delta_{\alpha \beta}\\
G_A &= f_{ABC}(\phi^m_B\pi^m_C - i\bar{\lambda}_B \lambda_C).
\end{align*}

It is claimed that

\begin{align*}
\{Q_{\alpha}, \bar{Q}_{\beta}\} &= 2 \delta_{\alpha \beta}H - 2g(\sigma^m)_{\alpha \beta} \phi^m_A G_A
\end{align*}

This can also be written as

\begin{align}
\{Q_{\alpha}, \bar{Q}_{\beta}\} &= \delta_{\alpha \beta}(\pi^m_A \pi^m_A + \frac{1}{2} g^2 (f_{ABC}\phi^m_B \phi^n_C)^2 + i2gf_{ABC}\bar{\lambda}_A \sigma^m\phi^m_B \lambda_C) - 2g(\sigma^m)_{\alpha \beta} \phi^m_A f_{ABC}(\phi^m_B\pi^m_C - i\bar{\lambda}_B \lambda_C).
\end{align}

I have spent many hours trying to confirm this, but unable so far to do so.

\begin{align*}
\{Q_{\alpha}, \bar{Q}_{\beta}\} &= \{(\sigma^m \lambda_A)_{\alpha}(\pi^m_A - iW^m_A) (\bar{\lambda}_B\sigma^n )_{\beta}(\pi^n_B + iW^n_B)\}\\
&=(\sigma^m_{\alpha \theta}\lambda_{A \theta})( \bar{\lambda}_{B \gamma}\sigma^n_{\gamma \beta})[\pi^m_A \pi^n_B + W^{n m}_{BA} + iW^n_B\pi^m_A - i W^m_A \pi^n_B + W^m_A W^n_B] \\
&+( \bar{\lambda}_{B \gamma}\sigma^n_{\gamma \beta})(\sigma^m_{\alpha \theta}\lambda_{A \theta})[\pi^n_B \pi^m_A - W^{m n}_{AB} - iW^m_A \pi^n_B + iW^n_B \pi^m_A + W^n_B W^m_A] \\
&= (\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})(\lambda_{A \theta} \bar{\lambda}_{B \gamma}) [\pi^m_A \pi^n_B + W^{n m}_{BA} + iW^n_B\pi^m_A - i W^m_A \pi^n_B + W^m_A W^n_B] \\
&+ (\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})( \bar{\lambda}_{B \gamma}\lambda_{A \theta})[\pi^n_B \pi^m_A - W^{m n}_{AB} - iW^m_A \pi^n_B + iW^n_B \pi^m_A + W^n_B W^m_A] \\
&= (\sigma^m_{\alpha \theta}\sigma^n_{\theta \beta})[\pi^m_A \pi^n_A + iW^n_A\pi^m_A - i W^m_A \pi^n_A + W^m_A W^n_A] \\
&+ (\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})[( \lambda_{A \theta}\bar{\lambda}_{B \gamma})-( \bar{\lambda}_{B \gamma}\lambda_{A \theta})]W^{m n}_{AB}\\
&= (\delta_{mn} \delta_{\alpha \beta} + i \epsilon_{mnp}\sigma^p_{\alpha \beta})[\pi^m_A \pi^n_A + iW^n_A\pi^m_A - i W^m_A \pi^n_A + W^m_A W^n_A] \\
&- 2(\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})( \bar{\lambda}_{B \gamma}\lambda_{A \theta})W^{m n}_{AB}\\
&= \delta_{\alpha \beta}(\pi^m_A \pi^m_A + W^m_AW^m_A) + 2\epsilon_{mnp}\sigma^p_{\alpha \beta} W^m_A \pi^n_A - 2(\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})( \bar{\lambda}_{B \gamma}\lambda_{A \theta})W^{m n}_{AB}\\
&=\delta_{\alpha \beta}(\pi^m_A \pi^m_A + \frac{1}{2} g^2 (f_{ABC}\phi^m_B \phi^n_C)^2) +g \epsilon _{mrs}f_{ABC}\phi^r_B \phi^s_C\epsilon_{mnp}\sigma^p_{\alpha \beta} W^m_A \pi^n_A - 2(\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})( \bar{\lambda}_{B \gamma}\lambda_{A \theta})W^{m n}_{AB}\\
&= \delta_{\alpha \beta}(\pi^m_A \pi^m_A + \frac{1}{2} g^2 (f_{ABC}\phi^m_B \phi^n_C)^2) - 2g(\sigma^m)_{\alpha \beta} \phi^m_A f_{ABC}\phi^m_B\pi^m_C - 2(\sigma^m_{\alpha \theta}\sigma^n_{\gamma \beta})( \bar{\lambda}_{B \gamma}\lambda_{A \theta})g \epsilon_{mnp}f_{ABC}\phi^p_C.
\end{align*}

The first three terms are correct. But the fourth term is wrong (it should instead be two different terms above). I have spent many hours on this. I think I must have some conceptual misunderstanding about these sorts of calculations, because I can't do it. I am hoping someone can help me out and clarify what I'm doing wrong please.
 
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So I think I have reduced the above to trying to conclude that:

$$
\sigma^{k \ \beta}_{\alpha} \delta_{\theta}^{\gamma} - \sigma^{k \ \gamma}_{\theta}\delta^{\beta}_{\alpha} = i \epsilon_{ijk}\sigma^{i \ \gamma}_{\alpha} \sigma^{j \ \beta}_{\theta}.
$$

If anyone has any suggestions, it would be appreciated.
 
I would try dropping in a commutator of Pauli matrices for ##\epsilon_{ijk}\sigma^{i \gamma}_{\alpha}## and see if you can get Kronecker's on the left hand side from products of the same Pauli matrices.
 
Gleeson said:
So I think I have reduced the above to trying to conclude that:

$$
\sigma^{k \ \beta}_{\alpha} \delta_{\theta}^{\gamma} - \sigma^{k \ \gamma}_{\theta}\delta^{\beta}_{\alpha} = i \epsilon_{ijk}\sigma^{i \ \gamma}_{\alpha} \sigma^{j \ \beta}_{\theta}.
$$

If anyone has any suggestions, it would be appreciated.
This should be
$$
\sigma^{k \ \gamma}_{\theta}\delta^{\beta}_{\alpha} - \sigma^{k \ \beta}_{\alpha} \delta_{\theta}^{\gamma} = i \epsilon_{ijk}\sigma^{i \ \gamma}_{\alpha} \sigma^{j \ \beta}_{\theta}.
$$.

Thanks for the suggestion, but I still couldn't show the two sides to be equal. I have checked various contractions, and they seem to be consistent at least.

If anyone else can see how to solve this, or to point out a mistake, please let me know.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

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