Problem evaluating an anticommutator in supersymmetric quantum mechanics

In summary, the evaluation of the anticommutator in supersymmetric quantum mechanics presents challenges due to the intricate interplay between bosonic and fermionic degrees of freedom. These difficulties arise from the need to maintain supersymmetry while performing calculations, as the anticommutator involves operators that do not commute in the traditional sense. Proper handling of these operators is essential for deriving physical quantities and ensuring the consistency of the supersymmetric framework. Techniques such as regularization and renormalization may be employed to address these issues and facilitate accurate computations.
  • #1
Gleeson
30
4
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
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