Problem evaluating an anticommutator in supersymmetric quantum mechanics

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SUMMARY

This discussion centers on the evaluation of the anticommutator in N = 4 gauge quantum mechanics, specifically verifying equation 5.31 from a referenced paper. The setup involves complex fermions represented by ##\lambda_{A \alpha}## and the use of Pauli matrices ##\sigma^i##. The user has struggled to confirm the correctness of the derived expressions for the anticommutator of the supercharges ##Q_{\alpha}## and ##\bar{Q}_{\beta}##, leading to a request for clarification on potential conceptual misunderstandings in the calculations.

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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.
 

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