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.
 

FAQ: Problem evaluating an anticommutator in supersymmetric quantum mechanics

What is an anticommutator in the context of supersymmetric quantum mechanics?

An anticommutator is a mathematical operation used in quantum mechanics, defined as the sum of the product of two operators in both possible orders. For operators A and B, the anticommutator is given by {A, B} = AB + BA. In supersymmetric quantum mechanics, anticommutators often involve fermionic operators and play a crucial role in defining supersymmetry algebra.

Why is evaluating an anticommutator important in supersymmetric quantum mechanics?

Evaluating an anticommutator is important because it helps in understanding the structure and properties of supersymmetric systems. Anticommutators are used to define supersymmetry transformations and to ensure that the algebra closes properly. They are also essential in deriving equations of motion and in analyzing the behavior of fermionic fields.

What are the common challenges encountered when evaluating an anticommutator?

Common challenges include handling the non-commutative nature of operators, dealing with the complexities of fermionic fields, and ensuring that the algebraic manipulations are consistent with the principles of supersymmetry. Additionally, ensuring that the resulting expressions respect the physical constraints and symmetries of the system can be mathematically demanding.

How do you approach solving an anticommutator problem in supersymmetric quantum mechanics?

To solve an anticommutator problem, one typically starts by identifying the relevant operators and their commutation or anticommutation relations. Next, the operators are expressed in a convenient basis, and the anticommutator is computed by carefully applying the definitions and properties of the operators. Symmetry considerations and consistency checks are also crucial to ensure the correctness of the solution.

Can you provide an example of evaluating an anticommutator in supersymmetric quantum mechanics?

Consider two fermionic operators, Q and Q†, which represent the supersymmetry generators. The anticommutator {Q, Q†} is often evaluated to find the Hamiltonian of the system. If {Q, Q†} = H, where H is the Hamiltonian, then the evaluation involves computing the product Q Q† + Q† Q and simplifying the expression using the known properties of Q and Q†. This result shows how the supersymmetry generators relate to the energy of the system.

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