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This question refers to equation (11) in the latest preprint of the following paper:

  • X. Han, S. A. Hartnoll and J. Kruthoff, "Bootstrapping Matrix Quantum Mechanics", Phys. Rev. Lett. 125, 041601 (2020), arXiv:2004.10212.

The authors give an expression for the generator of the ${\rm SU}(N)$ symmetry:

$$G=i[X, P]+N I$$

this form is explained by the sentence,

The final identity piece ensures that $\langle\operatorname{tr} G\rangle=0$, with the operator ordering $[X,P] =XP−PX$ in (11)

When I look at $G$, I see that the trace of the commutator $[X,P]$ will obviously be 0, as this is true for all commutators. Hence, I don't see why this additional $NI$ term is necessary, since it looks as if,

$$\langle\operatorname{tr}G\rangle = N\langle\operatorname{tr}I\rangle = N\langle N\rangle$$

what have I failed to understand here?

Urb
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1 Answers1

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You are confusing a matrix commutator with operator commutator.

$X$ and $P$ are matrix operators so $\text{tr}[X P] \neq \text{tr} [ P X ]$. More precisely, we note \begin{align} \text{tr} G &= i \text{tr} [ X , P ]+ N^2 \\ &= i \text{tr}[ X P ] - i \text{tr}[ P X ] + N^2\\ &= i X_{ij} P_{ji} - i P_{ji} X_{ij} + N^2 \\ &= - i [ P_{ij} , X_{ji} ] + N^2 \end{align} In the paragraph below equation (8) in your paper, we have $$ [ P_{ij} , X_{kl} ] = - i \delta_{il} \delta_{jk} $$ Thus, $$ [ P_{ij} , X_{ji} ] = - i \delta_{ii} \delta_{jj} = - i N^2 $$ which implies that $\text{tr} G = 0$.

Prahar
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