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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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    What space are $X$ and $P$ defined on? Please include all relevant information into your question so that answerers don't have to search the paper for it. If these are position and momentum operators, it is not at all obvious that the trace will be 0, let alone that it is well-defined to begin with, cf. e.g https://physics.stackexchange.com/a/149792/50583 – ACuriousMind Aug 11 '20 at 00:48

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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