Why is there an additional $NI$ term in this ${rm SU}(N)$ generator, from Matrix Quantum Mechanics?

Question

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 $langleoperatorname{tr} Grangle=0$,  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,
$$langleoperatorname{tr}Grangle = Nlangleoperatorname{tr}Irangle = Nlangle Nrangle$$
what have I failed to understand here?

Prahar Mitra · Answer

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 $text{tr} G = 0$.

Why is there an additional $NI$ term in this ${rm SU}(N)$ generator, from Matrix Quantum Mechanics?

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