Deriving Burger's equation for energy eigenvalues in $T\bar T$-deformed theories

Question

When doing $T \bar T$-deformation to 2d CFTs, it is interesting to ask how the original energy spectrum is shifted throughout the procedure. This is done as follows.

As mentioned in several papers/reviews: [1,2,3], the factorisation property of $T \bar T$ for a deformed CFT on a cylinder with radius $R$ reads (equation (35) in [1]) $$ \langle n|T \bar T|n\rangle = -\pi^2 \left(\langle n|T_{yy}|n\rangle\langle n|T_{xx}|n\rangle - \langle n|T_{xy}|n\rangle \langle n|T_{xy}|n\rangle \right).\tag{35} $$ Upon identifying $$ \langle n|T_{yy}|n\rangle = - \frac{1}{R} E_n(R;\alpha), \quad \langle n|T_{xx}|n\rangle = - \frac{\mathrm d}{\mathrm d R}E_n(R;\alpha), \quad \langle n|T_{xy}|n\rangle = \frac{\mathrm i}{R} P_n(R), \tag{36-37} $$ where $E_n(R;\alpha), P_n(R)$ are the energy/momentum eigenvalues of state $|n\rangle$ with deformation parameter $\alpha$, respectively, we get $$ \langle n| T \bar T|n\rangle = - \frac{\pi^2}{R}\left(E_n(R;\alpha) \frac{\mathrm d}{\mathrm dR} E_n(R;\alpha) + \frac{1}{R} P_n^2(R)\right).\tag{38} $$

The next step, which I don't understand properly, is to interpret $$ \langle n| T \bar T|n\rangle \stackrel{!}{=} \frac{\pi^2}{R} \frac{\partial}{\partial \alpha} E_n(R;\alpha), $$ which gives the Burgers' equation, describing the shift in energy spectrum.

I don't quite understand the last step. Could anyone elaborate why one can identify the expectation value of $T \bar T$ with $\partial E_n/\partial \alpha$?

Answer 1

That's a good question.

1. Refs. 1-3 are considering a scalar field theory where the 1+1D world-sheet (WS) is a cylinder with circumference $\ell$. The Lagrangian density is $T\bar{T}$-deformed $$ \frac{\partial {\cal L}_{(\alpha)}}{\partial \alpha} ~=~ \det(T^{\mu\nu}_{(\alpha)}), \tag{A}$$ cf. Ref. 2 eq. (5.1) and Ref. 3 eq. (2.1). Here $\alpha$ is a deformation parameter. The canonical SEM tensor is symmetric$^1$
   $$T^{\mu\nu}_{(\alpha)} ~=~ {\cal P}^{\mu}_{(\alpha)}\partial^{\nu}\phi -g^{\mu\nu}{{\cal L}_{(\alpha)}} ~=~(\mu\leftrightarrow \nu),\tag{B} $$ where the canonical momentum density is $$ {\cal P}^{\mu}_{(\alpha)}~:=~\frac{\partial {\cal L}_{(\alpha)}}{\partial (\partial_{\mu}\phi)}, \qquad {\cal P}^{\mu}_{(\alpha=0)}~=~\partial^{\mu}\phi.\tag{C}$$

2. The total energy and momentum are $$\begin{align} E_{(\alpha)}(\ell)~:=~&\int_0^{\ell}\! dx~T^{00}_{(\alpha)} ~\stackrel{(B)+(E)}{=}~P_{(\alpha)}(\ell)-\int_0^{\ell}\! dx~{\cal L}_{(\alpha)}, \tag{D}\cr P_{(\alpha)}(\ell)~:=~&\int_0^{\ell}\! dx~T^{01}_{(\alpha)} ~\stackrel{(B)}{=}~\int_0^{\ell}\! dx~{\cal P}^{1}_{(\alpha)}\partial_0\phi.\tag{E}\end{align}$$ Since $\langle F_{\rm ext} \rangle \delta\ell =$ external work = change in mechanical energy $$-\frac{1}{\ell}\int_0^{\ell}\! dx~T^{11}_{(\alpha)} ~\stackrel{(D)}{=}~\frac{dE_{(\alpha)}(\ell)}{d\ell},\tag{F}$$ cf. Ref. 1 eq. (36) and Ref. 3 eq. (3.23).

3. From periodicity of the cylinder the generator of translation must satisfy $$\exp\left(\frac{i}{\hbar}\ell \hat{P}_{(\alpha)}(\ell)\right)~=~\hat{\bf 1}\qquad\Rightarrow\qquad {\rm Spec}(\hat{P}_{(\alpha)}(\ell))~\subseteq~\frac{2\pi\hbar}{\ell}\mathbb{Z}\tag{G}$$ at the quantum-level. We conclude that the momentum is quantized and remains undeformed $$ \frac{\partial P_{(\alpha)}(\ell)}{\partial \alpha}~\stackrel{(G)}{=}~0,\tag{H}$$ cf. Ref. 3 below eq. (3.27).

4. Altogether, we obtain OP's sought-for equation $$-\frac{\partial E_{(\alpha)}(\ell)}{\partial \alpha} ~\stackrel{(A)+(D)+(H)}{=}~\int_0^{\ell}\! dx~ \det(T^{\mu\nu}_{(\alpha)}),\tag{I}$$ cf. Ref. 1 eq. (36) and Ref. 3 eq. (3.24).

5. For a proof of the factorizatization of $\langle n|\det(\hat{T}^{\mu\nu}_{(\alpha)})|n\rangle $ into diagonal matrix elements of the SEM tensor, see section 3.2 in Ref. 3.

References:

1. A.B. Zamolodchikov, arXiv:hep-th/0401146; eqs. (36)-(41).

2. F.A. Smirnov & A.B. Zamolodchikov, arXiv:1608.05499; section 5.

3. Y. Jiang, arXiv:1904.13376; section 3.3.

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$^1$ Conventions. We choose the WS metric to be $ds^2=dt^2-dx^2$ and the speed of light $c=1$.