Note on the nonrelativistic $T\bar{T}$ deformation

Next we add in a potential $V$ which is just a function of $\phi$ but does not contain the terms of the derivative of $\phi$. As before we expand the Lagrangian with respect to $\lambda$. The first few terms in the deformed Lagrangian are

$$\begin{split}\mathcal{L}_{0}=&-V+X+Y\\ \mathcal{L}_{1}=&-V^{2}+VX+Y(X+Y)\\ \mathcal{L}_{2}=&-V^{3}+V^{2}X-VY^{2}+Y(X^{2}+3XY+2Y^{2})\\ \mathcal{L}_{3}=&-V^{4}+V^{3}X-VY^{2}(3X+4Y)\\ &+Y(X^{3}+6X^{2}Y+10XY^{2}+5Y^{3})\\ \mathcal{L}_{4}=&-V^{5}+V^{4}X+2V^{2}Y^{3}-VY^{2}(6X^{2}+20XY+15Y^{2})\\ &+Y(X^{4}+10X^{3}Y+30X^{2}Y^{2}+35XY^{3}+14Y^{4}),\\ \end{split}$$

(8)

which can be rewritten in a general form as

$$\begin{split}\mathcal{L}_{n}&=-V^{n+1}+V^{n}X+\sum_{j=0}^{n}(-V)^{j}Y^{j+1}\sum_{k=0}^{n-2j}K(n,j,k),\\ K(n,j,k)&=X^{n-2j-k}Y^{k}C_{n+k}^{n-j-1}C_{n-j-1}^{k-1+j}C_{k-1-j}^{j}\frac{1}{k}.\end{split}$$

(9)

Substituting (9) into (3) and performing the summation we end up with the closed form of the deformed Lagrangian

$$\begin{split}\mathcal{L}=&-\frac{V}{1-\lambda V}\\ &-\frac{1}{2\lambda(1-\lambda V))}\left[-1-\lambda X+\sqrt{(1-\lambda X)^{2}-4\lambda(1-\lambda V)Y}\right].\end{split}$$

(10)

Alternatively the deformed Lagrangian can also be obtained by using the method of characteristics with the initial condition

$$\begin{split}\mathcal{L}_{0}=\phi\alpha+\beta^{2}-V,\quad\mbox{with}\hskip 8.61108pt\alpha=\partial_{0}\phi,\ \beta=\partial_{1}\phi.\end{split}$$

(11)

In terms of these new variables $\alpha$ and $\beta$ the flow equation (1) could be rewritten as

$$\begin{split}\frac{\partial\mathcal{L}}{\partial\lambda}=-\mathcal{L}^{2}+\beta\mathcal{L}\frac{\partial\mathcal{L}}{\partial\beta}+\alpha\mathcal{L}\frac{\partial\mathcal{L}}{\partial\alpha},\end{split}$$

(12)

which is equivalent to a set of equations:

$$\left\{\begin{aligned} &\frac{d\lambda}{ds}=1\\ &\frac{d\alpha}{ds}=-\alpha\mathcal{L}\\ &\frac{d\beta}{ds}=-\beta\mathcal{L}\\ &\frac{d\mathcal{L}}{ds}=-\mathcal{L}^{2}\\ &\mathcal{L}(s=0)\\ &=\phi\alpha(0)+\beta(0)^{2}-V\end{aligned}\right.\Rightarrow\hskip 8.61108pt\left\{\begin{aligned} &\lambda=s\\ &\alpha=\frac{C_{3}}{-s+C_{2}}\\ &\beta=\frac{C_{4}}{-s+C_{2}}\\ &\mathcal{L}=\frac{1}{s-C_{2}}\\ &-\frac{1}{C_{2}}=\phi\frac{C_{3}}{C_{2}}+(\frac{C_{4}}{C_{2}})^{2}-V.\end{aligned}\right.$$

(13)

Canceling the constants $C_{i}$’s in (13), we get

$$\begin{split}\mathcal{L}&=-\frac{V}{1-\lambda V}\\ &-\frac{1}{2\lambda(1-\lambda V))}\left[-1-\lambda X+\sqrt{(1-\lambda X)^{2}-4\lambda(1-\lambda V)Y}\right],\end{split}$$

(14)

which coincides with (10).

To include the modifications of the last two terms in (16), we make an ansatz on the exact Lagrangian and expand it with respect to $\lambda$ . By matching the expanded terms⁴⁴4The explicit forms of the Lagrangians $\mathcal{L}_{i}$ are quite involved, and we list them in Appendix A. to the first few orders we can fix our ansatz completely, and in the end we check our result with the flow equation. The final result we get is

$$\begin{split}\mathcal{L}&=-\frac{1}{2\lambda}\left[-1-\lambda X+\sqrt{(1-\lambda X)^{2}-4\lambda Y+8i\lambda^{2}A-4\lambda^{3}B}\right]\end{split}$$

(19)

where

$$\begin{split}&A=(\phi^{\ast}\partial_{1}\phi+\phi\partial_{1}\phi^{\ast})(\partial_{1}\phi\partial_{0}\phi^{\ast}-\partial_{0}\phi\partial_{1}\phi^{\ast}),\\ &B=\phi\phi^{\ast}(\partial_{1}\phi\partial_{0}\phi^{\ast}-\partial_{0}\phi\partial_{1}\phi^{\ast})^{2}.\\ \end{split}$$

(20)

In a similar way, we can find the deformed Lagrangian in the case that the complex scalar has a potential $V$,

$$\begin{split}&\mathcal{L}=-\frac{V}{1-\lambda V}-\frac{1}{2\lambda(1-\lambda V)}(-1-\lambda X+\\ &\sqrt{(1-\lambda X)^{2}-4\lambda(1-\lambda V)Y+8i\lambda^{2}(1-\lambda V)A-4\lambda^{3}(1-\lambda V)B}).\end{split}$$

(21)

The detailed derivation is straightforward but too tedious to be presented here.

The $T\bar{T}$-deformation is formally defined as a flow of the Lagrangian density of the theory:

$$\begin{split}\partial_{\lambda}\mathcal{L}=-\epsilon^{ab}\epsilon^{cd}T_{ac}^{(\lambda)}T_{bd}^{(\lambda)}\equiv-\mathcal{O}^{(\lambda)}_{T\bar{T}}\end{split}$$

(28)

where $T_{ab}^{(\lambda)}$ is the stress tensor of the theory with the flow parameter $\lambda$. It is generally believed that at least in the classical level the deformation can be equivalently defined as a flow of the Hamiltonian density of theory:

$$\begin{split}\partial_{\lambda}\mathcal{H}=\mathcal{O}^{(\lambda)}_{T\bar{T}}.\end{split}$$

(29)

The classical equivalence has been examined in the leading order in Kruthoff:2020hsi and more generally in Jorjadze:2020ili . However the equivalence is built on an implicit assumption that the Lagrangian is not singular or equivalently the Legendre transformation is invertible such that the $T\bar{T}$-deformation and the Legendre transformation are commutative. The subtlety of a singular Lagrangian is that it leads to a constrained Hamiltonian. In this section we want to show that in the Hamiltonian formalism the flow equation (29) itself is not enough to defined the $T\bar{T}$-deformation.

To illustrate our point, let us consider the simple model (22) with the Lagrangian

$$\begin{split}\mathcal{L}=&\mathcal{L}_{0}+\lambda\mathcal{L}_{1},\end{split}$$

(30)

where the leading-order deformation of the Lagrangian is

$$\begin{split}\mathcal{L}_{1}=\frac{i}{2}\left(\phi\partial_{t}\phi(\partial_{x}\phi^{\ast})^{2}-\phi^{\ast}\partial_{t}\phi^{\ast}(\partial_{x}\phi)^{2}\right)+\left(\partial_{x}\phi^{\ast}\partial_{x}\phi\right)^{2}.\end{split}$$

(31)

The energy density is now:

$$\begin{split}T_{00}=&\frac{\partial\mathcal{L}}{\partial(\partial_{t}\phi)}\partial_{t}\phi+\frac{\partial\mathcal{L}}{\partial(\partial_{t}\phi^{\ast})}\partial_{t}\phi^{\ast}-\mathcal{L}\\ =&\partial_{x}\phi\partial_{x}\phi^{\ast}-\lambda\left(\partial_{x}\phi^{\ast}\partial_{x}\phi\right)^{2}.\end{split}$$

(32)

This energy density is supposed to be identified with Hamiltonian density $\mathcal{H}$ after rewriting $\partial_{t}\phi,~{}\partial_{t}\phi^{\ast}$ in terms of $\phi,~{}\phi^{\ast}$ and the canonical momenta $\Pi,\Pi^{\ast}$, which are given by

$$\begin{split}&\Pi=\frac{\partial\mathcal{L}}{\partial(\partial_{t}\phi)}=\frac{i}{2}\phi^{\ast}+\frac{i\lambda}{2}\phi(\partial_{x}\phi^{\ast})^{2},\\ &\Pi^{\ast}=\frac{\partial\mathcal{L}}{\partial(\partial_{t}\phi^{\ast})}=-\frac{i}{2}\phi-\frac{i\lambda}{2}\phi^{\ast}(\partial_{x}\phi)^{2}.\end{split}$$

(33)

The problem is that the Lagrangian density $\mathcal{L}$ is singular so that the rewriting can not be performed. Since the energy density (32) does not depend on $\partial_{t}\phi,~{}\partial_{t}\phi^{\ast}$ explicitly, one tends to claim

$$\begin{split}\mathcal{H}=T_{00}=\partial_{x}\phi\partial_{x}\phi^{\ast}-\lambda\left(\partial_{x}\phi^{\ast}\partial_{x}\phi\right)^{2}.\end{split}$$

(34)

However clearly (34) is not consistent with the definition (29) even in the leading order

$$\begin{split}&\mathcal{H}=\mathcal{H}_{0}+\lambda\mathcal{O}^{(\lambda)}_{T\bar{T}}\\ &=\partial_{x}\phi\partial_{x}\phi^{\ast}-\lambda\left(\partial_{x}\phi^{\ast}\partial_{x}\phi\right)^{2}-\frac{i\lambda}{2}\left(\phi\partial_{t}\phi(\partial_{x}\phi^{\ast})^{2}-\phi^{\ast}\partial_{t}\phi^{\ast}(\partial_{x}\phi)^{2}\right).\end{split}$$

(35)

Strictly speaking, this is not the expression of Hamiltonian density because it contains the time-derivative terms $\partial_{t}\phi,~{}\partial_{t}\phi^{\ast}$ which come from $T_{ab}$. Using the equations of motion

$$\begin{split}i\partial_{t}\phi+\partial_{x}^{2}\phi=i\partial_{t}\phi^{\ast}-\partial_{x}^{2}\phi^{\ast}=0+\mathcal{O}(\lambda)\end{split}$$

(36)

one can rewrite (LABEL:HTt) as

$$\begin{split}&\mathcal{H}=\partial_{x}\phi\partial_{x}\phi^{\ast}-\lambda\left(\partial_{x}\phi^{\ast}\partial_{x}\phi\right)^{2}\\ &+\frac{\lambda}{2}\left((\partial_{x}\phi^{\ast})^{2}\partial_{x}^{2}\phi\phi+(\partial_{x}\phi)^{2}\partial_{x}^{2}\phi^{\ast}\phi^{\ast}\right).\end{split}$$

(37)

It is obvious that (34) and (LABEL:HT001) are in mismatch. This mismatch is problematic when one tries to compute the spectrum of the theory perturbatively. This problem does not appear in a relativistic theory, for example in a theory of the relativistic free boson. In the relativistic case, the canonical momentum depends on the flow parameter as well such that the deformed Hamiltonian densities are the same at the classical level in two different ways, as we show in the Appendix B. A more severe problem about the definition (29) is that it is not enough to define the deformation. For example if want to obtain the second-order correction of (LABEL:HT001) we need to first introduce the Lagrangian density

$$\begin{split}\mathcal{L}=\partial_{t}\phi\Pi+\partial_{t}^{\ast}\phi\Pi^{\ast}-\mathcal{H},\end{split}$$

(38)

with the primary constraints

$$\begin{split}\chi_{1}=\Pi-\frac{i}{2}\phi^{\ast}=0,\quad\chi_{2}=\Pi^{\ast}+\frac{i}{2}\phi=0,\end{split}$$

(39)

then derive the first-order canonical stress tensor through

$$\begin{split}T^{a}_{b}=\frac{\partial\mathcal{L}}{\partial(\partial_{a}\phi)}\partial_{a}\phi+\frac{\partial\mathcal{L}}{\partial(\partial_{a}\phi^{\ast})}\partial_{a}\phi^{\ast}-\delta^{a}_{b}\mathcal{L},\end{split}$$

(40)

and in the end substitute it into (29). Therefore besides the flow equation (29) one also needs to know how the constraints $\chi_{1}^{(\lambda)}$ and $\chi_{2}^{(\lambda)}$ change under the flow.

Assuming that we use (29) together with $\chi_{1}^{(\lambda)}$ and $\chi_{2}^{(\lambda)}$ to define the $T\bar{T}$-deformation of the non-relativistic free boson gas, we can use the Hellmann-Feynman theorem and factorization formula to derive the flow equation of the spectrum

$$\begin{split}\frac{d\langle n|H|n\rangle}{d\lambda}=\langle n|T_{00}|n\rangle\langle n|T_{11}|n\rangle-\langle n|T_{01}|n\rangle\langle n|T_{10}|n\rangle.\end{split}$$

(41)

But because the mismatch of (34) and (LABEL:HT001) we can not identify $\langle n|T_{00}|n\rangle$ with $E_{n}/R$ and $\langle n|H|n\rangle=E_{n}$ at the same time if we use (LABEL:HT001) as the Hamiltonian density, where $R$ is the size of the system.

It is not hard to check that the two constraints (39) are the second-class constraints so one may wonder how about to solve them to reduce the dimensions of the phase space first and then perform the $T\bar{T}$-deformation. The reduced Hamiltonian density is given in Gergely:2003zy

$$\begin{split}\mathcal{H}_{r}=-\frac{i}{2}\partial_{x}\tilde{\phi}\partial_{x}\tilde{\Pi},\quad\tilde{\phi}=\frac{\phi}{2}+i\Pi^{\ast},\quad\tilde{\Pi}=\Pi+i\frac{\phi^{\ast}}{2}.\end{split}$$

(42)

However this does not solve the problem: we still can not transform the Hamiltonian density into a Lagrangian density which only depends on $\tilde{\phi}$ and $\partial_{t}\tilde{\phi}$ so that the $T\bar{T}$-deformation can not be defined. A caveat of our analysis is that there may exist other ways to describe the reduced Hamiltonian system such that the velocity $\partial_{t}\phi$ could be solved in terms of the momentum.

The singular Lagrangian density can be cured by including the higher-order deformations. In particular the exact Lagrangian is not singular so in principle we can use the Legendre transformation to obtain a Hamiltonian density without any constraints. But it turns out the resulting Hamiltonian has a singularity at $\lambda=0$, indicating that it can not be expanded in a Taylor series of the flow parameter $\lambda$. Requiring the vanishing of these singularities introduces the constraints. Since there are different ways to introduce the constraints, there will be ambiguities in the constrained Hamiltonian. In short, the mismatch (34) and (LABEL:HT001) is not just a problem in the first-order deformation, instead it is a problem of the definition of the $T\bar{T}$-deformation in a system with singular Lagrangian density or equivalently in a constrained Hamiltonian system.

We conclude this section by stressing that the above argument is purely classical, without considering the problem of Feynman path integral and the ambiguities in the operator ordering.

The standard technique to solve the constrained Hamiltonian system is through the Dirac-Bergman Algorithm (DBA). In this section, we will focus on the first-order deformation and study how the deformation affects the commutation relations using DBA. First let us demonstrate the algorithm in the free theory with the action (22). The conjugate momentum of $\phi$ and $\phi^{\ast}$ are

$$\begin{split}\Pi=\frac{i}{2}\phi^{\ast},\quad\Pi^{\ast}=-\frac{i}{2}\phi,\end{split}$$

(43)

satisfying the standard equal-time commutation relation

$$\begin{split}[\phi(x),\Pi(y)]=i\delta(x-y),\quad[\phi^{\ast}(x),\Pi^{\ast}(y)]=i\delta(x-y).\end{split}$$

(44)

The Hamiltonian system has two constraints (39) satisfying

$$\begin{split}[\chi_{1},\chi_{2}]\equiv C_{1_{x},2_{y}}=\delta(x,y).\end{split}$$

(45)

These two constraints are the second class, so we can introduce the Dirac bracket which is defined by

$$\begin{split}&[\phi(x),\phi^{\ast}(y)]_{D}\\ &=[\phi(x),\phi^{\ast}(y)]-\int dx^{\prime}dy^{\prime}[\phi(x),\chi_{a}(x^{\prime})]C_{a_{x}^{\prime},b_{y}^{\prime}}^{-1}[\chi_{b}(y^{\prime}),\phi^{\ast}(y)]\\ &=0-\int dx^{\prime}dy^{\prime}\left(i\delta(x-x^{\prime})(-\delta(x^{\prime}-y^{\prime}))(-i\delta(y^{\prime}-y))\right)\\ &=\delta(x-y)\end{split}$$

(46)

as expected, recalling that the inverse of the matrix $C^{-1}_{a,b}$ is defined as

$$\begin{split}\int dx^{\prime}C_{a_{x},b_{x}^{\prime}}C^{-1}_{b_{x}^{\prime},c_{y}}=\delta_{ac}\delta(x-y).\end{split}$$

(47)

Now we consider the first-order deformed theory with the action

$$\begin{split}&\mathcal{L}=\mathcal{L}_{0}+\lambda\mathcal{L}_{1},\\ &\mathcal{L}_{1}=(\partial_{x}\phi^{\ast}\phi_{x}\phi)^{2}+\frac{1}{2}\left(i\phi\partial_{t}\phi(\partial_{x}\phi^{\ast})^{2}-i\phi^{\ast}\partial_{t}\phi^{\ast}(\partial_{x}\phi)^{2}\right).\end{split}$$

(48)

Correspondingly the two constraints become

$$\begin{split}&\chi_{1}=\Pi-\frac{i}{2}\phi^{\ast}-\frac{i\lambda}{2}\phi(\partial_{x}\phi^{\ast})^{2},\\ &\chi_{2}=\Pi^{\ast}+\frac{i}{2}\phi+\frac{i\lambda}{2}(\phi^{\ast}(\partial_{x}\phi)^{2}).\end{split}$$

(49)

From them we can compute

$$\begin{split}[\chi_{1}(x),\chi_{2}(y)]&\equiv C_{1_{x},2_{y}}=-C_{2_{y},1_{x}}\\ &=\delta(x-y)+\lambda G(x,y).\end{split}$$

(50)

where

$$\begin{split}G(x,y)=\partial_{y}\delta(x-y)\phi^{\ast}(y)\partial_{y}\phi(y)+\partial_{x}\delta(x-y)\partial_{x}\phi^{\ast}(x)\phi(x).\end{split}$$

(51)

The inverse of the matrix $C_{1_{x},2_{y}}$ up to the first order of $\lambda$ is simply

$$\begin{split}C_{1_{x},2_{y}}^{-1}=-\delta(x-y)+\lambda G(y,x).\end{split}$$

(52)

Then at the first order of $\lambda$ the non-vanishing commutation relations between $\phi$ and $\phi^{\ast}$ is

$$\begin{split}[\phi(x),\phi^{\ast}(y)]=0-C_{1_{x},2_{y}}^{-1}=\delta(x-y)-\lambda G(y,x).\end{split}$$

(53)

This implies that $\phi$ and $\phi^{\ast}$ are not canonical variables so that $\phi^{\ast}$ is not the creation operator to generate the Fock space. This fact is a reflection of the fact that $T\bar{T}$-deformation is equivalent to a dynamical coordinate transformation.

At the leading order we can introduce the canonical variables

$$\begin{split}\varphi=\phi+\frac{\lambda}{2}\phi^{\ast}(\partial_{x}\phi)^{2},\quad\varphi^{\ast}=\phi^{\ast}+\frac{\lambda}{2}(\partial_{x}\phi^{\ast})^{2}\phi,\end{split}$$

(54)

and conversely

$$\begin{split}\phi=\varphi-\frac{\lambda}{2}\varphi^{\ast}(\partial_{x}\varphi)^{2},\quad\phi^{\ast}=\varphi^{\ast}-\frac{\lambda}{2}\varphi(\partial_{x}\varphi^{\ast})^{2},\end{split}$$

(55)

such that

$$\begin{split}\langle x_{1}x_{2}|x_{3}x_{4}\rangle=\langle\varphi_{1}\varphi_{2}\varphi_{3}^{\ast}\varphi_{4}^{\ast}\rangle=\delta(x_{13})\delta(x_{24})+\delta(x_{14})\delta(x_{23}).\end{split}$$

(56)

The relations (54) are not canonical transformation and they are not the state flow which was introduced in Kruthoff:2020hsi . In terms of the new canonical variables the Hamiltonian density (34) up to the first order reads

$$\begin{split}T_{00}=\mathcal{H}\equiv H_{0}+H_{1}+\mathcal{O}(\lambda^{2})\end{split}$$

(57)

with

$$\begin{split}H_{0}=&\partial_{x}\varphi\partial_{x}\varphi^{\ast}\\ H_{1}=&-2\lambda(\partial_{x}\varphi\partial_{x}\varphi^{\ast})^{2}\\ &-\lambda\left(\varphi\partial_{x}\varphi\partial_{x}\varphi^{\ast}\partial_{x}^{2}\varphi^{\ast}+\varphi^{\ast}\partial_{x}\varphi^{\ast}\partial_{x}\varphi\partial_{x}^{2}\varphi\right).\end{split}$$

(58)

As a consistency check we also work out the first-order deformation of the number operator and momentum operator

$$\begin{split}N&=-\frac{i}{2}\left(\frac{\partial\mathcal{L}}{\partial\partial_{t}\phi}\phi-\frac{\partial\mathcal{L}}{\partial\partial_{t}\phi^{\ast}}\phi^{\ast}\right)\\ &=\phi^{\ast}\phi+\frac{\lambda}{2}\left({\phi^{\ast}}^{2}(\partial_{x}\phi)^{2}+(\partial_{x}\phi^{\ast})^{2}\phi^{2}\right)\\ &=\varphi^{\ast}\varphi+\mathcal{O}(\lambda^{2}),\end{split}$$

(59)

$$\begin{split}P&=\frac{\partial\mathcal{L}}{\partial\partial_{t}\phi}\partial_{x}\phi+\frac{\partial\mathcal{L}}{\partial\partial_{t}\phi^{\ast}}\partial_{x}\phi^{\ast}\\ &=\frac{i}{2}(\partial_{x}\phi\phi^{\ast}-\partial_{x}\phi^{\ast}\phi)(1-\lambda\,\partial_{x}\phi\partial_{x}\phi^{\ast}),\quad\\ &=\frac{i}{2}(\partial_{x}\varphi\varphi^{\ast}-\partial_{x}\varphi^{\ast}\varphi)+\text{total derivatives}+\mathcal{O}(\lambda^{2})\end{split}$$

(60)

which are indeed undeformed as expected. We expect that the spectrum of (57) can be derived by performing the second quantization and applying the standard quantum perturbation theory.

The explicit form of the Hamiltonian (57) implies that there is only two-to-two scattering, so without losing any generality we can focus on the two particle sector. When the total momentum is zero, the $T\bar{T}$-deformed spectrum satisfies a very simple equation BoseJ

$$\begin{split}E_{N}(R,\lambda)=E_{N}(R+\lambda E_{N},0).\end{split}$$

(61)

To illustrate the procedure of our calculation, we start from the two-particle sector with zero momentum, $u_{1}=-u_{2}=u>0$. In this case, a general two-particle eigenfunction is

$$\begin{split}\psi=\frac{\sqrt{2}}{R}\cos\left(u(x_{1}-x_{2})\right),\end{split}$$

(62)

with

$$\begin{split}u_{I}=\frac{2\pi}{R}I,\quad I\in\mathbf{Z}^{+}.\end{split}$$

(63)

First we compute the matrix elements of the normal ordered Hamiltonian (57) of the field theory in the position space

$$\begin{split}\langle\vec{y}|H_{0}+H_{1}|\vec{x}\rangle.\end{split}$$

(64)

For example the undeformed Hamiltonian density $\mathcal{H}_{0}$ is evaluated to be

$$\begin{split}\langle\vec{y}|\mathcal{H}_{0}|\vec{x}\rangle&=\langle 0|\varphi(y_{1})\varphi(y_{2})\partial_{x}\varphi^{\ast}\partial_{x}\varphi\varphi^{\ast}(x_{1})\varphi^{\ast}(x_{2})|0\rangle\\ &=\partial_{x}\delta(x-x_{1})\partial_{x}\delta(x-y_{2})\delta(y_{1}-x_{2})\\ &+\partial_{x}\delta(x-x_{2})\partial_{x}\delta(x-y_{2})\delta(y_{1}-x_{1})\\ &+\partial_{x}\delta(x-x_{1})\partial_{x}\delta(x-y_{1})\delta(y_{2}-x_{2})\\ &+\partial_{x}\delta(x-x_{2})\partial_{x}\delta(x-y_{1})\delta(y_{2}-x_{1}).\\ \end{split}$$

(65)

Then we use the wavefunction (26) to obtain the Hamiltonian

$$\begin{split}H_{0II}&\equiv\langle u_{I}|H_{0}|u_{I}\rangle=\frac{1}{2}\int d\vec{y\,}d\vec{x}\,dx\,\psi_{I}^{\ast}(\vec{y})\psi_{I}(\vec{x})\langle\vec{y}|H_{0}|\vec{x}\rangle\\ &=\int d\vec{x}\,\psi^{\ast}_{I}(\vec{x})(-\partial^{2}_{x_{1}}-\partial^{2}_{x_{2}})\psi_{I}(\vec{x})\\ &=2u_{I}^{2}\equiv\gamma I^{2},\end{split}$$

(66)

where $\gamma=\frac{8\pi^{2}}{R^{2}}$. Similarly we find

$$\begin{split}H_{1IJ}=-\frac{\lambda}{2}(\frac{\sqrt{2}}{R})^{2}\left(8u_{I}^{2}u_{J}^{2}R)\right)=-2\lambda\frac{\gamma^{2}I^{2}J^{2}}{R}.\end{split}$$

(67)

The result $H_{1II}=-2\lambda/RH_{0II}^{2}$ coincides with the subleading term in the $\lambda$ expansion of the exact spectrum:

$$\begin{split}&E_{N}(R,\lambda)=E_{N}^{(0)}-\frac{2\lambda}{R}{E_{N}^{(0)}}^{2}+\frac{7\lambda^{2}}{R^{2}}{E_{N}^{(0)}}^{3}-\frac{30\lambda^{3}}{R^{3}}{E_{N}^{(0)}}^{4}\\ &+\frac{143\lambda^{4}}{R^{4}}{E_{N}^{(0)}}^{5}-\frac{728\lambda^{5}}{R^{5}}{E_{N}^{(0)}}^{6}\dots\end{split}$$

(68)

which is solved from the flow equation in BoseJ .