Chapter 3 E2

the two particle operator can be defined at zero momentum

$$O_2=\phi(0)\phi(0)$$ We can also construct an operator that interpolates the same state

3.0.0.1 FT

we define the discrete Fourier Tansform in space

$$\phi(p,t)=\frac{1}{V}\sum_x e^{i2\pi \vec{p}\cdot \vec{x}/L}\phi(x,t)$$ with $\vec{p}=(n_x,n_y,n_z)$ and $n_i=0,1,...,L-1$ since the field $\phi(x,t)$ is real we have that $$\phi(-p,t)=\phi^*(p,t)$$

with back to back momentum. In the latter case to have an operator which projected on the $A1$ irrep of $O(24)$ we need to construct the combination $$O_2^{A1}=\frac{1}{\sqrt{3}}\left[\phi(p_x)\phi^*(p_x)+\phi(p_y)\phi^*(p_y)+\phi(p_z)\phi^*(p_z)\right]$$ where $p_x=(2\pi /L,0,0)$ and we use the fact that $\phi(-p)=\phi(p)^*$.

with these Operator we can construct the correlator $$\frac{1}{T}\sum_{t_1}\langle O_2^{A1}(t_1) O_2^{A1*}(t_1+t)\rangle = |A_{0\to2}| \left( e^{-E_2 t}+e^{-E_2 (T-t)}\right)+ |A_{1\to1}| e^{-E_1 t}$$

• $E_2$ is the energy of the two particle state at rest;

• $E_1$ is the energy of the single particle with momentum $|p|=2\pi/L$;

• $O_2^{A1*}=O_2^{A1}$ since it is real.

3.0.0.2 E irrep

The other irrep possible is $E$ which is two dimensional $$O_2^{E1}=\frac{1}{\sqrt{2}}\left[\phi(p_x)\phi(p_x)^*-\phi(p_y)\phi(p_y)^*\right]$$ which should interpolate an exited state, the other state of this irrep is $$O_2^{E2}=\frac{1}{\sqrt{2}}\left[\phi(p_x)\phi(p_x)^*+\phi(p_y)\phi(p_y)^*-2\phi(p_z)\phi(p_z)^*\right]$$

3.0.0.3 two particle with momentum operator

3.0.0.4 p=(1,0,0)

we can construct the two particle operator with momentum as $$O_{2p_x}=\phi(p_x)\phi(0) \,,\quad p_x=(1,0,0)$$ this operator is already in the irrep $A1$. The two particle correlator can be define as $$\langle O_{2p_x} O_{2p_x}^\dagger(t)\rangle = |A_{0\to2}|^2 \left( e^{-E_{2p_x} t}+e^{-E_{2p_x} (T-t)}\right)+ |A_{\phi(0)\to\phi(p_x)}|^2 \left(e^{-\omega_{p_x} T-m t+\omega_{p_x}t}+e^{-m T-\omega_{p_x} t+mt} \right)\,.$$ We can average over all the direction to have a more precise correlator $$\frac{1}{3}\sum_{i=x,y,z} \langle O_{2p_i} O_{2p_i}^\dagger(t)\rangle$$

3.0.0.5 p=(1,1,0)

$$O_{2p_{xy}}=\phi(p_{xy})\phi(0) \,,\quad p_{xy}=(1,1,0)$$ To construct the correlator we can average over all the direction to have a more precise correlator $$\frac{1}{3}\sum_{i=xy,yz,zx} \langle O_{2p_i} O_{2p_i}^\dagger(t)\rangle$$

3.0.0.6 p=(1,1,1)

The operator $$O_{2p_{xyx}}=\phi(p_{xyz})\phi(0) \,,\quad p_{xyz}=(1,1,1)\,,$$ and the correlator $$\langle O_{2p_{x,y,z}} O_{2p_{x,y,z}}^\dagger(t)\rangle$$