ELOG - 40m Phase maps loss estimation

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Sat May 9 17:01:08 2020, Yehonathan, Update, Loss Measurement, 40m Phase maps loss estimation

Wed May 13 15:13:11 2020, Yehonathan, Update, Loss Measurement, 40m Phase maps loss estimation

Thu May 14 12:21:56 2020, Yehonathan, Update, Loss Measurement, 40m Phase maps loss estimation

ETMX_Loss_Map_Perturbation_Theory(1).pdf

Thu May 14 19:00:43 2020, Yehonathan, Update, Loss Measurement, 40m Phase maps loss estimation

Tue May 19 15:39:04 2020, Yehonathan, Update, Loss Measurement, 40m Phase maps loss estimation $Mode_power_fraction1.pdf$

Message ID: 15333 Entry time: Thu May 14 19:00:43 2020 In reply to: 15332 Reply to this: 15338

Author:	Yehonathan
Type:	Update
Category:	Loss Measurement
Subject:	40m Phase maps loss estimation

Perturbation theory:

The cavity modes $\left|q\rangle_{mn}$ , where q is the complex beam parameter and m,n is the mode index, are the eigenmodes of the cavity propagator. That is:

$\hat{R}_{ITM}\hat{K}_L\hat{R}_{ETM}\hat{K}_L\left|q\rangle_{mn}=e^{i\phi_g}\left|q\rangle_{mn}$ ,

where $\hat{R}$ is the mirror reflection matrix. At the 40m, ITM is flat, so $\hat{R}_{ITM}=\mathbb{I}$ . ETM is curved, so $\hat{R}_{ETM}=e^{-i\frac{kr^2}{2R}}$ , where R is the ETM's radius of curvature.

$\phi_g$ is the Gouy phase.

$\hat{K}_L=\frac{ik}{2\pi L}e^{\frac{ik}{2L}\left|\vec{r}-\vec{r}'\right|^2}$ is the free-space field propagator. When acting on a state it propagates the field a distance L.

The phase maps perturb the reflection matrices slightly so:

$\hat{R}_{ITM}\rightarrow e^{ikh_1\left(x,y \right )}\approx 1+ikh_1\left(x,y \right )$

$\hat{R}_{ETM}\rightarrow e^{ikh_2\left(x,y \right )}e^{-i\frac{kr^2}{2R}}\approx\left[1+ikh_2\left(x,y \right )\right]e^{-i\frac{kr^2}{2R}}$ ,

Where h_12 are the height profiles of the ITM and ETM respectively. The new propagator is

$H = H_0+V$ , where $H_0$ is the unperturbed propagator and

$V=ikh_1\left(x,y \right )H_0+ik\hat{K}_Lh_2\left(x,y \right )e^{-i\frac{kr^2}{2R}}\hat{K}_L$

To find the perturbed ground state mode we use first-order perturbation theory. The new ground state is then

$|\psi\rangle=\textsl{N}\left[ |q\rangle_{00}+\sum_{m\geq 1,n\geq1}^{}\frac{{}_{mn}\langle q|V|q\rangle_{00}}{1-e^{i\left(m+n \right )\phi_g}}|q\rangle_{mn}\right]$

Where N is the normalization factor. The (0,1) and (1,0) modes are omitted because they can be zeroed by tilting the mirrors. Gouy phase of TEM00 mode is taken to be 0.

Some simplification can be made here:

${}_{mn}\langle q|V|q\rangle_{00}={}_{mn}\langle q|ikh_1\left(x,y \right )|q\rangle_{00}+{}_{mn}\langle q|\hat{K}_Likh_2\left(x,y \right )e^{-i\frac{kr^2}{2R}}\hat{K}_L|q\rangle_{00}$

${}_{mn}\langle q|\hat{K}_Likh_2\left(x,y \right )e^{-i\frac{kr^2}{2R}}\hat{K}_L|q\rangle_{00}={}_{mn}\langle q-L|ikh_2\left(x,y \right )e^{-i\frac{kr^2}{2R}}|q+L\rangle_{00}={}_{mn}\langle q+L|ikh_2\left(x,y \right )|q+L\rangle_{00}$

The last step is possible since the beam parameter q matches the cavity.

The loss of the TEM00 mode is then:

$L=1-\left|{}_{00}\langle q|\psi\rangle\right|^2$

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