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Michelson Interferometers, Modulations, and Transfer Functions - Homework 03

Syracuse University

This is the third homework assignment for Lasers and Optomechanics at Syracuse University.
It is due Monday, March 02, 2026 by 5 pm

You will need to complete the questions in this jupyter notebook and submit it via gitlab

1Homodyne Michelson

homodyne-michelson-interferometer.jpeg

Homodyne Michelson Interferometer Diagram

1.1Adjacency Matrix

Set up an Adjacency Matrix M\boldsymbol{M} for the Homodyne Michelson.
You may choose any electric fields vector you want, but you have to include the explicitly labeled electric fields.
The homodyne angle ϕHD\phi_\mathrm{HD} can be considered as the total phase accrued by the local oscillator field ELOE_\mathrm{LO} in the pick-off path:

ELO=rLOeiϕHDEin\begin{align} E_\mathrm{LO} = -r_\mathrm{LO} e^{-i \phi_\mathrm{HD}} E_\mathrm{in} \end{align}

1.2Electric Field Transfer Functions

Invert your Adjacency Matrix minus the identity (MI)1(\boldsymbol{M} - \boldsymbol{I})^{-1} to calculate the transfer functions from each electric field to every other electric field.
Pick out the field transfer functions from the input to the photodetectors PDA\mathrm{PD}_A and PDB\mathrm{PD}_B EPDAEin\dfrac{E_\mathrm{PDA}}{E_\mathrm{in}} and EPDBEin\dfrac{E_\mathrm{PDB}}{E_\mathrm{in}}.

1.3Substitutions

Apply the phase change of basis used in class ϕd=ϕxϕy2\phi_d = \dfrac{\phi_x - \phi_y}{2} and ϕc=ϕxϕy2\phi_c = \dfrac{\phi_x - \phi_y}{2} to your field transfer functions above.
You may also assume a perfect Michelson beamsplitter rBS=tBS=12r_\mathrm{BS} = t_\mathrm{BS} = \dfrac{1}{\sqrt{2}}.

1.4Power Transfer Functions

Calculate the input to power transfer functions PPDAPin\dfrac{P_\mathrm{PDA}}{P_\mathrm{in}} and PPDBPin\dfrac{P_\mathrm{PDB}}{P_\mathrm{in}}.

1.5Interpretation

How do PPDAPin\dfrac{P_\mathrm{PDA}}{P_\mathrm{in}} and PPDBPin\dfrac{P_\mathrm{PDB}}{P_\mathrm{in}} depend on the homodyne angle ϕHD\phi_\mathrm{HD}, and the common and differential Michelson phase ϕd\phi_d and ϕc\phi_c?
Can we manipulate the homodyne angle ϕHD\phi_\mathrm{HD} to detect the differential phase ϕd\phi_d?
Can you think of any problems you might run into if you tried to actually set up a homodyned Michelson?

2Asymmetric Michelson

Suppose you have a Michelson interferometer with the X-arm much longer than the Y-arm:

Lx>LyL_x > L_y

In this problem, we will calculate the full frequency response transfer functions of the Michelson

PasϕCARM(ω)andPasϕDARM(ω)\dfrac{P_\mathrm{as}}{\phi_\mathrm{CARM}}(\omega) \quad \mathrm{and} \quad \dfrac{P_\mathrm{as}}{\phi_\mathrm{DARM}}(\omega)

I recommend using an algebra helper like Mathematica or sympy for this problem. You may convert your answers to LaTeX using TeXform in Mathematica or some such similar function for sympy.

2.1Field tranfer functions

Derive the electric field transfer functions for the following:

ExEin,EyEin,EasEin\begin{align} \dfrac{E_\mathrm{x}}{E_\mathrm{in}}, \dfrac{E_\mathrm{y}}{E_\mathrm{in}}, \dfrac{E_\mathrm{as}}{E_\mathrm{in}} \end{align}

where EinE_\mathrm{in} is the input electric field E0eiω0tE_0 e^{i \omega_0 t},
ExE_\mathrm{x} is the field reflected from the X-arm mirror,
EyE_\mathrm{y} is the field reflected from the Y-arm mirror,
and EasE_\mathrm{as} is the field at the antisymmetric (or transmission) port.

Use rbs,tbsr_\mathrm{bs}, t_\mathrm{bs} for the beamsplitter reflection and transmission,
rx,ryr_\mathrm{x}, r_\mathrm{y} for the X and Y mirror reflections, and
ϕx,ϕy\phi_x, \phi_y for the single-pass phase accrued by the field as it propagates in the X-arm or Y-arm.

2.2End Mirror Modulation

Now, suppose we apply some common modulation to both end mirrors:

ϕc(t)=ϕx(t)=ϕy(t)=Γcos(ωt)\phi_c(t) = \phi_x(t) = \phi_y(t) = \Gamma \cos(\omega t)

Write what the new arm reflected transfer functions are for:

ExEin,EyEin\dfrac{E_\mathrm{x}}{E_\mathrm{in}}, \dfrac{E_\mathrm{y}}{E_\mathrm{in}}

Hint: Remember that you should end up with three distinct fields at ω0\omega_0 and ω0±ω\omega_0 \pm \omega

2.3Propagate to the antisymmetric port

Propagate your new transfer functions to the antisymmetric port, and compute a new EasEin\dfrac{E_\mathrm{as}}{E_\mathrm{in}}.

Be careful here.
Consider carefully what k=ω/ck = \omega/c is for each of your electric fields as they propagate back to the beamsplitter, accruing some phase ϕ=kL\phi = -k L.
Should each field accrue the same phase?

2.4Calculate the power response

Calculate the power response to the common motion PasPin\dfrac{P_\mathrm{as}}{P_\mathrm{in}}.

This is a lot of algebra, I recommend Mathematica or sympy to help you make sure you get things right here.
You may find it useful to express your arm phases in the common and differential bases again, since you have already incorporated the time-dependent term Γcos(ωt)\Gamma \cos(\omega t) as two additional sideband terms:

ϕx=ϕc+ϕd,ϕy=ϕcϕd\phi_x = \phi_c + \phi_d, \qquad \phi_y = \phi_c - \phi_d

You may assume an ideal beamsplitter

rbs=tbs=12r_\mathrm{bs} = t_\mathrm{bs} = \dfrac{1}{\sqrt{2}}

and balanced arms

rx=tyr_\mathrm{x} = t_\mathrm{y}

and small modulation from the end mirrors

Γ20\Gamma^2 \approx 0

Answer:

PasPin=rx2[sin2(2ϕd)2Γsin(4ϕd)sin(ωϕdω0)sin(ωt+ωϕcω0)]\dfrac{P_\mathrm{as}}{P_\mathrm{in}} = r_x^2 \left[ \sin^2(2 \phi_d) - 2 \Gamma \sin (4 \phi_d) \sin \left(\dfrac{\omega \phi_d}{\omega_0}\right) \sin \left(\omega t+\frac{\omega \phi_c}{\omega_0}\right) \right]

2.5Demodulate the power term

Calculate

PasDC,PasI,PasQP_\mathrm{as}^\mathrm{DC}, \, P_\mathrm{as}^I, \, P_\mathrm{as}^Q

by integrating PasP_\mathrm{as} over one cycle of ωt\omega t while multiplying by 1,cos(ωt)1, \cos(\omega t) and sin(ωt)\sin(\omega t), respectively.

Does your answer for PasDCP_\mathrm{as}^\mathrm{DC} seem familiar?

2.6Caluculate the Frequency Response to CARM motion in an Asymmetric Michelson

Calculate the frequency response

Pas(ω)=PasI(ω)+iPasQ(ω)P_\mathrm{as}(\omega) = P_\mathrm{as}^I(\omega) + i P_\mathrm{as}^Q(\omega)

to CARM motion Γ\Gamma.

Plot the magnitude and phase for PasΓ(ω)\dfrac{P_\mathrm{as}}{\Gamma}(\omega).

Assume

rx=1Lx=1000 mLy=1 mλ=1064 nm\begin{align} r_x &= 1\\ L_x &= 1000~\mathrm{m}\\ L_y &= 1~\mathrm{m}\\ \lambda &= 1064~\mathrm{nm} \end{align}