Michelson Interference with one mirror moving.

learning goals: 
(1) to understand how complex waves are used to predict a time average detector output 
(2) to understand both fixed time and fixed space calculations of the phase

(1) to understand how complex waves are used to predict a time average detector output,

Michelson interferometer with fixed mirrors but different path lengths x1 and x2. Add the NON-COMPLEX expressions for  two wave amplitudes at the output port

psi1=E0 Cos[k x1 -omega t];psi2=E0 Cos[k x2 -omega t];

Simplify[ExpToTrig[Together[TrigToExp[psi1+psi2]]]]

Sketch a phasor diagram of the same addition of harmonic waves. The Cos[(1/2) k (x1-x2)] term yields the interference or the magnitude of the phasor sum while the time dependent term is a consequence of the phasor rotation.  Show the phasor animation of a LC circuit.

The intensity is the square of the amplitude. This is what the photodiode measures. The working equation for the detector output is then

Vout=Simplify[ExpToTrig[Together[TrigToExp[(psi1+psi2) (psi1+psi2) ]]]]

Phasors are used to add waves not squares of waves.

These oscillations are at the wave frequency. What is that for a He-Ne laser?

Vout1=Vout/.E0->1/.k->1/.x1->1/2/.x2->1/.omega->3;Vout2=Vout/.E0->1/.k->1/.x1->1/2/.x2->2.5/.omega->3;

Plot[{Vout1,Vout2},{t,0,4}]

Can the detector measure these oscillations?

The voltage out of the detector is proportional to the time average.  The mathematics that does this time average uses complex expressions for the wave.

psi1=E0 Exp[I (k x1 -omega t)];
psi2=E0 Exp[I (k x2 -omega t)];
psi1conj=E0 Exp[-I (k x1 -omega t)];psi2conj=E0 Exp[-I (k x2 -omega t)];

Simplify[ExpToTrig[Together[TrigToExp[(psi1+psi2) (psi1conj+psi2conj)]]]]

Where have you seen a similar procedure?

If we launched an atom (in an eigenstate of momentum or a harmonic wavefunction) into this interferometer the probability density would be

Simplify[ExpToTrig[Together[TrigToExp[(psi1+psi2) (psi1conj+psi2conj)]]]]

What detector is time averaging to get this PDF?

Do you understand how complex waves are used to predict a time average detector output? 
Questions?

(2) to understand fixed time calculations of the phase (snapshot)

Let’s fix the time at t = 0 to calculate the phase difference of the waves between the two arms. This is the same as taking a snapshot of the waves in the interferometer and looking at the phase difference of the waves by counting crests

x2=x1+deltax;

psi1=E0 Cos[k x1 -omega t];psi2=E0 Cos[k x2 -omega t];

Simplify[ExpToTrig[Together[TrigToExp[psi1+psi2]]]]

The working equation involves measuring the period T between fringes seen by the detector and using lambda to find the velocity = deltax/T

Solve[Cos[((deltax k)/2)]^2==1/.deltax->2 v T,T]

Why is there a factor of 2 in deltax = 2 v T? 

The time between fringes is then

(π/(k v))/.k->2 Pi/lambda

amplitudes={psi1,psi2}/.E0->1/.k->1/.t->0/.omega->3

Plot[amplitudes/.deltax->1,{x1,4,8}]

Sketch a phasor diagram of the same addition of harmonic waves. The Cos[(1/2) k (x1-x2)] term yields the interference or the magnitude of the phasor sum while the time dependent term is a consequence of the phasor rotation.  Show the phasor animation of a LC circuit.

(2) to understand fixed space calculations of the phase (sit on a piling),

Let’s fix the our position in space by sitting on a piling in the water and watch wave go by us. Let the piling be located at x = 0 to calculate the phase difference of the waves between the two arms. The two waves returning from the different arms of the interferometer have different frequencies given by the Doppler shift in one arm and no shift in the other arm.
Let’s apply the complex notation to get the time averaged signal the detector actually measures

psi1=E0 Exp[I (k1 x1 -omega1 t)];
psi2=E0 Exp[I (k2 x2 -omega2 t)];
psi1conj=E0 Exp[-I (k1 x1 -omega1 t)];psi2conj=E0 Exp[-I (k2 x2 -omega2 t)];

Vdetector=Simplify[ExpToTrig[Together[TrigToExp[(psi1+psi2) (psi1conj+psi2conj)]]]]

Now k1 x1+k2 x2 is just a constant. The Doppler shift is given by

omega2=omega1 ( 1- 2 v/c);omega1=2 Pi c/lambda; Simplify[Vdetector]

How much time does it take for one crest to replace another?

Solve[Cos[(1/2) (((4 π t v)/lambda))]^2==1,t]

This is the same result obtained via the fixed time or snapshot method.

Clear["Global`*"];Remove["Global`*"]