The Lazy Laser Physicist: Part 2

Question

The Lazy Laser Physicist is quite shocked about what happened to his setup over night:

Almost all of his mirrors were taken away, but some other things have been put on his table. Besides these there is also a message:

Your setup was a total mess! I don't know how you could work there, but weren't all these unused mirrors totally in the way? It looked like you had one setup and then changed everything to a different path with possibly the minimal work. I hope you are aware of the depolarization if you use the mirrors at an angle of incidence other than 45°? Here are some things which you can use to reassemble your setup. Because I know you're lazy it also includes a variable phase shifter to change your beam path without moving any mirror.

Best wishes
Your supervisor

Well then, let's have a look what we have:

• 7 mirrors which have a reflective coating on one side (blue) and must be used with light incident at a 45° angle.
• 3 50:50 beamsplitters which reflect half of the incident's light intensity and transmit the other half.
• 1 variable phase shifter which multiplies a phase factor $e^{i \phi}$ to the transmitted electric field (see below). Unfortunately this phase shifter can only be tuned in the range $\phi \in \left[ -\frac{\pi}{8}, \frac{\pi}{8} \right]$. And it's so narrow that only one beam fits through it.

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Physics interlude

Interference

It is known that light can be described as waves of the electric field. When two laser beams are superimposed not the intensities, but the electric fields are added up: $E = E_1 + E_2$. Depending on the relative phase this can lead to constructive $$ E_1 = E_2 \qquad \Rightarrow \qquad E_1 + E_2 = 2E_1 $$ or destructive interference $$ E_1 = -E_2 \qquad \Rightarrow \qquad E_1 + E_2 = 0 $$ or anything in between as $E_1, E_2 \in \mathbb{C}$.

Phase change

The phase of the electric field changes in the following ways:

• Propagation through space. After a distance $L$ the electric field changes from $E$ to $E e^{i \cdot 2\pi \frac{L}{\lambda}}$. To make it easy we set the wavelength $\lambda$ to the length of one grid square.
• Reflection. When light is reflected from a mirror or beamsplitter it accumulates the phase $e^{i \frac{\pi}{2}} = i$.
• And of course the phase-shifter. When the light passes through the phase shifter it accumulates a phase of $e^{i \cdot 2\pi} = 1$ due to the length of one grid square, but the phase shifter can imprint an additional phase of $e^{i \phi}$ with $\phi \in \left[ -\frac{\pi}{8}, \frac{\pi}{8} \right]$ onto the electric field.

Beamsplitter

To clarify the action of the beamsplitter: Imagine 2 beams with $E_{\text{in, 1}}$ and $E_{\text{in, 2}}$ impinging on a beamsplitter in the following way:

Splitting the intensity in half means the electric field amplitude is divided by $\sqrt{2}$ as intensity is proportional to the square of the electric field. Hence, the $E_{\text{in, 1}}$ contributes $\frac{1}{\sqrt{2}} E_{\text{in, 1}}$ to $E_{\text{out, 1}}$ and $\frac{i}{\sqrt{2}} E_{\text{in, 1}}$ to $E_{\text{out, 2}}$. In an analoguous way one gets the contributions from $E_{\text{in, 2}}$, so that it total: $$ E_{\text{out, 1}} = \frac{1}{\sqrt{2}} \left( E_{\text{in, 1}} + i E_{\text{in, 2}} \right) \\ E_{\text{out, 2}} = \frac{1}{\sqrt{2}} \left( i E_{\text{in, 1}} + E_{\text{in, 2}} \right) $$ What happens if only one beam is impinging on the beamsplitter can be extracted if one of the incident fields is set to $0$.

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Finally the question

How does the physicist need to arrange the elements to direct the full intensity of the beam onto either detector A or B being able to switch between these two states by only changing the state of the phase-shifter?

Answer 1

Partial answer with terrible ASCII diagrams proper drawings

Arrange one part thus:

If the two beams are $90°$ out of phase one way, the entire beam goes to A, and if they are $90°$ out of phase the other way, the entire beam goes to B.

Arrange the rest such

that it forms a Mach-Zehnder interferometer:

So if our phase shifter had a $180°$ range,

then if we set up the path lengths right, then on one end of the phase shifter's range, the lower beam would be $-90°$ out of phase with the upper one; and, on the other end of the range, it would be $+90°$ out of phase with the other one. So, we could just connect this setup to the previous setup, and we'd be done.

But our phase shifter only has a $45°$ range. So arrange the remaining items

into a ring resonator:

This funky feedback loop should allow us to get a $180°$ phase difference using a phase shifter with only a $45°$ range. Replace the phase shifter in the previous setup with this contraption.

Put all together it looks like this:

Total: 3 beamsplitters , 7 mirrors , 1 phase shifter .

But, before I could finish the calculations to figure out exactly what path lengths we need to make this work, I looked into the laser beam!

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[EDIT TO ADD some explanations as to why the above works:]

To see that

when two beams of equal intensity and $90°$ phase difference hit a beam splitter, the beam will go in one direction only,

it is not necessary to do much mathematics:

the beam splitter splits an incoming beam into two outgoing beams, each outgoing part has equal amplitude, but the reflected outgoing part gets a $+90°$ phase shift.
Let's call the phase of the incoming beam going upwards on the page as $0°$. Suppose the incoming beam going rightwards on the page has phase $+90°$.

• The incoming beam going upwards splits into an outgoing beam going upwards, still $0°$ phase, and an outgoing beam going rightwards, with $+90°$ phase.
• The incoming beam going rightwards splits into an outgoing beam going rightwards, still with $+90°$ phase, and an outgoing beam going upwards, with $180°$ phase.

Therefore,

• The two outgoing beams going rightward both have $+90°$ phase, so they add.
• The two outgoing beams going upwards have $0°$ and $180°$ phases, so they cancel.
  And if, instead, the incoming beam going rightwards has phase $-90°$, the opposite happens.

To see that

the feedback loop contraption is a phase shifter

you might simply guess that

mirrors and splitters must conserve "energy"

and this guess actually works: the equations OP gave have the property that,

they conserve the sum of the squares of the amplitudes of the beams.

Physics interlude!

The physical power of an electromagnetic wave is indeed proportional to square amplitude of the electric field. The proportionality constant is the transmission line engineers' favourite constant, called the characteristic impedance of vacuum (well, of air, which is very close to vacuum as far as light is concerned).

That is how I arrived at the idea, but, I realize the puzzle isn't intended to be physically realistic. (Each grid square is one laser wavelength? That's really small!) So my answer won't rely on that. It will follow from the equations OP gave:

• For mirrors (and free space propagation), they just change phase and actually preserve amplitude. A fortiori, they preserve total squared amplitude.
• Beam splitters do not preserve amplitude. But, for a single incoming beam, each of the two outgoing beams has $1/\sqrt{2}$ the original amplitude, so the the squared amplitude is half. Thus the total outgoing square amplitude is equal to the incoming.
• It is less obvious that this remains true when there are two incoming beams on a splitter, since they interfere with each other. But it does!
• Call the incoming amplitudes $a$ and $b$ and the outgoing amplitudes $x$ and $y$, for the upwards and rightwards beams respectively.
• Call the incoming phases $\alpha$ and $\beta$.
• $x$ is going to be a combination of a beam of amplitude $a/\sqrt{2}$ and a beam of amplitude $b/\sqrt{2}$, and the phase difference between these component beams will be $\alpha - (\beta + 90^{\circ})$.
• $y$ is also going to be a combination of a beam of amplitude $a/\sqrt{2}$ and a beam of amplitude $b/\sqrt{2}$, and the phase difference between these component beams will be $(\alpha + 90^{\circ}) - \beta$.
• We can use the triangle cosine rule now to find the amplitudes $x$ and $y$: $$x^2 = \left( \frac{a}{\sqrt{2}} \right)^2 + \left( \frac{b}{\sqrt{2}} \right)^2 + 2 \left( \frac{a}{\sqrt{2}} \right) \left( \frac{b}{\sqrt{2}} \right) \cos(\alpha - (\beta + 90^{\circ}))$$ $$y^2 = \left( \frac{a}{\sqrt{2}} \right)^2 + \left( \frac{b}{\sqrt{2}} \right)^2 + 2 \left( \frac{a}{\sqrt{2}} \right) \left( \frac{b}{\sqrt{2}} \right) \cos((\alpha + 90^{\circ}) - \beta)$$
• But the two cosine terms differ by $180°$, so when you add the two equations together, they cancel: $$x^2 + y^2 = a^2 + b^2$$

So:

since all the component parts of the feedback loop contraption preserve total squared amplitude, and since overall the contraption has only one incoming and one outgoing beam, thus the outgoing beam must have the same amplitude as the incoming beam, so the only effect would be some sort of phase shift.

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[EDIT: previous answer fell into algebraic voodoo at this point, but it's not necessary]

To see that

$\pm 90°$ is in range of the feedback loop contraption

Using OP's original equations and notation (for the beamsplitter inside the ring resonator):

• Let's set $E_{in,1}=1$ and $E_{out,1}=+i$.
• First equation $E_{out,1}=(E_{in,1}+iE_{in,2})/\sqrt{2}$, so $i=(1+iE_{in,2})/\sqrt{2}$, so $$E_{in,2}=\sqrt{2}+i$$
• Second equation $E_{out,2}=(iE_{in,1}+E_{in,2})/\sqrt{2}$, so $E_{out,2}=(i+(\sqrt{2}+i))/\sqrt{2}$, so $$E_{out,2} = 1+i\sqrt{2}$$
• The outgoing upward beam goes around the loop and gets a phase shift, call it $\phi$, so $E_{in,2} = E_{out,2}e^{i\phi}$, so $e^{i\phi} = (\sqrt{2}+i)/(1+i\sqrt{2})$, so $$e^{i\phi} = \frac{2\sqrt{2}-i}{3}$$
• So finally $$\phi = \sin^{-1}(-\frac{1}{3}) = -19.4712^\circ$$
• And the same calculation with $E_{out,1}=-i$ gives the opposite sign for $\phi$
• But the phase shifter we are given has range $\pm 22.5°$ so it's in range.