# Photon states and interference in an Interferometer

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• zeronem
In summary, the Ubs operator for a beam splitter at a ratio of 50/50 is described by a matrix that is unitary and has a determinant of 1. It can be represented by a socalled Bogoliubov transformation and is related to the parameters ##\psi##, ##\varphi##, and ##\vartheta##. The specific values of these parameters depend on the type of beam splitter used.
zeronem
TL;DR Summary
Question on deriving the operator for a beam splitter of a specific ratio of split
Hello. I have a question. In the book I am reading, They derive the Ubs operator applied on a photon state with the beam splitter at a ratio of 50/50. A beam splitter that is used in the Mach-Zehnder interferometer.

I'm having a hard time deciphering whether the formula for beam splitter operator is for a beam splitter at 50/50 ratio only, or for any ratio of a photon state with amplitudes "a" and "b" of the photon going in from the left.

I feel this derivation was based off the a beam splitter at a ratio of 50/50 which means a and b would have to be 1/2.

If a and b of the photon state were not based on a 50/50 split, but rather a 70/30 split would that change the Ubs operator? Or would I go about an a = sqrt(7/10) and b = sqrt(3/10)? In the formula in the picture above?

It's not clear which book this is.

For me it was also quite a puzzle to understand the beam splitter until I came to think about it in terms of the most adequate formalism to describe (free) photons, which is QFT and annihilation and creation operators and interpret the beam splitter as some obstacle the photons (aka the electromagnetic waves) get scattered on. Now we assume a lossless beam splitter.

To get a complete description of the beam splitter you have to describe what hapens to a photon in each of the two in-channels, i.e., to which photons they are scattered. For a lossless beam splitterthis you can describe by a socalled Bogoliubov transformation, i.e., you have the annitilation operators for the two in-states ##\hat{a}_{j}##, ##j \in \{1,2 \}## with the bosonic commutation relations
$$[\hat{a}_j,\hat{a}_k^{\dagger}]=\delta_{jk}, \quad [\hat{a}_j,\hat{a}_k]=[\hat{a}_j^{\dagger},\hat{a}_k^{\dagger}].$$
Now the beam-splitter is a device in the realm of linear optics, i.e., the scattering of the photon modes in the in channel leads to photon modes in the out-channel
$$\hat{b}_k =\sum_j U_{jk} \hat{a}_j,$$
and the ##\hat{b}_k## must also fulfill the bosonic commutation relations as the in-channel annihilation and creation operators. Working this out leads to the conclusion that the matrix ##\hat{U}## must be unitary. Since an overall phase doesn't matter, you can assume that ##\hat{U} \in \mathrm{SU}(2)##.

Now we assume that ##\hat{b}_k## annihilates the state which describe the reflected photon of the state which is annihilated by ##\hat{a}_k##. Then our SU(2)-matrix reads
$$\hat{U}=\begin{pmatrix} r & t \\ t' & r'\end{pmatrix}.$$
Now we have the constraints
$$\hat{U}^{-1}=\hat{U}^{\dagger},$$
i.e., that ##\hat{U}## is unitary and our arbitrary choice of phase
$$\mathrm{det} U=1.$$
This implies
$$\hat{U}^{-1}=\frac{1}{\mathrm{det} \hat{U}} \begin{pmatrix} r' & -t \\ -t' & r \end{pmatrix} = \begin{pmatrix} r' & -t \\ -t' & r \end{pmatrix} = \begin{pmatrix} r^* & t^{\prime *} \\ t^* & r^{\prime *}\end{pmatrix}.$$
Thus we have
$$r'=r^*, \quad t'=-t^*.$$
Thus we have
$$\hat{U}=\begin{pmatrix}r & t \\ -t^* & r^* \end{pmatrix}.$$
Now writing
$$r=\rho \exp(\mathrm{i} \varphi), \quad t=\tau \exp(\mathrm{i} \vartheta)$$
with ##\rho,\tau>0## we have
$$\hat{U}=\begin{pmatrix} \rho \exp(\mathrm{i} \varphi) & \tau \exp(\mathrm{i} \vartheta) \\ -\tau \exp(-\mathrm{i} \vartheta) & \rho \exp(-\mathrm{i}\varphi) \end{pmatrix}.$$
From ##\mathrm{det} \hat{U}=1## we get
$$\rho^2+\tau^2=1 \; \Rightarrow \; \rho=\cos \psi, \quad \tau=\cos \psi, \quad \psi \in [0,\pi/2].$$
Thus
$$\hat{U}=\begin{pmatrix} \cos \psi \exp(\mathrm{i} \varphi) & \sin \psi \exp(\mathrm{i} \vartheta) \\ -\sin \psi \exp(-\mathrm{i} \vartheta) & \cos \psi \exp(-\mathrm{i} \varphi) \end{pmatrix}.$$
Of course the parameters depend on the specific realization of the beam splitter.

The meaning of the parameters in the above formalism is obviously as follows: If you have a photon 1 in the incoming channel, it's probability to be reflected is ##|r|^2=\cos^2 \psi## and being transmitted is ##|t'|^2=|t|^2=\sin^2 \psi##. The same probabilities hold for one photon entering in the state 2. You have a 50:50 chance for ##\psi=\pi/4##, i.e., ##\cos^2 \psi=\cos^2 \psi=1/2##.

The exponential factors are phase shifts, depending on the specific type of beam splitter. If you have a symmetric beam splitter, you have
$$r'=r, \quad t'=t.$$
Now above we have seen that
$$r'=r^*, \quad t'=-t^*.$$
This implies that a symmetric beam splitter is given if ##r^*=r##, i.e., ##r \in \mathbb{R}## and ##t=-t^*##, i.e., ##t \in \mathrm{i} \mathbb{R}##. This implies that a symmetric beam splitter has ##\varphi=0## and ##\vartheta =\pm \pi/2##, leading to
$$U=\begin{pmatrix} \cos \psi & \pm \mathrm{i} \sin \psi \\ \pm \mathrm{i} \sin \psi & \cos \psi \end{pmatrix}.$$
Another simple beam splitter is a glass plate with a thin metal coating on one side. Then for the input photon in state 1 the reflection is on the optical denser medium, giving an additional phase shift of ##\pi## relative to the reflected beam of input state 2, which is reflected on the optical less dense medium. This means that, because ##\text{arg} r= \varphi = \text{arg} r' + \pi=\pi-\varphi##, that ##\varphi=\pi/2##.

Then ##\vartheta## is determined by the relative phase shift between the reflected and the transmitted beam for the incoming photon in state 1. It is given by the $$\pi-\vartheta-\varphi=\pi/2-\vartheta =-(n-1) L/\lambda$$, where ##L## is the distance a light beam has to travel in the glass and ##\lambda## the wavelength. This gives ##\vartheta=\pi/2+(n-1)L/\lambda##.

Now let's see, how your book's matrix fits in. It is fine, but has another phase convention since ##\mathrm{det} U_{\text{BS}}=-1,##
i.e., it's related to the ##\hat{U}## in our convention by an additional phase factor ##\exp(\mathrm{i} \pi/2)=\mathrm{i},##
i.e.,
$$\hat{U}=\mathrm{i} \hat{U}_{\text{BS}} = \frac{1}{\sqrt{2}} \begin{pmatrix} \mathrm{i} & \mathrm{i} \\ \mathrm{i} & -\mathrm{i} \end{pmatrix}.$$
This is given by the parameters ##\psi=\pi/4##, ##\varphi=\pi/2##, and ##\vartheta=\pi/2##.

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weirdoguy and Mentz114
I believe the bean splitters they use in the book is the simple glass plate with coating inside. I’m working out an exercise problem in the book that explains that beam splitter this time around has a 70/30 ratio in reflection. Given that they formed a Ubs operator in chapter off a beam splitter with a reflection ratio of 50/50, i was thinking maybe I was going have to derive the beam splitter for a 70/30 ratio to multiply to a photon state with probabilistic amplitudes based off the 70/30 ratio. Meaning that in the interferometer with 2 detectors, the probability of a photon going through one detector and not the other is no longer 1/2, but 7/10 for detector 1 and 3/10 for detector 2. And of course I haven’t yet went through the phase shift operator and the second beam splitter since there is one.

In other words

$$|photon> = \sqrt{\frac{1}{2}}|right> + \sqrt{\frac{1}{2}}|down>$$

For a photon with a probability of 1/2.

And a beam splitter of 70/30 ratio this change the formula to $$|photon> = \sqrt{\frac{7}{10}}|right> + \sqrt{\frac{3}{10}}|down>$$?

They derive a complete formula for an

|out> = UbsUps Ubs|in> = Umz |in>

Where ps means phase shift. MZ mean Mach-Zehnder interferometer. But the formula they used directly above a as I think for the situation of a 50/50 beam splitter. That is where I’m having a hard time deciphering whether that formula could be used for a 70/30 ratio beam splitter and that I have to a re-derivation of the formula for 70/30 beam splitters.

Edit message:

I left out some of the formula for Umz specifically what it equals in terms of the phase shift and and whether it yields detector 1 or detector 2 given a specific phase shift. My confusion is that they derive that specific working with a 50/50 ratio and that the matrix formula would change under beam splitter ratio for 70/30 which would make while still being dependent on the phase shift, but different coefficients because of the 70/30 ratio. Meaning there would not be $$\frac {1}{\sqrt{2}}$$. But a different coefficient on the matrix operator.

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I know that my question is probably simple but I’m wording it too complicated.

My question is this, for two detectors in an interferometer there is a probability of exactly 1/2 per detector. So if there were a three detector setup on an interferometer would the photon probability be 1/3 per detector? Meaning a 33 percent chance one of the detectors will receive the photon?

And if that is the case, Ubs is dependent only on the beam splitter ratio and not the probability of the photon entering a detector in the interferometer. So if there were an n detector setup on an complex interferometer, the probably of a photon entering the nth detector would be 1/n? And the beam splitter ratio is independent of the probability of photon?

But now you now the precise form the unitary matrices ##\hat{U}## take and the meaning of the parameters (I gave some simple examples for concrete beam splitters too). The only thing you can deduce from saying "I've a 70/30 beamsplitter" is that ##\cos^2 \psi=0.7## (in my notation, if a photon gets reflected with probability 70% and transmitted with probability 30%; note that the conventions are different in different textbooks/papers!).

Ah, Ok. I understand. I’m working off a book authored by Pieter kok titled “A first introduction to Quantum Physics.”

So the probability of a photon is dependent on the ratio setup of the beam splitter?

The diagram is the type of interferometer setup I’m dealing with, a beam splitter and a few mirrors in place and a couple of looped QND(quantum non-demolition detectors). With the photon detectors at the ends.

This is the formula that I question the most. The a and b are probabilistic amplitudes of the photon going in, but the Ubs was clearly derived from a beam splitter of 50/50 ratio, but I suppose it can be applied to any situation of a beam splitter with a different ratio? Given the probabilistic amplitudes a and b of the photon are dependent on the ratio of said beam splitter?

I obviously don't understand where your problem is. As detailed in my yesterday's posting, the above given ##U_{\text{BS}}## is a valid description of a beam splitter. To understand where the matrix comes from, I'd need to know the specific construction of the beam splitter.

I also don't know what you mean by "probabilistic amplitudes". I think it describes simply that you have a single-photon state which is a superposition of the two input channels of the beam splitter.

I guess what I’m struggling with is whether the probability of a photon is dependent on the percent of reflection of a beam splitter, which is just a piece of glass with a semi-reflecting coating as stated in the book. I’m only focusing on the first beam splitter of the interferometer.

In the book they ascertain the quantum state of the photon before entering the beam splitter with 50/50 reflective ratio,
and then they derive the beam splitter operator in 50/50 reflective ratio case, to then be applied to the entering photon and this giving the result of a quantum state after the beam splitter. In the chapter they do it for a 50/50 ratio beam splitter. In the exercise problem they want to do it with 70/30 ratio beam splitter. So I was looking for the “universal” beam splitter operator formula to use for any given reflection ratio whether that be a beam splitter with a 60/40, 10/70, 80/20 reflective ratio.

As I said, you can have any ratio between the probabilities of reflection and transmission. In my notation it's given by
$$|r|^2=|r'|^2=\cos^2 \psi, \quad |t|^2=|t'|^2=\sin^2 \psi.$$
To understand, how to derive the unitary matrix, it helps a lot to learn classical electrodynamics and its application to optics (optics is just a subfield of electrodynamics since light is just an electromagnetic (wave) field). For reflection and transmission, look for "Fresnel equations". A start is Wikipedia:

https://en.wikipedia.org/wiki/Fresnel_equations

zeronem said:
suppose it can be applied to any situation of a beam splitter with a different ratio?

No, because, as @vanhees71 has been explaining in detail, for a different ratio the elements of the matrix ##U_{\text{BS}}## will be different.

zeronem said:
Given the probabilistic amplitudes a and b of the photon are dependent on the ratio of said beam splitter?

No, they're not, they're dependent on the state of the photon before it enters the beam splitter. (That state might be the result of previous operations with other beam splitters, but it's not related to the ratio of the beam splitter being described by the matrix in that equation in the textbook.)

vanhees71
PeterDonis said:
No, because, as @vanhees71 has been explaining in detail, for a different ratio the elements of the matrix ##U_{\text{BS}}## will be different.
No, they're not, they're dependent on the state of the photon before it enters the beam splitter. (That state might be the result of previous operations with other beam splitters, but it's not related to the ratio of the beam splitter being described by the matrix in that equation in the textbook.)

Thanks that’s what I figured. So in the exercise problem I must derive my own Ubs for a ratio of 70/30 and that is what I expected. I just wanted to make sure I was going the right route before I get any further with the derivation since the book doesn’t exactly throw in the universal formula for a beam splitter with variable ratios. And I kind of wanted the book to do that but that would be cheating, lol

Edit message:
Also thanks vahn I kind of gathered where you are coming from. With different ratios you’re going to have different angles of reflection. A 50/50 ratio gives you a pi/4 reflection. A 70/30 ratio would give you a reflection angle of .315pi if I calculated it correctly I hope. Noticed that you derived the Ubs for variable ratio haha. The book doesn’t take it that far it seems since it isn’t neccesarily textbook quality, and I was looking around the books chapter thinking that it did lol. I do plan on deriving myself the Ubs operator for a variable ratio in the future.

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Which "angles of reflection"? Nowhere did I mention any angles of reflection... I don't know the book you are using, but I get the impression that it's not very clearly explaining what a beam splitter is. There are of course very many different kinds of beam splitters, but the book should explain a few to give you a feeling, how those devices look in the quantum opticians' labs. A not too bad start is the Wikipedia page

https://en.wikipedia.org/wiki/Beam_splitter

vanhees71 said:
Which "angles of reflection"? Nowhere did I mention any angles of reflection... I don't know the book you are using, but I get the impression that it's not very clearly explaining what a beam splitter is. There are of course very many different kinds of beam splitters, but the book should explain a few to give you a feeling, how those devices look in the quantum opticians' labs. A not too bad start is the Wikipedia page

https://en.wikipedia.org/wiki/Beam_splitter
The book describes that the beam splitter is a half coated piece of glass as I’ve stated before. And I think I really meant a ratio between reflection and transmission. My bad wording it angle of reflection, not the right choice of words but when I think about a piece of coated glass being used I picture it that way. And I prefer to stay off Wikipedia. I’ll probably buy me more of textbook quality book here soon. Just wanted to self study from a much more smaller springer published book that’s attempting to take a different approach on things.

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vanhees71

## 1. What is a photon state in an interferometer?

A photon state in an interferometer refers to the specific quantum state of a photon as it travels through the interferometer. This state is determined by the properties of the photon, such as its energy, polarization, and direction of travel.

## 2. How does interference occur in an interferometer?

Interference in an interferometer occurs when two or more photon states overlap and interact with each other. This can result in constructive interference, where the photons reinforce each other, or destructive interference, where the photons cancel each other out.

## 3. What is the purpose of an interferometer in scientific research?

An interferometer is a powerful tool used in scientific research to measure small changes in the phase or amplitude of light waves. It is commonly used in fields such as astronomy, quantum optics, and metrology to make precise measurements and gather information about the properties of light and other electromagnetic waves.

## 4. How does an interferometer detect and measure changes in photon states?

An interferometer detects and measures changes in photon states by splitting a beam of light into two paths, recombining them, and observing the resulting interference pattern. Changes in the photon states, such as a phase shift or polarization change, will alter the interference pattern, allowing for precise measurements to be made.

## 5. What are some real-world applications of interferometers?

Interferometers have a wide range of applications in various fields of science and technology. They are used in the telecommunications industry for fiber-optic communication, in astronomy for measuring the size and distance of celestial objects, and in medical imaging for creating high-resolution images of internal structures. They are also used in the development of precision instruments such as atomic clocks and laser gyroscopes.

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