Polarization of light and probability

[rasp]

I Dont understand how light polarized along a certain axis in a prepared state can then pass through a second polarizer that is at an angle to the first. The experiment I’m trying to understand starts with unpolarized light which then passes through a first filter which vertically polarizes the light, such that all light that is passed through is vertically polarized. Then the light passes through a second 45 degree polarizer. My understanding is that 50% of the light will pass through. But why is that when none of the light was originally in the 45 degree plane? It is not that the light is oscillating at 45 degrees and has a component that oscillates vertically and another component that oscillates horizontally. It is only the measuring device that is in a 45 degree angle. The light was prepared in a vertical direction.

 

[tech99]

If you imagine a metal rod placed at 45 degrees rotation relative to vertical, you will notice that due to its vertical dimension it will partially respond to the vertical E-field. It will see a field strength of (cos 45) times the vertical field. It will then radiate a new wave in the 45 degree plane.
When we rotate a rod or slot around the polarisation axis, we see a cosine variation in the received voltage. We do not see a sudden stop when we rotate slightly.

 

[David Byrden]

QM is "linear".

Therefore we can break down the polarisation vector into two components, process them individually, and combine the results.

We can break down 1 unit of vertical into 0.7 of 45 degrees plus 0.7 of -45 degrees.

When we filter those two as you described, only the former passes through the filter.

 

[PeterDonis]

why is that when none of the light was originally in the 45 degree plane?

It's not true that "none of the light was originally in the 45 degree plane". See below.

It is not that the light is oscillating at 45 degrees and has a component that oscillates vertically and another component that oscillates horizontally.

But it is true that the light is oscillating vertically and has a component that oscillates at 45 degrees and a component that oscillates at 135 degrees. The only component it doesn't have is a component that oscillates horizontally.

You seem to be using a model of light that is basically a classical wave with a particular plane of oscillation. This model doesn't work when trying to explain quantum phenomena like how light interacts with polarizers.

 

[tech99]

You seem to be using a model of light that is basically a classical wave with a particular plane of oscillation. This model doesn't work when trying to explain quantum phenomena like how light interacts with polarizers.

I think that the classical approach does work. A rod sloping at 45 degrees has a vertical component of length, and sees a potential difference between its ends.

 

[vanhees71]

The problem with understanding this very important feature of quantum theory is that you seem to think of "light" as a stream of photons, misidentifying photons with quasi-classical bullet-like objects. This is the wrong picture!

If you want to use a classical picture about light you are much better off with thinking in terms of classical electromagnetic fields. Then it's very easy to understand why a horizontally polarized wave goes partially through a polarization filter oriented with an angle of $\pi/4$ relative to the horizontal direction (in the plane perpendicular to the wave vector $\vec{k}$ of the wave). The intensity behind this filter is indeed 1/2 of the original intensity (the "intensity" is a measure for the energy-density current, given by the Poynting vector, going through the area of your detector).

Now the correct quantum theory of single-photon states can be qualitatively understood by taking the classical intensity, normalized to one (i.e., you consider relative intensities like the 50% intensity in the above considered case relative to the intensity of the incoming light), as the probability for the photon, prepared in the given state, to go through the polarization filter in the given orientation. The only difference between QED and classical electrodynamics on this level of consideration simply is that a single photon cannot be split in parts having the same wave properties as the single photon you started with. Particularly you cannot split a photon within linear response theory, i.e., a linear optical device like a polarization filter won't split photons. Thus you can only have one whole photon go through the polarizer or have one whole photon completely absorbed by it. There's no way that you have half a photon going through and half a photon being absorbed, and that's the true quantum feature of single-photon states in such a setup, and it's pretty close to the way of thinking about em. waves within classical electrodynamics rather than thinking of photons as particle-like objects.

One whould note that single-photon are very much different from anything particle like you can think of. The reason is that due to relativistic QFT (which is the only consistent relativistic quantum theory we have today) a photon cannot be localized at all. There's not even a position operator for photons in the literal sense. The only thing you can say about a photon, knowing the quantum state it's preapred in, is the probability for finding it at a given place with some detector (the detector as a massive macroscopic object of course can be localized)!

 

[rasp]

Thank you. Am I then correct in interpreting PeterDonis comments to mean that a vertically polarized filter acting on unpolarized light emits all axis other than horizontal? It was my prior ‘misunderstanding’ that it blocks all axis other than pure vertical.

 

[vanhees71]

An ideal polarization filter let's through only em. waves polarized in one direction, i.e., behind the filter you have linearly polarized em. waves (e.g., light). If you put another polarization filter behind this polarized beam but rotated by an angle $\alpha$ with respect to the polarization direction of the incoming e.m. waves, you'll get again linearly polarized light, but now polarized in a direction rotated by the angle $\alpha$. The Intensity behind the second filter then turns out to be
$$I_1=I_0 \cos^2 \alpha,$$
which is known as Malus's Law.

Particularly if $\alpha=\pi/2$ the intensity behind the filter is 0, i.e., all light gets absorbed in this case.

But now think of two polarization filters. Let the incoming beam be horizontally polarized, i.e., defining the 0-degree direction. Then orient the first first filter in $45^{\circ}=\pi/4$ direction, and the 2nd in $90^{\circ}=\pi/2$ direction (all directions relative to the $0^{\circ}$-direction defined by the polarization direction of the incoming beam).

I only tell you that the result is not zero intensity as one might expect in an over-naive-photon picture. This is one of the most simple examples for the sometimes surprising quantum effects, but it's only surprising when wrongly using classical concepts about photons as particles, while in the "wave picture" it's very easy to understand. The finer details about photons however can indeed only be understood by the quite abstract formal description of relativistic quantum field theory.

 

[PeterDonis]

I think that the classical approach does work.

A classical field approach works, as @vanhees71 explains in post #6, as long as you don't try to detect individual photons at the output of the last polarization filter, but just measure the overall light intensity. But this approach does not model light as "a classical wave that is only oscillating in one plane", at least not as @rasp was trying to use that model in the OP.

 

[PeterDonis]

Am I then correct in interpreting PeterDonis comments to mean that a vertically polarized filter acting on unpolarized light emits all axis other than horizontal?

No. A vertically polarized filter acting on unpolarized light emits light that is linearly polarized in the vertical direction, as @vanhees71 said. Where you are going wrong is in thinking that linearly polarized light will only go through a filter oriented in exactly the same direction as the linear polarization. @vanhees71 explained the actual physical law involved (Malus' Law).

 

[rasp]

Pardon me for being dull. Can you try to explain it another way?. I still don’t understand why polarized light in a single direction would pass through with some probability amplitude any second filter that is not orthongal to the first. I understand the math of Malus law, what I don’t understand is how a singularly polarized light has any cosine component. In other words, it seems that the only way it would have a component cosine with it’s prepared axis would be if it the singularly polarized light is not actually singularly polarized but is made up (May I dare say, entangled?) with many components moving in all angles not orthongal to the prepared state and therefore the resulting intensity is just a function of the cosine squared of the angle of the prepared state to the measured state. But that can’t be the case as Peterson is and #vanhees71 explained the ideal polarizer leaves the photon stream in a single direction.

 

[PeterDonis]

what I don’t understand is how a singularly polarized light has any cosine component.

I don't know what you mean by "cosine component". The $\alpha$ in Malus' Law is the angle between the direction in which the light is linearly polarized and the direction in which the second filter is oriented. There is no "cosine component" of the light itself; the cosine is of $\alpha$, which is a property of the relationship between the light and the filter.

(May I dare say, entangled?)

There are no entangled states involved in any of this. Don't confuse yourself further than you already are.

Basically it seems like you want a model that satisfies your intuitions. There might not be one; you might need to retrain your intuitions. Malus' Law has plenty of experimental verification, so that's just the way polarized light works. If you can't understand why it works that way, welcome to science: every scientific explanation bottoms out somewhere in something that nobody can understand why it is the way it is. We just know it is that way because that's what experiments tell us.

 

[vanhees71]

Another very good advice is to first learn classical physics. In this case just look at classical electrodynamics (e.g., the book by Griffiths)!

 

[rasp]

Thank you all gentlemen or ladies. Please let me reiterate what I think I learned, but then again I ask to indulge your patience and to please correct me if I’m still confused. Starting with unpolarized light we pass it through an ideal polarizar of given angle, thereby creating a stream or wave of photons only along said angle. We then pass that stream through another ideal polarizer at angle theta to the first. The second polarizer will emit light with probability according to Maius law. What I still don’t understand is why the second polarizer allows the transmission of light that is not exactly aligned with its axis, according to probability of cosine theta squared. Didn’t the first polarizer restrict all emissions to only those aligned with its axis? Why does the second polarizer not also restrict according to the same mechanism. If I were to guess, maybe the reason that some component of light passes through the second polarizer is because it is a transverse light wave moving with some directions in the theta axis and not a steady stream of photons moving along the narrow and specified original direction. Maybe that portion of the wave that is aligned with the second olarizer passes through? Please comment. I think my problem in understanding is not so deep as relying on the intuition of classical logic vs quantum logic, but is simply not properly understanding how polarizers work.

 

[Nugatory]

What I still don’t understand is why the second polarizer allows the transmission of light that is not exactly aligned with its axis, according to probability of cosine theta squared

Before you take on this question, you may want to take the excellent advice above from @vanhees71 and learn how the classical wave model of electromagnetic radiation leads to Malus's Law without any quantum considerations. Googling for "Malus Law derivation" will get you started.

Starting with unpolarized light we pass it through an ideal polarizar of given angle, thereby creating a stream or wave of photons only along said angle.

You are allowing the (almost irrestibly natural) English-language description to mislead you. We use the words "polarized vertically" to describe the state of the photons that passed the filter just as we would say "held vertically" to describe a stick that we were carrying around.

However, it would be more accurate to say "in that quantum state in which it will pass a vertically oriented polarizer with 100% probability". Call that state $|V\rangle$; now with a fair amount of non-trivial math we can prove the identity $|V\rangle=\frac{\sqrt{2}}{2}(|45\rangle\pm|-135\rangle$ (where $|45\rangle$ is the state is that quantum state that will pass a polarizer at a 45 degree angle with 100% certainty).

Believe the math, not your mental model of a stick that is always unambigously oriented in a particular direction and won't pass sideways through a picket fence unless it is properly lined up with the pickets... and of course the fact that the quantum mechanical prediction for a large number of incdent photons matches the classical prediction above is a powerful sanity check on the math.

 

[rasp]

Thank you.

 

[rasp]

Another very good advice is to first learn classical physics. In this case just look at classical electrodynamics (e.g., the book by Griffiths)!

Thank you.

I thought I could get by without the basics, and truth be told, I don’t know classical electromagnetism, or even the derivation of Maxwell’s equations. I was caught up with a fascination with QM and made the mistake of trying to understand it without having the prerequisite background education.

 

[vanhees71]

There's no chance to understand quantum theory without a good grasp of classical mechanics and electrodynamics and, very important for all of physics but particularly for QT, the necessary level of mathematics, i.e., calculus, vector algebra and analysis. Some Lie-group and -algebra theory is very helpful but not mandatory. You learn about it in QT anyway.