# How do polarisers let a photon through

1. Sep 3, 2014

### Johan0001

So when it encounters a linear polariser, and is transmitted through , what actually happens to the photon.

Has it stopped vibrating in the plane perpendicular to the polariser, when it emerges?
This would be my obvious conclusion. Since putting a second perpendicular polariser after the first reveals no photons ever emerging.

Consider my analogy for simplicity.

If I hold a rope on one end, and tie the other end to a pole in the distance. Between myself and the pole the rope runs through a gate( having vertical beams evenly spaced) placed between myself and the pole.

1.If I induce a vertical vibration (Y- pulse) on the rope . Then the pulse goes through the gate without any , disturbance. The energy of the pulse is totally transmitted through to the pole.

2.If I induce a horizontal vibration ( X- pulse) on the rope. The energy of the pulse is totally absorbed by the gate. No pulse energy is transmitted to the pole.

3. If I induce a diagonal vibration 45% to the horizontal and vertical. From simple vector analysis , I can calculate that a certain % of the energy will be transmitted to the pole.
We would find a substantially smaller pulse being transmitted to the pole, and the rest being absorbed/reflected at the gate.

All this can be visualized in a classical , mathematical way. No problem there.

But now our original pulsating /vibrating ( in x-y plane) photon is not quite like this.
If it passes the polariser , why does it not lose most of its energy( or does it)?
If it is now only vibrating in 1 plane then it should be in the order of 1/360 its original energy?
Since it was vibrating in 360 degrees (unpolarised), its now down to one degree of freedom.

My understanding of the photon passing through a polariser must be incomplete.

OR

Is the vector superposition of states not then incorrect in describing the polarization of a photon?

2. Sep 3, 2014

### glance

One must be careful when talking of "photons".
When you consider a beam of light (i.e. lot of photons) crossing a polarizer, you can talk of a given percentage of it being absorbed/transmitted, and its intensity decreasing accordingly.
When you go down in intensity (i.e. in number of photons per unit time in the beam) so much that you start dealing with single photons things start getting tricky (i.e. quantum). Single photons are not partially absorbed or partially transmitted. They are absorbed or transmitted. It's a yes/no event. This is compatible with what you observe with a beam of light because you will have a certain probability of a photon with a certain polarization state being absorbed by a polarizer placed in a certain way, such that when you consider lot of photons you can safely talk of a "percentage of the beam being absorbed".

Another thing: when one says that the electromagnetic field carried by a beam of light "oscillates" in a certain direction, it does not mean that the photons composing the beam are actually "vibrating" in the classical sense. What oscillate are the electromagnetic fields, not some physical entity actually vibrating in space.

3. Sep 3, 2014

### Johan0001

Hi Glance

I am in agreement with your comments , I don't see any contradiction from what I have put forward?
Except
Surely a photon is exactly this , a small packet of oscillating electro - magnetic fields ?
Are you saying that an electromagnetic field can exist without photons, are they not one and the same?

4. Sep 3, 2014

### f95toli

No, photons are the quanta of the electromagnetic field; but that does not mean that they are "packets" of EM field. Even the concept of a single photon is a bit tricky since it implies that the field has specific number photons, i.e. it is in a so-called number state (in this case the number is one) and this field is extremely "non-classical".
I don't think there is a intuitive way of explaining this, it gets very complicated very quickly and all we can say that that math works extremely well.

5. Sep 3, 2014

### DrChinese

A photon does not gain or lose energy going through a polarizer. It will have the same frequency/momentum as before.

There is of course a chance that the photon will not pass through the polarizer. This chance is related to its original polarization, and is described by Malus' Law.

Although there are certain aspects of photon behavior that are susceptible to classical analysis, others defy such. As mentioned by f95toli, when the photon number is known (n=1 for example), you have what is usually called a Fock state. Such states have some properties that are complicated.

6. Sep 3, 2014

### Johan0001

May I simplify my questions, to establish a baseline for the understanding of a single photon,
when talking of a single photon passing through a polariser.

Does a electromagnetic field contain an ensemble of photons , yes or no.

what are the properties of the quanta. i.e. Do they involve electric and magnetic fields, however small?

7. Sep 3, 2014

### glance

As f95toli pointed out, tricky is the word here.
When I say that "there is no some physical entity vibrating in space" I mean that are not the "photons composing the beam" that somehow go right and then left and then right again and so on. What "oscillates" is an internal property of the photons.

I think a more well posed question would be "can the electromagnetic field be described without photons?", or even better "can the electromagnetic interactions be described without photons?".
The answer is: mostly, yes. As long as you are not looking close enough, which is in the vast majority of circumstaces, the classical description of light as an electromagnetic field is perfectly fine. A very important point here is that in the classical description of light (i.e. in classical electrodynamics) there is no such thing as a photon.

The concept of a photon becomes necessary when you start dealing with beams of very low intensity (when the electromagnetic field starts to show its discretized nature), or of very high intensity (e.g. effects of non linearity of the electromagnetic field, see photon-photon scattering) or in general in other very particular, and generally hard to achieve, circumstances.
Indeed, the concept of photon is so tricky that some even denied the usefullness of such a concept, see the very recent article of Willis Lamb (yes, the one of the shift): Antiphoton.

Of course, this does not mean that the concept of photon is not self-consistent, or incompatible with the classical limit (i.e. with the classical and "simpler" concept of electromagnetic fields). It just means that there is way more than what one would expect from its classical counterpart. For example, single photons carry orbital and spin angular momentum, with the latter being associated with the polarization state, and the former... let's just say trickier.
Even better: a photon with a definite frequency (hence with a definite energy) is completely delocalized, which means that it is really hard to think of it as an actual "particle"... at least as long as you don't detect it: then you will see it as a single, localized entity, appreciating its particle nature.

It depends. If the beam (or better saying, the quantum state under consideration) has a definite number of photons (i.e. you measure the intensity of the beam) it does not have a definite electromagnetic field. If the beam has a definite electromagnetic field (or in other words, the electromagnetic field as a specific phase) there is not a definite number of photons in it. This is an example of an uncertainty principle, just like the more commonly known position-momentum uncertainty principle.

Last edited: Sep 3, 2014
8. Sep 3, 2014

### Johan0001

If I use my description of the unpolarised photon

Then how can it maintain its energy after passing the polariser , if it now only oscillates in 1 plane , it has lost ""something" after being polarized , at least one degree of freedom ?

And light by definition is an electromagnetic wave that has at least 2 degrees of freedom
( x-y) for induced electro magnetic oscillations.

Somewhere the dots just don't connect.

9. Sep 3, 2014

### f95toli

Don't, because -as has been pointed out above- your description is not correct.

The energy of a photon only depends on the frequency,not its polarization. Moreover, photons do not "oscillate" in any plane.

In classical physics yes, but photons require QM; there is no such thing as a photon in classical physics and you can't mix the two pictures.

[/quote]

Because you are trying to use classical concepts for something that is very much non-classical.

10. Sep 3, 2014

### DrChinese

The description of light as an EM wave works sometimes. Sometimes a particle picture is a better description. I am fairly sure you are aware of the Heisenberg Uncertainty Principle (HUP) which more or less relates the two pictures. Quantum objects do not* have well-defined properties (such as spin/polarization) at all times. When a measurement collapses a photon into a spin eigenstate (a polarizer does that), its spin on its complementary basis becomes completely uncertain.

*Although this is somewhat interpretation dependent.

11. Sep 3, 2014

### atyy

There are many types of photons, so usually one has to be more specific and say what type of photon one is talking about. For example, there are photons that are not localized in space and which you cannot think of as a particle travelling at the speed of light, but there are other photons which you can roughly think of as a particle travelling at the speed of light.

If a certain type of single photon is unpolarized, one way you can think of it is that was polarized, but you did not know its polarization. Alternatively, you can think that each photon is polarized, but different photons have different polarizations. Here the concept is a "mixture" of polarizations.

Even if a photon has a definite polarization (45°), it can come out of a vertically oriented (90°) polarizer with a different polarization (90°). This is because a photon polarized at 45° is a superposition of a horizontally polarized photon and a vertically polarized photon. Since it is a superposition, half the time it will not pass through the vertical polarizer, while the other half of the times it will pass through the vertical polarizer and emerge with vertical polarization. This concept of "superposition" is not the same concept as the mixture above, since we are talking about a single photon with a definite polarization. The mixture means we are talking about many photons, each with different definite polarizations.

Last edited: Sep 3, 2014
12. Sep 3, 2014

### Johan0001

Lots of complicated descriptions of a photon , let me focus on those that sound uneccesarrily complicated , probabily for reasons I don't understand.

When an electron orbiting an atom , drops from a higher orbital to a lower , there is an exact amount of energy released from the atom in the form of a "Photon". This discrete amount of energy is the energy of the ""one"" photon. So if my laser consists of ,one atom's orbiting electron to fall from energy level 2 to its lowest energy level 1 , we get a photon with the minimum amount of energy that could have been emitted. So we have one photon traveling toward a polariser with energy hƒ.
If we had many atoms in the laser producing a beam then we divide the total energy by hƒ and we get the number of photons.(Assuming all the source atoms were the same element or atomic number), which should be the case in a laser.

But my specific question is in the case of one photon , I presume the energy of one photon is spread over some spacial distribution due to (HUP). But no matter.

What I am trying to establish is what actually changes in the "quanta" we call a photon , when it passes through a linear polariser.

Also if this energy packet has a electric and magnetic field oscilating in the perpendicular planes to its direction of motion, as described by wiki and many other references.

Because if a photon does have these orthogonal fields which induce each other , the polarization explained using sinusoidal waves make no sense to me. I was hoping my analogy, of the rope and gate , would simplify my question. But it seems like we can only describe a photon using QM and classical interpretations are incorrect.

If this is true why are there so many references using these sinusoidal fields, to describe the nature of light or photons.

Either a photon was polarized or not, why think of a polarized photon as unpolarised? Surely there is no merit in this line of thought , or is there a deeper reason?

13. Sep 3, 2014

### DrChinese

1. The quantum notion of spin (polarization) does not quite match up to classical notions of rotating components, as I am sure you are realizing. The more you try to marry up these views, the more difficult it is. Technically, the quantum view is always correct. But many times the classical view can give you an equivalent answer with greater ease.

2. I consider a photon to have the following polarization attributes.

a) Definite polarization, which can be known or unknown. A photon that has passed a polarizer is said to be of known polarization, whereas light from an ordinary light bulb is said to be of unknown polarization;

- or -

b) Indefinite polarization (a superposition of states). Polarization entangled photons are an example of this.

And you could technically say that a photon could be in a state in between a) and b) as well, although that case is usually ignored.

14. Sep 3, 2014

### atyy

Well, you were the one who mentioned unpolarized photons. I was just saying one way to make sense of that is to say that each photon is polarized, but you have a mixture of photons with different polarizations.

15. Sep 3, 2014

### atyy

One way to think about it is that its polarization changes. A photon polarized at 45° is a vector at 45°, which is a superposition of a vector at 0° and a vector at 90°. So if the polarizer is oriented at 90°, the photon will pass through with 50% chance, and after it has passed through it will have 90° polarization.

16. Sep 3, 2014

### vanhees71

Photons are among the most complicated notions of physics we know. It is impossible to give an adequate definition without referring to relativistic quantum field theory. The reason is that photons are described by massless quantized spin-1 fields, and any attempt to explain such beasts in terms of semiclassical particle analogues are flawed.

In most cases it's better to use a semiclassical description, where you treat the electromagnetic field as a classical field and the matter with quantum mechanics.

A laser never directly produces single-photon (Fock) states but something what's called a "coherent" state. If the intensity is not too low, this states are almost like classical wave fields. At low intensities you have large quantum fluctuations, but never a pure single-photon state. A coherent state consists of a superposition of photon states of any number. At low intensity it's dominated by the vacuum state ("no photons present") and the single-photon state, but there is also a non-zero probability to register more than one photon.

It's not so trivial to create single-photon states. Nowadays this is usually done using a laser, irradiating a birefringent chrystal, which leads to "parametric downconversion", in which process an entangled photon pair is created, i.e., a pure two-photon Fock state. Due to the properties of such entangled states, you can detect one of the two photons, and then be sure that you are left with a pure single-photon Fock state (socalled "heralded photons").

17. Sep 3, 2014

### johana

18. Sep 3, 2014

### atyy

As others have mentioned above, these classical sinusoidal fields do have their quantum counterpart, and are called coherent states with a large average number of photons. The funny thing about them is that although they are made of photons, they do not have a definite number of photons - they are superpositions of states with different numbers of photon. So every time you measure the number of photons in a coherent state, you will get a different answer, but the average answer will be well defined.

Anyway, since this wave is essentially classical you can think of it as a classical wave with orthogonal fields. In this wave the electric fields and magentic fields are orthogonal to each other. When people talk about polarization, they are, by convention, referring to the just the electric component. So if you just concentrate on the electric component, you will see that it is sinusoidal and polarized along one direction.

19. Sep 3, 2014

### Johan0001

thank you Johana , this is what I was using as my analogy , however it does seem like there is much more to a photon and its polarization.
Will need to take more time to have a better understanding.

20. Sep 4, 2014

### Johan0001

"How do polarisers let a photon through"

Well from what I have gathered , the answer is simply , we don't know.

What QM can tell us by gathering data from a repetitive measurements, is the Probability of making it through.
Once again the math works very well, but math and probability cannot determine all of reality we experience.
Its just a wave function we have carefully selected , so as not to bother with the "how it happened" because we still don't know.
We have moved passed this by saying "No matter how it happened" just accept that there is a probability that it will happen from our many observations we have made.

This wave function then becomes the subject of reality , because now we lose focus of the object it supposed to represent, in fact some now no longer believe the object is there until we collapse this wave function.

There needs to be a balance , to have agreement.