Circular Polarization of Light, Circularly polarized Light

Hi hexa,Mr homm,

I've got less clear since Mr Homm's first post - sorry I can't (yet) interpret l <L l ML l L> l^2 = 1 .. my problem not yours but .. being 'not wrong' is occasionally important. In Mr homm's first post (which I could understand)

Can I reasonably conclude from this that a single photon can have any polarisation from left through eliptical to vertical .. elliptical and so on to left? The mix exists within each individual photon? The impression my textbooks have given me that an equal mix of photons individually of (say) left and right polarisation would produce linear polarisation is actually wrong (possibly as a result of my failing to understand what is written). Confirmation .. please!

-C2.

Many thanks to hexa and Mr homm for their patience.

Hi Mr Homm,

Thanks again.

Do you mean to say:
1. True Right circular polarizer = QWP(1st) + LP + QWP(2nd)
2. True Left circular polarizer = LP + QWP
3. Right circular analyzer = QWP + LP
4. Left circular analyzer = LP +QWP

Yes. But what is stated in each box containing the polarizer is a left and a right circular polarizer.

Please also help to clarify Confused2 question.

Cheers.

Yes. But what is stated in each box containing the polarizer is a left and a right circular polarizer.

You're right, I meant to say ANALYZER there. Also, I made another mistake, due to forgetting how the angles would change sign when I turned the filters around. The correct statement is that a right circular analyzer is a right circular polarizer turned to face the other way (so that the light strikes the QWP), and a left circular polarizer similarly becomes a left circular analyzer. Right stays right and left stays left; they do not switch as I mistakenly said in the previous post.

Unless there was a mistake in packing the box, then what you have should be 1 left and 1 right circular polarizer. However, both of them should be used with the quarter wave plate in front and the linear polarizer in back. Otherwise, they will operate as linear polarizers.

Those are not quite right. Here is the correct list, but first I have to define my angles: In my description of constructing the true right and left circular polarizers two posts ago, I set up a coordinate system so that I could describe the orientations of the optic axes of the quarter wave plates and the linear polarizers. Referring back to that, the true left circular filter had the first QWP axis along the x axis, then the LP axis along the line y=x, and then the second QWP axis along the y axis. The relative orientations of the optic axes is the important thing, so notice that as you stand facing in the +z direction, the x axis is toward your left, the y axis points upward, and the line y=x is at a 45 degree angle upward and toward your left. This means that relative to the direction of the LP axis, the first QWP axis is 45 degrees counterclockwise and the second QWP axis is 45 degrees clockwise. Let's say counterclockwise is the positive rotation direction, as is usual, so QWP(+45) means a QWP with its axis 45 degrees counterclockwise relative to the LP axis, and QWP(-45) means the same thing but clockwise. Then:

True left circular filter = TLCF = QWP(+45) + LP + QWP(-45)

True right circular filter = TRCF QWP(-45) + LP + QWP(+45)

Photographic left circular polarizer = PLCP = LP + QWP(-45)

Photographic right circular polarizer = PRCP = LP + QWP(+45)

Photographic left circular analyzer = PLCA = QWP(+45) + LP

Photographic right circular analyzer = PRCA = QWP(-45) + LP.


To explain a bit further, suppose that the PRCP is oriented with the QWP axis on the y axis and the LP axis on the line y=x. Stand facing along the +z direction and hold the filter by the two points where the LP axis meets the circumference of the filter, and then flip the filter over so that the QWP side is in front. This rotates the filter around the line y=x. The QWP was on the back side, with its axis along the y axis, but when you rotate the filter around the line y=x, the QWP is not on the front side, and its axis now aligns with the x axis. This is 45 degrees clockwise from the line y=x, which is still the axis of the LP. So this filter now has become QWP(-45) + LP, which is a PRCA. Likewise, flipping the PLCP around the optic axis of the LP turns it into a PLCA.

Notice that turning over the true circular filters leaves them unchanged, which is the way a true filter is supposed to work. The first and second QWP's exchange positions and also their angles change sign, which leaves the overall structure of the filter exactly the same as before you flipped it over.

Further explanation about the analyzers: The purpose of a PRCA is to detect right circularly polarized light (which passes through it), and reject left circularly polarized light (which it absorbs). Let's look at how it does this. First, put some unpolarized light through a PRCP. When the light hits the LP it is filtered and only light polarized along the line y=x gets through. Then the QWP aligned at +45 degrees (which is the x axis) delays the x component by 1/4 wavelength to produced right circular polarized light.

Then let this light hit a PRCA. This has its QWP aligned at -45 degrees (which is the y axis), so it delays the y component of the light that hits it. Remember that the previous QWP had delayed the x component. Now the y component has been delayed too, which puts the x and y components back in phase. Since they are in phase, they now reach their maximum values in the x and y directions simultaneously, which means that the wave is now vibrating along the line y=x. Since this is how the LP of the analyzer is oriented, all the light goes through. Therefore, all the light from a PRCP will get through a PRCA.

What if you use a PRCP and a PLCA instead? Just as before, the light comes through the PRCP with an x component that is delayed by 1/4 wavelength relative to its y component. Now when this light hits the PLCA, the QWP of the analyzer is oriented along the x axis, so it delays the x component again. Now the x component has been delayed twice, so it is 1/2 wavelength behind the y component. But 1/2 wavelength of delay is the same as a sign change, so the wave now has its x component reach its maximum positive value just as the y component reaches its maximum negative value, so the wave is vibrating along the line y=-x. Since the LP in the analyzer is oriented along y=x, none of the light gets through it. Therefore, none of the light from a PRCP will get through a PLCA.

Likewise, you can check that the light from a PLCP will get through a PLCA but not a PRCA. One thing you should note about polarizers and analyzers: When light passes through a PRCP or PLCP, it is circularly polarizer afterward. Circularly polarized light is rotationally symmetric, so it DOES NOT MATTER how you rotate the analyzer, it will have no effect. All the light from a PRCP will get through a PRCA regardless of how you rotate it, and none of the light from a PRCP will get through a PLCA, regardless of how you rotate it. The same applies to light from a PLCP: regardless of the rotation angle, all of this light will pass through a PLCA and none of it will get through a PRCA.

To help you check out your equipment, here is what fraction of the original intensity to expect from all the different possible combinations of photographic polarizers:

The cases marked 0 to 1/2 depend on the orientation of the second stage. As you rotate it, the intensity will vary between these values. The other cases are independent of how the second stage is turned.

I am out of time now, so I will respond to confused_2's question in a separate post.

Hope this helps!

--Stuart Anderson

Hi Mr. Homm,

I sincerely must thank you for lifting the veil of confusion surrounding the construction of circular polarizer.
Your latest posting has finally provided me the answer that I have been seeking when I first raised this topic in this forum.
Before I comment on what you have stated, I would wait for your replies to the points raised by Confused2.

I truly cannot THANK YOU enough for sharing so generously your knowledge on how a true circular polarizer ought to be constructed as opposed to what constitute a photographic circular polarizer.
THANK YOU.

Cheers.

This post has been edited by hexa on Aug 5 2006, 10:36 AM

Hi Confused2,

Yes, this is correct according to standard QM. Every photon is always in some specific polarization state, and it can be any variety of circular, elliptical, or linear, with any orientation of the axis of polarization. However, it is not exactly right to think of a single photon as being in a mix of states. It is in ONE state, which may give the appearance of a mixture if you try to resolve it into components.

The state is mathematically a vector. A vector has a perfectly definite direction and magnitude, even prior to applying a coordinate system to it. If I kick a football, the velocity vector will have a definite speed and direction which I can directly see, and there is nothing indefinite or mixed about this. However, if I introduce a coordinate system, I can resolve this vector into components. At that point I can think of the vector as a mixture of those components. However, what if I introduce a different coordinate system, with the x axis not horizontal, for example? In the new coordinate system, the vector will resolve into different components. It is now a different mixture. But I haven't changed anything about the vector itself, I have only changed how I CHOOSE to take it apart. In fact, suppose I use a third coordinate system that has the x component aligned precisely along the actual velocity vector. In this system, I would say that the vector is not a mixture, but is pure x-component.

Therefore, for a vector, it is not correct to say that it is a mixture of components, and it is also not correct to say that it is not a mixture of components. In its own essential nature, it is neither a mixture nor a non-mixture. It is a vector. I can regard it as a mixture or as a non-mixture by choosing an appropriate coordinate system. Therefore a vector is not intrinsically mixed or unmixed, but it ALLOWS ITSELF to be thought of as a mixture or as a non-mixture, depending on your choice of coordinates.

Now a quantum state is exactly a vector, but the components can be complex numbers now, instead of just real numbers. It is a vector in a "complex vector space," which is just like a real vector space except that you can multiply vectors by complex numbers, take complex number linear combinations, and when you form the dot product, you must take the complex conjugate of one of the vectors. All the vector math works just the same as with real vector spaces. Because of this, the comments I made about vectors apply directly to quantum states:

Each photon has a polarization state. This state is perfectly definite, although we might not know what it is until we measure it. Because the state is a vector, we can think of it as a mixture if we choose one coordinate system or as a pure unmixed state if we choose another coordinate system. For example, if we use the linear polarization states |H> and |V> as the basis of our coordinates, then any polarization state can be written as a|H> + b|V>, including the circular polarization states |R> and |L>. As I said before, |R> = 1/√2*(-i|H> + |V>), so in this coordinate system it would be represented by the vector (-i/√2, 1/√2), as these are its components in this coordinate system. It looks like a mix. On the other hand, if I choose my coordinate system to be based on |L> and |R>, then of course |R> = 0*|L> + 1*|R>, so its vector component representation would be (0,1), as these are its components in this coordinate system. It looks like a pure unmixed state.

What is happening here is no more mysterious that what happened to the velocity vector when I chose a coordinate system with one of the coordinate axes lined up with the vector. OF COURSE the vector looks like a pure x-component, because we chose the coordinate system to make this happen. The point is, although you cannot see the quantum state vectors the way you can see velocity vectors, you have, just as with any vector, the freedom to choose any coordinate system you like, so you can regard the state as pure or mixed.

Now about what your textbooks say: People writing texts on classical physics often do not bother with the quantum explanations, or don't take enough care to make them clear. In a beam of polarized light, there are many photons, and each one of them is in the polarization state of the beam. If the beam is linear at a 20 degree angle, for example, then each photon in the beam is linear at at 20 degree angle. If the beam is right circularly polarized, so is each individual photon in it. It is definitely NOT true that an equal mixture of right and left circularly polarized photons produces a linearly polarized beam. Each photon individually is in a linear polarization state, which can be thought of as a combination of right and left circular polarization (if you choose to use that coordinate basis), or as a pure linear polarization (if you use that basis).

An easy way to see that this must be so is to think about each separate photon. All the left circularly polarized photons have a 1/2 chance of getting through a linear polarizer, regardless of its angle of orientation. The same is true for the right circularly polarized photons, so overall, a beam like this will have exactly 1/2 of the light getting through a linear polarizer at any angle, which is the definition of an unpolarized beam. This is clearly not the way a linearly polarized beam behaves.

In QM this is a dangerous argument to make, because this is the false argument that "proves" two slit interference cannot happen when one photon at a time goes through the apparatus. However, in this case, you have two separate populations of photons, and there is nothing tying their phases together, so the phase of the right circular photons will be randomly different from that of the left circular photons. This causes any interference effects to average away, and so you really do see always 1/2 of the photons getting through the linear polarizer.

The brief summary is:

Each photon is in a definite polarization state at all times.
This state may be thought of as pure or mixed, depending on your choice of basis states.
Polarized light is a beam of photons which all have the exact same polarization state, never a mixture of photons in different states.

Now about the notation you had trouble interpreting:

Something like this: |A> is a state vector. Think of it as a fancy notation for what in engineering you would write as an "A" with a vector arrow over the top.

Something like this: <A| is a state covector. It is the transpose complex conjugate of the vector |A>. You use it when you take the dot product. You probably know that one way to do a dot product is to take two vectors written as columns, turn the left one sideways (transpose it) to form a row vector, and them multiply them using the same rule you use for multiplying matrices. This is the same thing, only you also take the complex conjugate of every entry in the transposed vector, because that is a requirement for complex vector spaces. It makes the dot product of a vector with itself always come out positive and real, as it should, since this represents the square of the length of the vector, so it should always be positive and real.

Something like this: <A|B> is the dot product of A and B. It is just the transpose conjugate of A multiplied onto B by the matrix multiplication method.

Something like this: M|A> is the matrix M multiplying the vector A, just like in linear algebra.

Something like this: <A|M|B> is the conjugate of A as a row vector, then matrix M, then B as a column vector, all multiplied by the matrix multiplication method. The value of this is a complex number.

Hope this helps!

--Stuart Anderson

mr homm,hexa,
Working (slowly and painfully) backwards through 'section 2' (ongoing) I think my question was already answered. Very kind of you to give further time to ensure that I was not 'left out'.
Exceptional people .. both of you. Exemplary is a word that springs to mind.
Many thanks,
C2 (John).

Hi Mr Homm,

You had provided an explanation on how a true circular polarizer ought to be constructed as opposed to a photographic circular polarizer.
You had also provided a lucid explanation on the computation of probabilities using the state vector that define the physical state of a group of photons. You had done it with such vivid details that would put many authors on this topic to shame.

With the benefit of the explanation that you have provided, I will like to find out more on why the amplitude of the state vector is expressed as a cosine function rather than a sine function of the angle between the polarization axes of two linear polarizers to determine the probabilities of light passing through any two linear polarizers (i.e the basis of Malus Law)?

I would appreciate if you could share your insight on this topic.

Cheers.

This post has been edited by hexa on Aug 7 2006, 12:40 AM

@hexa

Glad to be able to finally answer your original question! There had been so much discussion already, and so many statements that were partially correct and partially wrong, that it took several posts to clear it all out and finally get down to the bedrock question. This kind of thing is very challenging to straighten out, I might add; it's easier when some people are clearly right and others are clearly wrong, but sorting out partial truths requires great care.

You're quite welcome. When I first started reading this thread, it struck me than the originally quite helpful and reasonable discussion had got seriously off track, and I decided to have a go at making the discussion more productive again. Basically, I can't resist trying to "fix" stuff that seems to be not quite right.

There is a classical way to look at it and a QM way to look at it. The classical way is to say that the E field is a vector and that the component of E aligned with the filter direction is passed through the filter and the component of E aligned perpendicular to the filter direction is absorbed by the filter. Therefore, the light exiting from the first filter consists entirely of the part that got through (of course!), and is therefore aligned with the direction of the first filter. Since the second filter is turned an angle θ relative to the first filter, the E field must be resolved into a new set of components parallel and perpendicular to the new filter direction. Since the E field vector makes an angle θ with the axis of the new filter, the component of E along the new axis is Ecosθ and the perpendicular component is Esinθ. Since only the parallel component gets through the second filter, the E field exiting the second filter is just Ecosθ.

Since the classical theory agrees with experiment (if it did not, it would have been abandoned long before QM ever began), QM must make the same prediction as classical theory. Otherwise QM would be wrong, and we would already have abandoned it, too. The question is, how does QM make this prediction? First, we know that the first filter selects only one polarization state to pass through. Let's assume the first filter axis is vertical, so the state |V> passes through and |H> is absorbed. Now every photon that hits the filter either is passed through or absorbed, so if you look at the probabilities of passing through or being absorbed, they add up to 1.0.

Now remember that in QM the photon is always in some particular polarization state originally, let's call it |O>, and if you perform a measurement on it, such as putting it through a filter, the photon is forced to jump into one of the eigenstates of the measurement operator. What are these eigenstates in the case of a linear polarizer? It is easy: LP|V> = 1|V> and LP|H> = 0|H>, so these are the eigenstates, and the eigenvalues are 1 and 0. Another basic postulate of QM is that after a measurement, probability of finding the photon in a particular eigenstate, such as |V> or |H>, is the square of the absolute value of the component of |O> projected onto that eigenstate. How do you calculate the projection? The same way as for ordinary vectors: you take the dot product of |O> with a unit vector along the desired direction. However, the eigenvectors are normalized, so they are already unit vectors. Therefore, the probability of finding the photon in state |V> after the measurement is |(<V|O>)^2|, and the probability for |H> is |(<H|O>)^2|.

Notice that since you already know that the probabilities add up to 1.0, and that |O> is normalized, so that <O|O>^2 = 1.0, this means that |(<V|O>)^2| + |(<H|O>)^2| = <O|O>^2. But this means, by the Pythagorean theorem, that |H> and |V> are perpendicular vectors in the space of states. (The Pythagorean theorem, being a part of pure mathematics, is independent of physics and is known to be true in abstract vector spaces of ANY dimension, so it certainly works here.) Technically, I didn't have to prove this here, because it is a theorem of Hilbert space theory (which is the mathematical theory behind the state spaces used in QM) that eigenvectors that have different eigenvalues are ALWAYS perpendicular. However, I wanted to show a simple way to see that it is true for the polarization states.

Therefore, when an unpolarized beam of light hits the first polarizer, since the beam consists of a random collection of different photons in different polarization states, on average exactly half of the photons get through. Some states will be closer to |V> and their probabilities of getting through the filter will be >1/2, and others closer to |H> and their probabilities will be <1/2, but it all averages out.

When light from the first filter strikes the second filter, all the photons are in the state |V>. The second filter has its own new polarization axis, turned at an angle θ relative to the first filter's axis. Let's say that the state that gets through the second filter is |V'> and the state that gets absorbed is |H'>. Just as with the first filter, these states are perpendicular to each other. Therefore, you can use them as a new basis for coordinates. What does the state |V> look like in this new basis? It must be some linear combination a|V'> + b|H'>, but what are a and b? From probability, you know that |a|^2 + |b|^2 = 1, and you know that when the second filter is aligned with the first filter, |V'> = |V>, so |a|^2 = 1.0 in that case, and |b|^2 = 0.0.

It is clear that if you turn the second filter by + or - 90 degrees, then |H'> = |V> and |V'> = |H>, possibly with some phase change due to the delay of traveling to the second filter. In that case, a and b clearly depend on θ, and so you can say that |a(0)| = 1.0, |b(0)| = 0.0, and |a(90)| = 0.0 and |b(90)| = 1.0. Also, notice that if you turn the second filter 360 degrees, everything returns to its original configuration, so a and b are periodic with period 360 degrees. This is enough information to pin down exactly what functions they are: a(θ) = cos(θ) and b(θ) = sin(θ). You can tell which one is cosine because |a(0)| = 1.0.

Therefore, when the second filter is turned at an angle θ, you get |V> = cos(θ)|V'> + sin(θ)|H'> (again, possibly multiplied by some phase factors, which have a magnitude of 1.0, so they don't affect the probability calculation). Since the second filter only passes through the component along |V'>, the probability of this photon getting through the second filter is |(<V'|V>)^2| = (cos(θ))^2. This is Malus' Law.

That's all for now, as I must go to bed. It's getting late here!

Hope this helps!

--Stuart Anderson

Hi hexa & Confused2,

I'm posting again to add something that I feel I neglected in my previous post. You might have been wondering why I took such a long way to show that the |V'> and |H'> states were expressed as combinations of |V> and |H> with coefficients cosθ and sinθ. Wouldn't it be simpler to just say that since the second filter is rotated by an angle θ, then the state vectors are rotated by θ also? Well there's a problem with that seemingly simple explanation, which is why I avoided it, but I forgot to point out what the problem was, which probably left the impression that I was doing everything the hard way for no reason.

The reason is that the vector space of states is an abstract space that describes the polarization of photons, not a geometric space that describes positions that are measurable with ruler and protractor. The fact that the polarization state space is 2 dimensional and that the space of directions perpendicular to the light path is 2 dimensional is just a coincidence. These spaces are unrelated. Other particles besides photons may have a different number of possible polarizations, so their polarization state space may have more than 2 dimensions. So you can't directly take angles from the geometric space and transfer them to the state space, because these spaces are completely separate from each other.

Now it does turn out that the angle is the same, but because you can't just assume that it must be so, you have to actually go to the trouble of proving it. The method I used was maybe a little awkward, so here is a somewhat better one that I thought of today:

You know that the state space is 2 dimensional, and than |H> and |V> are perpendicular and that |H'> and |V'> are also perpendicular. Therefore one of these is just a rotated version of the other, just as it any 2 dimensional vector space. You can imagine just like having a level coordinate system and a tilted coordinate system, such as are used in first year physics problems where blocks slide down inclines. One system is rotated relative to the other. Notice that this fact does NOT depend on there being any connection between the state space and the geometric space of x & y coordinates, because the reasoning here involves only known facts about the state space itself.

Since one set is a rotated version of the other set, it must be true that there is a matrix that does the rotation. If the state vectors are rotated through an angle θ' in the state space, then the matrix must be

Because the state vector space is complex, it is also possible that the new basis states have their phase changed. This would involve an extra matrix besides R multiplying them, but this would be a diagonal matrix with entries that looked like exp(i*phi), where phi is the phase constant. Since |exp(i*phi)| = 1, this phase matrix would not change any of the magnitudes of the components, and so the probabilities (which only depend on the squares of magnitudes of components) would not be affected. Therefore, it is safe to ignore this extra phase matrix for this problem. (As I said in an earlier post, absolute phase cannot be detected, only relative phase, and then only via interference, so it does not matter here.)

Now the only question that remains is, what is this angle θ'? It is easy to prove that it must be exactly the same as θ. Here is how. Suppose they were different, and that rotating the physical LP filter by θ caused the state vectors to rotate by θ'. If I rotate the filter by an infinitesimal amount dθ, then the state vector must rotate an infinitesimal amount dθ'. How do I know that it is infinitesimal? Because if the second filter is rotated infinitesimally, almost all the light must still get through, since it is still in virtually perfect alignment with the first filter. This means that |V'> must be almost exactly the same as |V>, because the projection of |V> onto |V'> is nearly 100%.

The R matrices have the nice property that R(θ1)*R(θ2) = R(θ1+θ2). This is obvious if you think about it, because R rotates the vectors, so rotating them and then rotating them some more, is the same as rotating them by the total angle all at once. Because of this property, you can think R(θ) as the product of N identical copies of R(dθ), rotating in tiny steps. But then R(θ') would be the product of N copies of R(dθ'). This means that θ = Ndθ and θ' = Ndθ', which means that θ'/θ = dθ'/dθ. Since the right hand side is a constant, it follows that θ' is proportional to θ. Therefore, all you need to find is the constant of proportionality.

This constant must be exactly 1. If it were not exactly 1, then let's say it is C. In that case θ' - θ = (C-1)θ. This means that if you rotate the LP filter by 360 degrees, the state vector will rotate by a different amount. In that case, it is either getting ahead or behind, so just keep rotating the LP filter 360 degrees at a time, and eventually θ'-θ will become very close to 90 degrees. But then |V> = |H'>, because |H'> is 90 degrees away from |V'>, so no light will get through the second filter. This is absurd, since you've been rotating the LP filter 360 degrees at a time, so it must be exactly in its original position, so you KNOW that all the light gets through it, since it was originally lined up with the first filter. Therefore it is impossible for θ' to be even microscopically different from θ. They are exactly the same.

Of course this means that R(θ') = R(θ), and so |V> = cos(θ)|V'> + sin(θ)|H'>. Therefore, as in my last post, the component along the |V'> direction has a magnitude of cos(θ), which means that the probability of a photon getting through the second filter is (cos(θ))^2, which is Malus' Law again.

This proof was somewhat more technical, but I hope it makes it clearer what is necessary to analyze this experiment.

Hope this helps!

--Stuart Anderson

Hi Mr Homm,

Thanks again.
You have again shown your finesse in providing the proof for Malus Law using Quantum Mechanics.
Notwithstanding the proof, I am uncomfortable that we cannot describe Malus Law from the geometrical space but must resort to the abstract Hilbert space which we cannot relate directly to physical reality.

Do you think it is possible to provide an account for this phenomenon that we can relate directly to physical reality?

Cheers.

Hi Mr Homm & Hexa,

I am enthralled by the depth of the discussion. I must thank Mr Homm for providing the answer to the same question that I too was seeking.

Your statement 1 to 4 are essentially paradoxical to one another.
Statement 5 requires the presence of an observer before we can know what state vector the photons are in prior to detection?

I hope Mr Homm can further elaborate on these paradox.

Yours,

Maltida

Hi Maltida,

This is not an 'explanation' .. hopefully my doubts will clarify yours.

The reference below is interesting. Although it only mentions photons in passing it does offer evidence of the way the electron spin behaves (or misbehaves.. depending on your POV).

http://www.upscale.utoronto.ca/GeneralInte...ernGerlach.html .

My interpretation at this point is that electron spin is always resolved (at the point of detection) into 'Head up/Tail down' OR 'Tail up/Head down' whereas photon spin is always resolved (at the point of detection) into 'Head first' or 'Tail first' . For a stream of identical particles in both cases there is a vector that can take any angle(s) but that angle can only be deduced from the ratio of heads and tails observed after many particles have been detected. The angle between the vector and the detector determines the mixture that will be detected but the vector is not in itself a mixture.

Considering the case of photons..
If we allowed the length of the vector to become anything other than 1 then we would not be representing the situation correctly .. as I understand it the 'representation' is to allow that the vector representing spin is a complex number with (always) a |magnitude| of 1 (**). It is the angle of this vector that determines the ratio of 'heads first' to 'tails first' at the point of detection.

As the vector approaches 90 degrees ( 0 + i ) to the direction of motion the vector (surprisingly?) acquires some of the same 'Heads up' or 'Tails up' property as was illustrated in the case of the electron. Hence (?) the behaviour of linearly polarised light is related to the angle between (say) a linear filter and this vector.

I hope this is at least 'useful'.

Comments, improvements (and criticism!) most welcome.

-C2.

(**) .. probably no real need to consider it as complex but it looked 'elegant' at the time .. best not to edit too much.

This post has been edited by Confused2 on Aug 12 2006, 01:16 PM

Hi everyone! I've been very busy for the last few days, but now I'm back for a while.

@ hexa:

Honestly, I don't think so. However, I have not investigated this very hard, because it wasn't a question that really bothered me. Perhaps it is possible, but here is why I think it is not:

First, it is just a coincidence that the state space and of the space perpendicular to the propagation direction are both two dimensional. If instead of the photon, we had some other particle with a different amount of spin, there might be more dimensions to the state space. In that case it would be obviously impossible to directly connect the two spaces in such a way that the space of geometric position was used instead of the state space.

Second, state space isn't necessarily more unnatural that geometric space, only less directly experienced. Our eyes don't perceive polarization, but bees and many fish do perceive it. If we had grown up with fish or bee eyes, we would have direct experience of polarization and how it behaves, and we would have a mental image of the state space that would be just as natural as the mental image we have of geometric space. Notice that this is really more a philosophical explanation than a physical one; that's because this is really a philosophical question. Since classical physics and QM both make the same predictions about beams of light, and both agree with experiment, there is no basis in physics for preferring one explanation over the other, at least based on experiments with strong beams of light. When two explanations make equivalent predictions, the only reasons for preferring one over the other are philosophical (Occam's razor) or emotional (familiarity and comfort). Therefore, I'm answering on the basis of philosophy right now.

Third, when you do look at experiments with very weak beams of light, so that there is only a single photon present at a time, then classical physics makes wrong predictions, but QM makes correct ones (although they are statistical in nature, they do not disagree with experimental results, and when the results from many photons are combined, QM results reproduce the predictions of classical physics). It is hard to see how to apply the geometric space to individual photons, since they appear to be point-like when they are detected, so they have no spatial extent. Therefore it is hard to associate a direction with them. On the other hand, the state space is designed to describe the polarization of individual photons, so it is always clear how to apply it to them.

BELOW IS A LONG RANT FULL OF MY PERSONAL OPINIONS:

Fourth, I tend to think about things from the opposite direction. I think the state space IS the physical reality, and the geometric space we perceive is a side-effect of the state space. Now this is very speculative, and it is just my opinion, but here it is: Really, all that exists is particles in quantum states. These states include many things, such as polarization, energy, etc. Among these things are position and momentum. A point in space is just an eigenstate of position, and a precise mass and velocity is just an eigenstate of momentum. These are related by the uncertainty relation, but for large objects, we don't notice it, because the uncertainty is too small for us to detect, and so we think that position and velocity can both be perfectly known.

In fact, I think that the uncertainty principle is really a bad way of thinking about the situation, because it leads to questions about how nature enforces the uncertainty and the possibility of hidden variables. The real situation (in my opinion) is that what we expect too much. The universe contains only half as much information as we think it does, and the reason that we think it contains more information is because we are spoiled by growing up in the luxury of a large scale, high energy part of the universe. With large objects, there are many many copies of the position and momentum information, so when we measure position, we destroy one copy of the momentum information, but there are lots more copies, so we don't notice. Similarly, when we measure momentum, we destroy one copy of the position information and don't notice.

It's like loading music into an mp3 player that has huge amounts of room and many copies of every song in it. If you download another song, it will overwrite something else, but you'll never notice because there are billions of backup copies. So you go along happily downloading songs whenever you want, and never notice a problem. But what if you had a truly tiny mp3 player that only had room for one song? Now when you download a new song, the previous song is gone, and you definitely notice. If you were used to having huge numbers of backup copies, you would probably thing something was wrong and that there was a song uncertainty principle. But from the point of view of information storage, nothing is wrong, we just have inflated expectations of how much information the player can store, because that's what we're used to. We are so used to having huge amounts of disposable information that we don't notice that we're destroying a little bit each time we measure something. However, with a single particle, there is NO backup copy of the information, so I will definitely notice if I destroy it.

So, in my opinion, there is no hidden variable theory because there is no uncertainty principle. There is only our inflated expectation of what we can ask of a particle. Because we can ask two questions (position and momentum) of an object with a trillion atoms in it, it doesn't follow that we can ask two questions of a single particle. A trillion atoms has plenty of storage space for lots of information, but a single particle can only store one piece of information. There is no missing information, we are just used to being able to "have our cake and eat it too," and then we are surprised when this is not possible with a single particle.

Another related idea is that the position and momentum states of a particle are Fourier transforms of each other. Mathematically, this means that the same information is contained in either description, since once can be turned into the other. Therefore, this is another way to see that there is only ONE piece of information (the state) which can be represented in two forms (position or momentum). If they are equivalent, why does position seem so much more basic to our perceptions? I think the answer is evolution. We have evolved in an environment where physical closeness correlates with relevance to our survival. Close prey can be caught, close plants eaten, close predators can get us, rather than distant ones. On the other hand, imagine a hypothetical type of creature that lives in huge interstellar gas clouds, and that the creatures are so gaseous that they pass through each other. The only way they can interact is to have nearly the same velocity so that their atoms are together long enough to do something. In that case, similar velocity would be the crucial thing correlated with survival, and not position. Such creatures would need to perceive the world in terms of velocity in order to survive, and so they would probably "see" velocity and "feel" distance, just as we "see" distance and "feel" velocity. For them, the momentum representation of a particle would seem to be the basic one, and the geometric position would seem like a strange abstraction.

Well, that's enough of my opinions for now. Later I'll tackle maltida and confused2's posts.

--Stuart Anderson

Hi Confused2,

Thanks for your response.
The link to Stern Gerlach Experiment is great.
However, it still does not address the question which I have raised to Mr Homm.
I am concern whether it is possible to describe the amplitude of a state vector as an entity in some geometrical space rather than in some abstract imaginery mathematical space. Afterall, physical reality must eventually be described in the 3D geometrical space.

I hope Mr Homm could address my question in greater depth.
Thanks.

Yours,

Maltida.

Hi Mr Homm,

Thanks for providing your comments to the question that I have raised.

Honestly, I don't think so. However, I have not investigated this very hard, because it wasn't a question that really bothered me. Perhaps it is possible, but here is why I think it is not:

Perhaps I should first raise the topic on kinetic theory of gas.
The temperature in the equation PV=RT is a macroscopic observation. The physical reality behind the phenomenon of temperature or pressure of a gas has a microscopic and Newtonian explanation using the kinetic theory. The description that each gas molecule has a distinct kinetic energy moving at a certain distinct velovity at a specific time provide the physical reality behind the phenomenon of temperature and pressure.

With the kinetic theory as a prelude, this led me to ask whether linear or circular polarized state of light has a more fundamental explanation based on the polarization axis of a photon and the angle θ that the axis of each photon makes with respect to a reference axis (arbitrary) define by the molecular axis of the polarizer.

The amplitude of a group of photon that define the base state is essentially a statistical representation of a group of photons based on its orientation with reference to a reference axis define by the polarizer.

Appreciate your comment on the above.

Cheers.