What is ¨unpolarized¨ light exactly?

• B
Gold Member
I learned that light can be linearly and circularly polarized. It can also be a mix of several simultaneously. Is there also a polarization state that is neither linear nor circular? (ie not polarized?) If so, what is it?

A. Neumaier
Yes, there is, for example ordinary thermal light. It is classically described by the stochastic (=fluctuating) Maxwell equations, and quantum mechanically by certain mixed states. See, e.g., my slides ''Classical models for quantum light''.

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Khashishi
Unpolarized light is a balanced mixture of polarized light. It could be a mixture of vertical and horizontally polarized light or a mixture of left circularly and right circularly polarized light or something else-- it doesn't matter. There's no way to tell the difference between the different mixtures.

From an email from Professor Giovanni Vignale sent to me December 2011:

An unpolarized photon is not in a pure state of polarization but in a mixture of different states, e.g. 50% V and 50% H, which is quite different from V+H (For example, V+H will pass with 100% probability through a polarizer at 45 degrees, but an unpolarized photon has 50% probability to pass) Mathematically a pure state is described by a vector in a "Hilbert space", but a mixed state is described by a density matrix.

Gold Member
From an email from Professor Giovanni Vignale sent to me December 2011:
An unpolarized photon is not in a pure state of polarization but in a mixture of different states, e.g. 50% V and 50% H, which is quite different from V+H (For example, V+H will pass with 100% probability through a polarizer at 45 degrees, but an unpolarized photon has 50% probability to pass) Mathematically a pure state is described by a vector in a "Hilbert space", but a mixed state is described by a density matrix.
I will take this opportunity to have so light shed on something that isn´t clear to me, and this context is a good one to me: How do you prepare a mixed state (in this case, a photon that has probability H to be horizontally polarized, and a probability V to be vertically polarized, and how is this physically possible??
Unpolarized light is a balanced mixture of polarized light. It could be a mixture of vertical and horizontally polarized light or a mixture of left circularly and right circularly polarized light or something else-- it doesn't matter. There's no way to tell the difference between the different mixtures.
So, since the trace of the density matrix must be 1, this means that all four polarization modes must be prepared with a contribution ¼?

A. Neumaier
How do you prepare a mixed state
Just take thermal light, the sort that enlightens you, literally. It is already unpolarized, hence prepared as desired.

Its splitting into (2 or 4 or any number of) polarized states is physically meaningless.

By mixing it with completely polarized light (obtained by passing it through a polarizer) you can prepare light in arbitrary partial polarization states.

Gold Member
Is thermal light a mixture of polarized photons? Eg, do there even exist unpolarized photons?? Or is a mixed state a physically distinct state per photon?

@A. Neumaier: I dropped out reading your slides because I couldn´t follow the math steps... But I´ll try again!

stevendaryl
Staff Emeritus
I will take this opportunity to have so light shed on something that isn´t clear to me, and this context is a good one to me: How do you prepare a mixed state (in this case, a photon that has probability H to be horizontally polarized, and a probability V to be vertically polarized, and how is this physically possible??

Mixed states aren't hard to understand: they can be interpreted as just ignorance as to the true state. Imagine a polarizing filter whose orientation was rapidly changing in an unpredictable way. The light that passes through it would have a random polarization, which would be described as a mixture. That's different from a superposition, because a superposition of H-polarized light and V-polarized is light with a definite polarization at an angle between horizontal and vertical.

entropy1
Gold Member
Mixed states aren't hard to understand: they can be interpreted as just ignorance as to the true state. Imagine a polarizing filter whose orientation was rapidly changing in an unpredictable way. The light that passes through it would have a random polarization, which would be described as a mixture. That's different from a superposition, because a superposition of H-polarized light and V-polarized is light with a definite polarization at an angle between horizontal and vertical.
So each photon behaves like it is not polarized, because their polarizations are uniformly randomly distributed?

stevendaryl
Staff Emeritus
Is thermal light a mixture of polarized photons? Eg, do there even exist unpolarized photons?? Or is a mixed state a physically distinct state per photon?

Mixed states can be understood through the ignorance interpretation of probabilities. If there is a probability of $p_H$ of producing a horizontally polarized photon, and a probability of $p_V$ of producing a vertically polarized photon, then the situation would be described using the mixed state:

$\rho = p_H |H\rangle \langle H| + p_V |V\rangle \langle V|$

Being mixed is a subjective fact about a photon. There is no experiment that can distinguish between a mixed photon and a pure-state photon.

stevendaryl
Staff Emeritus
So each photon behaves like it is not polarized, because its polarization is uniformly randomly distributed?

Yes, there is no difference between an unpolarized photon and a photon with an unknown polarization.

entropy1
A. Neumaier
Mixed states aren't hard to understand: they can be interpreted as just ignorance as to the true state.
This works only in certain cases. If you have a beam of unpolarized light, you cannot know its true pure state - it has none! The preparation procedure decides the state, and it is a mixed state from the very beginning.

vanhees71
stevendaryl
Staff Emeritus
This works only in certain cases. If you have a beam of unpolarized light, you cannot know its true pure state - it has none! The preparation procedure decides the state, and it is a mixed state from the very beginning.

Okay, but if you pass that light through a polarizing filter, one photon at a time, there is no experimental difference between unpolarized photons and photons of unknown polarization.

A. Neumaier
Being mixed is a subjective fact about a photon. There is no experiment that can distinguish between a mixed photon and a pure-state photon.
Being unpolarized is an objective fact about sunlight. If it were polarized (though unknown to an observer with less than maximal knowledge) it could be found out by passing it through various polarization filters and observe how much the intensity decreases. But the intensity decreases by the same amount no matter how you orient it. This proves it is unpolarized.

A. Neumaier
Okay, but if you pass that light through a polarizing filter, one photon at a time, there is no experimental difference between unpolarized photons and photons of unknown polarization.
If you make a counting statistics there is a difference.

If you don't make a counting statistics the state makes no difference at all since each photon that passed is polarized in the direction of the polarizer!

vanhees71 and entropy1
stevendaryl
Staff Emeritus
Being unpolarized is an objective fact about sunlight. If it were polarized (though unknown to an observer with less than maximal knowledge) it could be found out by passing it through various polarization filters and observe how much the intensity decreases.

That only proves that the photons from the sun don't have the SAME (unknown) polarization. If a process generatef randomly polarized photons, the output would be indistinguishable from unpolarized light. Wouldn't it?

stevendaryl
Staff Emeritus
If you make a counting statistics there is a difference.

You seem to be assuming that there are only two possibilities: (1) All photons have the same unknown polarization. (2) The light is unpolarized.

The other possibility is that the photons all have different unknown polarizations.

Gold Member
The other possibility is that the photons all have different unknown polarizations.
...and is that the same (also) as a mixed state?

stevendaryl
Staff Emeritus
...and is that the same (also) as a mixed state?

Yes, that's what a mixed state is...well, usually.

entropy1
A. Neumaier
That only proves that the photons from the sun don't have the SAME (unknown) polarization. If a process generated randomly polarized photons, the output would be indistinguishable from unpolarized light. Wouldn't it?
The assumption is meaningless.

How would you test whether a process generated (a) uniformly randomly polarized photons or (b) randomly left and right polarized photons with equal probability? There is no way to tell, hence this level of detail is physically meaningless. What is meaningful is only what one can indeed check - that the source prepared unpolarized light.

This is why people in quantum optics always work with density operators - except in idealized discussions or very simple situations, where states can be assumed to be pure.

entropy1
Gold Member
Yes, that's what a mixed state is...well, usually.
You make me curious! What do you mean by ¨usually¨?

A. Neumaier
The other possibility is that the photons all have different unknown polarizations.
To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state - they have no polarization at all. There is no other possible operational meaning to your statement. Pretending that they have a definite polarization without being able to check it is irrelevant unscientific talk (Ockham's razor).

entropy1
Gold Member
To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state - they have no polarization at all. There is no other possible operational meaning to your statement. Pretending that they have a definite polarization without being able to check it is irrelevant unscientific talk (Ockham's razor).
But if that was the case, we would effectively have the same situation, right? (with respect to what we (can) know?)

stevendaryl
Staff Emeritus
To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state

Then you're agreeing with me, since that's what I was saying.

A. Neumaier
But if that was the case, we would effectively have the same situation, right? (with respect to what we (can) know?)
If it were the case it would look the same. But it would also look the same if it were only left- and right-polarized photons, or only (3|left>+4|right>)/5 and (4|left>-3|right>)/5 polarized photons, or many other possibilities. These are just different ways to prepare unpolarized light if you need to make it from polarized light.

But the state is always defined by the preparation procedure, and it makes no operational sense to ascribe a different state to each single photon if in fact a stream of photons is prepared.

entropy1
A. Neumaier
Then you're agreeing with me, since that's what I was saying.
No. I was arguing against your use of meaningless phrases. I am saying that the only operational sense one can make of your statement is that each of the prepared photons is unpolarized - it makes no sense to assign a random polarization to each single photon, only the prepared non-pure state. Unknowable means ''non-pure'' - this is not the same as ''definitely pure but unknown and different state in each instance''.

Gold Member
But the state is always defined by the preparation procedure, and it makes no operational sense to ascribe a different state to each single photon if in fact a stream of photons is prepared.
So, in case of thermal light, we can´t know if the individual photons have polarizations or not, for QM does not answer that question, and therefore it is not relevant?

Can we even not make out from the way we prepare the light (the photons) if they in fact are polarized?

vanhees71
Gold Member
2021 Award
Unpolarized photons are described by the statistical operator (in polarization space) ##\hat{\rho}=1/2 \hat{1}## by definition. That's it. There's nothing else you need to know about what unpolarized photons are.

entropy1
stevendaryl
Staff Emeritus
You make me curious! What do you mean by ¨usually¨?

Well, there are two different ways that mixed states arise in quantum mechanics.

Proper mixed states: You have some process that puts a system into an unknown pure state. In that case, you can describe the system using a mixed state: $\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |$, where $p_j$ is the probability that it is in state $|\psi_j\rangle$. The "mixed state" reflects our ignorance about the state.

Improper mixed states: If you have two systems that are entangled, then they would be described as a superposition of composite pure states of the form:
$|\Psi\rangle = \sum_{i \alpha} C_{i\alpha} |\psi_i\rangle |\phi_\alpha \rangle$, where $|\psi_i\rangle$ is a complete set of basis states for the first system and $|\phi_\alpha\rangle$ is a complete set of basis states for the second. Now, if the second system is unobservable, then it's inconvenient to include it in the description. In that case, we can get rid of the second system by tracing over its degrees of freedom.

entropy1
stevendaryl
Staff Emeritus
No. I was arguing against your use of meaningless phrases.

And by those phrases, I meant the same thing that you are saying: There is no difference between unpolarized light and light where the photons have different unknown polarizations. I think you are just being argumentative for no reason. You're not clarifying anything, you're picking fights where there is none.

A. Neumaier
So, in case of thermal light, we can´t know if the individual photons have polarizations or not, for QM does not answer that question, and therefore it is not relevant?
Operationally, we cannot collect more information that the coefficients in the density matrix. Thus the shut-up-and-calculate mode in which predictions are made and experiments are interpreted only speaks about that information. Nothing more can be predicted and tested, so everything else is irrelevant (and beyond science).

However, different interpretations of quantum mechanics may make additional (uncheckable) assertions beyond that. They are operationally meaningless but can be used as props for the intuition. For example, the Copenhagen interpretation (on which stevendaryl bases his arguments) assigns a pure state to each single state and then has to interpret the density matrix as something subjective due to lack of knowledge. But from what is operationally verifiable it is the opposite: the density matrix contains the full knowable information while the additional information in the assumed pure states - being unobservable - is subjective and spurious.

The situation is slightly different if the light is very dim, so that photons are created so slowly that one can change the polarizer settings before each new photon arrives. Then one can test the state of each photon separately. (This works only if one knows the state from the preparation procedure since one can make only one measurement per photon. Thus one needs ''photons on demand'' or ''heralded photons''.) But even in this case, the state of each prepared photon will typically still be a mixed state. See the photons on demands analysis in the slides mentioned in post #2.

vanhees71
Gold Member
2021 Award
Well, there are two different ways that mixed states arise in quantum mechanics.

Proper mixed states: You have some process that puts a system into an unknown pure state. In that case, you can describe the system using a mixed state: $\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |$, where $p_j$ is the probability that it is in state $|\psi_j\rangle$. The "mixed state" reflects our ignorance about the state.

Improper mixed states: If you have two systems that are entangled, then they would be described as a superposition of composite pure states of the form:
$|\Psi\rangle = \sum_{i \alpha} C_{i\alpha} |\psi_i\rangle |\phi_\alpha \rangle$, where $|\psi_i\rangle$ is a complete set of basis states for the first system and $|\phi_\alpha\rangle$ is a complete set of basis states for the second. Now, if the second system is unobservable, then it's inconvenient to include it in the description. In that case, we can get rid of the second system by tracing over its degrees of freedom.
Any statistical operator can be written in the first way, because it's self-adjoint and thus you can express it in terms of its eigenstates. Then the ##|\psi_j \rangle## are even a complete orthonormal system.

In the second example you simply have a reduced density operator of a pure state of a composite system, i.e., you use the partial trace
$$\hat{\rho}=\mathrm{Tr}_B |\Psi \rangle \langle \Psi|.$$
It's again a self-adjoint trace-class operator with ##\mathrm{Tr} \hat{\rho}## in the Hilbert space of the partial system ##A##.

stevendaryl
Staff Emeritus
Unknowable means ''non-pure'' - this is not the same as ''definitely pure but unknown and different state in each instance''.

I think you're mixing your philosophy up with your physics. There are times when a mixed state is a necessity, as when you are observing one subsystem of an entangled composite system. If that's what you mean, then it would be more useful to say that explicitly.

stevendaryl
Staff Emeritus