# B What is ¨unpolarized¨ light exactly?

Tags:
1. Apr 19, 2016

### entropy1

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?

2. Apr 19, 2016

### 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''.

Last edited: Apr 19, 2016
3. Apr 19, 2016

### 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.

4. Apr 19, 2016

### StevieTNZ

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

5. Apr 20, 2016

### entropy1

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??
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?

6. Apr 20, 2016

### A. Neumaier

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.

7. Apr 20, 2016

### entropy1

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!

8. Apr 20, 2016

### stevendaryl

Staff Emeritus
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.

9. Apr 20, 2016

### entropy1

So each photon behaves like it is not polarized, because their polarizations are uniformly randomly distributed?

10. Apr 20, 2016

### stevendaryl

Staff Emeritus
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.

11. Apr 20, 2016

### stevendaryl

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

12. Apr 20, 2016

### A. Neumaier

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.

13. Apr 20, 2016

### stevendaryl

Staff Emeritus
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.

14. Apr 20, 2016

### A. Neumaier

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.

15. Apr 20, 2016

### A. Neumaier

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!

16. Apr 20, 2016

### stevendaryl

Staff Emeritus
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?

17. Apr 20, 2016

### stevendaryl

Staff Emeritus
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.

18. Apr 20, 2016

### entropy1

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

19. Apr 20, 2016

### stevendaryl

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

20. Apr 20, 2016

### A. Neumaier

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.