# Which properties of the quantum are random?

1. Jun 17, 2011

### San K

Which properties of the photon/quantum are random?

Spin - Yes
Position (within the "range/orbital") - Yes
Momentum - Yes

Phase - No
Polarization - No
Coherence (derivative of phase) - No

2. Jun 17, 2011

### Demystifier

Polarization of a photon is essentially the same as as spin of a photon. In that sense, polarization is also "random", i.e. subject to probabilistic laws.

Phase of a single photon is not even measurable.

3. Jun 17, 2011

### San K

interesting......then why do we have left and right polarizers?

i mean the left polarizer would allow only left-photons to pass through....but if polarization is random then the left polarizers would keep changing?

4. Jun 20, 2011

### Demystifier

But you can also have a photon in the vertical (or horizontal) polarization, which is a superposition of left and right polarization. If you transmit such a photon through a left polarizer, there is a 50% chance that it will pass. So, it's probabilistic too.

5. Jun 20, 2011

### Zarqon

Just a point: You can't really say that a specific property is always random or not, it depends on the context, and how you have prepare a particular state. For example, a photon in a Fock state has a known photon number, but an unknown (random) phase, whereas a photon prepared in a coherent state may have a well known phase, but instead containing an unknown (random) number of photons.

6. Jun 21, 2011

### San K

interesting. so the same property can be random/non-random depending upon the state?

i.e. can we lock/unlock the randomness, in properties, of a photon?

7. Jun 21, 2011

### Goldstone1

Sure, all you need to know is all the eigenstates of a particles. But knowing two complimentary observables however, makes it an issue that ''randomness'' is a word which replaces a ''lack of knowledge'' on the system. Randomness is an illusion of our semantic attachment for the need to know everything.

8. Jun 21, 2011

### San K

I had similar ideas, subject to verification.

how? can you give an example/experiment where two complimentary observables were known?

9. Jun 21, 2011

### Goldstone1

Well, you can only know one with great certainty, but the other becomes increasingly unknowable, which is what was meant by the post.

10. Jun 22, 2011

### Goldstone1

That doesn't mean it is random though.

We seem to also use the radiation of ripe systems as a perfect example of random systems. I don't see why... 1) If there is a mechanism, then it isn't random 2) you can freeze the system using the zeno effect, then your system of radiating particles are completely knowable over large periods of time.

11. Jun 22, 2011

### San K

I did not understand the "photon number" part.

If we have two photons in a coherent state, don't we have a well know phase as well as a known number of photons (i.e. two)?

12. Jun 22, 2011

### Goldstone1

To say it was random, you would need to explain a large part of decoherence physics.

$$|\psi> = \sum_i |i><i|\psi>$$

would be our state of the system also knowing that the $$|i>$$'s form the Einselection basis. If one says the initial state of the system was $$\epsilon$$, then one can make a before and after equation based on what has happened in it's evolution. In effect, a system can either loose information or gain information. If we are talking about radiative systems, giving off radiation, then the after equation

$$|A>= \sum_i |\epsilon_i><i| \psi>$$

If one can ultimately know how strongly $$|i>|\epsilon>$$ evolves into $$\epsilon_i$$ is completely knowable, because there is nothing which dictates in the equations that it cannot be knowable.

13. Jun 24, 2011

### Zarqon

photon number = number of photons

A coherent state is per definition a superposition of photon number states. As soon as you have one well defined photon number, then it must also have an unknown phase, because you cannot even define a phase for only one photon number, since a phase is a relative concept.

Simple example:

A state with exactly two photons can be written |2>, and there is only one possibility. However, a state with randomly either 1 or 2 photons could be written in a infinite number of ways, for example, |1> + |2> or |1> - |2>. The latter ones are coherent states and the phase is the difference in sign between the individual photon number states.

14. Jun 24, 2011

### San K

well put, this and the previous post. thanks for info

15. Jun 24, 2011

### Goldstone1

But their post does not prove a true random system. No part of it actually says random, it says ''lack of knowledge''. This is a priori of the arguement solicitated, but understand lack of knowledge is just a part of ignorance. No system is incomplete and beyond the reach of human evolution.