# How do photons “decide”?

I was reading that when horizontally polarized light hits a vertical Polaroid all the light is blocked out. But when the Polaroid is off the vertical, some but not all photons "decide" to jump into the new plane of polarization. Could this be a "road less traveled" kind of effect?

If a run of two or three photons make the jump then conditions are affected in such a way, that the next photon is less likely to make the jump. Then as one or more photons get blocked, conditions cool down a bit increasing the likelihood that another run of jumps will occur: a mechanism of so called "deciding".

There is no such mechanism. The probability for a photon to pass through a polarizer at an angle $\theta$ is $\cos^2(\theta)$, regardless of what has happened before, and regardless of how many photons "at once" try to pass through it. As Bell's theorem tells us, the quantum world is really random (or non-local).

• Here, here, young thin Edwin. People should be though QM with the last sentence engraved above a blackboard. :) – Bubble Sep 21 '14 at 21:48
• Doesn't the very fact that the angle of a physical apparatus affects the angle of polarization violate this theorem. After all the photons are not coming out of the Polaroid with random angles of polarization. Certainly conditions apply that are affecting outcomes at the quantum mechanical level. I know that's probably the second in a run of two dumb questions. – user56930 Sep 21 '14 at 21:52
• @user56930: Ah, with random I mean that the outcome of a specific measurement is not predetermined (or only influenced by what happened before), but, in this case, always distributed according to $\cos^2(\theta)$. That it is random does not mean that it is uniformly random (which would mean equal probability for every possible outcome). – ACuriousMind Sep 21 '14 at 21:59

If that happened, we would be able to detect it by looking at correlations between successive photons' "decisions."

That is, suppose you represent each pair of consecutive photons (1 and 2, 2 and 3, 3 and 4, etc.) with $+1$ if they both made the same "decision" or $-1$ if one went through the polarizer and the other didn't. Take the average of these numbers for all pairs and call it $C$. If a run of photons making the same decision changed the probability for the following photons, $C$ would be greater than zero. In reality, it comes out to zero, meaning that each photon doesn't change its behavior depending on what the one before it did. So we have clear experimental evidence that this modification of probabilities doesn't happen.

• I understand the simplicity of probabilities reigning at the quantum mechanical level. But I wonder how it is that a physical manipulation, a manual adjustment of the angle of a Polaroid can have an effect on a probability. Descartes wondered how an idea can move a muscle. How does a classical physical manipulation transmit a measurable effect on a probability if not by a mechanism? – user56930 Sep 21 '14 at 22:09
• I realize that I'm asking two questions. One is how do photons "decide"? The other, how is Bell's theorem not violated when a physical twist of an apparatus has an effect on outcomes? Thank you. – user56930 Sep 21 '14 at 22:19
• @user56930 how much do you know about wavefunctions? These last couple comments suggest you might not be familiar with the idea, but I don't think I can give you a sensible answer that doesn't involve them. The gist is that physical manipulations change the potential energy field, which in turn changes how the wavefunction evolves in time, and the wavefunction determines the probabilities. – David Z Sep 22 '14 at 0:15
• If you care to explain in terms of wave functions briefly of course I'll view it as a chance to read up. I'm trying to get calculus. Thanks – user56930 Sep 22 '14 at 13:16
• @user56930 The (position-space) wavefunction $\Psi(\vec{x},t)$ is a function of position and time that, in basic quantum mechanics, represents all the information that can be known about a particle. The probability for a particle to be found within a region $R$ at time $t$ is $\int_R \lvert\Psi(\vec{x},t)\rvert^2\mathrm{d}^3\vec{x}$. It changes in time in a manner given by the Schroedinger equation. You might want to follow up with our QM book recommendations but note that they assume knowledge of multivariable calculus and linear algebra. – David Z Sep 22 '14 at 20:25