I was thinking about the question I posted yesterday, and I thought of a better way to ask it.

I'm trying to figure out why QM necessitates "pure randomness". Assume you have a photon that has a hidden variable. This hidden variable is a pseudorandom number generator $f(t) \in \mathbb{R}$ such that $0 \leq f(t) \leq 1$. If $f(t) > 0.5$, the photon passes through the polarizer, and if $f(t) \leq 0.5$, it does not. If the experimenter could figure out what this PRNG is, he could predict the result of every measurement, which is more than QM can predict.

In other words, the photon has a local hidden variable that if known would remove the possibility of "true" randomness, while still reproducing the probability distribution predicted by QM.

However, Bell's theorem rules out this possibility. That's not what I have a problem with — giving up locality is fine with me. So consider this:

The PRNG is no longer a hidden variable of each photon, but a hidden variable of an entangled two photon system. I'm sure this can be done with one PRNG, but for simplicity of explanation, let's say there are two individual PRNG's associated with the entire system: $g_1(t)$ and $g_2(t)$.

The photons are entangled and separated. Photon 1 heads toward polarizer 1 with angle $\theta_1$ and photon 2 heads toward polarizer 2 with angle $\theta_2$. It's well known that the probability that each photon gives the same measurement is given by:

$$P(\theta_1, \theta_2) = \cos^2(\theta_1 - \theta_2)$$

and this has been experimentally verified. It's clear to see that because the angles of each polarizer can be altered while each photon is still in flight, there must be an instantaneous connection between the measurement results.

However, to me, this still doesn't imply true randomness.

Suppose photon 1 gets to its polarizer first at time $t_1$. Whether it passes through the polarizer is simply given by the boolean $X_1 = g_1(t_1) > 0.5$. Now define another boolean

$$Y = g_2(t_2) < \cos^2 (\theta_1 - \theta_2)$$

, where $t_2$ is the time that photon 2 arrives at its polarizer. Whether photon 2 passes through the polarizer is then given by:

$$ \overline{X}_1 \overline{Y} + X_1 Y$$

As far as I can tell, this doesn't violate any of the postulates of QM or any kind of no-go theorem, and it's deterministic. Where did I go wrong?

  • $\begingroup$ The problem here is that you've explicitly stated that photon 2 will not pass through the polarizer if photon 1 does and the angle between them is $\pi/2$. This is not true; photon 2 can still pass through polarizer 2, however the measurements are no longer correlated. The point of saying that it is random now is that there is no way to determine whether it will pass through the polarizer without interacting with the photon. In essence, it is impossible to figure out the PRNG before measuring $\endgroup$
    – Jim
    Apr 3, 2014 at 15:28

2 Answers 2


There are published deterministic alternatives to the indeterministic Copenhagen interpretation of Quantam Mechanics.

De Broglie–Bohm theory is the most well known.

David Bohm wrote a book The Undivided Universe: An Ontological Interpretation of Quantum Theory just before he died in 1992, with Basil Hiley. The book explains Bohm's deterministic interpretation and compares it to indeterministic interpretations such as the Copenhagen interpretation and many worlds interpretation. Link to book review. Quoting from the review: "Thus, in Bohmian mechanics, the configuration of a system of particles evolves via a deterministic motion choreographed by the wave function. In particular, when a particle is sent into a double-slit apparatus, the slit through which it passes and where it arrives on the photographic plate are completely determined by its initial position and wave function."

The determinstic interpretation has not been disproven.

There is a extentsive article in the Stanford University Encylopedia of Philosophy on Bohmian Mechanics.

  • $\begingroup$ When they can make predictions, they both yield the same experimental results. The advantage of the Copenhagen view is that it naturally conduces to QFT. The experts I asked were not aware of any Bohmian interpretation of QFT. $\endgroup$
    – Davidmh
    Apr 2, 2014 at 13:08
  • $\begingroup$ @Davidmh That's really suprising that the experts you asked weren't aware of the interpretation. Maybe they know it as "pilot wave" model or another term. The Bohmian interpretation is certainly referred to in peer reviewed literature starting with Bohm, D., 1952, “A Suggested Interpretation of the Quantum Theory in Terms of ‘Hidden’ Variables, I and II,” Physical Review, 85: 166–193. A list of papers is availible here: bohmian-mechanics.net/research_papers.html . Prof. Detlef Dürr of LMU (Munich) and Prof. Sheldon Goldstein of Rutgers are active in this area. $\endgroup$
    – DavePhD
    Apr 2, 2014 at 15:53
  • $\begingroup$ There are about 6 Bohmian Mechanics questions at Physics Stack Exchange including: "Why Do People Still Talk about Bohmian Mechanics" physics.stackexchange.com/questions/7112/… $\endgroup$
    – DavePhD
    Apr 2, 2014 at 16:00
  • $\begingroup$ @Davidmh There is "Bohmian Mechanics and Quantum Field Theory" Phys.Rev.Lett. 93 (2004) 090402 arxiv.org/abs/quant-ph/0303156 $\endgroup$
    – DavePhD
    Apr 2, 2014 at 18:26

If you want to insist that a classical theory of hidden variables can reproduce the predictions of QM, no one can say you are wrong (just stubborn). But if you admit one more thing, that physics ought to be local, then we've got you!

So without thinking about your model in detail, I can say that such a model might reproduce the predictions of QM, but it cannot be local.

You haven't found a contradiction, so you needn't conclude that your workings or assumptions are wrong.

  • 1
    $\begingroup$ I'm not trying to insist on a classical theory -- I learn best by trying to come up with a counterexample and then figuring out what went wrong (I suppose you could call it learning by contradiction?) $\endgroup$
    – Nick
    Mar 29, 2014 at 22:51
  • $\begingroup$ @Nick once you convince yourself of Bell's theorem there should be no ambiguities to what can, and cannot be done here. I'd say that's the way to go. $\endgroup$
    – Danu
    Mar 29, 2014 at 22:59
  • $\begingroup$ Good then. That's exactly what I'm doing here. $\endgroup$
    – Nick
    Mar 29, 2014 at 23:11

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