Let suppose you roll a die, and it falls into a hidden place, for example under furniture.
Then although the experiment has already been made (the die already has a number to show), that value can not be known, so the experiment was not fully realized.
Then till you see the die's top side, the probability remain p = 1/6.
I see no difference between this and the wave function collapse, at least as an analogy.
Can someone explain a deeper difference?



6 Answers 6


A somewhat similar (yet inverted) question was posed, infamously, by Einstein, Podolsky, and Rosen as an argument against the same phenomenon you are challenging.

The basic argument was that such quantum wave collapses are indistinguishable from dice falling under couches, but under opposite grounds as yours -- both must be "determined" by what were later named "hidden variables". That a quantum wave collapse, like a falling dice, is completely deterministic by the nature of its initial conditions, and its apparent randomness is due to our current inability to measure these initial conditions.

It was John S. Bell who thought up of an experiment to test this.

The basic idea was statistical, and involved quantum entanglement -- if hidden variables existed (and the particles already having a determined, yet unknown state at their inception), statistics should yield a certain correlation between entangled particles. If there are no hidden variables, then this correlation would not show up.

Surprisingly, this correlation did not manifest by even a large margin of experimental error. This proves an inherent difference in the randomness of quantum wave collapse as opposed to simply not knowing the landing of a dice.

The result culminated in Bell's theorem.

  • $\begingroup$ I am tempted to think that any value showed just after a measure, was "a hidden -value-" before measurement, perhaps no hidden -variables- could explain measured values, but themselves could be "hidden values", why we can't think that way? I mean that they were already "defined" by "universe" before the measurement, I see no difference, as hidden dice, "true random" could be defined at measurement or billion years before, I see no difference. I've put two dice sum and an entanglement examples as a comment in Luboš Motl answer, then he guided me to Hardy's paradox, I am trying to understand it $\endgroup$
    – HDE
    Feb 24, 2011 at 11:55
  • $\begingroup$ Luboš said (as I understood) that "hidden value" supposition can't be done, because realism fails in QM, this is a very surprising to me, more examples of this are welcome, @Justin L. perhaps bell theorem is speaking exactly about this, but I think hidden variables to explain values is not the same that hidden values, thanks for any further comment $\endgroup$
    – HDE
    Feb 24, 2011 at 12:03
  • $\begingroup$ @HDE I guess my answer didn't directly address the question at hand. Indeed, isolated quantum randomness (without entanglement), is more or less indistinguishable. What I was meaning to say was that there is something fundamentally, inherently different about determination in the context of quantum mechanics than just a simple dice roll. $\endgroup$
    – Justin L.
    Feb 24, 2011 at 23:43

You're absolutely right: the probabilities predicted by quantum mechanics are conceptually fully analogous to probabilities predicted by classical statistical mechanics, or statistical mechanics with a somewhat undetermined initial state - just like your metaphor with dice indicates. In particular, the predicted probability is a "state of our knowledge" about the system and no object has to "collapse" in any physical way in order to explain the measurements.

There are two main differences between the classical and quantum probabilities which are related to one another:

  1. In classical physics - i.e. in the case of dice assuming that it follows classical mechanics - one may imagine that the dice already has a particular value before we look. This assumption isn't useful to predict anything but we may assume that the "sharp reality" exists prior to and independently of any observations. In quantum mechanics, one is not allowed to assume this "realism". Assuming it inevitably leads to wrong predictions.

  2. The quantum probabilities are calculated as $|c|^2$ where $c$ are complex numbers, the so-called probability amplitudes, which may interfere with other contributions to these amplitudes. So the probabilities of outcomes, whenever some histories may overlap, are not given as the sum over probabilities but the squared absolute value of the sum of the complex probability amplitudes: in quantum mechanics, we first sum the complex numbers, and then we square the result to get the total probability. On the other hand, there is no interference in classical physics; in classical physics, we would surely calculate the probabilities of individual histories, by any tools, and then we would sum the probabilities.

Of course, there is a whole tower of differences related to the fact that the observable (quantities) in quantum mechanics are given by operators that don't commute with each other: this leads to new logical relationships between statements and their probabilities that would be impossible in classical physics.

A closely related question to yours:

Why is quantum entanglement considered to be an active link between particles?

The reason why people often misunderstand the analogy between the odds for dice and the quantum wave function is that they imagine that the wave function is a classical wave that may be measured in a single repetition of the situation. In reality, the quantum wave function is not a classical wave and it cannot be measured in a single case of the situation, not even in principle: we may only measure the values of the quantities that the wave function describes, and the result is inevitably random and dictated by a probability distribution extracted from the wave function.

  • $\begingroup$ @Luboš Motl, question understood +1 good answer, however I still can't see difference, rolling two dice and taking sum as result, the distribution is a triangle with the peak at 7, and you can't directly measure that "wave" neither, and if you unhide (and know the value) of one dice, then the final sum result is altered, the whole experiment is altered, there is some kind of interference. $\endgroup$
    – HDE
    Feb 23, 2011 at 14:37
  • $\begingroup$ @Luboš Motl comment II, why do you say realism drive to wrong predictions? imagine a pair of entangled particles A/B, measuring one, the other state is determined, but if no one else knows the measure was done for A, the B state is still indetermined till a measure is done, then the realism suposition(or even having a Real [but unknown] deterministic value!), doesn't change the non-determinism $\endgroup$
    – HDE
    Feb 23, 2011 at 14:40
  • 1
    $\begingroup$ Dear HDE, well, I am agreeing that it is analogous. Just concerning your first comment above, it's not true that there is any interference of probability amplitudes in classical physics. Classical physics only knows the real, non-negative probabilities and they may only be added in the normal way which means that there's no interference. In quantum mechanics, probability amplitudes may be added both constructively and destructively and they may even cancel, which is impossible for classical probabilities that are non-negative. $\endgroup$ Feb 23, 2011 at 15:19
  • $\begingroup$ Concerning your second comment, nope, it is not just determinism that fails in quantum mechanics: it's realism that fails. All the fun experiments with entanglement - such as quantum eraser, Bell's inequalities, GHZM state, and Hardy's paradox - see physics.stackexchange.com/questions/4040/… for the latter - imply that the very assumption of realism fails in the real world (governed by quantum mechanics). Just by assuming that the objects have properties before they're measured, you inevitably deduce predictions that are experimentally refuted. $\endgroup$ Feb 23, 2011 at 15:21
  • $\begingroup$ @Luboš Motl, Wow! (despite the fact that I am deep against realism), It would surprise me that there is already a way to show that it fails!, I will try hard to understand what Hardy's paradox says, thanks for the reference! $\endgroup$
    – HDE
    Feb 23, 2011 at 15:49

The other answers provide an account of quantum theory as a source of the answer. But I would like to consider the question from a more general perspective: that of the meaning of "probability" especially in the classical context.

Specifically what is meant by "probability" of the dice value in this question and where has the value 1/6 come from? This may seem like a trivial point, but there are (at least) three different perspectives on the answer:

(1) Frequentist. This is the view that probability is a number determined empirically by a series of tests. So this view says that such tests have been conducted on the dice and in 1/6 of cases (subject to some statistical factor) a given number was returned.

(2) Absolutist. This is the view (although I am not sure anyone will promote it) that the probability (with its value here of 1/6) is a derived physical property of the dice (like temperature, say). As a physical property it has been determined by some detailed analysis. One possible such analysis might be on the phase space of all possible motions, which has been appropriately partitioned and it is found that 1/6 of the volume is associated with each resultant value. This analysis might be done for that dice, or for some idealised dice. So the absolutist would claim that physics determines (classical) probabilities, and would view the frequentist data as simply experimental (dis-)confirmation of a specific model.

(3) Relativist. This is really the Bayesian viewpoint as espoused by Jaynes. This is using conditional probability $P(A|C)$ to define a probability in a relative way, with Bayes Theorem - which also allows us to construct $P(C|A)$ - this allows discussion of Inference and Prior Probabilities. The Bayesian view (especially as promoted by Jaynes) would be that there is no absolute probability, but only levels of knowledge, "observer inferencing" and (changing) prior assumptions. So the 1/6 value would be determined subjectively, as a measure of your "expectation" for example. It would have nothing to do with any intrinsic physics of the dice.

So to return to Quantum Mechanics. It too deals with probabilities, which are calculated from $|\Psi(x)*\Psi(x)|$. So what do these probabilities mean now that we have done some analysis?

Jaynes liked to promote his Bayesian view (ie 3 above) that he claimed reconciled Bohr with Einstein, thus interpreting $\Psi$ as somewhat epistemological. Phrases I have seen in other Stack answers like "the $\Psi$ collapse is all in your head" when discussing $\Psi$ might suggest that some physicists share an aspect of this view.

However an alternative view might be that $\Psi$ really solves the Absolutists problem: namely it provides a well validated physical basis for calculating probabilities in an absolute sense. If this were true then the only physical way to get to 1/6 in the dice would be to do some giant quantum calculation from its atoms!


In QM one can also deal with probabilities instead of probability amplitudes when one describes the prepared beam in terms of density matrix. For example (I assume there is six states $|n>$):

$\rho = (1/6)\sum_n {|n><n|}$. (1)

Mixture is generally different from a pure (superposition) state with the same probabilities:

$\psi = (1/6)^{1/2}\sum_n {|n>}$. (2)

The dice experiment corresponds to (1). It would be nice if you could read something about difference between mixtures and superpositions in QM (see sources of coherent and non coherent light, for example).


The deeper difference between the classical and quantum dice is this - quantum dice obey the superposition principle. Consequently you can create entangled states of the form:

$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|x \rangle_{D1} |y \rangle_{D2} \pm |y \rangle_{D1} |x \rangle_{D2} ) $$

where the subscripts denote the particular die (1 or 2), $|x\rangle_{D}$ and $|y\rangle_{D}$ are two different possible states of the die. The simplest case is when the role of the "die" is played by qubits. Then the above expression becomes:

$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|\uparrow \rangle_{D1} |\downarrow \rangle_{D2} \pm |\downarrow \rangle_{D1} |\uparrow \rangle_{D2} ) $$

Entanglement appears to have no classical analog and it actually plays the role of a usable information resource in quantum computation.

  • $\begingroup$ Entanglement has a classical analog. Given two quantum systems with state spaces S and T, the joint system has state space S⊗T. Any joint state that isn't a product x⊗y is entangled. Similarly, if two probabilistic systems have PDFs in vector spaces P and Q, the joint distribution lives in P⊗Q. You can then have states analogous to entanglement that are linear combinations of products of vectors in P and Q. But QM allows linear combinations with complex coefficients and that is what has no classcial analogue. (en.wikipedia.org/wiki/Quantum_entanglement#Pure_states) $\endgroup$
    – Dan Piponi
    Aug 25, 2011 at 0:03

I was going to make this a comment but here it comes, since maybe it is a simpler formulation of a fundamental difference than all the rest of the answers:

The difference is that the value of the classical die under the couch can be seen/measured by many people.They will not tell you, and you still will not know, but the value that was thrown will not change. In quantum mechanics each observation/measurement changes the system, from one measurement to the other . The die's throw will not change from one look of a hidden to you observer, to the next.

  • 1
    $\begingroup$ @anna your wording in the second para seems odd. I don't see how having $N$ observers, rather than just one, helps explain the difference between classical and quantum ... on second thoughts, I get your point. Perhaps you could clean up the language a bit. $\endgroup$
    – user346
    Feb 24, 2011 at 6:43
  • $\begingroup$ @Deepak Vaid Thanks, I tried to make it clearer. $\endgroup$
    – anna v
    Feb 24, 2011 at 7:03
  • $\begingroup$ @anna v, observation is part of the experiment, if you add more observation those are different experiments (unless that with observation you mean to re read already made results) then different experiments are like roll dice again. $\endgroup$
    – HDE
    Feb 24, 2011 at 13:12
  • $\begingroup$ Classically, observation is not part of the experiment. Something is or is not. You can read off what the face of the die says, as many times as you want and it will not change if you do not throw it again. Quantum mechanically, each "reading" will change the value, and yes observation becomes part of the experiment. So one could rephrase what I said more simply : the difference in probability measurements between classical and qm view points depends on the fact that the observer is part of the experiment in qm and not in cm. $\endgroup$
    – anna v
    Feb 24, 2011 at 13:30
  • $\begingroup$ a second observer would only measure something different if you sequentially measured two values of non-commuting operators. If you leave the system in a diagonal state instead, the measurement remains constant $\endgroup$ Feb 25, 2011 at 14:04

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