2
$\begingroup$

In the proof of Bell's theorem of 1964, referenced e.g. here, the definition of a hidden variable seems to be any variable from which we can derive the correlation between the detectors, by calculating the integral over the values.

Isn't this what quantum mechanics already does? The wavefunction provides a spectrum that can then be integrated to obtain the correlation, where does one assume in the proof that the variable $\lambda$ is of some other "non-quantum" nature?

$\endgroup$
5
  • 4
    $\begingroup$ Note that Bell's theorem is concerned with local hidden variables, while the wavefunction describes the whole configuration space at once. $\endgroup$
    – Ruslan
    Commented May 31 at 22:02
  • 1
    $\begingroup$ Yes -- maybe a better way to put it is that Bell's theorem is concerned with localized variables, variables which are associated with locations in space and time. That way one would calculate the local result from an integral over local values. Multiparticle wavefunctions are famously "unlocalized" -- they are functions on a higher dimensional configuration space, not spacetime. So they can't serve this role. $\endgroup$ Commented May 31 at 23:58
  • $\begingroup$ But the violations of Bell's inequality also occur for two qubits in one point of space. So unlocalized wavefunctions may not be the real essence of the explanation. $\endgroup$ Commented Jun 1 at 6:26
  • $\begingroup$ "Any variable from which we can derive the correlation between the detectors" is not the correct definition. See what local beables are in the words of Bell himself: The theory of local beables $\endgroup$ Commented Jun 1 at 8:55
  • $\begingroup$ John Stewart Bell, the leading Bohmian until the 1990s, exclaimed: “The absurdity is that these theories are called “hidden variable” theories. This is an absurdity because here, it is not in the wave function that we find an image of the visible world, and of the results of experiments, but in these “hidden”(!) complementary variables. (...) The most hidden of the variables, in this image of the pilot wave, is the wave function, which only manifests itself to us through its influence on the complementary variables. » fr.wikipedia.org/wiki/Th%C3%A9orie_de_De_Broglie-Bohm $\endgroup$
    – The Tiler
    Commented Jun 4 at 10:27

4 Answers 4

1
$\begingroup$

Bell's theorem assumes that the outcomes of the two measurements are $A(\vec a,λ)$ and $B(\vec b,λ)$ where $\vec a$ and $\vec b$ are the measurement settings, $λ$ is arbitrary "data" shared by the two particles, and $A$ and $B$ are arbitrary functions.

If you take $λ$ to be the wave function (plus a source of random bits to decide the outcomes of probabilistic measurements), and $A$ and $B$ to implement the measurement postulate of QM, you'll get the wrong answer. The reason is that you have to modify ("collapse") the wave function after each measurement, in a way that depends on the measurement, and feed the modified wave function into the next measurement. To fit that into Bell's framework, supposing wlog that you consider measurement A first, you'd have to say the measurement outcomes are $A(\vec a,λ)$ and $B(\vec b,λ')$ where $λ'$ is a function of $\vec a$ and $λ$, or equivalently that the outcomes are $A(\vec a,λ)$ and $B(\vec a,\vec b,λ)$. You can't prove Bell's inequality any more with that change.

$\endgroup$
1
  • $\begingroup$ Just saying that "if you take $\lambda$ to be the wave function" then "you'll get the wrong answer", does not really answer OP's question what goes wrong. It merely asserts that something does go wrong, which OP probably had heard already, but is not willing to accept without more evidence... $\endgroup$ Commented Jun 1 at 6:08
0
$\begingroup$

the definition of a hidden variable seems to be any variable from which we can derive the correlation between the detectors,

The hidden variables referred to in the context of Bell's inequalities are usually local. A hidden variable theory limits itself to information available locally to avoid any requirement for FTL knowledge of states.

Let's say Alice and Bob are two spacelike separated observers with analysers whose orientation are defined by the vectors $\vec{a}$ and $\vec{b} $ respectively. The probability for the expected correlations' of their measurements is given by:

$$ P(\vec{a}, \vec{b}) = -\vec{a} \cdot \vec{b} $$ This QM prediction includes non local variables such as the orientation of the other observers analyser. A hidden variable theory has to make a prediction for the correlation $(-\vec{a} \cdot \vec{b}) \ $ measured by, for example, Bob without knowledge of the orientation of Alice's detector $\vec{a}$ at the instant Bob makes his measurement, which is impossible. The equation effectively becomes
$$ P(\vec{a}, \vec{b}) = -\text{?} \cdot \vec{b} $$

Bob could conspire with Alice to gain advance knowledge of the configuration of Alice's analyser. For example Alice could agree to always measure up spin. Now Bob can predict that if he make a simultaneous measurement for up spin, that the correlation should always be be zero, (i.e. always the opposite of what Alice measures at the instant Bob make his measurement). So the hidden local variables are the the orientation of Bob's analyser and the knowledge in his head of the orientation of Alice's analyser, but he still cannot predict the correct outcome for what he should measure without knowledge of the simultaneous measurement outcome made by Alice, which is classically impossible if they are spacelike separated. (By classical, I mean non-quantum behaviour).

$\endgroup$
0
-1
$\begingroup$

"where does one assume in the proof that the variable λ is of some other "non-quantum" nature?" One assumes in the proof that any pair of observable quantities has a well-defined correlation, which is essentially tantamount to assuming that any pair of observable quantities can be simultaneously observed. This assumption is true of classical random variables, but false of quantum mechanical observables. Bell assumes that the hidden variables (such as $\lambda$) are classical (hence that the correlations are always well defined) and derives a contradiction with quantum mechanics (and with experiment).

$\endgroup$
7
  • $\begingroup$ What you say here only agrees with OP's question but does not answer it. Because any pair of observable quantities does have a well-defined correlation (it's just higher than classical physics can explain). And any pair of observable quantities can be simultaneously observed (in the sense that Alice and Bob can just do their measurements). Saying that these assumptions are false in QM is incorrect. $\endgroup$ Commented Jun 1 at 6:12
  • $\begingroup$ @JosBergervoet: Alice cannot simultaneously observe spin in two directions. If she and Bob could both do that, you'd have well-defined correlations for all four of the random variables Alice's observed spin in direction A, Alice's in direction B, Bob's in direction A, and Bob's in direction B. These correlations would then have to satisfy Bell's inequality, but in fact they don't. So the failure to satisfy Bell depends on the non-observability of everything simultaneously. $\endgroup$
    – WillO
    Commented Jun 1 at 13:25
  • $\begingroup$ Alice and Bob definitely can measure what they want for the two entangled particles in the usual setup discussed with EPR (maybe lightyears away from each other, is then added!) So that simply cannot be the reason why their results cannot be described by hidden variables. OP's question is not about certain things that cannot be measured, but about things that can be measured and behave nonclassically. (The other problem would also be interesting, admittedly...) $\endgroup$ Commented Jun 1 at 13:50
  • 1
    $\begingroup$ @JosBergervoet: We have four observables. We discover that $Prob(A\neq D)< Prob(A\neq B)+Prob(B\neq C)+Prob(C\neq D)$. This is contradictory, because whenever $A\neq D$, we expect at least one of $A\neq B,B\neq C,$, or $C\neq D$. We avoid the contradiction because not all of these pairs can be simultaneously measured. If you are suggesting that Alice can simultaneously measure spin in two directions (observables $B$ and $C$, say) then you are plain wrong. If you are suggesting that she can always measure whichever she wants to, that is true but misses the point entirely. (CTD) $\endgroup$
    – WillO
    Commented Jun 1 at 14:15
  • 1
    $\begingroup$ (CTD) The point is that because Alice can't measure $B$ and $C$ simultaneously, we do not need to assign a probability to the event $B\neq C$, so we avoid the contradiction. That's exactly how Bell's argument works. $\endgroup$
    – WillO
    Commented Jun 1 at 14:17
-1
$\begingroup$

In classical physics the evolution of each measurable quantity $x$ is described by a function $x(t)$ whose value is the same as the result you would get by measuring $x$.

I'm going to discuss quantum theory in the Heisenberg picture rather than the Schrodinger picture because the Heisenberg picture describes the evolution of physical quantities and individual systems more clearly than does the Schrodinger picture. I should also note that the account I will give here supposes that the equations of motion of quantum theory are an accurate description of how the world actually works and aren't modified by collapse or extra particles as in pilot wave theory. Collapse and extra particle assumptions contradict the equations of motion of quantum theory so I won't include them.

In quantum physics the evolution of a physical quantity $y$ is described by an observable $\hat{Y}(t)$ whose value is a Hermitian operator where the possible results of a measurement of $y$ are the eigenvalues of $\hat{Y}(t)$. Most observables have more than one eigenvalue and more than one possible measurement result. The evolution of $\hat{Y}(t)$ describes a process involving all of the possible values of $\hat{Y}(t)$. To make predictions one needs another Hermitian operator $\rho$ called the state which is a record of what measurements happened in the past and what outcomes you saw. The expectation value of measurements of $\hat{Y}(t)$ in the state $\rho$ is

$$\langle\hat{Y}(t)\rangle=tr(\rho\hat{Y}(t))$$

Suitable experiments such as single particle interference experiments can involve manipulating the projectors of $\hat{Y}(t)$ corresponding to each of the possible values in such a way that the expectation values depend on what happened to each of the possible values. For an example, see section 2 of this paper:

https://arxiv.org/abs/math/9911150

Note that the value of $\hat{Y}(t)$ at a given time is a matrix and the possible measurement results are the eigenvalues so the value of $\hat{Y}(t)$ is not the same as any of the possible measurement results, they aren't even the same kind of mathematical object.

A hidden variable would be a quantity $y(t)$ whose value at any given time time was the same as the outcome of measuring $y$. For something like the spin of an electron $y(t)$ would be either $+\tfrac{\hbar}{2}$ or $-\tfrac{\hbar}{2}$. The corresponding quantum observable would be $+\tfrac{\hbar}{2}$ times a matrix from a set of matrices satisfying the Pauli algebra, such as $$\frac{\hbar}{2}\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$

Unsurprisingly if you try to represent a quantity that is described by a Hermitian matrix by a single number instead the resulting theories contradict one another:

https://arxiv.org/abs/quant-ph/9906007

https://arxiv.org/abs/1109.6223

After you have done a measurement you only see a single outcome but that isn't because the outcome is described by a hidden variable. Rather, it is because interactions that produce copyable records of a measured quantity suppress interference, this is called decoherence:

https://arxiv.org/abs/quant-ph/0306072

This suppression of interference causes the different versions of the measurement device after the measurement to evolve independently of one another to a good approximation, this is often called the many worlds interpretation:

https://arxiv.org/abs/1111.2189

https://arxiv.org/abs/quant-ph/0104033

It should be noted that this approximation isn't perfect and the Bell correlations illustrate one way that this approximation can break down. The correspondence between measurement results on different systems only comes about after their measurement results are compared. Spatially separated systems are not yet divided into non-interfering relative states with respect to one another. For a description of local relative states see:

https://arxiv.org/abs/2008.02328

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.