I'm trying to understand the motivation for local hidden variable theory. The EPR paradox considers the following thought experiment, where we can express a state $|\psi \rangle \in H_{Alice} \otimes H_{Bob}$ as perfectly "opposite" eigenvectors of a measurement $A$ and another measurement $B$

$|\psi \rangle = \sum |u_{n} \rangle | \psi_{n} \rangle = \sum |v_{n} \rangle | \varphi_{n} \rangle$

The original example was formulated in the case $A$ was the momentum operator and $B$ was the position operator. A convenient example, is the one due to Bohm, where $|\psi \rangle = \frac{1}{\sqrt{2}}|01 \rangle - \frac{1}{\sqrt{2}}|10 \rangle $ is the singlet state and $A = \sigma_{z}$ is measurement along the $z$-axis and $B = \sigma_{x}$ is measurement along the $x$-axis. In the case of the singlet state, we can express $|\psi \rangle $ as either in terms of the eigenvectors $|0 \rangle$ and $|1 \rangle$ of $\sigma_{z}$ with values $\pm 1$ respectively or in terms of the eigenvectors $\frac{1}{2}( |0 \rangle +|1 \rangle)$, $\frac{1}{2}( |0 \rangle -|1 \rangle)$ of $\sigma_x$ with values $\pm 1$ respectively.The paradox occurs with the assumption of local causality but with the observations that Alice's measurements on her system collapse Bob's system to the eigenvector with the exact opposite eigenvalue. This leads Alice to be able to predict with certainty the values of two non-commuting measurements.

I was wondering how the local variable system, by defining two random variables $A(a, \lambda)$ and $B(b, \lambda)$ on a probability space $(\Lambda, p(\lambda))$ denoting the values taken by the measurements $A$ and $B$. Rectify the above paradox.

The idea that you could explain QM by local hidden variables was eventually proven false by Bell's theorem, but I wanted to understand why this was the attempt to complete QM with respect to this example in the first place.


1 Answer 1


If we have local hidden variables, then there's no real "collapse". When the particles were created, they had definite "plans" for what they would do if they encountered a detector aligned along any conceivable axis. What we call a "singlet state" is just a conspiracy that the two particles entered into when they were created, comparing each other's "plans" and making sure that if particle A came out of a detector spin-up, then particle B would come out of the same kind of detector spin-down.

Standard quantum mechanics, under this interpretation, is just incomplete. The probabilistic interpretation of quantum mechanics is just a sign that we don't have all the information about the "real" state of the particle; there are variables determining the state of the particle that are hidden from us.

  • $\begingroup$ So formally, do we think of the conspiracy as the hidden variable? $\endgroup$
    – user135520
    Nov 17, 2021 at 3:24
  • 2
    $\begingroup$ @user135520 : I think the biggest difference between your proposal, "two random variables $A(a,\lambda)$ and $B(b,\lambda)$ on a probability space $(\Lambda,p(\lambda))$ denoting the values taken by the measurements $A$ and $B$" and what is suggested here, is that hidden variables theory doesn't give two random variables, it gives one non-random variable on the same space (i.e., the "conspiratorial" deterministic measurement pair correlations). We can't measure the hidden variables, so these measurements seem random to us. $\endgroup$ Nov 17, 2021 at 7:06
  • $\begingroup$ I think actually in the case of perfect correlation such as the singlet state, we have that $B(p,\lambda)=-A(p, \lambda)$ where $p$ is the measurement setting, or direction we are measuring in, so computing the correlation $E[AB] = -E[A^{2}]=-1$ $\endgroup$
    – user135520
    Nov 17, 2021 at 20:52
  • $\begingroup$ The EPR paradox says only one side (local) should be measured, so should the measurement operator $A\otimes\mathbb{1}_2$ be considered ? I tried to introduce a measurement matrix whose colums are made of multiples of an eigenvector. Applying this matrix at the place A gives a parametrized probability. This cannot be above 1/2 since a subuniverse of probabilities is considered. But it can be 0 for some parameter, which gives in turn info about the other side. $\endgroup$ Feb 14, 2022 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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