Jilang said:
I understand that the probabilities don’t factorise, but I am curious as to whether the probability amplitudes do. Is there a consensus on that?
I posted a note about this a year or so ago. Here's the summary:
In a spin-1/2 anti-correlated EPR experiment, the joint probability that Alice will measure spin-up at angle \alpha and Bob will measure spin-up at angle \beta is given by:
P(A \hat B | \alpha, \beta) = \frac{1}{2} sin^2(\frac{\beta - \alpha}{2})
According to Bell, to explain this via local hidden variables would require a parameter \lambda and three probability distributions:
- P(\lambda)
- P(A | \alpha, \lambda)
- P(B | \beta, \lambda)
such that:
P(A \hat B | \alpha, \beta) = \sum_\lambda P(\lambda) P(A | \alpha, \lambda) P(B | \beta, \lambda)
(where the sum might be an integral, if \lambda takes on continuous values). If there were such probability distributions, then there would be a straight-forward local explanation of the EPR result:
- Each pair of particles has a corresponding "hidden variable" \lambda selected according to the probability distribution P(\lambda)
- When Alice measures her particle, she gets spin-up with probability P(A|\alpha, \lambda)
- When Bob measures his particle, he gets spin-up with probability P(B|\beta, \lambda)
- These probabilities are independent.
Alas, Bell showed that there were no such probability distributions. Now, you can ask the analogous question about amplitudes:
Let \psi(A \hat B | \alpha, \beta) be the amplitude (square-root of the probability, basically) that Alice and Bob will get spin-up at their respective angles. So \psi(A \hat B | \alpha, \beta) = \frac{1}{\sqrt{2}} sin(\frac{\beta - \alpha}{2}) (times a phase factor). Then the analogous question for amplitudes is:
Does there exist a parameter \lambda and amplitudes \psi(\lambda), \psi(A | \alpha, \lambda), \psi(B | \beta, \lambda) such that:
\psi(A \hat B | \alpha, \beta) = \sum_\lambda \psi(\lambda) \psi(A | \alpha, \lambda) \psi(B | \beta, \lambda) ?
The answer is straightforwardly "yes".
- Let \lambda have two possible values, \lambda_u and \lambda_d.
- Let the amplitude for these values be: \psi(\lambda_u) = \frac{1}{\sqrt{2}} and \psi(\lambda_d) = - \frac{1}{\sqrt{2}}
- Let the conditional amplitudes be:
- \psi(A | \alpha, \lambda_u) = cos(\frac{\alpha}{2})
- \psi(A | \alpha, \lambda_d) = sin(\frac{\alpha}{2})
- \psi(B | \beta, \lambda_u) = sin(\frac{\alpha}{2})
- \psi(B | \beta, \lambda_d) = cos(\frac{\beta}{2})
Then \psi(A \hat B | \alpha, \beta) = \sum_\lambda \psi(\lambda) \psi(A |\alpha, \lambda) \psi(B |\beta, \lambda)
= \psi(\lambda_u) \psi(A |\alpha, \lambda_u) \psi(B|\beta, \lambda_u) + \psi(\lambda_d) \psi(A |\alpha, \lambda_d) \psi(B|\beta, \lambda_d)
= \frac{1}{\sqrt{2}} (cos(\frac{\alpha}{2}) sin(\frac{\beta}{2}) - sin(\frac{\alpha}{2}) cos(\frac{\beta}{2})
= \frac{1}{\sqrt{2}} sin(\frac{\beta - \alpha}{2})
I don't know physically what it means that amplitudes, rather than probabilities factor, but it shows that quantum problems are often a lot simpler in terms of amplitudes.