Recent content by pines-demon

This is getting very cryptic. I take ##S_\theta=S_x\cos\theta+S_p\sin\theta=\alpha S_x+\beta S_p##, I do what then? ##S_\theta^2## or ##S_\theta S_\theta^\dagger##? Either way I get interference term right? Is not that different to your calculation. What am I missing?

Isn’t ##\alpha\beta=\cos\theta\sin\theta## different than 0 for most angles?

Is this related to the state? to the observables or to what? I am lost here. There are clearly interference terms in whatever you are doing there.

I think you misunderstood, see the edit to post #37, are you telling me that you can write $$|\psi\rangle=\int \mathrm d x |x\rangle |x\rangle$$ as the product of two states? How? Spin zero fermions?

Can you be more explicit? What do you mean that it does not contain interferences?

We were writing in short form, the initial state has an integral ##\int \mathrm d x\, |x\rangle |\pm x\rangle##. (I will edit it) Also it is not clear that you have fermions or not with your state because we don't know what is going on with spin. Not that it matters.

Ok lets try. Prepare your state in whatever version you prefer ##\int \mathrm d x\, |x\rangle|\pm x\rangle##. Now you have two options either you keep continuous variables and then you need a continuous variables extension of the Bell inequalities (there are various approaches). Or you bin, by...

Fine take ##|p\rangle|-p\rangle## what is the problem of being in the same position? We can figure out the spin or use a boson.

Experiment.

What is the issue with that? just take a finite time after that? Why are you focusing on that point? You could instead start with some weird potential where that is its prepared ground-state, at least to first approximation. Also why take a different sign convention to mine?

Thanks for stepping in. I hoped that there was something stronger that I was missing. Trying to perform Bell tests on the original EPR set up is a hard task, but there is no argument to discard it right out of the bat.

Does it matter? just prepare the state. Agreed

Sorry if we are speaking past each other, I wanted an answer to a previous statement of another user, I am not even following the rest of the conversation if that has anything to do with your question. Let's say a particle detector, the position is where we detect one of the entangled...

Take some state like ##|\psi\rangle=\int \mathrm d x\, |x\rangle|-x\rangle## (if you are worried that this is not normalized correctly, limit it to a given space), in momentum it looks something like ##|\psi\rangle=\int \mathrm d p\, |p\rangle|p\rangle## The position of detection, keep it 1D...

What is the issue? There are Bell inequalities for continuous variables or you can discretize your space. I mean you measure the particle at three angles (mixtures of position and momentum). I am not saying it is easy but I see no argument against it.