Jump to content

Recommended Posts

Posted

It is said, that entanglement of two particles can make them expose correlated results of measured observable. For example, we can entangle two electrons to have contrary oriented spins. Also we can entangle them to have spins co-oriented. We can entangle other observables, not only a spin.

 

The question is can we entagle time also?

 

For example, can we entangle two nuclei to decay simultaneously? Or can we make them to decay at specific moments?

 

If we can, then how this time-relation entanglement will correspond with special relativity with it's relativism of a time?

Posted

But every measurement is inherently random. For example, if we have 45 degree polarized photon, then it will randomly bypass vertical polarizer or stop. I. e. the answer for the question "is it 90 degree polarized or 0 degree" is inherent random.

 

But we CAN entangle two photons.

Posted

I don't pretend to know much about this, but...

 

you can entangle the spin because it's an intrinsic value of the photon. Can time be considered an intrinsic value also?

Posted

You can induce transitions from excited states; that's what lasers do (stimulated emission)

 

But time is not an observable - not a property of the particle. You entangle properties that are governed by some conservation law.

Posted

Is it such a perfect thing? The closest I've come in my readings, and also sitting and figuring the random angle of two polarized detectors affecting coincidence, is that rather than depending on the cosine of the relative angle, the probabilty has a linear term also, not contained in a cosine expansion. It even seems to me that we are speaking of an information loss, not gain.

Posted

Fredrik educates me as I look up "Bayes rule". I'll be damned it sounds like just what I'm talking of here, but I don't grok it yet. What I worked out is the probability of two polarized detectors at relative angle phi to flash in coincidence, given a randomly oriented pair of equal and opposite photons. This yields P= 1/2 + (1/4)cos(2phi). Is this the right starting point?

  • 2 weeks later...
Posted

Look at the case where you maintain polarized detectors "at the same angle in space". The relative angle is zero; why is the answer only 3/4? Since we have offered a "monte carlo" of random polarizations, some of those detected in one counter do not make it in the other.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.