Recent content by gentzen

You are missing Your sign measurement operators ##S_x## and ##S_p## are no longer representable by a single state, hence you have to represent them as matrices/operators. But forming the superposition of matrices doesn't give you interference. You only get interference, if you form the...

Yes, of course. I guess what you didn't understand is that your ##|\alpha|^2S_x+|\beta|^2S_p## is missing those interference terms. That was my point, why I showed you an observable where those terms are not missing.

This is an observable. Your ##S_x## and ##S_p## were observables too. Maybe to clarify, we have ##\alpha=\cos(\theta)## and ##\beta=\sin(\theta)## in your case.

Suppose you start with the state ##|\psi_{\alpha,\beta}\rangle=\alpha |z+\rangle+\beta|z-\rangle##. Then there is a measurement that will always give +1, and never -1. I want to realize that measurement. When you measure the state ##|\psi\rangle=|z+\rangle## with that measurement, you will get...

The state is entangled, but pines-demon tried to get rid of the continuous measurement outcomes, to be closer to typical Bell measurements. I told him before So when pines-demon came up with a sufficiently well specified experiment with binning, I tried to explain to him why that specific...

You mean, because they are not normalizable? But they are well defined in a suitable space of distributions, and one can form their complex superposition in that space. Less sure about the ##|x\rangle\langle x|## part. Products of distributions can be nasty, and taking the dual is also...

$$(\alpha|x\rangle+\beta|p\rangle)(\alpha^*\langle x|+\beta^*\langle p|)=|\alpha|^2|x\rangle\langle x|+|\beta|^2|p\rangle\langle p|+\text{interference terms}$$ $$\text{interference terms} = \alpha\beta^*|x\rangle\langle p| +\alpha^*\beta|p\rangle\langle x|$$

If you define it like this, then ##S_x## and ##S_p## are already matrices (or rather operators), basically ##S_x=\int_0^\infty dx\, |x\rangle\langle x|## and ##S_p=\int_0^\infty dp\, |p\rangle\langle p|##. And your ##S_{\theta}## then doesn't contain interferences between ##|x\rangle## and...

It just makes it easy for me to see that the state is not time invariant. With the signs of your state, I would have been a bit less sure, even so I guess that it won't be time invariant either. No, I had no problem with being in the same position in that sense. I just wanted to nail down the...

Because the spin in the normal "measurements showing Bell's inequality violations" is constant in time. The position is not, which on the one hand gives the possibility to measure "some mixture between momentum and position at creation time". And on the other hand, it constitutes an important...

Because starting with |p>|-p> corresponds to the EPR description of the experiment (the total momentum is zero). In principle, one should be able to compute the position representation from that. I just guessed that it would be the integral over |x>|x>, because your position and moment...

It measures the spin of the particle in some specific direction, as it was initially before the bending of the beam. (I am still a bit confused about the number of parameters. An arbitrary direction in 3D space would still only be two continuous parameters. Maybe I should try to look up the...

If I have two entangled particles in "two beams" with known fixed direction, then I want to measure their spin, not the spin of two other particles corresponding to different "two beams" with different known fixed directions. I could try to bent a given beam with some combination of electric...

Now I understand what you wanted to say: The initial state ##|\psi\rangle## for the EPR experiment can be assumed to be ψ... I guess |x>|-x> as starting point is "unusual" for the EPR experiment, I would rather have started with the integral over |p>|-p> (possibly restricted by some max energy)...

Maybe read what you wrote exactly, see whether it makes any sense to you. And then ask yourself, how anybody who wants to have a serious discussion focused on physical details should respond to that. Wow, you still ignore the issue of when the detection happens, compared to the moment of...