On Tuesday, November 5, 2024 at 6:45:43 AM UTC-7 smitra wrote:
On 03-11-2024 06:16, Alan Grayson wrote: > The wf under consideration is 1/sqrt (2)[ |spin Up> + |spin Dn> ], and > my question is this: > Why PRECISELY is the Ignorance Interpretation false? Brent says it is > "exactly wrong". I'd like to know his reasoning, or anyone's > reasoning. TY, AG This can most clearly be seen by considering a certain measurement on the state: |psi> = 1/sqrt(2) [|u1, u2, u3> - |d1, d2, d3>] where ui denotes spin up of the ith particle and di denotes spin down for the ith particle relative to a defined z-axis. We then consider 3 observers where observer i is going to measure spin nr. i. *"We then consider 3 observers where observer i is going to measure spin nr. i." ? AG* And we then consider the observable Ai which corresponds to observer i measuring the x-component of spin i while the two observers measure the y-components of the spins allocated to them. And we then assign a value ±1 for measuring spin op or down, and Ai multiples the results of the measurements for the 3 observers. How does Ai act on psi? It's easy to see that: sigma_x|u> = |d> sigma_x|d> = |u> sigma_y|u> = i |d> sigma_y|d> = -i |u> If we then let Ai act on |psi>, then the up and down states are reversed, and the two sigma_y operators yield a minus sign for both the first and the second term. So, we see that the two terms in |psi> get interchanged and there is an overall minus sign, so the ent effect is that the state is left unchanged: Ai|psi> = |psi> So, the outcome of measuring two y components and one x component and multiplying the results with each other is always going to yield 1 as the answer. But suppose that the measurement results of the spins are locally predetermined. When we measure spin up then there was some set of local hidden variable that made the outcome to be spin up when deciding to measure the spin in the direction it was measured in, and the variables would also tell us what would have happened had we measured a different component of the spin. With the outcome of all possible measurements being predetermined in the hidden variable scenario and given that we know that measuring each Ai is always going to yield, we have that the value of A1 A2 A3 equals 1. While we can only measure 1 of the Ai, all 3 are well defined independent of what we choose to measure if we assume hidden variables. If we denote the unknown value of the ith spin in the x and y directions by Six and Siy, respectively, then we have: 1 = A1 A2 A3 = (S1x S2y S3y) (S1y S2x S3y) (S1y S2y S3x)= S1x S2x S3x S1y^2 S2y^2 S3y^2 = S1x S2x S3x because the square of a spin is always 1. So, if we assume local hidden variables, we predict that if all observers measure the x component that then the product of their results always equals 1. But, with sigma_x flipping up and down states we have: sigmax1 sigmax2 sigmax3|psi> = -|psi> So, measuring S1x S2x S3x will always yield minus 1, not plus 1. Saibal -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/707e16cf-475a-41ca-b8d0-c67c7898d8b7n%40googlegroups.com.
