On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote: > > > > On 1/16/2019 7:25 PM, agrays...@gmail.com <javascript:> wrote: > > > > On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote: >> >> >> >> On 1/13/2019 9:51 PM, agrays...@gmail.com wrote: >> >> This means, to me, that the arbitrary phase angles have absolutely no >> effect on the resultant interference pattern which is observed. But isn't >> this what the phase angles are supposed to effect? AG >> >> >> The screen pattern is determined by *relative phase angles for the >> different paths that reach the same point on the screen*. The relative >> angles only depend on different path lengths, so the overall phase angle is >> irrelevant. >> >> Brent >> > > > *Sure, except there areTWO forms of phase interference in Wave Mechanics; > the one you refer to above, and another discussed in the Stackexchange > links I previously posted. In the latter case, the wf is expressed as a > superposition, say of two states, where we consider two cases; a > multiplicative complex phase shift is included prior to the sum, and > different complex phase shifts multiplying each component, all of the form > e^i (theta). Easy to show that interference exists in the latter case, but > not the former. Now suppose we take the inner product of the wf with the > ith eigenstate of the superposition, in order to calculate the probability > of measuring the eigenvalue of the ith eigenstate, applying one of the > postulates of QM, keeping in mind that each eigenstate is multiplied by a > DIFFERENT complex phase shift. If we further assume the eigenstates are > mutually orthogonal, the probability of measuring each eigenvalue does NOT > depend on the different phase shifts. What happened to the interference > demonstrated by the Stackexchange links? TIA, AG * > > Your measurement projected it out. It's like measuring which slit the > photon goes through...it eliminates the interference. > > Brent >

*But if that's the case, won't the probability density of the eigenvalue being measured (by Born's rule) be the value in the absence of interference, which I presume is the classical value? AG * -- 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.