On Friday, January 18, 2019 at 3:49:10 AM UTC, Brent wrote: > > > > On 1/17/2019 5:23 PM, [email protected] <javascript:> wrote: > > > > On Thursday, January 17, 2019 at 8:33:21 AM UTC, [email protected] > wrote: >> >> >> >> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote: >>> >>> >>> >>> On 1/16/2019 7:25 PM, [email protected] wrote: >>> >>> >>> >>> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote: >>>> >>>> >>>> >>>> On 1/13/2019 9:51 PM, [email protected] 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 >>> >> >> *That's what I suspected; that going to an orthogonal basis, I departed >> from the examples in Stackexchange where an arbitrary superposition is used >> in the analysis of interference. Nevertheless, isn't it possible to >> transform from an arbitrary superposition to one using an orthogonal basis? >> And aren't all bases equivalent from a linear algebra pov? If all bases are >> equivalent, why would transforming to an orthogonal basis lose >> interference, whereas a general superposition does not? TIA, AG* >> > > *I don't get it. If it's easy to show the existence of interference for a > general superposition where the components have different phase shifts, why > would the interference disappear for a special case using orthonormal basis > components? TIA, AG * > > > But taking the inner product with the *ith* eigenstate is not > transforming to a different basis. > > Brent >
*I know. I meant that from a general superposition used in the Stackexchange articles, I wrote that general form as a superposition of eigenstates, and this is where there was an implicit transformation to a different, specific basis. 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 [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
