> On 17 Jan 2019, at 09:33, agrayson2...@gmail.com wrote: > > > > 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 > > 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 donâ€™t understand this. All the bases we have used all the time are supposed to be orthonormal bases. We suppose that the scalar product (e_i e_j) = delta_i_j, when presenting the Born rule, and the quantum formalism. Bruno > > -- > 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 > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to everything-list@googlegroups.com > <mailto:everything-list@googlegroups.com>. > Visit this group at https://groups.google.com/group/everything-list > <https://groups.google.com/group/everything-list>. > For more options, visit https://groups.google.com/d/optout > <https://groups.google.com/d/optout>. -- 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.