> On 18 Jan 2019, at 18:50, [email protected] wrote: > > > > On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote: > >> On 17 Jan 2019, at 14:48, [email protected] <javascript:> wrote: >> >> >> >> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote: >> >>> On 17 Jan 2019, at 09:33, [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 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 >> >> Generally, bases in a vector space are NOT orthonormal. > > Right. But we can always build an orthonormal base with a decent scalar > product, like in Hilbert space, > > > >> For example, in the vector space of vectors in the plane, any pair of >> non-parallel vectors form a basis. Same for any general superposition of >> states in QM. HOWEVER, eigenfunctions with distinct eigenvalues ARE >> orthogonal. > > Absolutely. And when choosing a non degenerate observable/measuring-device, > we work in the base of its eigenvectors. A superposition is better seen as a > sum of some eigenvectors of some observable. That is the crazy thing in QM. > The same particle can be superposed in the state of being here and there. Two > different positions of one particle can be superposed. > > This is a common misinterpretation. Just because a wf can be expressed in > different ways (as a vector in the plane can be expressed in uncountably many > different bases), doesn't mean a particle can exist in different positions in > space at the same time. AG
It has a non null amplitude of probability of being here and there at the same time, like having a non null amplitude of probability of going through each slit in the two slits experience. If not, you can’t explain the inference patterns, especially in the photon self-interference. > > Using a non orthonormal base makes only things more complex. >> I posted a link to this proof a few months ago. IIRC, it was on its >> specifically named thread. AG > > But all this makes my point. A vector by itself cannot be superposed, but can > be seen as the superposition of two other vectors, and if those are > orthonormal, that gives by the Born rule the probability to obtain the "Eigen > result” corresponding to the measuring apparatus with Eigen vectors given by > that orthonormal base. > > I’m still not sure about what you would be missing. > > You would be missing the interference! Do the math. Calculate the probability > density of a wf expressed as a superposition of orthonormal eigenstates, > where each component state has a different phase angle. All cross terms > cancel out due to orthogonality, ? Sin(alpha) up + cos(alpha) down has sin^2(alpha) probability to be fin up, and cos^2(alpha) probability to be found down, but has probability one being found in the Sin(alpha) up + cos(alpha) down state, which would not be the case with a mixture of sin^2(alpha) proportion of up with cos^2(alpha) down particles. Si, I don’t see what we would loss the interference terms. > and the probability density does not depend on the phase differences. What > you get seems to be the classical probability density. AG I miss something here. I don’t understand your argument. It seems to contradict basic QM (the Born rule). Bruno > > 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 [email protected] <>. >>> To post to this group, send email to [email protected] <>. >>> 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 [email protected] <javascript:>. >> To post to this group, send email to [email protected] >> <javascript:>. >> 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 [email protected] > <mailto:[email protected]>. > To post to this group, send email to [email protected] > <mailto:[email protected]>. > 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 [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.
