On Sunday, January 20, 2019 at 11:54:43 AM UTC, [email protected] wrote: > > > > On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote: >> >> >> 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] 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). >> > > *Suppose we want to calculate the probability density of a superposition > consisting of orthonormal eigenfunctions, each multiplied by some amplitude > and some arbitrary phase shift. If we take the norm squared using Born's > Rule, don't all the cross terms zero out due to orthonormality? Aren't we > just left with the SUM OF NORM SQUARES of each component of the > superposition? YES or NO? If YES, the resultant probability density doesn't > depend on any of the phase angles. AG* >
*YES or NO? AG * > > >> 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. For more options, visit https://groups.google.com/d/optout.
