> On 24 Jan 2019, at 09:29, agrayson2...@gmail.com wrote: > > > > On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com > <http://gmail.com/> wrote: > > > On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote: > >> On 18 Jan 2019, at 18:50, agrays...@gmail.com <> wrote: >> >> >> >> On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote: >> >>> On 17 Jan 2019, at 14:48, agrays...@gmail.com <> wrote: >>> >>> >>> >>> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote: >>> >>>> On 17 Jan 2019, at 09:33, agrays...@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 <> 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 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,
Distinct eigenvalue correspond to orthonormal vector, so I tend to always superpose only orthonormal functions, related to those eigenvalue. > each multiplied by some amplitude and some arbitrary phase shift. like (a up + b down), but of course we need a^2 + b^2 = 1. You need to be sure that you have normalised the superposition to be able to apply the Born rule. > If we take the norm squared using Born's Rule, don't all the cross terms zero > out due to orthonormality? ? The Born rule tell you that you will find up with probability a^2, and down with probability b^2 > Aren't we just left with the SUM OF NORM SQUARES of each component of the > superposition? YES or NO? If you measure in the base (a up + b down, a up -b down). In that case you get the probability 1 for the state above. > If YES, the resultant probability density doesn't depend on any of the phase > angles. AG > > YES or NO? AG Yes, if you measure if the state is a up + b down or a up - b down. No, if you measure the if the state is just up or down 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 everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to email@example.com > <mailto:firstname.lastname@example.org>. > 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 email@example.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.