> On 24 Jan 2019, at 09:29, [email protected] wrote: > > > > On Sunday, January 20, 2019 at 11:54:43 AM UTC, [email protected] > <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, [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,
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 [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.
