> On 17 Jan 2019, at 14:48, agrayson2...@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 <javascript:> 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. 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.

Bruno




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