On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote:
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> On 17 Jan 2019, at 09:33, agrays...@gmail.com <javascript:> wrote:
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>
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> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote:
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>>
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>> On 1/16/2019 7:25 PM, agrays...@gmail.com wrote:
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>> 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. 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. I posted a link to 
this proof a few months ago. IIRC, it was on its specifically named thread. 
AG*

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