On Friday, January 18, 2019 at 4:26:25 AM UTC, agrays...@gmail.com wrote:
>
>
>
> On Friday, January 18, 2019 at 3:49:10 AM UTC, Brent wrote:
>>
>>
>>
>> On 1/17/2019 5:23 PM, agrays...@gmail.com wrote:
>>
>>
>>
>> On Thursday, January 17, 2019 at 8:33:21 AM UTC, 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 get it. If it's easy to show the existence of interference for a 
>> general superposition where the components have different phase shifts, why 
>> would the interference disappear for a special case using orthonormal basis 
>> components? TIA, AG *
>>
>>
>> But taking the inner product with the *ith* eigenstate is not 
>> transforming to a different basis.
>>
>> Brent
>>
>
> *I know. I meant that from a general superposition used in the 
> Stackexchange articles, I wrote that general form as a superposition of 
> eigenstates, and this is where there was an implicit transformation to a 
> different, specific basis. AG *
>

*I suppose you could start with a superposition of eigenstates and get, or 
not get interference depending on the type of mathematical operation is 
performed. So I am unclear what's going on here; why taking the inner 
product is tantamount to looking at which slit the particle is going 
through.  AG *

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