# Re: Coherent states of a superposition

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On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote:
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> On 1/16/2019 7:25 PM, agrays...@gmail.com <javascript:> wrote:
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> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote:
>>
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>> 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
>>
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> *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
>```
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*But if that's the case, won't the probability density of the eigenvalue
being measured (by Born's rule) be the value in the absence of
interference, which I presume is the classical value? AG *

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