# Re: Coherent states of a superposition

```> On 17 Jan 2019, at 09:33, agrayson2...@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 <javascript:> 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 donâ€™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

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