# Re: Coherent states of a superposition

```
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 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
>```
```
*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*

>
> --
> You received this message because you are subscribed to the Google Groups
> "Everything List" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> To post to this group, send email to everyth...@googlegroups.com
> <javascript:>.
> Visit this group at https://groups.google.com/group/everything-list.
> For more options, visit https://groups.google.com/d/optout.
>
>
>

--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email