# Re: Coherent states of a superposition

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On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com wrote:
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>
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> On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote:
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>>
>> On 18 Jan 2019, at 18:50, agrays...@gmail.com wrote:
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>>
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>> On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote:
>>>
>>>
>>> On 17 Jan 2019, at 14:48, agrays...@gmail.com wrote:
>>>
>>>
>>>
>>> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote:
>>>>
>>>>
>>>> On 17 Jan 2019, at 09:33, 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 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. *
>>>
>>>
>>> Right. But we can always build an orthonormal base with a decent scalar
>>> product, like in Hilbert space,
>>>
>>>
>>>
>>> *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.*
>>>
>>>
>>> Absolutely. And when choosing a non degenerate
>>> observable/measuring-device, we work in the base of its eigenvectors. A
>>> superposition is better seen as a sum of some eigenvectors of some
>>> observable. That is the crazy thing in QM. The same particle can be
>>> superposed in the state of being here and there. Two different positions of
>>> one particle can be superposed.
>>>
>>
>> *This is a common misinterpretation. Just because a wf can be expressed
>> in different ways (as a vector in the plane can be expressed in uncountably
>> many different bases), doesn't mean a particle can exist in different
>> positions in space at the same time. AG*
>>
>>
>> It has a non null amplitude of probability of being here and there at the
>> same time, like having a non null amplitude of probability of going through
>> each slit in the two slits experience.
>>
>> If not, you can’t explain the inference patterns, especially in the
>> photon self-interference.
>>
>>
>>
>>
>>
>> Using a non orthonormal base makes only things more complex.
>>>
>> * I posted a link to this proof a few months ago. IIRC, it was on its
>>>
>>>
>>> But all this makes my point. A vector by itself cannot be superposed,
>>> but can be seen as the superposition of two other vectors, and if those are
>>> orthonormal, that gives by the Born rule the probability to obtain the
>>> "Eigen result” corresponding to the measuring apparatus with Eigen vectors
>>> given by that orthonormal base.
>>>
>>> I’m still not sure about what you would be missing.
>>>
>>
>> *You would be missing the interference! Do the math. Calculate the
>> probability density of a wf expressed as a superposition of orthonormal
>> eigenstates, where each component state has a different phase angle. All
>> cross terms cancel out due to orthogonality,*
>>
>>
>> ?  Sin(alpha) up + cos(alpha) down has sin^2(alpha) probability to be fin
>> up, and cos^2(alpha) probability to be found down, but has probability one
>> being found in the Sin(alpha) up + cos(alpha) down state, which would not
>> be the case with a mixture of sin^2(alpha) proportion of up with
>> cos^2(alpha) down particles.
>> Si, I don’t see what we would loss the interference terms.
>>
>>
>>
>> * and the probability density does not depend on the phase differences.
>> What you get seems to be the classical probability density. AG *
>>
>>
>>
>> I miss something here. I don’t understand your argument. It seems to
>> contradict basic QM (the Born rule).
>>
>
> *Suppose we want to calculate the probability density of a superposition
> consisting of orthonormal eigenfunctions, each multiplied by some amplitude
> and some arbitrary phase shift. If we take the norm squared using Born's
> Rule, don't all the cross terms zero out due to orthonormality? Aren't we
> just left with the SUM OF NORM SQUARES of each component of the
> superposition? YES or NO? If YES, the resultant probability density doesn't
> depend on any of the phase angles. AG*
>```
```
*YES or NO? AG *

>
>
>> Bruno
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