# Re: Coherent states of a superposition

```> On 24 Jan 2019, at 09:29, agrayson2...@gmail.com wrote:
>
>
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> On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com
> <http://gmail.com/> wrote:
>
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> On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote:
>
>> On 18 Jan 2019, at 18:50, agrays...@gmail.com <> wrote:
>>
>>
>>
>> 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,```
```
Distinct eigenvalue correspond to orthonormal vector, so I tend to always
superpose only orthonormal functions, related to those eigenvalue.

> each multiplied by some amplitude and some arbitrary phase shift.

like  (a up + b down), but of course we need a^2 + b^2 = 1. You need to be sure
that you have normalised the superposition to be able to apply the Born rule.

> If we take the norm squared using Born's Rule, don't all the cross terms zero
> out due to orthonormality?

?

The Born rule tell you that you will find up with probability a^2, and down
with probability b^2

> Aren't we just left with the SUM OF NORM SQUARES of each component of the
> superposition? YES or NO?

If you measure in the base (a up + b down, a up -b down). In that case you get
the probability 1 for the state above.

> If YES, the resultant probability density doesn't depend on any of the phase
> angles. AG
>
> YES or NO? AG

Yes, if you measure if the state is a up + b down or a up - b down.
No, if you measure the if the state is just up or down

Bruno

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> Bruno
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