# Re: Coherent states of a superposition

```> On 18 Jan 2019, at 18:50, agrayson2...@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 <javascript:> 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
>> specifically named thread. AG
>
> 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).

Bruno

>
> Bruno
>
>
>
>
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