Re: Coherent states of a superposition

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On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote:
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> On 17 Jan 2019, at 14:48, agrays...@gmail.com <javascript:> wrote:
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> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote:
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>> On 17 Jan 2019, at 09:33, agrays...@gmail.com wrote:
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>> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote:
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>>> On 1/16/2019 7:25 PM, agrays...@gmail.com wrote:
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>>> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote:
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>>>> On 1/13/2019 9:51 PM, agrays...@gmail.com wrote:
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>>>> 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
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> *Generally, bases in a vector space are NOT orthonormal. *
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> Right. But we can always build an orthonormal base with a decent scalar
> product, like in Hilbert space,
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> *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.*
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> 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.
>```
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*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*

Using a non orthonormal base makes only things more complex.
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* I posted a link to this proof a few months ago. IIRC, it was on its
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> 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.
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> 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, and the probability density
does not depend on the phase differences.  What you get seems to be the
classical probability density. AG *

>
> Bruno
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