# Re: Coherent states of a superposition

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On 1/17/2019 5:23 PM, agrayson2...@gmail.com wrote:
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On Thursday, January 17, 2019 at 8:33:21 AM UTC, agrays...@gmail.com wrote:
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On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote:

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:

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
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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
*

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```        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*

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*I don't get it. If it's easy to show the existence of interference for a general superposition where the components have different phase shifts, why would the interference disappear for a special case using orthonormal basis components? TIA, AG *
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But taking the inner product with the /ith/ eigenstate is not transforming to a different basis.
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Brent

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