On 1/17/2019 5:23 PM, agrayson2...@gmail.com wrote:

On Thursday, January 17, 2019 at 8:33:21 AM UTC, 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


        *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.


    *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 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 *

But taking the inner product with the /ith/ eigenstate is not transforming to a different basis.


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