> On 30 Jan 2019, at 02:59, agrayson2...@gmail.com wrote:
> 
> 
> 
> On Tuesday, January 29, 2019 at 4:37:34 AM UTC-7, Bruno Marchal wrote:
> 
>> On 28 Jan 2019, at 22:50, agrays...@gmail.com <javascript:> wrote:
>> 
>> 
>> 
>> On Friday, January 25, 2019 at 7:33:05 AM UTC-7, Bruno Marchal wrote:
>> 
>>> On 24 Jan 2019, at 09:29, agrays...@gmail.com <> wrote:
>>> 
>>> 
>>> 
>>> On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com 
>>> <http://gmail.com/> wrote:
>>> 
>>> 
>>> 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 
>>>>> 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). 
>>> 
>>> 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
>> 
>> I assume orthNORMAL eigenfunctions. I assume the probability densities sum 
>> to unity. Then, using Born's rule, I have shown that multiplying each 
>> component by e^i(theta) where theta is arbitrarily different for each 
>> component, disappears when the probability density is calculated, due to 
>> orthonormality.
> 
> 
> That seems to violate elementary quantum mechanics. If e^I(theta) is 
> different for each components, Born rule have to give different probabilities 
> for each components---indeed given by the square of e^I(theta).
> 
> The norm squared of e^i(thetai) is unity, except for the cross terms which is 
> zero due to orthonormality. AG 
> 
>> What you've done, if I understand correctly, is measure the probability 
>> density using different bases, and getting different values.
> 
> The value of the relative probabilities do not depend on the choice of the 
> base used to describe the wave. Only of the base corresponding to what you 
> decide to measure. 
> 
> 
> 
>> This cannot be correct since the probability density is an objective value, 
>> and doesn't depend on which basis is chosen. AG
> 
> Just do the math. Or read textbook.
> 
> Why don't YOU do the math ! It's really simple. Just take the norm squared of 
> a superposition of component eigenfunctions, each multiplied by a probability 
> amplitude, and see what you get !  No need to multiply each component by 
> e^i(thetai).  Each amplitude has a phase angle implied. This is Born's rule 
> and the result doesn't depend on phase angles, contracting what Bruce wrote 
> IIUC. If you would just do the simple calculation you will see what I am 
> referring to! AG


Bruce is right. Let us do the computation in the simple case where e^i(theta) = 
-1. (Theta = Pi)

Take the superposition (up - down), conveniently renormalised. If I multiply 
the whole wave (up - down) by (-1), that changes really nothing. But if I 
multiply only the second term, I get the orthogonal state up + down, which 
changes everything. (up +down) is orthogonal to (up - down).

Bruno



> 
> 
> The probabilities dos not depend on the choice of the base, but they are 
> different when the components are different, given that the probabilities are 
> given by the quake of those components, in the base corresponding to what you 
> decide to measure, and this in all base used to describe the wave.
> 
> OK, AG 
> 
> Bruno
> 
> 
> 
> 
>> 
>> 
>> 
>> 
>>> 
>>> 
>>> Bruno
>>> 
>>> 
>>> 
>>> 
>>> 
>>> 
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