Dear Poojary, I will have to disappoint you by saying that your arguments in your paper are false. The meaning of [x,p] = i (for simplicity of notation, I identify h-bar with 1), which you have used in your paper, is not that [x,p] ψ = ι ψ for some ψ, but for all ψ in the single-particle Hilbert space of the problem at hand. Two notions are very relevant here. Firstly, the notion of Hilbert space (a complete normed space) and arbitrariness of ψ in this Hilbert space. Therefore, two important questions that you will have to address are the following: 1. What is the Hilbert space to which your "proof" applies? 2. Having specified this Hilbert space, is your function, ψ, in this Hilbert space, and, if so, is it the most general function in this space? Having very briefly looked through the pdf file of your paper that you have provided here above, my definite conclusion is that your "proof" is no proof at all. It merely amounts to some exercise in the application of the complex coordinates on the two-dimensional plane.

Dear Poojary, I will have to disappoint you by saying that your arguments in your paper are false. The meaning of [x,p] = i (for simplicity of notation, I identify h-bar with 1), which you have used in your paper, is not that [x,p] ψ = ι ψ for some ψ, but for all ψ in the single-particle Hilbert space of the problem at hand. Two notions are very relevant here. Firstly, the notion of Hilbert space (a complete normed space) and arbitrariness of ψ in this Hilbert space. Therefore, two important questions that you will have to address are the following: 1. What is the Hilbert space to which your "proof" applies? 2. Having specified this Hilbert space, is your function, ψ, in this Hilbert space, and, if so, is it the most general function in this space? Having very briefly looked through the pdf file of your paper that you have provided here above, my definite conclusion is that your "proof" is no proof at all. It merely amounts to some exercise in the application of the complex coordinates on the two-dimensional plane.

Dr Behnam Farid,

Thank you for your answer, I am not disappointed.

I used already exisiting proof for HUP using metric mechanics [x.p-p.x = ih].

I am using complete metric space (hilbert space).

I had one proof of HUP which used metric mechanics to prove the position and movementum do not comute (which means it has non zero output) and thats where uncertainity comes into picture.

I argued if we take position and movementum to complete metric space , position and movementum comute (which is equal to zero).

Hence I take it has a proof for proving HUP is incomplete picture of what we actually observe.

Kindly let me know if you want to know anything more.

Really appreciate that you took your time to read my paper.

Regards,

Bhushan Poojary

You are welcome Poojary. Could you please explicitly say whether you agree with me or not? This issue is not clear to me.

Incidentally, I would recommend you to work with a finite-dimensional matrix model of a quantum-mechanical problem whereby you can directly verify that indeed the matrix representations of x and p do not commute. Working in a finite-dimensional Hilbert space, issues are very transparent. Chapter 6 (Matrix Formulation of Quantum Mechanics) of the book by Leonard Schiff (Quantum Mechanics, 3rd edition, McGraw-Hill, New York, 1968) has a good discussion of this subject matter. The matrix representations of x and p for a one-dimensional quantum harmonic oscillator are explicitly presented in this chapter (beginning on page 183). Using these matrices, you can directly verify the operator relationship [x,p] = i I, where x and p are the relevant matrices and I the unit matrix.   

I fully agree with Behnam Farid, and believe that you should withdraw this paper forthwith.

Dr Behnam Farid,

Thank you , I will go through the articles you have mentioned and get back to you with more details on it.

Regards,

Bhushan Poojary

You are welcome Bhushan.

who can prove actually your ideas on the uncertainty principle by experiments ?

This is a rather fundamental question but I do not know the answer to it. Sorry !

No problem Olszewski,

Even single slit data would be helpfull.

Regards,

Bhushan Poojary

Dear colleagues!

Experiments prove nothing, they can only disprove.

Dr Eugene,

I  want data for single slit experiment (wavelength used, slit length and spread).

I wished some institute would help me with this.

Regards,

Bhushan Poojary

Institutes make money, Bhushan.

I agree with you Eugene.

But I am not looking for any money nor I want to take any credit from them to prove me right or wrong.

But it seems very difficult to find one.

Regards,

Bhushan Poojary

Nothing can be proved (exept tautology), Bhushan.

Any, Manuel.