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Heisenberg Uncertainty Principle scientific prove

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Heisenberg's uncertainty principle states that: if the x-component of the momentum of a particle is measured with an uncertainty $$\Delta \vec p_x$$ then its x-position cannot, at same time, be measured more accurately than $$\Delta\vec x=\frac {\hbar}{2\Delta\vec p_x }$$ $$\Delta\vec x\Delta\vec p_x \ge \frac {\hbar}{2}$$ what is scientific prove of this principle?

This post has been migrated from (A51.SE)
Closed as per community consensus as the post is High school level question; too low-level for PhysicsOverflow
asked Apr 20, 2012 in Closed Questions by Bad Boy (-10 points) [ no revision ]
recategorized Apr 22, 2014 by dimension10
This is a standard textbook question, so I am moving it to physics.SE. Moreover, when refer to a component of a vector then you drop the arrow.

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1 Answer

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This is, as mentioned by Piotr Migdal, in the comments, standard textbook material. 

In Matrix Mechanics, observables like position and momentum are promoted to operators. Obviously, multiplication of operators need not commute. Specifically, the difference is given by the commutator bracket:    

\(\left[ X, P\right] = XP-PX=i\hbar I \)

where I is the identity matrix.  

Therefore, there have to be uncertainties (standard deviation after taking an infinite number of measurements) in X and P and it should be intuitive that:  

\( {\sigma _x}{\sigma _p} \ge k\hbar \)  

For some scalar (constant) value of k. So far, everything applies to any pair of such dynamical variables, like position and momentum. It happens to be, however, that specifically for position and momentum, k is \(\frac12 \). This can be derived from special cases, c.f. Feynman, Hibbs and Styer, who derive this in the first chapter through special cases.       

One may also derive it through the wave interpretation, through Fourier transforms, and stuff; see Wikipedia.      

answered Feb 22, 2014 by dimension10 (1,950 points) [ no revision ]




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