There are several derivations of the uncertainty principle. The popular version of this principle found in many college textbooks, usually says, that you cannot measure position and momentum with certainty because if you try to exactly measure position, u would need a higher resolution microsope and an higher resolution microscope, means, more smaller lamda, and hence a larger frequency. However,

From the debroglie wavelength equation

2 pi / lamda = p / hbar

where p is the momentum and lamda is the wavelength.

Suppose we are tryin to measure an electron.

The incoming photon has momentum h / lamda. As a result of the collision, the photon transfers part of all of its momentum to the electron. The uncertainty in momentum is Dp = h / lamda. We also know that the photon is also a wave. So, the uncertainty in position is about Dx = lamda.

so Dp * Dx = h

Well, it's ok, at the high school or fresh man college level, to say, that this derivation is sensible and so acceptable.

However, is it? IMHO, no, it is in fact misleading and leads to a lot of confusion and sometimes even creates a havoc between your high school physics and you in an truthfullness of uncertainty principle.

I agree that that uncertainty principle is correct, but the question is if the above explanation is sufficient, necessary or in fact, admissible to exactly describe the truth of the uncertainty principle.

I am a stronger believer in Einstien. You can even say, that i even worship him. After all, i am indian. And i am allowed to do ascestry worship, although he is not an ancestor of mine. In fact, there is a famous picture of Bose yes, the Bose-Einstien condensation guy, with a picture of Einstien's picture hung among the God's and with Bose, paying homage to Einstien, by putting his hands in a position,as people do, when they pray. Yes, Einstien's photo even has the dot on it.

I will tell you, why I am a strong believer of Einstien. First and foremost, in my opnion, Einstien wasnt wrong about what he felt, in fact, he was right, more right than anyone, of his time. His gut feeling is inscapbable. In fact, the famous EPR paradox after all, is in fact, as real, as me or you.

When srchodinger developed wave mechanics, (anectodally, passed to me, from many internet sites), Einstien did believe that it was a successful description of nature. However, when Von Neuman showed that schrodinger's picture and hisenberg's picture, einstien started raising his doubts.

Basically, here's the deal with the uncertainty principle. The most important thing to note about the uncertainty principle is that the so called uncertainty does not apply to a single particle, but to an ensemble of particle.

The Dx, is just another 'lousy' way of writing the standard deviation of x. sigma_x = sqrt(< (x - <x> )^2 >). The second to note is that, the standard deviation is taken over distribution, not a single particle. It makes all the difference in the world, to say that something is a statistically true and something is always true. As much difference as walking into a feminist colony and proclaming that since the mean of SAT maths scores of female students is 30 points lower than males, then all women should be poor at math compared to any man. This will also get you killed, if you add a racist twist to this statistics.

First and foremost, it is important to understand the concept of statiscal 'proof' , what it means to have a correlation and what the notion of average means. The average, basically, takes a distribution of numbers and turns it into one number. And from that 1 number, you cannot say what the other numbers are. The standard deviation, is another number from the average, that takes a distribution of numbers and turns it into a single number.

sigma_p = sqrt(< (x -

)^2> ) sigma_x = sqrt(< (x - <x> )^2> ) So, basically, it says something like this. Suppose you have your wave equation W(x) sigmap^2 intx-infty,infty ( W*(x) ( p -

)^2 W(x) ) sigmax^2 intx-infty,infty ( W*(x) ( x - <x> )^2 W(x) ) Then sigmax * sigmap >= hbar/2 Basically, the uncertainty principle in its pureform is < (deltaA)^2> < (deltaB)^2 > >= 1/4 ||<A,B>|| ^2 It's very, very important to not that, the uncertainty principle, does not say anything about a single particle and it's also important to pay attention to the expected value in that expression. In fact, the debroglie formula lamda = h / p has a new dimension in wave mechanices. This formula does not literally translate into wave mechanics. I am sure, you are familar with taylor serier. f(x+ a) = f(x) + a f'(x) + a^2/2==== f(x) + a^3/3! f'(x) + ....

==

Now, intresting you can also write the above expression as

f(x+ a) = exp(a D/Dx) f(x)

       = 1 + a D/Dx + a^2 D^2/Dx^2  +  a^3 D^3/D^3 x f(x)
       = f(x) + a Df(x)/Dx + a^2 D^2 f(x)/Dx^2 + ...

Notice that apply exp( a D/Dx) translates f(x) to f(x+a). In Quantum mechanics, momentum is a generator of translation. as we saw in the above expression

Suppose i claim that exp( a D/Dx) we say above can be written as exp( i a O ) where a is the translation distance and O is some kind of operator, related to D/Dx and i the famous complex number. Now, you may say, why we need an i in there. First, we know that momentum causes translation. that means

exp( i a p) f(x) = f(x+a)

However, the units is all wrong. The expresion a*p is required to be unitless . The above expression is a classical wave expression. exp( i a p) = cos(a p) + i sin(a p)

So, p must be some kind of frequency, taken over space, not over time.

Now, we go back to our debroglie relation.

2*pi / lamda = p/ h

1 / lamda = p / (h/2*pi) 1/ lamda = p / (hbar)

Our p / (hbar) expression will statisfy both the frequency aspect of p and units of m ^-1.

so exp( i a p / hbar) should be our expression

so if i apply exp( i a p /hbar) f(x), i should get f(x+ a). So, now, we claimed that p is an operator so, it related to D/Dx.

if exp( i a p / hbar) f(x) = f(x+a)

then p must be (hbar/i D/Dx)

so exp( i a (hbar/i D/Dx) p / bar) = exp( a D/Dx)

Now, you may be inclined to ask, why we thru all the trouble, of putting the complex i and hbar in there, after all, it cancels itself out. The prensence of hbar in p, gives it the correct units as momentum. The presence of i is tricky to understand. (even i have trouble admitting to myself that it is needed). Until, i found a nice page on the web,

http://www.umassd.edu/1Academic/CArtsandSciences/physics/Research/Quantum/Quantum.html

which explains why i is there. Although, i remain unconvinced, because by inducing the i into the expression, we have introduced wave like behaviour into nature of things. However the expression,

exp( i a (hbar/i D/Dx) p / bar) = exp( a D/Dx)

itself does not need the i.

Now, our commutation expression < (deltaA)^2> < (deltaB)^2 > >= 1/4 ||<A,B>|| ^2

A = x B = ih D/Dx

A,B = AB - BA

that is to

x-ihbar D/Dx  - (-i) hbar D/Dx  x

Now, the expresion may be even confusing. The way to look at it to say that commuation actions on some function

(x -ihbar D/Dx + ihbarD/Dx *x) f(x)

= -xihbar D f(x) /Dx + ihbar D/Dx (x* f(x) )

D/Dx ( x f(x) ) = x D(f(x))/Dx + f(x) * Dx/Dx

D/Dx ( x f(x) ) - x Df(x)/Dx = f(x)

ih D/Dx ( x ''' f(x) ) - x ''' Df(x) / Dx = ih f(x)

So x,p = ihbar

taking ||<x,p||>||^2 ||ihbar||^2 hbar^2

So, we have

< (deltax)^2 > < (deltap)^2 > >= hbar^2/4

About the relation < (deltaA)^2> < (deltaB)^2 > >= 1/4 ||<A,B>|| ^2

Now, if you wonder, where i came up with the above formula, i would say that you consult an advanced physics book. A very good book, probably found in your college library is 'Introduction to Quantum Mechanics' by Griffiths. Now, if you are lookin for a much more advanced text, 'Modern Quantum Mechanics' By J.J. Sakurai, IMHO, i can tell, is much more than a book. It's an art. This is book used in a graduate Introduction to QM, at Rutgers. Ofcourse, for an beginer, following the book, is not easy. If cannot afford a book, or goin to the library, there are many websites, decidated to this. However, IMHO, whether you skim thru a book, or goto a website, its more important to seek a sense of statisfaction with an explanation, then accept an explanation to be true. If you not satisfied, then, do more research until you have convincing explanation for yourself of the genuine meaning of the uncertainty principle. Then, you can explain it to your friends over dinner. Or even, to your first born, on his first birthday.

-suresh