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 - <p> )^2> )
sigma_x = sqrt(< (x - <x> )^2> )
So, basically, it says something like this. Suppose you have your wave
equation W(x)
sigma_p^2 = int_x=[-infty,infty] ( W*(x) ( p - <p> )^2 W(x) )
sigma_x^2 = int_x=[-infty,infty] ( W*(x) ( x - <x> )^2 W(x) )
Then sigma_x * sigma_p >= hbar/2
Basically, the uncertainty principle in its pureform is
< (delta_A)^2> < (delta_B)^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.
