Heisenberg's uncertainty principle says that (Delta x) [Delta (mv)] >= h/4 pi
Delta x represents the uncertainty in position and delta (mv) represents the uncertainty in momentum.

An example problem in my textbook is:
The mass of an electron is 9.11x10^(-31) kg and its velocity is 2.2x10^6 m/s (known to within 10%, or 0.2x10^6 m/s), then

(Delta x) >= h/ {4 pi[Delta (mv)]}
(Delta x) >= 6.6x10^(-34) / {(4)(pi)[9.11x10^(-31)](0.2x10^6)}

The part that brings me into trouble is, how come delta (mv) = m * delta v = [9.11x10^(-31)](0.2x10^6)? How come you can pull the "m" out of the delta? I can't understand this operation at all...

My second question: the problem says the uncertainty of velocity of the electron is 0.2x10^6m/s. Does this mean (2.2x10^6 +/- 0.2x10^6) m/s or (2.2x10^6 +/- 0.1x10^6) ? I am not sure about this...

Also, delta x in physics means CHANGE in position, but here the delta x means the UNCERTAINTY in position. Why are they both given the same symbol? Aren't they different things?

Delta typically symbolized change, not necessarily in velocity or any other unit.

-AntonK

The following are SWAG's (Sophisticated Wild Ass Guesses), which are slightly more reliable than WAG's.When the Uncertainty Principle was first developed, there was no standard name for an amount of uncertainty. They might have chosen some standard name for an amount of error, which would be some what misleading. They could have made up a new term. They arbitrarily decided to use Delta, which was a common name for an increment in calculus [As in F(X + DeltaX) = . . . ].

I just shared a bottle of wine at dinner and some had several cups of coffee with Irish Cream liquor, so I am not quite up to analysing the equation you posted. The m (for mass) might have been replaced by the value of mass for some particle being discussed, allowing Velocity Uncertainty to be expressed as a function of Position Uncertainty & vice versa for some patrticular particle.BTW: Einstein once said that he considered momentum to be a more fundamental quantity than either velocity or mass. I think Newton expressed some of his laws of motion in terms of change in momentum due to a force, rather than using the equation Force = Mass*Acceleration. Both of these brilliant men viewed momentum as a fundamental quantity.

The part that brings me into trouble is, how come delta (mv) = m * delta v = [9.11x10^(-31)](0.2x10^6)? How come you can pull the "m" out of the delta? I can't understand this operation at all...

Delta(mv) means (mv)_{highest value} minus mv_{lowest value}

If the mass m is constant, then this is the same as:

m(v_{highest value} - v_{lowest value})
= m Delta(v)

My second question: the problem says the uncertainty of velocity of the electron is 0.2x10^6m/s. Does this mean (2.2x10^6 +/- 0.2x10^6) m/s or (2.2x10^6 +/- 0.1x10^6) ? I am not sure about this...

It means plus or minus 0.2 x 10^-6.

Also, delta x in physics means CHANGE in position, but here the delta x means the UNCERTAINTY in position. Why are they both given the same symbol? Aren't they different things?

A little, but they are similar things.

The effect of the delta x in the uncertainty principle is to change the value of the position from x to (x plus or minus delta x), in an unknown way, at the moment of measurement of x.

That's a bit different to starting x at one value, then measuring its final value later on. However, in both cases we are talking about a change in the value, so it makes sense to use the same symbol.

thank you very much for your explanation