# Expectation Velues

## Main Question or Discussion Point

Every quantum mechanical operator has an observable in classical mechanics

<x> - position
...
<x^2> - ?
<p^2> - ?

What is the meaning on these expectation values?

v^2 = <x^2> - <x>^2

What is the meaning of this? edit: It looks to me like uncertainty in position. Is it the average uncertainty in position?

Admin Please move me to the right forum, if i'm not in eet.

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jtbell
Mentor
v^2 = <x^2> - <x>^2
This is the square of the uncertainty in position. That is,

$$\Delta x = \sqrt {<x^2> - <x>^2}$$

This is the square of the uncertainty in position. That is,

$$\Delta x = \sqrt {<x^2> - <x>^2}$$
That's what i figures too after some digging in the kinetics section of my chem book

cheers.

i still don't get what <x^2> means. What i found was "the average of the square of molecular speeds" but i still don't entirely get why its operator is X^2.

edit: Also is there a ways to represent it geometrically or graphically

jtbell
Mentor
i still don't get what <x^2> means.
It's simply the average value of the square of the position, in the limit as the number of measurements goes to infinity. If you have N measurements of the position, then

$$<x^2> = \frac{1}{N} \sum_{i=1}^N {x_i^2}$$

Actually this is only an approximation. It becomes exact as $N \rightarrow \infty$.

The "average of the square of molecular speeds" would be $<v^2>$, calculated similarly, but with v replacing x.

It's simply the average value of the square of the position, in the limit as the number of measurements goes to infinity. If you have N measurements of the position, then

$$<x^2> = \frac{1}{N} \sum_{i=1}^N {x_i^2}$$

Actually this is only an approximation. It becomes exact as $N \rightarrow \infty$.

The "average of the square of molecular speeds" would be $<v^2>$, calculated similarly, but with v replacing x.
i get it now.

thanks.

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