Meaning of Expectation Values for <x^2> and <p^2> in Classical Mechanics

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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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13characters said:
v^2 = <x^2> - <x>^2

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

\Delta x = \sqrt {&lt;x^2&gt; - &lt;x&gt;^2}
 
jtbell said:
This is the square of the uncertainty in position. That is,

\Delta x = \sqrt {&lt;x^2&gt; - &lt;x&gt;^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
 
13characters said:
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

&lt;x^2&gt; = \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 &lt;v^2&gt;, calculated similarly, but with v replacing x.
 
jtbell said:
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

&lt;x^2&gt; = \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 &lt;v^2&gt;, calculated similarly, but with v replacing x.

i get it now.

thanks.
 
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