# Standard deviation of expectation values

1. Apr 18, 2007

### neu

Very basic question which has confused me:

if the variance of an expectation value <A> is:

uncertainty of $$A=<(A-<A>)^2>^0.5$$

how is this equal to:

$$(<A^2>-<A>^2)^0.5$$

??

Last edited: Apr 18, 2007
2. Apr 18, 2007

### Staff: Mentor

Expand it out:
$$<(A-<A>)^2> = < A^2 - 2A<A> + <A>^2 > = < A^2 - <A>^2 >$$

3. Apr 18, 2007

### Staff: Mentor

Start by expanding the squared term in parentheses:

$(A - <A>)^2 = A^2 - 2<A>A + <A>^2$

Note that <A> is simply a number and can be manipulated like any other numeric constant. Simplify the resulting expectation value by taking advantage of general properties of expectation values, i.e.

$<A+B> = <A> + <B>$

$<cA> = c<A>$

where c is a numeric constant.