# Quantum mechanics operator manipulation

1. Apr 4, 2013

### Tom_12

1. The problem statement, all variables and given/known data
consider operator defined as $\hat{O_A} = \hat{A} -<\hat{A}>$
show that $(ΔA)^2=<\hat{O_A}^2>$

2. Relevant equations
$(ΔA)^2=<\hat{A}^2>-<\hat{A}>^2$

3. The attempt at a solution
$(ΔA)^2=<\hat{A}^2>-<\hat{A}>^2$
$= <\hat{A}^2> - (\hat{A} -\hat{O_A})^2$
$= <\hat{A}^2> - \hat{A}^2 + 2\hat{A}<\hat{O_A}> - \hat{O_A}^2$

or

$\hat{O_A} = \hat{A} -<\hat{A}>$

$\hat{O_A}^2 = (\hat{A} -<\hat{A}>)^2$

$\hat{O_A}^2 = \hat{A}^2-2\hat{A}<\hat{A}>+\hat{A}^2$

But I don't know how to convert the operator $\hat{O_A}$ into the expectation value $<\hat{O_A}>$...?

2. Apr 4, 2013

### George Jones

Staff Emeritus
What happens when the expectation of both sides of the last line is taken?

3. Apr 4, 2013

### Tom_12

Does this happen:

$<\hat{O_A}^2> = <\hat{A}^2>-2<\hat{A}><\hat{A}>+<\hat{A}>^2$

$<\hat{O_A}^2> = <\hat{A}^2>-2<\hat{A}>^2+<\hat{A}>^2$

$<\hat{O_A}^2> = (ΔA)^2$....?

I am not sure what happens when you take the expectation value of a term that is already an expectation value, does it just remains unchanged?

4. Apr 4, 2013

### vela

Staff Emeritus
An expectation value is just a number, so yes, it remains unchanged.

5. Apr 5, 2013

### Tom_12

Oh yes, of course. Thank you.

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