Quantum mechanics operator manipulation

[Tom_12]

Homework Statement

consider operator defined as \hat{O_A} = \hat{A} -<\hat{A}>
show that (ΔA)^2=<\hat{O_A}^2>

Homework Equations

(ΔA)^2=<\hat{A}^2>-<\hat{A}>^2

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}>...?

 

[George Jones]

\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}>...?

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

 

[Tom_12]

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

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?

 

[vela]

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

 

[Tom_12]

Oh yes, of course. Thank you.