Understanding the Disappearance of 2*A^*<A> and 2*B^*<B> in Derivations

[M. next]

Hello guys,

Check the attachment please.
How is it what's written after "HENCE"?
Where did 2*A^*<A> term go?
the same thing for 2*B^*<B>??

 

[dextercioby]

It was absorbed generating a term similar to the one already there, namely <A>^2 or <B>^2.

<A>^2 - 2<A><A> + <A^2> = <A^2> + <A>^2 - 2 <A>^2 = <A^2> - <A>^2 and the same for B.

 

[M. next]

thanks for your reply,

but you wrote: <A>^2 - 2<A><A> + <A^2>
it is supposed to be: <A>^2 - 2<A>A + A^2

see figure please

 

[Cthugha]

thanks for your reply,

but you wrote: <A>^2 - 2<A><A> + <A^2>
it is supposed to be: <A>^2 - 2<A>A + A^2

If you have a look at the figure you have posted, you will see that you are supposed to take the expectation value of that second line you just gave (<A>^2 - 2<A>A + A^2). Taking the expectation value of that line gives you the first line you just gave (<A>^2 - 2<A><A> + <A^2>).

 

[M. next]

how come?
i didn't get it, mind elaborating?

 

[Chopin]

The key is to remember that $\langle A \rangle \equiv \langle \psi | A | \psi \rangle$.

So we have:
$$\langle \psi | (\Delta A)^2 | \psi\rangle = \langle \psi | \hat{A}^2 - 2\hat{A}\langle \hat{A} \rangle + \langle \hat{A} \rangle ^ 2 | \psi \rangle \\ = \langle \psi | \hat{A}^2 | \psi \rangle - 2\langle \hat{A} \rangle \langle \psi | \hat{A} | \psi \rangle + \langle \hat{A} \rangle ^2 \langle \psi | \psi \rangle \\ = \langle \hat{A}^2\rangle - 2\langle \hat{A}\rangle \langle \hat{A} \rangle + \langle \hat{A} \rangle^2 \\ = \langle \hat{A}^2\rangle - \langle \hat{A}\rangle ^2$$

Does that make sense?

 

[M. next]

Yes Very!
Thank you very much Chopin.