Something in the derivation

1. Oct 22, 2012

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>??

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2. Oct 22, 2012

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.

3. Oct 22, 2012

M. next

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

4. Oct 22, 2012

Cthugha

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>).

5. Oct 22, 2012

M. next

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

6. Oct 22, 2012

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?

7. Oct 23, 2012

M. next

Yes Very!
Thank you very much Chopin.