-1
$\begingroup$

Whilst working on a project I kept stumbeling across two different expressions for the standard deviation $\Delta{X}^2 = <(X - <X>)^2 >$ and the other $\Delta{X}^2 = <X^2> - <X>^2$. In one of my books I found the following "derivation" $$\Delta{X}^2 = \langle(X - \langle X\rangle)^2 \rangle$$ $$ = \langle X^2 - 2X\langle X \rangle + \langle X \rangle ^2 \rangle $$ $$= (\langle X^2 \rangle- \langle X \rangle ^2) $$

It is the jump from the second to the last line that doesnt make any sense to me because the way I understand it this would imply:

$$ X\langle \psi|X|\psi\rangle = \langle\psi|XX|\psi\rangle$$ And this cannot be the case since X is an operator, no?

And on an unrelated note, how do I use the braket notation in LaTeX?

$\endgroup$
  • $\begingroup$ Look closely at your brackets in that derivation again. In LaTeX, you can use the '\langle' and '\rangle' commands for braket notation. $\endgroup$ – Wouter Nov 5 '14 at 10:10
  • $\begingroup$ When you've sorted out the brackets in the derivation, note that the second term in the second line is not $-2X\langle X \rangle$. (if you don't see that, look carefully at that line and remember that taking the expectation value is linear, i.e. $\langle a A + b B \rangle = a\langle A \rangle + b\langle B \rangle$ with $a,b$ constants and $A,B$ operators) $\endgroup$ – Wouter Nov 5 '14 at 10:16
  • $\begingroup$ @Wouter I think I get it now: I have to treat the expectation value as a constant then it works perfectly. This is possible because the expectation value is always a real number in the end. Thank you, this was bothering me! $\endgroup$ – SandraK Nov 5 '14 at 10:43
  • $\begingroup$ Exactly. You're welcome :) $\endgroup$ – Wouter Nov 5 '14 at 10:58
2
$\begingroup$

So let's remember that $X$ is an operator, and $\langle X \rangle$ is just a number, and we can use the definition of the expectation value $\langle O \rangle = \langle \psi | O | \psi \rangle$ to work this out.

\begin{eqnarray}\Delta X^2 =& \langle X^2 - 2X\langle X \rangle + \langle X \rangle ^2 \rangle \\ =& \langle X^2 \rangle -\langle 2X\langle X \rangle \rangle + \langle X \rangle ^2\end{eqnarray}

That middle term is dealt with easily \begin{eqnarray}\langle 2X\langle X \rangle \rangle = \langle \psi|X \langle X \rangle | \psi \rangle =\langle X \rangle \langle \psi|X | \psi \rangle = \langle X \rangle ^2\end{eqnarray}

Substituting this back in, we get

\begin{eqnarray}\Delta X^2 =& \langle X^2 \rangle -\langle 2X\langle X \rangle \rangle + \langle X \rangle ^2 \\ =& \langle X^2 \rangle -2\langle X \rangle^2 + \langle X \rangle ^2 \\ =& \langle X^2 \rangle- \langle X \rangle ^2 \end{eqnarray}

$\endgroup$
0
$\begingroup$

There is a very big "expectation value" bracket all around the expression, and from the 2nd to the 3rd line, you have to use its properties. So, let's write it the long way:

$$ \langle (X^2 - 2X\langle X\rangle + \langle X\rangle^2) \rangle = \langle X^2\rangle - \langle 2X\langle X \rangle \rangle + \langle X \rangle^2 $$

(now use that the expectation value of the expectation value of something is the expection value of something. That sentence sounds weird, but it is actually correct)

$$...= \langle X^2\rangle - \langle 2X^2\rangle + \langle X\rangle^2$$ (now use $$X^2 - 2X^2 = -X^2$$)

$$ ... = - \langle X^2 \rangle + \langle X\rangle^2$$

I think I messed up the signs somewhere, but this should explain how to get from the 2nd to the 3rd line.

$\endgroup$
  • $\begingroup$ This is incorrect (probably just due to going too quickly in your enthusiasm to answer though). And more importantly (in my opinion): we don't like to just give the answer to this kind of question here. It's much better to let the asker come to the answer themselves, for their own benefit. $\endgroup$ – Wouter Nov 5 '14 at 10:25
  • $\begingroup$ You're right, especially with the enthusiasm. Finally a question I know the answer to! ;-) $\endgroup$ – Gully Nov 5 '14 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.