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?

  • $\begingroup$ 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

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}


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.

  • $\begingroup$ 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

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