# Rewrite standard deviation for operators ($\Delta A$)

According to my notes the Standard deviation in QM is defined as:

$$(\Delta A)^2 = \langle \psi| A^2|\psi\rangle -\langle\psi|A|\psi\rangle^2$$

Now I stumbled on the following claim: $$(\Delta A)^2 = \langle\psi|(A-\langle A\rangle)^2|\psi\rangle$$

How is that possible? I recognize some parallelism to the standardized standard deviation though:

$$s^2 = \dfrac{1}{n}\sum_{i = 1}^n\left(x_i-\bar{x}\right)^2 = \dfrac{1}{n}\sum_{i = 1}^n{x_i}^2-\bar{x}^2$$

But I do see no accordance while writing it out:

\begin{align}(\Delta A)^2 &= \langle\psi|(A-\langle A\rangle)^2|\psi\rangle = \langle\psi|A^2-2A\langle A\rangle + \langle A\rangle^2| \psi\rangle\\[24pt] & = \langle \psi|A^2|\psi\rangle - 2\langle\psi|A|\psi\rangle\,\langle A\rangle+\langle\psi|\langle A^2\rangle | \psi\rangle\end{align}

What's wrong with that?

• Can you tell us the difference between $\langle \psi | A | \psi \rangle$ and $\langle A \rangle$? Jun 1, 2022 at 12:48
• I've got it, it's the same.
– Leon
Jun 1, 2022 at 19:47

Picking up from where you left, \begin{align} (\Delta A)^2 &= \langle \psi|A^2|\psi\rangle - 2\langle\psi|A|\psi\rangle\,\langle A\rangle+\langle\psi|\langle A\rangle | \psi\rangle, \\ &= \langle A^2 \rangle - 2\langle A\rangle\,\langle A\rangle+\langle A\rangle\langle\psi | \psi\rangle, \\ &= \langle A^2 \rangle - 2\langle A\rangle\,\langle A\rangle+\langle A\rangle, \\ &= \langle A^2 \rangle - \langle A\rangle\, \end{align} as desired.

It is worth recalling that $$\langle A\rangle$$ is a common shorthand for $$\langle\psi|A|\psi\rangle$$.

• Where I left I had a little typo. I mean to write $\mathbf{\langle\psi|\langle A^2\rangle | \psi\rangle}$ But I got the key idea from your indication, many thanks.
– Leon
Jun 1, 2022 at 19:46

Start with $$(\Delta Q)^2 = {\langle(Q-\langle Q\rangle )^2\rangle}_{\Psi}$$

Expanding the square gives: $${\langle Q^2+\langle Q\rangle ^2 -2\langle Q\rangle Q\rangle}_{\Psi}$$

Given that $$\langle\langle Q\rangle \rangle_{\Psi} = \langle Q\rangle _{\Psi}$$ we have:

$${\langle Q^2 \rangle} _{\Psi} +{\langle Q\rangle^2} _{\Psi} -2{\langle Q\rangle^2}_{\Psi}$$

Hence the result...