As per this video, if $\hat{A}$ is a quantum operator, the uncertainty is given by

$$\Delta \hat{A}=\sqrt{\langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2}$$

I understand what this expression means in a purely mathematical sense, but I have no physical intuition to it.

How should I interpret the terms $\langle\hat{A}^2\rangle$ and $\langle\hat{A}\rangle^2$, physically? And why, physically, should uncertainty be the square root of their difference?

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    $\begingroup$ Quantum mechanics is a little bit of a red-herring here. This is just the standard definition of standard deviation (which we often use as a measure of uncertainty in measurements), where $\langle \hat{A} \rangle$ is the average value of the physical observable associated with $\hat{A}$. $\endgroup$
    – march
    Feb 14 '19 at 21:19
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    $\begingroup$ Once you accept that $\langle \cdot\rangle$ denotes an expectation value of a random variable in the ordinary sense of statistics, this is just the standard expression for standard deviation. Do you know that, and so are effectively asking why $\langle \cdot\rangle$ is the expectation value, or does this already answer your question? $\endgroup$
    – ACuriousMind
    Feb 14 '19 at 21:20

Admittedly that expression is somewhat un-intuitive

$$ \Delta \hat{A}=\sqrt{\langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2} $$

But you can rewrite the term below the square root in a mathematically equivalent way, and get

$$ \Delta \hat{A}=\sqrt{\left<(\hat{A}-\langle\hat{A}\rangle)^2\right>} $$

Now, here $(\hat{A}-\langle\hat{A}\rangle)$ is obviously an operator with mean value $0$, and the whole expression is quite intuitively the standard deviation of $\hat{A}$.

  • $\begingroup$ Okay, I see the analogy with standard deviation, but since $\hat{A}$ is an operator, I don't see why that analogy should hold for operators in general. Can you justify the analogy? $\endgroup$
    – spraff
    Feb 14 '19 at 23:27
  • $\begingroup$ @spraff It just follows from the definition of expectation value $\langle\hat{A}\rangle$. $\endgroup$ Feb 14 '19 at 23:45

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