# What's the intuitive interpretation of quantum uncertainty $\Delta \hat{A}=\sqrt{\langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2}$?

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?

• 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 $\hat{A}$. – march Feb 14 at 21:19
• 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? – ACuriousMind Feb 14 at 21:20

## 1 Answer

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 $$\hat{A}$$.

• 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? – spraff Feb 14 at 23:27
• @spraff It just follows from the definition of expectation value $\langle\hat{A}\rangle$. – Thomas Fritsch Feb 14 at 23:45