# Proof of quantum mechanical position uncertainty

How can you prove the uncertainty for position is:

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

$\Delta{x}$, taken to be the root mean square of x.

$$\Delta{x} =\sqrt{\langle \left(x-\langle x\rangle\right)^2\rangle}$$

$$\Delta{x} =\sqrt{\langle \left(x-\langle x\rangle\right) \left(x-\langle{x}\rangle\right)\rangle}$$

$$\Delta{x} =\sqrt{\langle x^2-2x\langle x\rangle +\langle x \rangle^2\rangle}$$

This is the bit which I am not sure about and why I can do it (taking the outer braket and acting it on the inner x values:

$$\Delta{x} =\sqrt{\langle x^2\rangle -2\langle x \rangle \langle x\rangle +\langle x \rangle^2}$$

$$\Delta{x} =\sqrt{\langle x^2\rangle -2\langle x\rangle^2 +\langle x \rangle^2}$$

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

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– Qmechanic♦ Oct 16 '12 at 23:49

2
I'm afraid your step is incorrect (the last formula). Expanding $\langle(x-\langle x \rangle)^2\rangle$ you obtain $\langle x^2 -2x \langle x \rangle x - \langle x \rangle^2\rangle$. From here you only need to use that $\langle x \rangle$ is a number and that expectation value is linear. Since this looks like a homework, I won't work it all out for you (important part of the learning process in physics is to calculate things for yourself). But hopefully this is enough of a hint to get you to the right answer. – SMeznaric Oct 16 '12 at 23:55
@SMeznaric that could be a good answer – David Zaslavsky♦ Oct 16 '12 at 23:56
You're right, here goes. – SMeznaric Oct 16 '12 at 23:57
By the way, the title of your question seems to be in no relation to the body... – Fabian Oct 17 '12 at 10:08
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## 1 Answer

I'm afraid your step is incorrect (the last formula). Expanding $\langle(x−\langle x \rangle)^2\rangle$ you obtain $\langle x^2−2x\langle x\rangle + \langle x\rangle^2\rangle$. From here you only need to use that $\langle x\rangle$ is a number and that expectation value is linear. Since this looks like a homework, I won't work it all out for you (important part of the learning process in physics is to calculate things for yourself). But hopefully this is enough of a hint to get you to the right answer.

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yeah sorry, that was a mistake. I will correct that bit. I had done that expansion but it did not make sense to me that it would work. Still doesn't. – Magpie Oct 17 '12 at 14:12
Now that you have the expansion you can use the linearity of the expectation value, i.e. $\langle a \hat{x} + b \hat{y} \rangle = a \langle \hat{x} \rangle + b \langle \hat{y} \rangle$, where I marked the operators with hats and $a,b$ are numbers. In your expression the operators are $x$ and $x^2$, and $\langle x \rangle$ is a number. You are very close to the answer. – SMeznaric Oct 18 '12 at 15:39
Also, as an aside, your last formula is still incorrect. You need $⟨\rangle$ around the value inside the square root, otherwise you are taking the square root of the position operator (not what you want) – SMeznaric Oct 18 '12 at 17:57
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