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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:

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

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

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

Magpie
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1 Answers1

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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.

SMeznaric
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