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The difference between the two expressions $\langle \hat p^2 \rangle_{\psi}$ and $\langle \hat p \rangle_{\psi}^2$ is defined by the squared uncertainty: $$\Delta p^2 = {\langle \hat p^2 \rangle_{\psi} - \langle \hat p \rangle_{\psi}^2}$$.

On the other hand, the variance of a random variable is $$var(X)=E[(X-E[X])^2]=E[X^2]-E[X]^2.$$

How can one represent the uncertainty on the momentum operator through a variance?

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Why "on the other hand", they are the same idea. – Ron Maimon Jul 14 '12 at 3:36

Defining the operator $$\Delta A\equiv A-\langle A\rangle$$ the dispersion of an operator A is $$\langle(\Delta A)^{2}\rangle\equiv \langle A^{2}\rangle-\langle A\rangle^{2}$$ The expectation value $\langle(\Delta A)^{2}\rangle$ is called the variance of the operator A. The quantity you call the "squared uncertainty" is actually the variance of the momentum operator.

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