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Well, the difference between the two expressions $\langle \hat p^2 \rangle_{\psi}$ and $\langle \hat p \rangle_{\psi}^2$ is exactly $\Delta p^2$ , i.e. the squared uncertainty (variance) of the momentum, where the $\hat p$ is momentum operator and uncertainty on the momentum operator is defined by: $$\Delta p= \sqrt {\langle \hat p^2 \rangle_{\psi} - \langle \hat p \rangle_{\psi}^2}$$. why the difference between $\langle \hat p^2 \rangle_{\psi}$ and $\langle \hat p \rangle_{\psi}^2$ is NOT zero?

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The fact that the quantities $\langle p^2\rangle$ and $\langle p\rangle^2$ are different is not something specific to quantum mechanics, but exists in any context where one can define an average value. Let's take a simple example and suppose that $p$ is a balanced binary random variable which can take the values $+1$ and $-1$. Since the variable is balanced, one has $\langle p\rangle=0$ and $\langle p\rangle^2=0$. On the other hand $(+1)^2=1=(-1)^2$, so $p^2=1$ in all cases and $\langle p ^2\rangle = 1 \neq \langle p \rangle^2$.

The difference between the two quantities is called the variance and its square root is the standard deviation $\sigma$. This quantity is the most commonly used in statistics to measure the width of a probability distribution. It is always positive, and $\sigma=0$ if and only if the random variable is constant.

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Well, the difference between the two expressions is exactly $(\Delta p)^2$, i.e. the squared uncertainty (variance) of the momentum, as the very question correctly says.

To be sure, the real question is why it's not zero. It's not zero. Just write the function $\psi$ in the momentum representation. Then there is a probabilistic distribution $$ \rho(p)=|\tilde \psi(p)|^2 $$ and the expectation values of $p$ and $p^2$ are simply $$ \langle p \rangle = \int dp\,p\,\rho(p) $$ and $$ \langle p^2 \rangle = \int dp\,p^2\,\rho(p) $$ There is no reason why the latter should be the square of the former. If you first square a function and then integrate it, it's something else than if you first integrate it and then square the result.

For example, $\langle p\rangle$ is zero for a huge family of functions $\rho(p)$. It's enough if the distribution is correctly shifted in the horizontal $p$ direction. For every function $\rho(p)$, you may shift it to $\rho'(p) = \rho(p+P)$ so that the new $\rho'(p)$ has $\langle p \rangle = 0$.

On the other hand, if you want $\langle p^2 \rangle = 0 $, you need $\rho(p)\sim \delta (p)$: it must be a delta-function and the probability of nonzero values of the momentum has to be zero, otherwise the expectation value would clearly be positive as it would be a positive linear combination of non-negative terms and some of them would be strictly positive.

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the real question is why it's not zero. this is exactly what i want to ask, thanks!!. – user8784 Jun 13 '12 at 11:13

A simple calculation shows that $\langle p^2\rangle-\langle p\rangle^2= \left<\left(p-\langle p\rangle\right)^2\right>$, hence this is always nonnegative, and positive unless $p$ has a sharp, deterministic value. This holds both for classical variables $p$ and for quantum variables.

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