Time independent Schrodinger equation While studying Introduction to Quantum Mechanics by D. J. Griffiths, in the time independent Schrodinger equation chapter, the author provided 3 arguments, first one being:
Every expectation value is constant in time, which makes sense because the time dependent part is eliminated from the integration. Further in the text, he mentioned that," $\langle x \rangle$   is constant, hence $\langle p \rangle = 0$. Nothing ever happens in a stationary state" 
My queries:
1. Does that mean every stationary state will have $\langle p \rangle = 0$? 
2. If expectation value of momentum is zero, then how come  $\langle p^2 \rangle$ is not zero?( I did encounter a problem with $\langle p\rangle = 0$, but $\langle p^2 \rangle$ was not. It was a solved numerical.)
 A: The easiest way to think of this is through Ehrenfest's theorem which states that
$$
\langle p\rangle = m\frac{d}{dt}\langle x\rangle  \tag{1}
$$
Here, $p=mv=m\frac{d}{dt}x$; taking the average on both sides gives (1).
Since $\langle x\rangle$ is not a function of $t$ in a stationary state, $\langle p\rangle =0$ follows immediately.  
Ehrenfest's theorem is mathematical statement that reflects the observation that, on average quantum mechanics should give the same results as classical mechanics.
To see how $\langle p^2\rangle \ne 0$ even though $\langle p\rangle=0$, first note that $\langle p^2\rangle \ne \langle p\rangle^2$: since $p^2$ is a positive only function, its average cannot be $0$ for a non-negative probability density $\vert\psi(x)\vert^2$.  Instead, think of the kinetic energy $T=\frac{1}{2m}p^2$ and
indeed $\frac{1}{2m}\langle p^2\rangle=\langle T\rangle $.  In this sense, $\langle p^2\rangle$ is related (up to a factor of $2m $) to the (non-negative) average kinetic energy of the particle.   Classically, this kinetic energy is certainly on average greater than $0$ since it is $0$ only at the turning points of the motion, and positive everywhere else.  Averaging a bunch of non-negative number gives a average greater than $0$.
A: The mean value of $p$ is $0$ because the $-|p|$ contribution cancels out the $|p|$ one. For $p^2$, both contributions add up. Another way to see it is that $p$ is an odd function while $p^2$ is even.
