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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.)

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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 $0$.

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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.

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  • $\begingroup$ Thanks, that makes sense. Does this mean that every stationary state will have $\langle p \rangle = 0$? $\endgroup$ – Ashutosh Singh Sep 5 '17 at 21:52
  • $\begingroup$ Well, this mean value operator $\langle \mathcal{O} \rangle$ is related to the trace of the operator $\mathcal{O}$ and the basis of eigenfunctions with which you do the trace can be the set of stationary states. Those do not depend on $p$ so $\langle p \rangle = 0$. Talking about the operator $\langle \rangle$ applied on state do not make much sense, is it rather applied to operators, which acts on the space of states. $\endgroup$ – gingras.ol Sep 5 '17 at 22:03

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