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In Shankar's Principles of Quantum Mechanics, he applies the HUP to estimate the ground state energy of the Hydrogen atom. In the proof, it is said that the first step to minimizing the expectation value of energy $\langle H \rangle$ is to consider states where all the expectation values of momentum are 0, i.e. $\langle P_i \rangle = 0$ where $i\in\{x,y,z\}$. I am having some trouble understanding why this is so, as $\langle H \rangle$ depends on $\langle P_i^2 \rangle$.

Is this because $\langle P_i^2 \rangle = (\Delta P_i)^2+\langle P_i \rangle^2$? Why can't we consider a 'frame of reference' in momentum space where $\langle P_i \rangle = 0$? Why does this mean that $\langle P_i \rangle = 0$ implies a minimum $\langle H \rangle$?

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  • $\begingroup$ Is it possible to see all the construction somewhere? HUP connects E and t, p and x... $\endgroup$ – jaromrax Feb 7 '17 at 12:32
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Shankar mentions the answer while discussing the estimation of a ground state of a particle in a one-dimensional potential well using the HUP; similar logic applies to a hydrogen atom and a three-dimensional system. I do not recall seeing an estimation of the minimum $\left<H\right>$ for a hydrogen atom through this approach.

Now $\left<P\right>=0$ for the following reason. Since a bound state is a stationary state, $\left<P\right>$ is time independent. If this $\left<P\right>\neq0$, the particle must (in average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.

That seems pretty self-explanatory... so it's not that the proof only considers states with $\left<P\right>=0$: the fact is that there are no valid candidate states to consider wherein $\left<P\right>$ is either positive or negative.

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