mjhjmtu

My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(Delta x)^2= big<x^2 big>+big<xbig>^2$ for a harmonic oscillator in the ground state energy.

We have established for a harmonic oscillator
$hat x=sqrt hbarover 2momega(hat a+hat a^dagger)$ so $big<xbig>=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>$ which gives us $big<x^2big>=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig>$.
And in a similar fashion $big<xbig>^2=big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$.

To me this means that $(Delta x)^2=big<nbig|hbarover 2momega(hat a+hat a^dagger)^2big|nbig> + big<n big|sqrt hbarover 2momega(hat a+hat a^dagger) big|nbig>^2$. However, the book seems to drop (with no explanation) $big<xbig>^2$ and comes up with $(Delta x)^2=big<0big|hbarover 2momega(hat a+hat a^dagger)^2big|0big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $big<x^2big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?

::chuckles::

I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.

Three facts:

• 
  $hata$ and $hata^dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.


• The numbered states are a set of eigenstates, so they are orthogonal to one another.


• Expand $left( hata + hata^dagger right)^2$, and see why it has a very different character than $left( hata + hata^dagger right)$.