Why does the position uncertainty of a harmonic oscillator not have the expectation value squared term?

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 $$(\Delta x)^2= \big+\big^2$$ for a harmonic oscillator in the ground state energy.

We have established for a harmonic oscillator $$\hat x={\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger})$$ so $$\big=\big$$ which gives us $$\big=\big$$. And in a similar fashion $$\big^2=\big^2$$.

To me this means that $$(\Delta x)^2=\big + \big^2$$. However, the book seems to drop (with no explanation) $$\big^2$$ and comes up with $$(\Delta x)^2=\big<0\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|0\big>$$. 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 $$\big$$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?

• $$\hat{a}$$ and $$\hat{a}^\dagger$$ 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.
• Expand $$\left( \hat{a} + \hat{a}^\dagger \right)^2$$, and see why it has a very different character than $$\left( \hat{a} + \hat{a}^\dagger \right)$$.
• I just expanded them and I see exactly what you mean, I need to be more careful about how I use operators. Looking at this I can't see any way $<x>^2$ would ever contribute to the uncertainty, no matter what energy state it is in because of the lowering and raising ladder operations. Is that the case? – matryoshka Oct 13 '18 at 22:11
• Yup. Or at least, "Yup for the harmonic oscillator". The situation where $\hat{x}$ is composed of a sum of simple raising and lowering operators is special to the SHO. – dmckee Oct 13 '18 at 22:14