# 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? Oct 13, 2018 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. Oct 13, 2018 at 22:14